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Main Date: 23 Oct 2007 11:25:59 From: joseph.santaniello@gmail.com Subject: Gyroscopic forces revisited Hi All, A while ago there was a thread discussing no-handed riding and the role gyroscopic forces play in bike turning. I hypothesized that gyroscopic forces were not that important, and several things were pointed out to me that almost convinced me. A good dose of insomnia allowed me to think about this for several hours and I am now convinced that gyroscopic forces are virtually irrelevant for riding bikes, hands or no-hands. It was postulated that a clown-bike with tiny wheel would be difficult to ride no hands due to the small gyroscopic forces. Since I don't have a clown-bike to play with, this remains a mystery. But I have observed little kids riding small wheel (10") bikes at below walking pace. I have also spun some 12" wheels in my hand a various speeds to feel what sort of gyroscopic forces are there. Not much. A little kid with a sense of balance not as developed as an adult can ride one of these bikes. I do not believe a kid could keep one of these bikes upright by manual correction alone. These small bikes are stable by themselves, and since the gyroscopic forces are so low, there must be something else at work here. This isn't proof or anything, this is just what got me thinking. I finally pulled out a big plywood board (much to the chagrin of my wife who imagines there are myriad things I could be better spending my time on) and propped it up at an angle and put a bike on it to simulate what happens when a bike is in a turn. It was suggested that in a turn a pendulm hung from the top-tube would hang parallel to the seat tube. When riding no hands in a constant radius turn, this cannot be the case, and I suspect that it is not the case with hands on the bars either. Experiments with my plywood board show that when force is applied straight down through the bike the steering remains straight no matter what the angle of the board (simulated angle of lean). In a turn, the steering cannot be straight, otherwise it wouldn't be called a turn, it would be called a crash. So in a turn (a no-handed one in particular) something has to be holding the steering at a non-straight angle. The only thing it can be is that the center of mass is moved to the side of the plane that is the centerline of the bike. This makes the steering flop into the turn. The combined force of gravity and the acceleration of the turn act from the center of mass through the contact patches of the tires. Since the center of mass is not in the plane of the bike's centerline, this means that the combined force is not parallel to the seat tube and thus a pendulum hanging from the top- tube could not be parallel with the seat tube. Riding with hands on the bars I suspect is the same. But a rider could force the bike to be in the same plane, but then they would need to hold the steering at the proper angle manually. This would no doubt require quite a bit of skill, and I believe in practice to be virtually impossible. But perhaps it is just this skill which separates the good from the great. So what does all that mean? It means that "flop" from an off-center center of mass is what makes a bike turn, and thus while gyroscopic forces make help the initial turn of the steering due to induced lean, it is an unnecessary component that is ultimately irrelevant to turing a bike. The whole COM argument was brought about by thinking about how a radio- controlled motorcycle I used to have woked. Joseph   Date: 31 Oct 2007 21:02:47 From: Joe Riel Subject: Re: Gyroscopic forces revisited Tom Sherman <sunsetss0003@REMOVETHISyahoo.com > writes: > jobst.brandt@stanfordalumni.org aka Jobst Brandt wrote: >> ... >> Centrifugal force is a product of arc radius and speed." > Poor word choice. I usually associate product with multiplication, not the more general sense (production) that is being used here. > And mass. Yes. Centripetal force is m*v^2/r. -- Joe Riel   Date: 30 Oct 2007 07:15:39 From: Subject: Re: Gyroscopic forces revisited On Oct 23, 7:54 pm, jobst.bra...@stanfordalumni.org wrote: > Joseph Santaniello writes: > > A while ago there was a thread discussing no-handed riding and the > > role gyroscopic forces play in bike turning. > > I hypothesized that gyroscopic forces were not that important, and > > several things were pointed out to me that almost convinced me. A > > good dose of insomnia allowed me to think about this for several > > hours and I am now convinced that gyroscopic forces are virtually > > irrelevant for riding bikes, hands or no-hands. > > It was postulated that a clown-bike with tiny wheel would be > > difficult to ride no hands due to the small gyroscopic forces. > > Since I don't have a clown-bike to play with, this remains a > > mystery. But I have observed little kids riding small wheel (10") > > bikes at below walking pace. I have also spun some 12" wheels in my > > hand a various speeds to feel what sort of gyroscopic forces are > > there. Not much. A little kid with a sense of balance not as > > developed as an adult can ride one of these bikes. I do not believe > > a kid could keep one of these bikes upright by manual correction > > alone. These small bikes are stable by themselves, and since the > > gyroscopic forces are so low, there must be something else at work > > here. This isn't proof or anything, this is just what got me > > thinking. > > You should have seen the demo at InterBike where an engineer built a > front wheel with a forward rotating flywheel (brass disk) between > the spokes driven by a small motor at about the speed you expect from > a 27" wheel at 10-15mph. He rode this bicycle no-hands at below 2mph, > steady as a rock, up and down the isles. Without the flywheel > turning independently, the bicycle was as difficult to ride no-hands > as any other bicycle with wheels that size. > Ah, but was the disc spinning in the same direction as the wheel or in the opposite direction? > > > > I finally pulled out a big plywood board (much to the chagrin of my > > wife who imagines there are myriad things I could be better spending > > my time on) and propped it up at an angle and put a bike on it to > > simulate what happens when a bike is in a turn. It was suggested > > that in a turn a pendulum hung from the top-tube would hang parallel > > to the seat tube. When riding no hands in a constant radius turn, > > this cannot be the case, and I suspect that it is not the case with > > hands on the bars either. Experiments with my plywood board show > > that when force is applied straight down through the bike the > > steering remains straight no matter what the angle of the board > > (simulated angle of lean). In a turn, the steering cannot be > > straight, otherwise it wouldn't be called a turn, it would be called > > a crash. So in a turn (a no-handed one in particular) something has > > to be holding the steering at a non-straight angle. The only thing > > it can be is that the center of mass is moved to the side of the > > plane that is the centerline of the bike. This makes the steering > > flop into the turn. The combined force of gravity and the > > acceleration of the turn act from the center of mass through the > > contact patches of the tires. Since the center of mass is not in > > the plane of the bike's centerline, this means that the combined > > force is not parallel to the seat tube and thus a pendulum hanging > > from the top- tube could not be parallel with the seat tube. Riding > > with hands on the bars I suspect is the same. But a rider could > > force the bike to be in the same plane, but then they would need to > > hold the steering at the proper angle manually. This would no doubt > > require quite a bit of skill, and I believe in practice to be > > virtually impossible. But perhaps it is just this skill which > > separates the good from the great. > > Please discover why one should use paragraphs. You must have come > across this in school. > > > So what does all that mean? It means that "flop" from an off-center > > center of mass is what makes a bike turn, and thus while gyroscopic > > forces make help the initial turn of the steering due to induced > > lean, it is an unnecessary component that is ultimately irrelevant > > to turning a bike. > > The whole COM argument was brought about by thinking about how a > > radio controlled motorcycle I used to have worked. > > I think your research came up with the wrong result. > > An easily repeatable exercise of coasting down a smooth road at more > than 20mph riding no-hands, is to shake one knee from side to side > while resting the other one against the top tube for stability. I > think there is where you will see the effect the best. In addition, > shimmy on a bicycle cannot occur without gyroscopic forces of the > front wheel. > > Jobst Brandt   Date: 29 Oct 2007 17:22:00 From: Jon_C Subject: Re: Gyroscopic forces revisited On Oct 29, 10:32 am, jobst.bra...@stanfordalumni.org wrote: > Jon Crouch writes: > > So there are other forces at play too'. As I said earlier, if > > camber thrust acts on a vertical wheel then it would pull the bottom > > of the wheel up the slope causing the wheel to now lean down the > > slope, countering the camber thrust to some degree'. Also as the > > contact patch of the wheel when vertical is off center and on the > > uphill side of the wheel while the CoG is still on the centerline of > > the wheel, gravity will tend to tip the wheel in the downhill > > direction'. Do these mechanisms make sense? > > Please describe your rolling wheel experiment in detail'. I'm > > interested to know what speeds yo rolled the wheel at and how steel > > the camber was. > > I think the test is described unambiguously. I didn't notice that > "camber thrust" is speed dependent. Because we don't all have a > bicycle race track available, an ordinary crowned street will do. > Roll the wheel parallel to the center line on a crowned street at your > favorite speeds and not that it does not climb toward the high part of > the road as has been proposed by "camber thrust". Proposed by who??? You just quoted me saying that there are obvious forces that might counteract camber force and cause the wheel to arc downhill and then totally ignore that in your response. Please don't put words into my mouth. If you must deny the effect of camber thrust, please let us know what forces cause a lone wheel to roll in a circle.     Date: 29 Oct 2007 18:57:53 From: Subject: Re: Gyroscopic forces revisited Jon Crouch writes: >>> So there are other forces at play too'. As I said earlier, if >>> camber thrust acts on a vertical wheel then it would pull the >>> bottom of the wheel up the slope causing the wheel to now lean >>> down the slope, countering the camber thrust to some degree'. >>> Also as the contact patch of the wheel when vertical is off center >>> and on the uphill side of the wheel while the CoG is still on the >>> centerline of the wheel, gravity will tend to tip the wheel in the >>> downhill direction'. Do these mechanisms make sense? >>> Please describe your rolling wheel experiment in detail'. I'm >>> interested to know what speeds yo rolled the wheel at and how >>> steel the camber was. >> I think the test is described unambiguously. I didn't notice that >> "camber thrust" is speed dependent. Because we don't all have a >> bicycle race track available, an ordinary crowned street will do. >> Roll the wheel parallel to the center line on a crowned street at >> your favorite speeds and not that it does not climb toward the high >> part of the road as has been proposed by "camber thrust". > Proposed by who??? You just quoted me saying that there are obvious > forces that might counteract camber force and cause the wheel to arc > downhill and then totally ignore that in your response. Please > don't put words into my mouth. For who the bell tolls? I didn't say that you made that claim but rather the proponents of "camber thrust" claim that the wheel will turn toward the side where the axis of rotation pierces the road. > If you must deny the effect of camber thrust, please let us know > what forces cause a lone wheel to roll in a circle. The same force that causes a penny flop (in place). I notice there was no explanation how a stunt rider can ride around the wall in a vertical barrel. His speed and angle to the wall are closely linked while his traction on the wall has nothing to do with "camber thrust" unless we see a re-definition of what that is. Jobst Brandt   Date: 29 Oct 2007 09:09:31 From: Joe Riel Subject: Re: Gyroscopic forces revisited jobst.brandt@stanfordalumni.org writes: > I think the test is described unambiguously. I didn't notice that > "camber thrust" is speed dependent. Because we don't all have a > bicycle race track available, an ordinary crowned street will do. > Roll the wheel parallel to the center line on a crowned street at your > favorite speeds and not that it does not climb toward the high part of > the road as has been proposed by "camber thrust". Have you tried this test? I have. It proves nothing---I am unable to roll the wheel with sufficient precision/repeatibility. Sometimes it goes up, sometimes it goes down. Camber thrust is well described in Milliken's "Race Car Vehicle Dynamics". The included graphs, alas, aren't particularly suited for bicycle wheels. It does mention, however, that for motorcycle tires, which have round cross sections, camber thrust provides significant lateral force for lean angles up to 50 degrees. -- Joe Riel     Date: 29 Oct 2007 18:49:26 From: Subject: Re: Gyroscopic forces revisited Joe Riel writes: >> I think the test is described unambiguously. I didn't notice that >> "camber thrust" is speed dependent. Because we don't all have a >> bicycle race track available, an ordinary crowned street will do. >> Roll the wheel parallel to the center line on a crowned street at >> your favorite speeds and not that it does not climb toward the high >> part of the road as has been proposed by "camber thrust". > Have you tried this test? I have. It proves nothing---I am unable > to roll the wheel with sufficient precision/repeatability. > Sometimes it goes up, sometimes it goes down. I have fairly steep crown on the road in front of my house and can roll a front wheel briskly down the road. It gradually veers to the curb with no visible loss of speed in the few seconds it takes to get to its destination. > Camber thrust is well described in Milliken's "Race Car Vehicle > Dynamics". The included graphs, alas, aren't particularly suited > for bicycle wheels. It does mention, however, that for motorcycle > tires, which have round cross sections, camber thrust provides > significant lateral force for lean angles up to 50 degrees. I find they have practically no relevance to bicycles. Motorcycles have a substantially wider contact patch and the farther they lean the more they approach a rotary scrub brush of a street cleaner. There are two reasons for hiking out on fast motorcycle turns, one is to keep hardware from grounding, the other is to reduce scrub and subsequent loss of traction. When leaned at near maximum angle the contact patch has significantly different tracking from inside to outside, the inside scrubbing forward, the outside scrubbing rearward, the neutral point lying somewhere outside the center of the contact patch. It is much like a bearing ball in a race, where the ball is sliding forward at the edge of the contact patch and rearward in the center with only a line somewhere in between that tracks. Jobst Brandt   Date: 29 Oct 2007 14:19:24 From: Jon_C Subject: Re: Gyroscopic forces revisited On Oct 27, 6:02 pm, jobst.bra...@stanfordalumni.org wrote: > Jon Crouch writes: > >>>> With a tire cross section of 0", what forces cause the tire to > >>>> turn about point B? > >>> Good question :) I guess in theory a wheel of zero thickness on an > >>> infinitely hard flat surface won't experience any camber > >>> thrust. Thats kinda hypothetical tho :) > >> Well lets approach the limit in reality. Do you believe that a > >> slick 60mm tire corners differently from a 23mm cross section tire? > >> Where does the difference enter as an effect and is it a step > >> function? > > I'd imagine there'd be a difference due to the way the tyres slip > > during a turn. It's highly unlikely that they'd behave identically > > but I don;t really see how it's relevant. > >> A 23mm tire is essentially a zero width tire for the turn radii of > >> interest. > > No it's not. Regardless of turn radii, it still creates camber > > thrust relative to it's lean angle. > > Yes and the Coriolis effect gets in there too, but the effect you > propose is in the noise and isn't detectable for bicycles. If it is > significant, then you should be able to deflate a tire enough so that > when rolled on a banked straight course (straight crowned road in > front of your house) it climbs toward the road center instead of > turning toward the fall line. > > Jobst Brandt How would deflating the tyre increase camber thrust? You don't seem to have grasped the concept and yet you're going to great lengths to deny it's validity.     Date: 29 Oct 2007 14:34:52 From: Subject: Re: Gyroscopic forces revisited Jon Crouch writes: >>>>>> With a tire cross section of 0", what forces cause the tire to >>>>>> turn about point B? >>>>> Good question :) I guess in theory a wheel of zero thickness on >>>>> an infinitely hard flat surface won't experience any camber >>>>> thrust. Thats kinda hypothetical tho :) >>>> Well lets approach the limit in reality. Do you believe that a >>>> slick 60mm tire corners differently from a 23mm cross section >>>> tire? Where does the difference enter as an effect and is it a >>>> step function? >>> I'd imagine there'd be a difference due to the way the tyres slip >>> during a turn. It's highly unlikely that they'd behave >>> identically but I don;t really see how it's relevant. >>>> A 23mm tire is essentially a zero width tire for the turn radii >>>> of interest. >>> No it's not. Regardless of turn radii, it still creates camber >>> thrust relative to it's lean angle. >> Yes and the Coriolis effect gets in there too, but the effect you >> propose is in the noise and isn't detectable for bicycles. If it >> is significant, then you should be able to deflate a tire enough so >> that when rolled on a banked straight course (straight crowned road >> in front of your house) it climbs toward the road center instead of >> turning toward the fall line. > How would deflating the tyre increase camber thrust? You don't seem > to have grasped the concept and yet you're going to great lengths to > deny it's validity. It increases the width of the contact patch from which some torque about the vertical axis of the wheel is supposedly acting. Jobst Brandt   Date: 29 Oct 2007 14:14:58 From: Jon_C Subject: Re: Gyroscopic forces revisited On Oct 27, 5:54 pm, jobst.bra...@stanfordalumni.org wrote: > Also "camber thrust" is not magically created because someone believes > in it. If it is so clearly defined, you might develop what the forces > are knowing the tire cross section and its major diameter... And how would that very complex calculation help you? The force generated by a camber thrust depends on tyre shape/size/rigidity, friction between road surface/tyre surface, speed and lean angle. If you're going to deny the effect of camber thrust, can you provide an alternative theory as to what causes a lone wheel to turn when it leans? >If you roll a lone wheel straight ahead on a banked surface, its gradual turn > toward the fall line will occur gradually and is in opposition to your > claim. Try it and see how the "camber thrust" acts. So there are other forces at play too. As I said earlier, if camber thrust acts on a vertical wheel then it would pull the bottom of the wheel up the slope causing the wheel to now lean down the slope, countering the camber thrust to some degree. Also as the contact patch of the wheel when vertical is off center and on the uphill side of the wheel while the CoG is still on the centerline of the wheel, gravity will tend to tip the wheel in the downhill direction. Do these mechanisms make sense? Please describe your rolling wheel experiment in detail. I'm interested to know what speeds yo rolled the wheel at and how steel the camber was. > >>> Now as you (and I) have said, a bicycle will take a much larger > >>> arc than described above when riding at speed but that doesn't > >>> mean that camber force isn't present. In fact it's still camber > >>> force that's making your bike turn but at the same time the bike's > >>> tyres are 'slipping' (as described above) causing the bile to take > >>> a much larger radius. > > When is this force present and how large is it? If you tried the > above test, you'll note that it doesn't interfere with the wheel > turning to the opposite side predicted by your proposal. > Which force? > >> The lean angle is defined by the centrifugal acceleration and the > >> curvature, not the converse. That is where I believe you have the > >> cart before the horse. > > You keep quoting me and then make totally unrelated responses. > > Anyhow, lean angle (required to balance) is determined solely by > > centrifugal force. Centrifugal force is a product of arc radius and > > speed. I'm not sure where I said the converse of that. > > Your "camber thrust" claims to have an effect on that but maybe you > didn't make clear how. I'm not sure what you;re saying here. > > >>> Google some more my friend. These are all commonly accepted > >>> principles. > >> Just because they are repeated among certain groups, does not make > >> them correct or true. It's like the rotating mass on a bicycle is > >> more important than static mass on the frame. That one survived > >> many years in these newsgroups only because it was repeated so > >> often > > Try a physics text book then. > > http://en.wikipedia.org/wiki/Special:Search?search=camber+thrust > > I don't see any relevance there and other references only repeat > without evidence or magnitude, the same as what you have said. > > Jobst Brandt     Date: 29 Oct 2007 14:32:22 From: Subject: Re: Gyroscopic forces revisited Jon Crouch writes: >> Also "camber thrust" is not magically created because someone >> believes in it. If it is so clearly defined, you might develop >> what the forces are knowing the tire cross section and its major >> diameter... > And how would that very complex calculation help you? The force > generated by a camber thrust depends on tyre shape/size/rigidity, > friction between road surface/tyre surface, speed and lean angle'. > If you're going to deny the effect of camber thrust, can you provide > an alternative theory as to what causes a lone wheel to turn when it > leans? >> If you roll a lone wheel straight ahead on a banked surface, its >> gradual turn toward the fall line will occur gradually and is in >> opposition to your claim'. Try it and see how the "camber thrust" >> acts. > So there are other forces at play too'. As I said earlier, if > camber thrust acts on a vertical wheel then it would pull the bottom > of the wheel up the slope causing the wheel to now lean down the > slope, countering the camber thrust to some degree'. Also as the > contact patch of the wheel when vertical is off center and on the > uphill side of the wheel while the CoG is still on the centerline of > the wheel, gravity will tend to tip the wheel in the downhill > direction'. Do these mechanisms make sense? > Please describe your rolling wheel experiment in detail'. I'm > interested to know what speeds yo rolled the wheel at and how steel > the camber was. I think the test is described unambiguously. I didn't notice that "camber thrust" is speed dependent. Because we don't all have a bicycle race track available, an ordinary crowned street will do. Roll the wheel parallel to the center line on a crowned street at your favorite speeds and not that it does not climb toward the high part of the road as has been proposed by "camber thrust". >>>>> Now as you (and I) have said, a bicycle will take a much larger >>>>> arc than described above when riding at speed but that doesn't >>>>> mean that camber force isn't present'. In fact it's still >>>>> camber force that's making your bike turn but at the same time >>>>> the bike's tyres are 'slipping' (as described above) causing the >>>>> bile to take a much larger radius. >> When is this force present and how large is it? If you tried the >> above test, you'll note that it doesn't interfere with the wheel >> turning to the opposite side predicted by your proposal. > Which force? "Camber thrust" or is it not a force or vector quantity? >>>> The lean angle is defined by the centrifugal acceleration and the >>>> curvature, not the converse'. That is where I believe you have >>>> the cart before the horse. >>> You keep quoting me and then make totally unrelated responses. >>> Anyhow, lean angle (required to balance) is determined solely by >>> centrifugal force'. Centrifugal force is a product of arc radius >>> and speed'. I'm not sure where I said the converse of that. >> Your "camber thrust" claims to have an effect on that but maybe you >> didn't make clear how. > I'm not sure what you're saying here. Where does "camber thrust" enter into the: "Anyhow, lean angle (required to balance) is determined solely by centrifugal force'. Centrifugal force is a product of arc radius and speed." >>>>> Google some more my friend'. These are all commonly accepted >>>>> principles. >>>> Just because they are repeated among certain groups, does not >>>> make them correct or true'. It's like the rotating mass on a >>>> bicycle is more important than static mass on the frame'. That >>>> one survived many years in these newsgroups only because it was >>>> repeated so often >>> Try a physics text book then. http://en.wikipedia.org/wiki/Special:Search?search=camber+thrust >> I don't see any relevance there and other references only repeat >> without evidence or magnitude, the same as what you have said. Jobst Brandt       Date: 31 Oct 2007 21:36:34 From: Tom Sherman Subject: Re: Gyroscopic forces revisited jobst.brandt@stanfordalumni.org aka Jobst Brandt wrote: > ... > Centrifugal force is a product of arc radius and speed." And mass. -- Tom Sherman - Holstein-Friesland Bovinia When did ignorance of biology become a "family value"?         Date: 01 Nov 2007 14:00:19 From: Subject: Re: Gyroscopic forces revisited Tom Sherman writes: >> Centrifugal force is a product of arc radius and speed." > And mass. In the context, the mass was constant (for a given rider), so don't quibble. Jobst Brandt   Date: 27 Oct 2007 20:29:28 From: Jon_C Subject: Re: Gyroscopic forces revisited On Oct 27, 1:14 am, jobst.bra...@stanfordalumni.org wrote: > Jon Crouch writes: > >> With a tire cross section of 0", what forces cause the tire to turn > >> about point B? > > Good question :) I guess in theory a wheel of zero thickness on an > > infinitely hard flat surface won't experience any camber > > thrust. Thats kinda hypothetical tho :) > > Well lets approach the limit in reality. Do you believe that a slick > 60mm tire corners differently from a 23mm cross section tire? Where > does the difference enter as an effect and is it a step function? I'd imagine there'd be a difference due to the way the tyres slip during a turn. It's highly unlikely that they'd behave identically but I don;t really see how it's relevant. > A 23mm tire is essentially a zero width tire for the turn radii of > interest. No it's not. Regardless of turn radii, it still creates camber thrust relative to it's lean angle. > Jobst Brandt     Date: 27 Oct 2007 22:02:13 From: Subject: Re: Gyroscopic forces revisited Jon Crouch writes: >>>> With a tire cross section of 0", what forces cause the tire to >>>> turn about point B? >>> Good question :) I guess in theory a wheel of zero thickness on an >>> infinitely hard flat surface won't experience any camber >>> thrust. Thats kinda hypothetical tho :) >> Well lets approach the limit in reality. Do you believe that a >> slick 60mm tire corners differently from a 23mm cross section tire? >> Where does the difference enter as an effect and is it a step >> function? > I'd imagine there'd be a difference due to the way the tyres slip > during a turn. It's highly unlikely that they'd behave identically > but I don;t really see how it's relevant. >> A 23mm tire is essentially a zero width tire for the turn radii of >> interest. > No it's not. Regardless of turn radii, it still creates camber > thrust relative to it's lean angle. Yes and the Coriolis effect gets in there too, but the effect you propose is in the noise and isn't detectable for bicycles. If it is significant, then you should be able to deflate a tire enough so that when rolled on a banked straight course (straight crowned road in front of your house) it climbs toward the road center instead of turning toward the fall line. Jobst Brandt   Date: 27 Oct 2007 20:21:23 From: Jon_C Subject: Re: Gyroscopic forces revisited On Oct 27, 1:10 am, jobst.bra...@stanfordalumni.org wrote: > That's a nice picture but if you roll a wheel out on a parking lot, > you'll notice that it tightens the curve from practically straight > ahead to finally doing a wobble flop onto its side. Those are a > continuum of curves defines by speed and resultant lateral > acceleration. They are not defined by the lean angle but rather the > lean angle is defined by the speed which equals lateral acceleration. All I'm saying is that at a certain angle of lean a wheel will _try_ to follow a certain arc. That's CAMBER THRUST. Obviously speed will affect lean angle which will effect camber thrust. But that's exactly what I said in the paragraph below. > > > Obviously the wheel needs to be at a certain speed to achieve this > > or else centrifugal force or gravity will change the angle of the > > wheel and throw it into a weave or a different arc.. Anyhow, that's > > the natural turning radius of a wheel at a given angle to the ground > > and you can test it very easily with a kids hoop or a bike wheel. > > We aren't riding a single wheel alone but operating a bicycle that > does not act as a single wheel. Obviously. I was explaining camber thrust as requested by you. Atatching a wheel to a bicycle doesn't magically make it immune to camber thrust. > > > Now as you (and I) have said, a bicycle will take a much larger arc > > than described above when riding at speed but that doesn't mean that > > camber force isn't present. In fact it's still camber force that's > > making your bike turn but at the same time the bike's tyres are > > 'slipping' (as described above) causing the bile to take a much > > larger radius. > > The lean angle is defines by the centrifugal acceleration and the > curvature, not the converse. That is where I believe you have the > cart before the horse. You keep quoting me and then make totally unrelated responses. Anyhow, lean angle (required to balance) is determined solely by centrifugal force. Centrifugal force is a product of arc radius and speed. I;m not sure where I said the converse of that. > > Google some more my friend. These are all commonly accepted > > principles. > > Just because they are repeated among certain groups, does not make > them correct or true. It's like the rotating mass on a bicycle is more > important than static mass on the frame. That one survived many years > in these newsgroups only because it was repeated so often. > > Jobst Brandt Try a physics text book then.     Date: 27 Oct 2007 21:54:48 From: Subject: Re: Gyroscopic forces revisited Jon Crouch writes: >> That's a nice picture but if you roll a wheel out on a parking lot, >> you'll notice that it tightens the curve from practically straight >> ahead to finally doing a wobble flop onto its side. Those are a >> continuum of curves defines by speed and resultant lateral >> acceleration. They are not defined by the lean angle but rather >> the lean angle is defined by the speed which equals lateral >> acceleration. > All I'm saying is that at a certain angle of lean a wheel will _try_ > to follow a certain arc. That's CAMBER THRUST. Obviously speed > will affect lean angle which will effect camber thrust. But that's > exactly what I said in the paragraph below. >>> Obviously the wheel needs to be at a certain speed to achieve this >>> or else centrifugal force or gravity will change the angle of the >>> wheel and throw it into a weave or a different arc.. Anyhow, >>> that's the natural turning radius of a wheel at a given angle to >>> the ground and you can test it very easily with a kids hoop or a >>> bike wheel. >> We aren't riding a single wheel alone but operating a bicycle that >> does not act as a single wheel. > Obviously. I was explaining camber thrust as requested by you. > Attaching a wheel to a bicycle doesn't magically make it immune to > camber thrust. Also "camber thrust" is not magically created because someone believes in it. If it is so clearly defined, you might develop what the forces are knowing the tire cross section and its major diameter. If you roll a lone wheel straight ahead on a banked surface, its gradual turn toward the fall line will occur gradually and is in opposition to your claim. Try it and see how the "camber thrust" acts. >>> Now as you (and I) have said, a bicycle will take a much larger >>> arc than described above when riding at speed but that doesn't >>> mean that camber force isn't present. In fact it's still camber >>> force that's making your bike turn but at the same time the bike's >>> tyres are 'slipping' (as described above) causing the bile to take >>> a much larger radius. When is this force present and how large is it? If you tried the above test, you'll note that it doesn't interfere with the wheel turning to the opposite side predicted by your proposal. >> The lean angle is defined by the centrifugal acceleration and the >> curvature, not the converse. That is where I believe you have the >> cart before the horse. > You keep quoting me and then make totally unrelated responses. > Anyhow, lean angle (required to balance) is determined solely by > centrifugal force. Centrifugal force is a product of arc radius and > speed. I'm not sure where I said the converse of that. Your "camber thrust" claims to have an effect on that but maybe you didn't make clear how. >>> Google some more my friend. These are all commonly accepted >>> principles. >> Just because they are repeated among certain groups, does not make >> them correct or true. It's like the rotating mass on a bicycle is >> more important than static mass on the frame. That one survived >> many years in these newsgroups only because it was repeated so >> often > Try a physics text book then. http://en.wikipedia.org/wiki/Special:Search?search=camber+thrust I don't see any relevance there and other references only repeat without evidence or magnitude, the same as what you have said. Jobst Brandt   Date: 27 Oct 2007 02:09:53 From: Jon_C Subject: Re: Gyroscopic forces revisited On Oct 26, 7:18 pm, "Kerry Montgomery" <kamon...@teleport.com > wrote: > With a tire cross section of 0", what forces cause the tire to turn about > point B? > Thanks, > Kerry Good question :) I guess in theory a wheel of zero thickness on an infinitely hard flat surface won't experience any camber thrust. Thats kinda hypothetical tho :)     Date: 28 Oct 2007 23:20:11 From: Michael Press Subject: Re: Gyroscopic forces revisited In article <1193450993.812759.320110@o80g2000hse.googlegroups.com > , Jon_C <JonCCrouch@gmail.com > wrote: > On Oct 26, 7:18 pm, "Kerry Montgomery" <kamon...@teleport.com> wrote: > > > With a tire cross section of 0", what forces cause the tire to turn about > > point B? > > Thanks, > > Kerry > > Good question :) I guess in theory a wheel of zero thickness on an > infinitely hard flat surface won't experience any camber thrust. Thats > kinda hypothetical tho :) The camber thrust occurs even with zero thickness tires. It is a function of the radius of curvature of a wheel track of a leaned over wheel. -- Michael Press       Date: 29 Oct 2007 00:04:41 From: Subject: Re: Gyroscopic forces revisited Michael Press writes: >>> With a tire cross section of 0", what forces cause the tire to >>> turn about point B? >> Good question :) I guess in theory a wheel of zero thickness on an >> infinitely hard flat surface won't experience any camber >> thrust. Thats kinda hypothetical tho :) > The camber thrust occurs even with zero thickness tires. It is a > function of the radius of curvature of a wheel track of a leaned > over wheel. How is this force expressed (as a vector and units) and how does it occur when, for instance, rolling a pizza (disk) knife horizontally in a straight line across a tilted surface. This quantity and its effect on bicycle cornering seems hard to define. I am not understanding what this force is and where it applies, especially in the example I mention above. Jobst Brandt         Date: 28 Oct 2007 22:01:49 From: Michael Press Subject: Re: Gyroscopic forces revisited In article <47252399$0$14061$742ec2ed@news.sonic.net >, jobst.brandt@stanfordalumni.org wrote: > Michael Press writes: > > >>> With a tire cross section of 0", what forces cause the tire to > >>> turn about point B? > > >> Good question :) I guess in theory a wheel of zero thickness on an > >> infinitely hard flat surface won't experience any camber > >> thrust. Thats kinda hypothetical tho :) > > > The camber thrust occurs even with zero thickness tires. It is a > > function of the radius of curvature of a wheel track of a leaned > > over wheel. > > How is this force expressed (as a vector and units) and how does it > occur when, for instance, rolling a pizza (disk) knife horizontally in > a straight line across a tilted surface. > > This quantity and its effect on bicycle cornering seems hard to > define. I am not understanding what this force is and where it > applies, especially in the example I mention above. A hoop rolling with its axis oblique to the roadway will turn in an arc. To turn there must be a reaction force at the contact patch. -- Michael Press           Date: 29 Oct 2007 13:55:30 From: Subject: Re: Gyroscopic forces revisited Michael Press writes: >>>>> With a tire cross section of 0", what forces cause the tire to >>>>> turn about point B? >>>> Good question :) I guess in theory a wheel of zero thickness on >>>> an infinitely hard flat surface won't experience any camber >>>> thrust. Thats kinda hypothetical tho :) >>> The camber thrust occurs even with zero thickness tires. It is a >>> function of the radius of curvature of a wheel track of a leaned >>> over wheel. >> How is this force expressed (as a vector and units) and how does it >> occur when, for instance, rolling a pizza (disk) knife horizontally in >> a straight line across a tilted surface. >> This quantity and its effect on bicycle cornering seems hard to >> define. I am not understanding what this force is and where it >> applies, especially in the example I mention above. > A hoop rolling with its axis oblique to the roadway will turn in an > arc. To turn there must be a reaction force at the contact patch. You'll notice a sphere in that test will do the same thing, also changing direction to roll down the fall line. I see no explanation of how this "camber force" force is involved. Jobst Brandt             Date: 29 Oct 2007 12:18:43 From: Michael Press Subject: Re: Gyroscopic forces revisited In article <4725e652$0$14138$742ec2ed@news.sonic.net >, jobst.brandt@stanfordalumni.org wrote: > Michael Press writes: > > >>>>> With a tire cross section of 0", what forces cause the tire to > >>>>> turn about point B? > > >>>> Good question :) I guess in theory a wheel of zero thickness on > >>>> an infinitely hard flat surface won't experience any camber > >>>> thrust. Thats kinda hypothetical tho :) > > >>> The camber thrust occurs even with zero thickness tires. It is a > >>> function of the radius of curvature of a wheel track of a leaned > >>> over wheel. > > >> How is this force expressed (as a vector and units) and how does it > >> occur when, for instance, rolling a pizza (disk) knife horizontally in > >> a straight line across a tilted surface. > > >> This quantity and its effect on bicycle cornering seems hard to > >> define. I am not understanding what this force is and where it > >> applies, especially in the example I mention above. > > > A hoop rolling with its axis oblique to the roadway will turn in an > > arc. To turn there must be a reaction force at the contact patch. > > You'll notice a sphere in that test will do the same thing, also > changing direction to roll down the fall line. I see no explanation > of how this "camber force" force is involved. Do not complicate the matter with spheres yet. A hoop tilted to the roadway and rolling such that its track is an arc of a circle must have a reaction force at the contact patch to accelerate the hoop. This is the camber force. -- Michael Press               Date: 29 Oct 2007 19:51:17 From: Subject: Re: Gyroscopic forces revisited Michael Press writes: >>>>>>> With a tire cross section of 0", what forces cause the tire to >>>>>>> turn about point B? >>>>>> Good question :) I guess in theory a wheel of zero thickness on >>>>>> an infinitely hard flat surface won't experience any camber >>>>>> thrust. Thats kinda hypothetical tho :) >>>>> The camber thrust occurs even with zero thickness tires. It is >>>>> a function of the radius of curvature of a wheel track of a >>>>> leaned over wheel. >>>> How is this force expressed (as a vector and units) and how does >>>> it occur when, for instance, rolling a pizza (disk) knife >>>> horizontally in a straight line across a tilted surface. >>>> This quantity and its effect on bicycle cornering seems hard to >>>> define. I am not understanding what this force is and where it >>>> applies, especially in the example I mention above. >>> A hoop rolling with its axis oblique to the roadway will turn in >>> an arc. To turn there must be a reaction force at the contact >>> patch. >> You'll notice a sphere in that test will do the same thing, also >> changing direction to roll down the fall line. I see no >> explanation of how this "camber force" force is involved. > Do not complicate the matter with spheres yet. > A hoop tilted to the roadway and rolling such that its track is an > arc of a circle must have a reaction force at the contact patch to > accelerate the hoop. This is the camber force. By that you seem to say that the steering angle has nothing to do with it but only the lean. Meanwhile cars with essentially no camber go around curves apparently without this effect. My point is that cornering on a single track vehicle that leans is no different from a non leaning vehicle. The steering angle of the front wheel is what is giving the lateral force in both instances. The conical effect does not make sense. If it does, someone must be able to show the vector forces. The diagram is static and does not represent the physical reality. Jobst Brandt                 Date: 29 Oct 2007 23:38:06 From: Michael Press Subject: Re: Gyroscopic forces revisited In article <472639b5$0$14073$742ec2ed@news.sonic.net >, jobst.brandt@stanfordalumni.org wrote: > Michael Press writes: > > >>>>>>> With a tire cross section of 0", what forces cause the tire to > >>>>>>> turn about point B? > > >>>>>> Good question :) I guess in theory a wheel of zero thickness on > >>>>>> an infinitely hard flat surface won't experience any camber > >>>>>> thrust. Thats kinda hypothetical tho :) > > >>>>> The camber thrust occurs even with zero thickness tires. It is > >>>>> a function of the radius of curvature of a wheel track of a > >>>>> leaned over wheel. > > >>>> How is this force expressed (as a vector and units) and how does > >>>> it occur when, for instance, rolling a pizza (disk) knife > >>>> horizontally in a straight line across a tilted surface. > > >>>> This quantity and its effect on bicycle cornering seems hard to > >>>> define. I am not understanding what this force is and where it > >>>> applies, especially in the example I mention above. > > >>> A hoop rolling with its axis oblique to the roadway will turn in > >>> an arc. To turn there must be a reaction force at the contact > >>> patch. > > >> You'll notice a sphere in that test will do the same thing, also > >> changing direction to roll down the fall line. I see no > >> explanation of how this "camber force" force is involved. > > > Do not complicate the matter with spheres yet. > > > A hoop tilted to the roadway and rolling such that its track is an > > arc of a circle must have a reaction force at the contact patch to > > accelerate the hoop. This is the camber force. > > By that you seem to say that the steering angle has nothing to do > with it but only the lean. Meanwhile cars with essentially no camber > go around curves apparently without this effect. I am _not_ saying that. I am not drawing inferences. > My point is that cornering on a single track vehicle that leans is no > different from a non leaning vehicle. The steering angle of the front > wheel is what is giving the lateral force in both instances. The > conical effect does not make sense. If it does, someone must be able > to show the vector forces. The diagram is static and does not > represent the physical reality. -- Michael Press           Date: 29 Oct 2007 08:30:22 From: Tim McNamara Subject: Re: Gyroscopic forces revisited In article <rubrum-15BE59.22014928102007@newsclstr03.news.prodigy.net >, Michael Press <rubrum@pacbell.net > wrote: > In article <47252399$0$14061$742ec2ed@news.sonic.net>, > jobst.brandt@stanfordalumni.org wrote: > > > Michael Press writes: > > > > >>> With a tire cross section of 0", what forces cause the tire to > > >>> turn about point B? > > > > >> Good question :) I guess in theory a wheel of zero thickness on > > >> an infinitely hard flat surface won't experience any camber > > >> thrust. Thats kinda hypothetical tho :) > > > > > The camber thrust occurs even with zero thickness tires. It is a > > > function of the radius of curvature of a wheel track of a leaned > > > over wheel. > > > > How is this force expressed (as a vector and units) and how does it > > occur when, for instance, rolling a pizza (disk) knife horizontally > > in a straight line across a tilted surface. > > > > This quantity and its effect on bicycle cornering seems hard to > > define. I am not understanding what this force is and where it > > applies, especially in the example I mention above. > > A hoop rolling with its axis oblique to the roadway will turn in an > arc. To turn there must be a reaction force at the contact patch. How about gravity?             Date: 29 Oct 2007 12:16:45 From: Michael Press Subject: Re: Gyroscopic forces revisited In article <timmcn-212E9A.08302229102007@news.iphouse.com >, Tim McNamara <timmcn@bitstream.net > wrote: > In article <rubrum-15BE59.22014928102007@newsclstr03.news.prodigy.net>, > Michael Press <rubrum@pacbell.net> wrote: > > > In article <47252399$0$14061$742ec2ed@news.sonic.net>, > > jobst.brandt@stanfordalumni.org wrote: > > > > > Michael Press writes: > > > > > > >>> With a tire cross section of 0", what forces cause the tire to > > > >>> turn about point B? > > > > > > >> Good question :) I guess in theory a wheel of zero thickness on > > > >> an infinitely hard flat surface won't experience any camber > > > >> thrust. Thats kinda hypothetical tho :) > > > > > > > The camber thrust occurs even with zero thickness tires. It is a > > > > function of the radius of curvature of a wheel track of a leaned > > > > over wheel. > > > > > > How is this force expressed (as a vector and units) and how does it > > > occur when, for instance, rolling a pizza (disk) knife horizontally > > > in a straight line across a tilted surface. > > > > > > This quantity and its effect on bicycle cornering seems hard to > > > define. I am not understanding what this force is and where it > > > applies, especially in the example I mention above. > > > > A hoop rolling with its axis oblique to the roadway will turn in an > > arc. To turn there must be a reaction force at the contact patch. > > How about gravity? Gravity acts perpendicular to a flat road surface. The reaction force to make the hoop roll on an arc of a circle acts parallel to the road surface. -- Michael Press               Date: 29 Oct 2007 15:56:43 From: Tim McNamara Subject: Re: Gyroscopic forces revisited In article <rubrum-E98E0B.12164429102007@newsclstr03.news.prodigy.net >, Michael Press <rubrum@pacbell.net > wrote: > In article <timmcn-212E9A.08302229102007@news.iphouse.com>, > Tim McNamara <timmcn@bitstream.net> wrote: > > > In article > > <rubrum-15BE59.22014928102007@newsclstr03.news.prodigy.net>, > > Michael Press <rubrum@pacbell.net> wrote: > > > > > In article <47252399$0$14061$742ec2ed@news.sonic.net>, > > > jobst.brandt@stanfordalumni.org wrote: > > > > > > > Michael Press writes: > > > > > > > > >>> With a tire cross section of 0", what forces cause the tire > > > > >>> to turn about point B? > > > > > > > > >> Good question :) I guess in theory a wheel of zero thickness > > > > >> on an infinitely hard flat surface won't experience any > > > > >> camber thrust. Thats kinda hypothetical tho :) > > > > > > > > > The camber thrust occurs even with zero thickness tires. It > > > > > is a function of the radius of curvature of a wheel track of > > > > > a leaned over wheel. > > > > > > > > How is this force expressed (as a vector and units) and how > > > > does it occur when, for instance, rolling a pizza (disk) knife > > > > horizontally in a straight line across a tilted surface. > > > > > > > > This quantity and its effect on bicycle cornering seems hard to > > > > define. I am not understanding what this force is and where it > > > > applies, especially in the example I mention above. > > > > > > A hoop rolling with its axis oblique to the roadway will turn in > > > an arc. To turn there must be a reaction force at the contact > > > patch. > > > > How about gravity? > > Gravity acts perpendicular to a flat road surface. The reaction force > to make the hoop roll on an arc of a circle acts parallel to the road > surface. But hoops with extremely low coefficients of friction, and hence little reaction force (e.g., "scrub"), display this behavior. It looks to me that it's due to gravity not a reaction force against the ground. If the hoop can't tilt, as in a two track vehicle such as a car, then a reaction force is obviously necessary. But for a hoop such as a coin or a tire, it's not clear to me that a scrub effect or reaction force against the ground is necessary to cause a curving trajectory when rolling across a flat surface.                 Date: 29 Oct 2007 23:35:34 From: Michael Press Subject: Re: Gyroscopic forces revisited In article <timmcn-D6ADDA.15563829102007@news.iphouse.com >, Tim McNamara <timmcn@bitstream.net > wrote: > In article <rubrum-E98E0B.12164429102007@newsclstr03.news.prodigy.net>, > Michael Press <rubrum@pacbell.net> wrote: > > > In article <timmcn-212E9A.08302229102007@news.iphouse.com>, > > Tim McNamara <timmcn@bitstream.net> wrote: > > > > > In article > > > <rubrum-15BE59.22014928102007@newsclstr03.news.prodigy.net>, > > > Michael Press <rubrum@pacbell.net> wrote: > > > > > > > In article <47252399$0$14061$742ec2ed@news.sonic.net>, > > > > jobst.brandt@stanfordalumni.org wrote: > > > > > > > > > Michael Press writes: > > > > > > > > > > >>> With a tire cross section of 0", what forces cause the tire > > > > > >>> to turn about point B? > > > > > > > > > > >> Good question :) I guess in theory a wheel of zero thickness > > > > > >> on an infinitely hard flat surface won't experience any > > > > > >> camber thrust. Thats kinda hypothetical tho :) > > > > > > > > > > > The camber thrust occurs even with zero thickness tires. It > > > > > > is a function of the radius of curvature of a wheel track of > > > > > > a leaned over wheel. > > > > > > > > > > How is this force expressed (as a vector and units) and how > > > > > does it occur when, for instance, rolling a pizza (disk) knife > > > > > horizontally in a straight line across a tilted surface. > > > > > > > > > > This quantity and its effect on bicycle cornering seems hard to > > > > > define. I am not understanding what this force is and where it > > > > > applies, especially in the example I mention above. > > > > > > > > A hoop rolling with its axis oblique to the roadway will turn in > > > > an arc. To turn there must be a reaction force at the contact > > > > patch. > > > > > > How about gravity? > > > > Gravity acts perpendicular to a flat road surface. The reaction force > > to make the hoop roll on an arc of a circle acts parallel to the road > > surface. > > But hoops with extremely low coefficients of friction, and hence little > reaction force (e.g., "scrub"), display this behavior. It looks to me > that it's due to gravity not a reaction force against the ground. If > the hoop can't tilt, as in a two track vehicle such as a car, then a > reaction force is obviously necessary. But for a hoop such as a coin or > a tire, it's not clear to me that a scrub effect or reaction force > against the ground is necessary to cause a curving trajectory when > rolling across a flat surface. This is a simple example of Newton's laws. To change the direction of a body in motion requires force. A hoop rolling on a curve on a flat plane has a force acting on it. I maintain that the force is a friction based reaction force at the contact patch. A hoop rolling on a frictionless surface will not roll in curve. If it is tilted the center of gravity will fall straight down. -- Michael Press     Date: 27 Oct 2007 06:14:30 From: Subject: Re: Gyroscopic forces revisited Jon Crouch writes: >> With a tire cross section of 0", what forces cause the tire to turn >> about point B? > Good question :) I guess in theory a wheel of zero thickness on an > infinitely hard flat surface won't experience any camber > thrust. Thats kinda hypothetical tho :) Well lets approach the limit in reality. Do you believe that a slick 60mm tire corners differently from a 23mm cross section tire? Where does the difference enter as an effect and is it a step function? A 23mm tire is essentially a zero width tire for the turn radii of interest. Jobst Brandt   Date: 27 Oct 2007 01:59:09 From: Jon_C Subject: Re: Gyroscopic forces revisited On Oct 26, 3:52 pm, jobst.bra...@stanfordalumni.org wrote: > Jon Crouch writes: > >> Tire slip is minuscule or we would see black marks on the road and > >> tires would wear out in a few hundred miles. Tire tread is only a > >> couple of millimeters thick at best. In contrast 25mm cross > >> section tires last for as much as 3000 miles with plenty of fast > >> descending. > > It's not that miniscule. A lone 29" wheel rolling at 45 degrees > > would naturally follow a 23" radius arc. > > I think you'll have to explain that in greater detail. When riding no > hands down high speed curves at around 40 degrees lean to the road, > there is no such steering force. a two foot (~23") radius is a pretty > severe curvature. OK I'll explain. A wheel on it's own rolling while at an angle to the ground will turn in a circle centered about the point on the ground where the line of it's axle intersects the ground, as in my pic: http://picasaweb.google.com/JonCCrouch/UntitledAlbum/photo#5125629114602972546 The reason the wheel turns that it's contact patch is at an angle to the wheel's axle so basically the wheel is acting as a cone. (the red part in my pic). This concept is called Camber Thrust. Obviously the wheel needs to be at a certain speed to achieve this or else centrifugal force or gravity will change the angle of the wheel and throw it into a weave or a different arc.. Anyhow, that's the natural turning radius of a wheel at a given angle to the ground and you can test it very easily with a kids hoop or a bike wheel. Now as you (and I) have said, a bicycle will take a much larger arc than described above when riding at speed but that doesn't mean that camber force isn't present. In fact it's still camber force that's making your bike turn but at the same time the bike's tyres are 'slipping' (as described above) causing the bile to take a much larger radius. > > > A rider might manage a 200 foot radius arc leaning at 45 degrees if > > he goes fast enough. Unless you want to deny the principal of > > camber thrust you have to accept that there;s a lot of slip > > happening there. > > I don't see where you get that. Please explain how this fits with > riding no-hands on a straight banked surface such as the edge of a > crowned road. I do this regularly and don't compensate for slope. > I think you do compensate. As I already said, I'm not sure how but unless you deny the effect of camber force you must acknowledge that the rider is doing something to compensate. > > Now slip is a slightly misleading word because the tyre isn't > > sliding as in a skid. It's actually 'walking' across road surface > > as it spins. The part of the tyre on the road stretches and then > > the wheel rolls on to an unstretched bit of tyre that then stretches > > too, letting the wheel creep outward without actually letting go of > > the road. > > That is how a slime filled flat tire does it, but it isn't sliding on > the road but rather internally like a tracked vehicle. The lateral > deformation of tread rubber is minuscule and recoverable minus > hysteretic loss in its rebound. There isn't a lot of scrubbing taking > place in a contact patch that is about a cm wide. Well the lateral deformation isn't recoverable because the tyre rolls onto an undeformed section which then deforms too and so on and so forth. I'm repeating myself tho. > > >>> Steering flop and contact patch drag will also come into play on a > >>> leaning no-handed bike but they're countered to a large degree by > >>> trail (shopping cart wheel effect). > > That is unclear what you mean by that. You bring effects in and > discard them as fast. What is happening? I simply mentioned that two well known steering forces are fairly negligible. Did that really need an explanation?? > > > >>> Moving the CoG of the rider will change the angle of the wheels > >>> and so alter the amount of camber thrust and contact patch > >>> offset/drag. That's about all the forces involved I think. None > >>> of them rely on gyroscopic inertia but as you say, inertia will > >>> give the wheel a tendency to remain stable and resist sudden > >>> changes in direction. > >> From this one might conclude that vehicles that don't lean can't > >> corner. Riding a bicycle no-hands on a straight crowned road > >> requires no lean and the "camber thrust" has no effect. This > >> "camber thrust" seems to be a new concept that came from the people > >> who recently discovered countersteer or ones who believe pushing > >> down on the inside handlebar improves cornering. > > I can't see how you could conclude that. I only mentioned the > > forces involved in a bicycle, I didn't say there were no other > > forces in the universe. Oh and camber thrust isn't a new concept by > > any means. Ask any penny-farthing rider. > > I can and he doesn't know anything about it. Maybe its a long lost > effect. The conclusion arises because you mentioned the forces that > steer the wheel in curves. The same effects should apply to non > leaning wheels and if they don't there must be a basic difference in > tracking curves for the two types. By the way, formerly cars had > positive camber on front wheels while today negative camber is used, > front and rear, so what are these differences? > Well I described turning forces that are caused by the lean of a wheel. Why should they apply to non-leaning wheels??? Yes there are basic differences in their tracking curves. Non-leaning (car) wheels have to be turned much more than a bike wheel travellig the same arc and speed. [ Positive/negative camber describes whether pairs of (car) wheels sharing the same axle point inwards or outwards at the top. It's not applicable to bicycles. ] > > As for riding straight on cambered surfaces, I'm still not sure how > > we manage that. The laws of physics tell us there has to be some > > degree of camber thrust but how us humans counter it is a hard one > > to answer. I'm still leaning toward subtle weaving and weight > > shifting. It might be something that feels so instinctive that we > > don;t even notice doing it. > > Here's a good article on steering forces: > > http://www.tonyfoale.com/Articles/Tyres/TYRES.htm > > "When Newton first expounded to the world his theories of mechanics, > no doubt he had on his mind, things other than the interaction of > motorcycle tyres with the road surface." > > That's not a good start and doesn't explain the "less obvious" parts > of this treatise. He doesn't add anything to what you claim in this > thread. > > Jobst Brandt Google some more my friend. these are all commonly accepted principles.     Date: 27 Oct 2007 06:10:07 From: Subject: Re: Gyroscopic forces revisited Jon Crouch writes: >>>> Tire slip is minuscule or we would see black marks on the road >>>> and tires would wear out in a few hundred miles. Tire tread is >>>> only a couple of millimeters thick at best. In contrast 25mm >>>> cross section tires last for as much as 3000 miles with plenty of >>>> fast descending. >>> It's not that minuscule. A lone 29" wheel rolling at 45 degrees >>> would naturally follow a 23" radius arc. >> I think you'll have to explain that in greater detail. When riding >> no hands down high speed curves at around 40 degrees lean to the >> road, there is no such steering force. a two foot (~23") radius is >> a pretty severe curvature. > OK I'll explain. A wheel on it's own rolling while at an angle to > the ground will turn in a circle centered about the point on the > ground where the line of it's axle intersects the ground, as in my > pic: http://picasaweb.google.com/JonCCrouch/UntitledAlbum/photo#5125629114602972546 > The reason the wheel turns that it's contact patch is at an angle to > the wheel's axle so basically the wheel is acting as a cone. (the > red part in my pic). This concept is called Camber Thrust. That's a nice picture but if you roll a wheel out on a parking lot, you'll notice that it tightens the curve from practically straight ahead to finally doing a wobble flop onto its side. Those are a continuum of curves defines by speed and resultant lateral acceleration. They are not defined by the lean angle but rather the lean angle is defined by the speed which equals lateral acceleration. > Obviously the wheel needs to be at a certain speed to achieve this > or else centrifugal force or gravity will change the angle of the > wheel and throw it into a weave or a different arc.. Anyhow, that's > the natural turning radius of a wheel at a given angle to the ground > and you can test it very easily with a kids hoop or a bike wheel. We aren't riding a single wheel alone but operating a bicycle that does not act as a single wheel. > Now as you (and I) have said, a bicycle will take a much larger arc > than described above when riding at speed but that doesn't mean that > camber force isn't present. In fact it's still camber force that's > making your bike turn but at the same time the bike's tyres are > 'slipping' (as described above) causing the bile to take a much > larger radius. The lean angle is defines by the centrifugal acceleration and the curvature, not the converse. That is where I believe you have the cart before the horse. >>> A rider might manage a 200 foot radius arc leaning at 45 degrees >>> if he goes fast enough. Unless you want to deny the principal of >>> camber thrust you have to accept that there's a lot of slip >>> happening there. >> I don't see where you get that. Please explain how this fits with >> riding no-hands on a straight banked surface such as the edge of a >> crowned road. I do this regularly and don't compensate for slope. > I think you do compensate. As I already said, I'm not sure how but > unless you deny the effect of camber force you must acknowledge that > the rider is doing something to compensate. >>> Now slip is a slightly misleading word because the tyre isn't >>> sliding as in a skid. It's actually 'walking' across road surface >>> as it spins. The part of the tyre on the road stretches and then >>> the wheel rolls on to an unstretched bit of tyre that then >>> stretches too, letting the wheel creep outward without actually >>> letting go of the road. >> That is how a slime filled flat tire does it, but it isn't sliding >> on the road but rather internally like a tracked vehicle. The >> lateral deformation of tread rubber is minuscule and recoverable >> minus hysteretic loss in its rebound. There isn't a lot of >> scrubbing taking place in a contact patch that is about a cm wide. > Well the lateral deformation isn't recoverable because the tyre > rolls onto an undeformed section which then deforms too and so on > and so forth. I'm repeating myself tho. >>>>> Steering flop and contact patch drag will also come into play on >>>>> a leaning no-handed bike but they're countered to a large degree >>>>> by trail (shopping cart wheel effect). >> That is unclear what you mean by that. You bring effects in and >> discard them as fast. What is happening? > I simply mentioned that two well known steering forces are fairly > negligible. Did that really need an explanation?? >>>>> Moving the CoG of the rider will change the angle of the wheels >>>>> and so alter the amount of camber thrust and contact patch >>>>> offset/drag. That's about all the forces involved I think. >>>>> None of them rely on gyroscopic inertia but as you say, inertia >>>>> will give the wheel a tendency to remain stable and resist >>>>> sudden changes in direction. >>>> From this one might conclude that vehicles that don't lean can't >>>> corner. Riding a bicycle no-hands on a straight crowned road >>>> requires no lean and the "camber thrust" has no effect. This >>>> "camber thrust" seems to be a new concept that came from the >>>> people who recently discovered countersteer or ones who believe >>>> pushing down on the inside handlebar improves cornering. >>> I can't see how you could conclude that. I only mentioned the >>> forces involved in a bicycle, I didn't say there were no other >>> forces in the universe. Oh and camber thrust isn't a new concept >>> by any means. Ask any penny-farthing rider. >> I can and he doesn't know anything about it. Maybe its a long lost >> effect. The conclusion arises because you mentioned the forces >> that steer the wheel in curves. The same effects should apply to >> non leaning wheels and if they don't there must be a basic >> difference in tracking curves for the two types. By the way, >> formerly cars had positive camber on front wheels while today >> negative camber is used, front and rear, so what are these >> differences? > Well I described turning forces that are caused by the lean of a > wheel. Why should they apply to non-leaning wheels??? Yes there are > basic differences in their tracking curves. Non-leaning (car) > wheels have to be turned much more than a bike wheel traveling the > same arc and speed. > [Positive/negative camber describes whether pairs of (car) wheels > sharing the same axle point inward or outward at the top. It's not > applicable to bicycles.] >>> As for riding straight on cambered surfaces, I'm still not sure >>> how we manage that. The laws of physics tell us there has to be >>> some degree of camber thrust but how us humans counter it is a >>> hard one to answer. I'm still leaning toward subtle weaving and >>> weight shifting. It might be something that feels so instinctive >>> that we don;t even notice doing it. >>> Here's a good article on steering forces: >> http://www.tonyfoale.com/Articles/Tyres/TYRES.htm >> "When Newton first expounded to the world his theories of >> mechanics, no doubt he had on his mind, things other than the >> interaction of motorcycle tyres with the road surface." >> That's not a good start and doesn't explain the "less obvious" >> parts of this treatise. He doesn't add anything to what you claim >> in this thread. > Google some more my friend. These are all commonly accepted > principles. Just because they are repeated among certain groups, does not make them correct or true. It's like the rotating mass on a bicycle is more important than static mass on the frame. That one survived many years in these newsgroups only because it was repeated so often. Jobst Brandt   Date: 27 Oct 2007 01:05:51 From: Jon_C Subject: Re: Gyroscopic forces revisited On Oct 26, 2:09 pm, "Kerry Montgomery" <kamon...@teleport.com > wrote: > Jon, > How did you calculate that a lone 29" wheel at 45 degrees naturally follows > a 23" arc? Did you assume a particular tire cross section? If taken to an > extreme, I think a sphere mounted on an axle will always roll straight > ahead, as long as the axle angle doesn't reach 90 degrees. > Thanks, > Kerry Sorry, 20.5" to be exactish. It's the distance from the contact patch to the point where the acle line bisects the ground. [ 29 * sin(45) ] I think you';re wrong about the sphere. It will roll in circles around the point at which it's axle line bisects the ground too.   Date: 26 Oct 2007 12:25:11 From: joseph.santaniello@gmail.com Subject: Re: Gyroscopic forces revisited On Oct 26, 8:51 pm, Tim McNamara <tim...@bitstream.net > wrote: > In article <dabac.2z2...@no-mx.forums.cyclingforums.com>, > > > > dabac <dabac.2z2...@no-mx.forums.cyclingforums.com> wrote: > > Joseph Santaniello Wrote: > > > > > I have witnessed descending in a similar fashion, but with both > > > > hands behind the rider's butt. This position is significantly > > > > more aerodynamic, as evidenced by the riders marked speed > > > > increase upon moving the arms back like that. > > > jobst.bra...@stanfordalumni.org Wrote: > > > > I doubt that this does what you say it does because frontal area is > > > what causes drag, the closing area behind the rider being a bluff > > > body and has no streamlining (as a long pointed tail). Frontal > > > area is smaller with the shoulders pulled in in front of the rider > > > with arms in chest, elbows on knees. > > > Then why does the tuck used by ski jumpers and speed skaters look the > > way it does? According to your line of reasoning they should be > > faster if they kept their arms against the chest. How strange that > > none of those guys have discovered such an easy way to gain an > > edge... > > Never having speed skated or ski jumped, I am speculating here. My > understanding about ski jumping is that the position they adopt is one > intended to allow small corrections in midair. My understanding about > the speed skating position is that it's about ergonomics as well as > aerodynamics. > > Aerodynamics is only part of the equation. I recall reading about > Miguel Indurain's trip to the wind tunnel. They found a position that > would be much faster than the one he usually used. He then pointed out > that he couldn't actually pedal the bike in that position. Ski jumping has the ramp where speed is built up, and flying where lift is needed. Maybe a super-man position would allow greater speed on the ramp. That's actually how they did it in the old days. Once in the air they would windmill their arms for stability. I speed skate, but I suck, so take this for what it's worth: The arms back allows the CG to be a little back. With the CG too far forward, the tips dig in and it is hard to keep a nice low crouch. The arms do come forward alternatingly when you are really trying to give it some stick, but only a little. Joseph   Date: 26 Oct 2007 12:17:44 From: joseph.santaniello@gmail.com Subject: Re: Gyroscopic forces revisited On Oct 26, 7:52 pm, Jon_C <JonCCro...@gmail.com > wrote: > On Oct 26, 12:26 pm, jobst.bra...@stanfordalumni.org wrote: > > > Tire slip is minuscule or we would see black marks on the road and > > tires would wear out in a few hundred miles. Tire tread is only a > > couple of millimeters thick at best. In contrast 25mm cross section > > tires last for as much as 3000 miles with plenty of fast descending. > > It's not that miniscule. A lone 29" wheel rolling at 45 degrees would > naturally follow a 23" radius arc. A rider might manage a 200 foot > radius arc leaning at 45 degrees if he goes fast enough. Unless you > want to deny the principal of camber thrust you have to accept that > there;s a lot of slip happening there. Now slip is a slightly > misleading word because the tyre isn't sliding as in a skid. It's > actually 'walking' across road surface as it spins. The part of the > tyre on the road stretches and then the wheel rolls on to an > unstretched bit of tyre that then stretches too, letting the wheel > creep outwards without actually letting go of the road. > > > > > > Steering flop and contact patch drag will also come into play on a > > > leaning no-handed bike but they're countered to a large degree by > > > trail (shopping cart wheel effect). > > > Moving the CoG of the rider will change the angle of the wheels and > > > so alter the amount of camber thrust and contact patch offset/drag. > > > That's about all the forces involved I think. None of them rely on > > > gyroscopic inertia but as you say, inertia will give the wheel a > > > tendency to remain stable and resist sudden changes in direction. > > > From this one might conclude that vehicles that don't lean can't > > corner. Riding a bicycle no-hands on a straight crowned road requires > > no lean and the "camber thrust" has no effect. This "camber thrust" > > seems to be a new concept that came from the people who recently > > discovered countersteer or ones who believe pushing down on the inside > > handlebar improves cornering. > > > Jobst Brandt > > I can't see how you could conclude that. I only mentioned the forces > involved in a bicycle, I didn't say there were no other forces in the > universe. Oh and camber thrust isn't a new concept by any means. Ask > any penny-farthing rider. > > As for riding straight on cambered surfaces, I'm still not sure how we > manage that. The laws of physics tell us there has to be some degree > of camber thrust but how us humans counter it is a hard one to answer. > I'm still leaning towards subtle weaving and weight shifting. It might > be something that feels so instinctive that we don;t even notice doing > it. > > Here's a good article on steering forces:http://www.tonyfoale.com/Articles/Tyres/TYRES.htm Riding straight on cambered surfaces is possible because the steering is straight due to the force of gravity working straight down through the bike centerline, and this with trail makes there be no tendency for the steering to flop one way or the other. When both wheels are straight ahead (no steering) in the same line, the two camber-thrust arcs work such that the forces they generate are parallel, pointing up the slope, canceling each other out in a way. Joseph   Date: 26 Oct 2007 17:52:43 From: Jon_C Subject: Re: Gyroscopic forces revisited On Oct 26, 12:26 pm, jobst.bra...@stanfordalumni.org wrote: > Tire slip is minuscule or we would see black marks on the road and > tires would wear out in a few hundred miles. Tire tread is only a > couple of millimeters thick at best. In contrast 25mm cross section > tires last for as much as 3000 miles with plenty of fast descending. It's not that miniscule. A lone 29" wheel rolling at 45 degrees would naturally follow a 23" radius arc. A rider might manage a 200 foot radius arc leaning at 45 degrees if he goes fast enough. Unless you want to deny the principal of camber thrust you have to accept that there;s a lot of slip happening there. Now slip is a slightly misleading word because the tyre isn't sliding as in a skid. It's actually 'walking' across road surface as it spins. The part of the tyre on the road stretches and then the wheel rolls on to an unstretched bit of tyre that then stretches too, letting the wheel creep outwards without actually letting go of the road. > > Steering flop and contact patch drag will also come into play on a > > leaning no-handed bike but they're countered to a large degree by > > trail (shopping cart wheel effect). > > Moving the CoG of the rider will change the angle of the wheels and > > so alter the amount of camber thrust and contact patch offset/drag. > > That's about all the forces involved I think. None of them rely on > > gyroscopic inertia but as you say, inertia will give the wheel a > > tendency to remain stable and resist sudden changes in direction. > > From this one might conclude that vehicles that don't lean can't > corner. Riding a bicycle no-hands on a straight crowned road requires > no lean and the "camber thrust" has no effect. This "camber thrust" > seems to be a new concept that came from the people who recently > discovered countersteer or ones who believe pushing down on the inside > handlebar improves cornering. > > Jobst Brandt I can't see how you could conclude that. I only mentioned the forces involved in a bicycle, I didn't say there were no other forces in the universe. Oh and camber thrust isn't a new concept by any means. Ask any penny-farthing rider. As for riding straight on cambered surfaces, I'm still not sure how we manage that. The laws of physics tell us there has to be some degree of camber thrust but how us humans counter it is a hard one to answer. I'm still leaning towards subtle weaving and weight shifting. It might be something that feels so instinctive that we don;t even notice doing it. Here's a good article on steering forces: http://www.tonyfoale.com/Articles/Tyres/TYRES.htm     Date: 26 Oct 2007 19:52:08 From: Subject: Re: Gyroscopic forces revisited Jon Crouch writes: >> Tire slip is minuscule or we would see black marks on the road and >> tires would wear out in a few hundred miles. Tire tread is only a >> couple of millimeters thick at best. In contrast 25mm cross >> section tires last for as much as 3000 miles with plenty of fast >> descending. > It's not that miniscule. A lone 29" wheel rolling at 45 degrees > would naturally follow a 23" radius arc. I think you'll have to explain that in greater detail. When riding no hands down high speed curves at around 40 degrees lean to the road, there is no such steering force. a two foot (~23") radius is a pretty severe curvature. > A rider might manage a 200 foot radius arc leaning at 45 degrees if > he goes fast enough. Unless you want to deny the principal of > camber thrust you have to accept that there;s a lot of slip > happening there. I don't see where you get that. Please explain how this fits with riding no-hands on a straight banked surface such as the edge of a crowned road. I do this regularly and don't compensate for slope. > Now slip is a slightly misleading word because the tyre isn't > sliding as in a skid. It's actually 'walking' across road surface > as it spins. The part of the tyre on the road stretches and then > the wheel rolls on to an unstretched bit of tyre that then stretches > too, letting the wheel creep outward without actually letting go of > the road. That is how a slime filled flat tire does it, but it isn't sliding on the road but rather internally like a tracked vehicle. The lateral deformation of tread rubber is minuscule and recoverable minus hysteretic loss in its rebound. There isn't a lot of scrubbing taking place in a contact patch that is about a cm wide. >>> Steering flop and contact patch drag will also come into play on a >>> leaning no-handed bike but they're countered to a large degree by >>> trail (shopping cart wheel effect). That is unclear what you mean by that. You bring effects in and discard them as fast. What is happening? >>> Moving the CoG of the rider will change the angle of the wheels >>> and so alter the amount of camber thrust and contact patch >>> offset/drag. That's about all the forces involved I think. None >>> of them rely on gyroscopic inertia but as you say, inertia will >>> give the wheel a tendency to remain stable and resist sudden >>> changes in direction. >> From this one might conclude that vehicles that don't lean can't >> corner. Riding a bicycle no-hands on a straight crowned road >> requires no lean and the "camber thrust" has no effect. This >> "camber thrust" seems to be a new concept that came from the people >> who recently discovered countersteer or ones who believe pushing >> down on the inside handlebar improves cornering. > I can't see how you could conclude that. I only mentioned the > forces involved in a bicycle, I didn't say there were no other > forces in the universe. Oh and camber thrust isn't a new concept by > any means. Ask any penny-farthing rider. I can and he doesn't know anything about it. Maybe its a long lost effect. The conclusion arises because you mentioned the forces that steer the wheel in curves. The same effects should apply to non leaning wheels and if they don't there must be a basic difference in tracking curves for the two types. By the way, formerly cars had positive camber on front wheels while today negative camber is used, front and rear, so what are these differences? > As for riding straight on cambered surfaces, I'm still not sure how > we manage that. The laws of physics tell us there has to be some > degree of camber thrust but how us humans counter it is a hard one > to answer. I'm still leaning toward subtle weaving and weight > shifting. It might be something that feels so instinctive that we > don;t even notice doing it. > Here's a good article on steering forces: http://www.tonyfoale.com/Articles/Tyres/TYRES.htm "When Newton first expounded to the world his theories of mechanics, no doubt he had on his mind, things other than the interaction of motorcycle tyres with the road surface." That's not a good start and doesn't explain the "less obvious" parts of this treatise. He doesn't add anything to what you claim in this thread. Jobst Brandt     Date: 26 Oct 2007 11:09:16 From: Kerry Montgomery Subject: Re: Gyroscopic forces revisited "Jon_C" <JonCCrouch@gmail.com > wrote in message news:1193421163.811684.5450@v3g2000hsg.googlegroups.com... > On Oct 26, 12:26 pm, jobst.bra...@stanfordalumni.org wrote: > >> Tire slip is minuscule or we would see black marks on the road and >> tires would wear out in a few hundred miles. Tire tread is only a >> couple of millimeters thick at best. In contrast 25mm cross section >> tires last for as much as 3000 miles with plenty of fast descending. > > It's not that miniscule. A lone 29" wheel rolling at 45 degrees would > naturally follow a 23" radius arc. A rider might manage a 200 foot > radius arc leaning at 45 degrees if he goes fast enough. Unless you > want to deny the principal of camber thrust you have to accept that > there;s a lot of slip happening there. Now slip is a slightly > misleading word because the tyre isn't sliding as in a skid. It's > actually 'walking' across road surface as it spins. The part of the > tyre on the road stretches and then the wheel rolls on to an > unstretched bit of tyre that then stretches too, letting the wheel > creep outwards without actually letting go of the road. > >> > Steering flop and contact patch drag will also come into play on a >> > leaning no-handed bike but they're countered to a large degree by >> > trail (shopping cart wheel effect). >> > Moving the CoG of the rider will change the angle of the wheels and >> > so alter the amount of camber thrust and contact patch offset/drag. >> > That's about all the forces involved I think. None of them rely on >> > gyroscopic inertia but as you say, inertia will give the wheel a >> > tendency to remain stable and resist sudden changes in direction. >> >> From this one might conclude that vehicles that don't lean can't >> corner. Riding a bicycle no-hands on a straight crowned road requires >> no lean and the "camber thrust" has no effect. This "camber thrust" >> seems to be a new concept that came from the people who recently >> discovered countersteer or ones who believe pushing down on the inside >> handlebar improves cornering. >> >> Jobst Brandt > > I can't see how you could conclude that. I only mentioned the forces > involved in a bicycle, I didn't say there were no other forces in the > universe. Oh and camber thrust isn't a new concept by any means. Ask > any penny-farthing rider. > > As for riding straight on cambered surfaces, I'm still not sure how we > manage that. The laws of physics tell us there has to be some degree > of camber thrust but how us humans counter it is a hard one to answer. > I'm still leaning towards subtle weaving and weight shifting. It might > be something that feels so instinctive that we don;t even notice doing > it. > > Here's a good article on steering forces: > http://www.tonyfoale.com/Articles/Tyres/TYRES.htm > Jon, How did you calculate that a lone 29" wheel at 45 degrees naturally follows a 23" arc? Did you assume a particular tire cross section? If taken to an extreme, I think a sphere mounted on an axle will always roll straight ahead, as long as the axle angle doesn't reach 90 degrees. Thanks, Kerry       Date: 26 Oct 2007 22:18:43 From: Michael Press Subject: Re: Gyroscopic forces revisited In article <13i4bai49lol74@corp.supernews.com >, "Kerry Montgomery" <kamontgo@teleport.com > wrote: > "Jon_C" <JonCCrouch@gmail.com> wrote in message > news:1193421163.811684.5450@v3g2000hsg.googlegroups.com... > > On Oct 26, 12:26 pm, jobst.bra...@stanfordalumni.org wrote: > > > >> Tire slip is minuscule or we would see black marks on the road and > >> tires would wear out in a few hundred miles. Tire tread is only a > >> couple of millimeters thick at best. In contrast 25mm cross section > >> tires last for as much as 3000 miles with plenty of fast descending. > > > > It's not that miniscule. A lone 29" wheel rolling at 45 degrees would > > naturally follow a 23" radius arc. A rider might manage a 200 foot > > radius arc leaning at 45 degrees if he goes fast enough. Unless you > > want to deny the principal of camber thrust you have to accept that > > there;s a lot of slip happening there. Now slip is a slightly > > misleading word because the tyre isn't sliding as in a skid. It's > > actually 'walking' across road surface as it spins. The part of the > > tyre on the road stretches and then the wheel rolls on to an > > unstretched bit of tyre that then stretches too, letting the wheel > > creep outwards without actually letting go of the road. > > > >> > Steering flop and contact patch drag will also come into play on a > >> > leaning no-handed bike but they're countered to a large degree by > >> > trail (shopping cart wheel effect). > >> > Moving the CoG of the rider will change the angle of the wheels and > >> > so alter the amount of camber thrust and contact patch offset/drag. > >> > That's about all the forces involved I think. None of them rely on > >> > gyroscopic inertia but as you say, inertia will give the wheel a > >> > tendency to remain stable and resist sudden changes in direction. > >> > >> From this one might conclude that vehicles that don't lean can't > >> corner. Riding a bicycle no-hands on a straight crowned road requires > >> no lean and the "camber thrust" has no effect. This "camber thrust" > >> seems to be a new concept that came from the people who recently > >> discovered countersteer or ones who believe pushing down on the inside > >> handlebar improves cornering. > >> > >> Jobst Brandt > > > > I can't see how you could conclude that. I only mentioned the forces > > involved in a bicycle, I didn't say there were no other forces in the > > universe. Oh and camber thrust isn't a new concept by any means. Ask > > any penny-farthing rider. > > > > As for riding straight on cambered surfaces, I'm still not sure how we > > manage that. The laws of physics tell us there has to be some degree > > of camber thrust but how us humans counter it is a hard one to answer. > > I'm still leaning towards subtle weaving and weight shifting. It might > > be something that feels so instinctive that we don;t even notice doing > > it. > > > > Here's a good article on steering forces: > > http://www.tonyfoale.com/Articles/Tyres/TYRES.htm > > > Jon, > How did you calculate that a lone 29" wheel at 45 degrees naturally follows > a 23" arc? Did you assume a particular tire cross section? If taken to an > extreme, I think a sphere mounted on an axle will always roll straight > ahead, as long as the axle angle doesn't reach 90 degrees. Right triangle ABC with the right angle at A. C is the contact patch A is the center of the axle. B is the intersection of the hub axis with the road surface. CA = 29"/2. angle ABC = 45 deg. Therefore the radius of curvature of the turn is BC = CA/sin(angle ABC) = (29"/2)/(sin(45 deg)) = 20" 1/2. I do not know where 23" comes from. -- Michael Press         Date: 26 Oct 2007 16:18:33 From: Kerry Montgomery Subject: Re: Gyroscopic forces revisited "Michael Press" <rubrum@pacbell.net > wrote in message news:rubrum-F5FD56.15184426102007@newsclstr02.news.prodigy.com... > In article <13i4bai49lol74@corp.supernews.com>, > "Kerry Montgomery" <kamontgo@teleport.com> wrote: > >> "Jon_C" <JonCCrouch@gmail.com> wrote in message >> news:1193421163.811684.5450@v3g2000hsg.googlegroups.com... >> > On Oct 26, 12:26 pm, jobst.bra...@stanfordalumni.org wrote: >> > >> >> Tire slip is minuscule or we would see black marks on the road and >> >> tires would wear out in a few hundred miles. Tire tread is only a >> >> couple of millimeters thick at best. In contrast 25mm cross section >> >> tires last for as much as 3000 miles with plenty of fast descending. >> > >> > It's not that miniscule. A lone 29" wheel rolling at 45 degrees would >> > naturally follow a 23" radius arc. A rider might manage a 200 foot >> > radius arc leaning at 45 degrees if he goes fast enough. Unless you >> > want to deny the principal of camber thrust you have to accept that >> > there;s a lot of slip happening there. Now slip is a slightly >> > misleading word because the tyre isn't sliding as in a skid. It's >> > actually 'walking' across road surface as it spins. The part of the >> > tyre on the road stretches and then the wheel rolls on to an >> > unstretched bit of tyre that then stretches too, letting the wheel >> > creep outwards without actually letting go of the road. >> > >> >> > Steering flop and contact patch drag will also come into play on a >> >> > leaning no-handed bike but they're countered to a large degree by >> >> > trail (shopping cart wheel effect). >> >> > Moving the CoG of the rider will change the angle of the wheels and >> >> > so alter the amount of camber thrust and contact patch offset/drag. >> >> > That's about all the forces involved I think. None of them rely on >> >> > gyroscopic inertia but as you say, inertia will give the wheel a >> >> > tendency to remain stable and resist sudden changes in direction. >> >> >> >> From this one might conclude that vehicles that don't lean can't >> >> corner. Riding a bicycle no-hands on a straight crowned road requires >> >> no lean and the "camber thrust" has no effect. This "camber thrust" >> >> seems to be a new concept that came from the people who recently >> >> discovered countersteer or ones who believe pushing down on the inside >> >> handlebar improves cornering. >> >> >> >> Jobst Brandt >> > >> > I can't see how you could conclude that. I only mentioned the forces >> > involved in a bicycle, I didn't say there were no other forces in the >> > universe. Oh and camber thrust isn't a new concept by any means. Ask >> > any penny-farthing rider. >> > >> > As for riding straight on cambered surfaces, I'm still not sure how we >> > manage that. The laws of physics tell us there has to be some degree >> > of camber thrust but how us humans counter it is a hard one to answer. >> > I'm still leaning towards subtle weaving and weight shifting. It might >> > be something that feels so instinctive that we don;t even notice doing >> > it. >> > >> > Here's a good article on steering forces: >> > http://www.tonyfoale.com/Articles/Tyres/TYRES.htm >> > >> Jon, >> How did you calculate that a lone 29" wheel at 45 degrees naturally >> follows >> a 23" arc? Did you assume a particular tire cross section? If taken to an >> extreme, I think a sphere mounted on an axle will always roll straight >> ahead, as long as the axle angle doesn't reach 90 degrees. > > Right triangle ABC with the right angle at A. > C is the contact patch > A is the center of the axle. > B is the intersection of the hub axis with the road surface. > CA = 29"/2. > angle ABC = 45 deg. > Therefore the radius of curvature of the turn is > BC = CA/sin(angle ABC) = (29"/2)/(sin(45 deg)) = 20" 1/2. > > I do not know where 23" comes from. > > -- Michael, Thanks for the description of a diagram of the wheel leaned over. I get the your same value for BC, assuming a tire cross section of 0". If the tire has a cross section larger than 0", the diagram gets more complicated. As the tire cross section grows, no single point C can satisfy the conditions of 1) being at 29"/2 from A and forming a 90 degree angle CAB 2) being the point at which the tire touches the road. Starting with the drawing from your description, adding a circle whose center is on segment AC and the circumference of which passes through point C, a portion of that circumference lies below the surface of the road. With a tire cross section of 0", what forces cause the tire to turn about point B? Thanks, Kerry           Date: 28 Oct 2007 23:17:22 From: Michael Press Subject: Re: Gyroscopic forces revisited In article <13i4tefqks8lb77@corp.supernews.com >, "Kerry Montgomery" <kamontgo@teleport.com > wrote: > "Michael Press" <rubrum@pacbell.net> wrote in message > news:rubrum-F5FD56.15184426102007@newsclstr02.news.prodigy.com... > > In article <13i4bai49lol74@corp.supernews.com>, > > "Kerry Montgomery" <kamontgo@teleport.com> wrote: > > > >> "Jon_C" <JonCCrouch@gmail.com> wrote in message > >> news:1193421163.811684.5450@v3g2000hsg.googlegroups.com... > >> > On Oct 26, 12:26 pm, jobst.bra...@stanfordalumni.org wrote: > >> > > >> >> Tire slip is minuscule or we would see black marks on the road and > >> >> tires would wear out in a few hundred miles. Tire tread is only a > >> >> couple of millimeters thick at best. In contrast 25mm cross section > >> >> tires last for as much as 3000 miles with plenty of fast descending. > >> > > >> > It's not that miniscule. A lone 29" wheel rolling at 45 degrees would > >> > naturally follow a 23" radius arc. A rider might manage a 200 foot > >> > radius arc leaning at 45 degrees if he goes fast enough. Unless you > >> > want to deny the principal of camber thrust you have to accept that > >> > there;s a lot of slip happening there. Now slip is a slightly > >> > misleading word because the tyre isn't sliding as in a skid. It's > >> > actually 'walking' across road surface as it spins. The part of the > >> > tyre on the road stretches and then the wheel rolls on to an > >> > unstretched bit of tyre that then stretches too, letting the wheel > >> > creep outwards without actually letting go of the road. > >> > > >> >> > Steering flop and contact patch drag will also come into play on a > >> >> > leaning no-handed bike but they're countered to a large degree by > >> >> > trail (shopping cart wheel effect). > >> >> > Moving the CoG of the rider will change the angle of the wheels and > >> >> > so alter the amount of camber thrust and contact patch offset/drag. > >> >> > That's about all the forces involved I think. None of them rely on > >> >> > gyroscopic inertia but as you say, inertia will give the wheel a > >> >> > tendency to remain stable and resist sudden changes in direction. > >> >> > >> >> From this one might conclude that vehicles that don't lean can't > >> >> corner. Riding a bicycle no-hands on a straight crowned road requires > >> >> no lean and the "camber thrust" has no effect. This "camber thrust" > >> >> seems to be a new concept that came from the people who recently > >> >> discovered countersteer or ones who believe pushing down on the inside > >> >> handlebar improves cornering. > >> >> > >> >> Jobst Brandt > >> > > >> > I can't see how you could conclude that. I only mentioned the forces > >> > involved in a bicycle, I didn't say there were no other forces in the > >> > universe. Oh and camber thrust isn't a new concept by any means. Ask > >> > any penny-farthing rider. > >> > > >> > As for riding straight on cambered surfaces, I'm still not sure how we > >> > manage that. The laws of physics tell us there has to be some degree > >> > of camber thrust but how us humans counter it is a hard one to answer. > >> > I'm still leaning towards subtle weaving and weight shifting. It might > >> > be something that feels so instinctive that we don;t even notice doing > >> > it. > >> > > >> > Here's a good article on steering forces: > >> > http://www.tonyfoale.com/Articles/Tyres/TYRES.htm > >> > > >> Jon, > >> How did you calculate that a lone 29" wheel at 45 degrees naturally > >> follows > >> a 23" arc? Did you assume a particular tire cross section? If taken to an > >> extreme, I think a sphere mounted on an axle will always roll straight > >> ahead, as long as the axle angle doesn't reach 90 degrees. > > > > Right triangle ABC with the right angle at A. > > C is the contact patch > > A is the center of the axle. > > B is the intersection of the hub axis with the road surface. > > CA = 29"/2. > > angle ABC = 45 deg. > > Therefore the radius of curvature of the turn is > > BC = CA/sin(angle ABC) = (29"/2)/(sin(45 deg)) = 20" 1/2. > > > > I do not know where 23" comes from. > > Michael, > Thanks for the description of a diagram of the wheel leaned over. I get the > your same value for BC, assuming a tire cross section of 0". > If the tire has a cross section larger than 0", the diagram gets more > complicated. As the tire cross section grows, no single point C can satisfy > the conditions of 1) being at 29"/2 from A and forming a 90 degree angle CAB > 2) being the point at which the tire touches the road. Starting with the > drawing from your description, adding a circle whose center is on segment AC > and the circumference of which passes through point C, a portion of that > circumference lies below the surface of the road. > With a tire cross section of 0", what forces cause the tire to turn about > point B? The tire width is about 1 or 2 cm. The change in the radius of curvature will be about one tire width plus about dR/sin(45 deg), where R is the radius of the wheel. It looks like the change in radius of curvature will be about -2 cm. -- Michael Press   Date: 26 Oct 2007 14:25:31 From: Jon_C Subject: Re: Gyroscopic forces revisited On Oct 26, 9:48 am, "joseph.santanie...@gmail.com" <joseph.santanie...@gmail.com > wrote: > On Oct 26, 3:18 pm, Jon_C <JonCCro...@gmail.com> wrote: > > > > > On Oct 24, 11:04 am, "joseph.santanie...@gmail.com" > > > <joseph.santanie...@gmail.com> wrote: > > > On Oct 24, 2:40 pm, Jon_C <JonCCro...@gmail.com> wrote: > > > > > On Oct 23, 10:51 pm, jobst.bra...@stanfordalumni.org wrote: > > > > > > Jon Crouch writes: > > > > > > How about a wheel on it's own? If you roll a wheel and it leans= over > > > > > > it turns in that direction. The reason is that the contact patc= h of > > > > > > a leaning tyre is parallel to the ground and so at an angle to = the > > > > > > wheel's axle so the wheel acts as a rolling cone and turns into= the > > > > > > corner (which maintains it's balance). I'm pretty sure that's t= he > > > > > > main steering force involved in no-handed riding. > > > > > > (I seem to remember reading about someone testing this on a > > > > > > motorbike and finding that on a prolonged corners the handlebar= s are > > > > > > actually turned slightly outward instead of inward as you;d > > > > > > expect. I think that was due to rake/trail but I don;t have tim= e to > > > > > > think about that right now :) > > > > > > I think you ought to go to a velodrome. These tracks have sloped > > > > > straights and more steeply banked circular curves at each end. R= iders > > > > > have no problem riding no-hands on these tracks, either on the > > > > > straight sections or in the banked curve (if they are going fast > > > > > enough to not strike a pedal). > > > > > > If you don't have a track, try a crowned paved street and notice = how > > > > > the angle with the pavement has essentially no effect on steering= , the > > > > > centerline of the contact patch moving less than 1/8" off center = for a > > > > > 10=B0 side slope. > > > > > > Jobst Brandt > > > > > To go in a straight line on an angled surface you'd have to keep the > > > > steering pointed slightly down the slope (ie. left on a velodrome > > > > straight) to counter-act the steering force from the slope. Maybe a > > > > slight weave-like cycle occurs where the rightward steering force f= rom > > > > the slope causes the rider to 'fall' left (centrifugal force) which > > > > causes the bike to steer back to the left and so on. Aslo, shifting > > > > the rider's center of gravity to the right will lean the bike frame= to > > > > the left slightly, causing the front wheel to 'flop' (and steer) le= ft > > > > countering camber thrust from the slope. > > > > > I don't think the 1/8" offset of the contact patch is too relevant > > > > btw. Camber thrust is due the angle between the wheel's axle and the > > > > ground (cone effect) not contact patch offset. > > > > I don't think you do need have the steering anything but straight on a > > > sloped surface. Not having a velodrome handy, that's why I broke out > > > my trusty plywood board. I too thought the angle of the wheels on the > > > surface had something to do with it, but it doesn't. I put my bike on > > > the board and fiddled with it quite a bit, and the angle of contact > > > alone does not appear to matter. What does matter is where the center > > > of mass is relative to the contact patches and the combined gravity > > > and turn forces. When the center of mass is along the centerline of > > > the bike and these forces are acting through it to the patches, the > > > steering satys straight no matter the angle of the wheels to the road. > > > > Using a sloped board you can simulate the forces on a riderless bike > > > during a turn. (On Mars...) > > > > Perhaps some minimum speed is necessary to not fall over riding on a > > > sloped velodrome, but I suspect this is from requiring a certain > > > amount of momentum to be able to "climb" the banking everytime you > > > make a steering correction up-track. > > > > Joseph > > > Joseph, your experiment fails because your bike is stationary on the > > board. Camber thrust is a turning force caused by a wheel _rolling_ on > > a surface it's not perpendicular to (rolling cone effect). It's > > impossible to create camber thrust with a stationary wheel.http://picas= aweb.google.com/JonCCrouch/UntitledAlbum/photo#5125629114... > > > Also by being stationary you're eliminating the the effect of wheel > > drag which, combined with trail (shopping cart wheel effect) gives the > > wheel a tendency to point straight ahead and combined with an offset > > contact patch (when leaning) give the wheel a tendency to turn inwards. > > Very interesting. Do you think these things contribute significantly > to turning a leaned bike? Or are they just what accounts for the > scrubbing sound? > > Joseph Yup camber thrust is the main mechanism that generates cornering force. It's countered to a large degree by tyre slip and by the fact that you have 2 wheels held relatively parallel by the trail of the steering geometry (two cones held parallel wont want to roll in an arc. Infact they wont want to roll much at all). The countering forces cause a leaning bike to take a much wider arc than a lone wheel at the same speed and lean angle would). Steering flop and contact patch drag will also come into play on a leaning no-handed bike but they're countered to a large degree by trail (shopping cart wheel effect). Moving the CoG of the rider will change the angle of the wheels and so alter the amount of camber thrust and contact patch offset/drag. That's about all the forces involved I think. None of them rely on gyroscopic inertia but as you say, inertia will give the wheel a tendency to remain stable and resist sudden changes in direction.     Date: 26 Oct 2007 16:26:34 From: Subject: Re: Gyroscopic forces revisited Jon Crouch writes: >>> Also by being stationary you're eliminating the the effect of >>> wheel drag which, combined with trail (shopping cart wheel effect) >>> gives the wheel a tendency to point straight ahead and combined >>> with an offset contact patch (when leaning) give the wheel a >>> tendency to turn inwards. >> Very interesting. Do you think these things contribute >> significantly to turning a leaned bike? Or are they just what >> accounts for the scrubbing sound? > Yup camber thrust is the main mechanism that generates cornering > force. It's countered to a large degree by tyre slip and by the > fact that you have 2 wheels held relatively parallel by the trail of > the steering geometry (two cones held parallel wont want to roll in > an arc. Infact they wont want to roll much at all). The countering > forces cause a leaning bike to take a much wider arc than a lone > wheel at the same speed and lean angle would). Tire slip is minuscule or we would see black marks on the road and tires would wear out in a few hundred miles. Tire tread is only a couple of millimeters thick at best. In contrast 25mm cross section tires last for as much as 3000 miles with plenty of fast descending. > Steering flop and contact patch drag will also come into play on a > leaning no-handed bike but they're countered to a large degree by > trail (shopping cart wheel effect). > Moving the CoG of the rider will change the angle of the wheels and > so alter the amount of camber thrust and contact patch offset/drag. > That's about all the forces involved I think. None of them rely on > gyroscopic inertia but as you say, inertia will give the wheel a > tendency to remain stable and resist sudden changes in direction. From this one might conclude that vehicles that don't lean can't corner. Riding a bicycle no-hands on a straight crowned road requires no lean and the "camber thrust" has no effect. This "camber thrust" seems to be a new concept that came from the people who recently discovered countersteer or ones who believe pushing down on the inside handlebar improves cornering. Jobst Brandt       Date: 26 Oct 2007 22:06:05 From: Michael Press Subject: Re: Gyroscopic forces revisited In article <4722153a$0$14068$742ec2ed@news.sonic.net >, jobst.brandt@stanfordalumni.org wrote: > ones who believe pushing down on the inside > handlebar improves cornering. 'tis'nt pushing down on the inside; rather 'tis unloading the outside bar that eases cornering. -- Michael Press         Date: 26 Oct 2007 23:15:07 From: Subject: Re: Gyroscopic forces revisited Michael Press writes: >> ones who believe pushing down on the inside handlebar improves >> cornering. > 'tis'nt pushing down on the inside; rather 'tis unloading the > outside bar that eases cornering. xplain! Jobst Brandt           Date: 27 Oct 2007 00:42:17 From: Michael Press Subject: Re: Gyroscopic forces revisited In article <472274fb$0$14096$742ec2ed@news.sonic.net >, jobst.brandt@stanfordalumni.org wrote: > Michael Press writes: > > >> ones who believe pushing down on the inside handlebar improves > >> cornering. > > > 'tis'nt pushing down on the inside; rather 'tis unloading the > > outside bar that eases cornering. > > xplain! Explain effective cornering to you?! I evolved a way of cornering the bicycle that takes little effort from me. Fine adjustments in the corner are quick, precise, and accurate. * Get out of the saddle. * Unload the inside pedal. * Load the outside pedal * Unload the outside bar. Result is the bicycle and me floating around corners, without having to "muscle" it. -- Michael Press             Date: 27 Oct 2007 01:19:56 From: Subject: Re: Gyroscopic forces revisited Michael Press writes: >>>> ones who believe pushing down on the inside handlebar improves >>>> cornering. >>> 'tis'nt pushing down on the inside; rather 'tis unloading the >>> outside bar that eases cornering. >> xplain! > Explain effective cornering to you?! Explain what this and the other things you list below have any effect on cornering, both for traction and ease of operation. > I evolved a way of cornering the bicycle that takes little effort > from me. Fine adjustments in the corner are quick, precise, and > accurate. Sounds like the garden of eden. > * Get out of the saddle. > * Unload the inside pedal. > * Load the outside pedal > * Unload the outside bar. > Result is the bicycle and me floating around corners, without having > to "muscle" it. xplain! How do these operations affect the handling or control in a curve. That was the question. I don't know any fast descenders who "get out of the saddle" if the road isn't too rough to sit, and then they don't go as fast because the tires hop. What does unloading the inside pedal do? This sounds like it came from the same folks who believe in pushing down on the inside (curve) end of the handlebar. To be blunt, these are religious tenets, nothing more. I think you have seen the reason for the position in the cornering picture I posted. They are outlined in the FAQ item on Descending. Jobst Brandt               Date: 28 Oct 2007 23:02:43 From: Michael Press Subject: Re: Gyroscopic forces revisited In article <4722923c$0$14129$742ec2ed@news.sonic.net >, jobst.brandt@stanfordalumni.org wrote: > Michael Press writes: > > >>>> ones who believe pushing down on the inside handlebar improves > >>>> cornering. > > >>> 'tis'nt pushing down on the inside; rather 'tis unloading the > >>> outside bar that eases cornering. > > >> xplain! > > > Explain effective cornering to you?! > > Explain what this and the other things you list below have any effect > on cornering, both for traction and ease of operation. I do not have to explain anything to you. You asked, you do not like the answer you get, and you deride me. You knew you would not like the answer, but you asked anyway. I think you are looking for an excuse to say nasty things. > > I evolved a way of cornering the bicycle that takes little effort > > from me. Fine adjustments in the corner are quick, precise, and > > accurate. > > Sounds like the garden of eden. > > > * Get out of the saddle. > > * Unload the inside pedal. > > * Load the outside pedal > > * Unload the outside bar. > > > Result is the bicycle and me floating around corners, without having > > to "muscle" it. > > xplain! How do these operations affect the handling or control in a > curve. That was the question. I don't know any fast descenders who > "get out of the saddle" if the road isn't too rough to sit, and then > they don't go as fast because the tires hop. What does unloading the > inside pedal do? This sounds like it came from the same folks who > believe in pushing down on the inside (curve) end of the handlebar. It came from nobody else. I made it up all by myself. > To be blunt, these are religious tenets, nothing more. Therefore it is not religion. > I think you have seen the reason for the position in the cornering > picture I posted. They are outlined in the FAQ item on Descending. Please, explain it all to us again. -- Michael Press                 Date: 29 Oct 2007 02:58:22 From: Ben C Subject: Re: Gyroscopic forces revisited On 2007-10-28, Michael Press <rubrum@pacbell.net > wrote: > In article <4722923c$0$14129$742ec2ed@news.sonic.net>, > jobst.brandt@stanfordalumni.org wrote: > >> Michael Press writes: [...] >> > * Get out of the saddle. >> > * Unload the inside pedal. >> > * Load the outside pedal >> > * Unload the outside bar. >> >> > Result is the bicycle and me floating around corners, without having >> > to "muscle" it. FWIW I remember Liggett and Sherwen commenting during a descent on the TdF that the way to go round a corner was to put your weight on the inside hand and outside leg. I think that is good advice although I think it's sort of what I was doing anyway before I heard it.                   Date: 29 Oct 2007 14:06:15 From: Subject: Re: Gyroscopic forces revisited Ben C? writes: >>>> * Get out of the saddle. >>>> * Unload the inside pedal. >>>> * Load the outside pedal >>>> * Unload the outside bar. >>>> Result is the bicycle and me floating around corners, without >>>> having to "muscle" it. > FWIW I remember Liggett and Sherwen commenting during a descent on > the TdF that the way to go round a corner was to put your weight on > the inside hand and outside leg. > I think that is good advice although I think it's sort of what I was > doing anyway before I heard it. This is classic bicycle science at work. A strong riders who can win races becomes scientists who explain how they win through technique. No question about the mecahnism for how this works is asked. This is religion. Jobst Brandt                     Date: 29 Oct 2007 15:53:06 From: Sandy Subject: Re: Gyroscopic forces revisited Dans le message de news:4725e8d7$0$14138$742ec2ed@news.sonic.net, jobst.brandt@stanfordalumni.org <jobst.brandt@stanfordalumni.org > a réfléchi, et puis a déclaré : > Ben C? writes: > >>>>> * Get out of the saddle. >>>>> * Unload the inside pedal. >>>>> * Load the outside pedal >>>>> * Unload the outside bar. > >>>>> Result is the bicycle and me floating around corners, without >>>>> having to "muscle" it. > >> FWIW I remember Liggett and Sherwen commenting during a descent on >> the TdF that the way to go round a corner was to put your weight on >> the inside hand and outside leg. > >> I think that is good advice although I think it's sort of what I was >> doing anyway before I heard it. > > This is classic bicycle science at work. A strong riders who can win > races becomes scientists who explain how they win through technique. > No question about the mecahnism for how this works is asked. This is > religion. > > Jobst Brandt Your envy is showing.                       Date: 31 Oct 2007 21:42:45 From: Tom Sherman Subject: Re: Gyroscopic forces revisited Sandy Leurre wrote: > Dans le message de news:4725e8d7$0$14138$742ec2ed@news.sonic.net, > jobst.brandt@stanfordalumni.org <jobst.brandt@stanfordalumni.org> a > réfléchi, et puis a déclaré : >> Ben C? writes: >> >>>>>> * Get out of the saddle. >>>>>> * Unload the inside pedal. >>>>>> * Load the outside pedal >>>>>> * Unload the outside bar. >>>>>> Result is the bicycle and me floating around corners, without >>>>>> having to "muscle" it. >>> FWIW I remember Liggett and Sherwen commenting during a descent on >>> the TdF that the way to go round a corner was to put your weight on >>> the inside hand and outside leg. >>> I think that is good advice although I think it's sort of what I was >>> doing anyway before I heard it. >> This is classic bicycle science at work. A strong riders who can win >> races becomes scientists who explain how they win through technique. >> No question about the mecahnism for how this works is asked. This is >> religion. >> >> Jobst Brandt > > Your envy is showing. Obviously engineering school teaches a much more realistic explanation of existence than law school. -- Tom Sherman - Holstein-Friesland Bovinia When did ignorance of biology become a "family value"?                   Date: 29 Oct 2007 08:29:06 From: Tim McNamara Subject: Re: Gyroscopic forces revisited In article <slrnfib4me.m9c.spamspam@bowser.marioworld >, Ben C <spamspam@spam.eggs > wrote: > On 2007-10-28, Michael Press <rubrum@pacbell.net> wrote: > > In article <4722923c$0$14129$742ec2ed@news.sonic.net>, > > jobst.brandt@stanfordalumni.org wrote: > > > >> Michael Press writes: > [...] > >> > * Get out of the saddle. * Unload the inside pedal. * Load > >> > the outside pedal * Unload the outside bar. > >> > >> > Result is the bicycle and me floating around corners, without > >> > having to "muscle" it. > > FWIW I remember Liggett and Sherwen commenting during a descent on > the TdF that the way to go round a corner was to put your weight on > the inside hand and outside leg. > > I think that is good advice although I think it's sort of what I was > doing anyway before I heard it. Whereas ISTR Bernard Hinault's book saying to carry your weight on the outside hand and outside leg (with photo of him taking his inside hand off the bars in a turn). And then you have Davis Phinney telling you to steer to the outside of the turn to tighten your radius. Etc. Sean Yates's advice is the same as Liggett/Sherwin. FWIW that's the grip I use on the bike in corners (although who really knows, because normally I am not paying attention to it, I'm just riding my bike).               Date: 26 Oct 2007 18:35:44 From: jim beam Subject: Re: Gyroscopic forces revisited jobst.brandt@stanfordalumni.org wrote: > Michael Press writes: > >>>>> ones who believe pushing down on the inside handlebar improves >>>>> cornering. > >>>> 'tis'nt pushing down on the inside; rather 'tis unloading the >>>> outside bar that eases cornering. > >>> xplain! > >> Explain effective cornering to you?! > > Explain what this and the other things you list below have any effect > on cornering, both for traction and ease of operation. > >> I evolved a way of cornering the bicycle that takes little effort >> from me. Fine adjustments in the corner are quick, precise, and >> accurate. > > Sounds like the garden of eden. > >> * Get out of the saddle. >> * Unload the inside pedal. >> * Load the outside pedal >> * Unload the outside bar. > >> Result is the bicycle and me floating around corners, without having >> to "muscle" it. > > xplain! How do these operations affect the handling or control in a > curve. That was the question. I don't know any fast descenders who > "get out of the saddle" if the road isn't too rough to sit, and then > they don't go as fast because the tires hop. What does unloading the > inside pedal do? This sounds like it came from the same folks who > believe in pushing down on the inside (curve) end of the handlebar. > > To be blunt, these are religious tenets, nothing more. stop being so dismissive and start being analytical jobst. ever thought what contribution frame flex may have on geometry? you /do/ know that frames flex when loaded on only one pedal, right? > > I think you have seen the reason for the position in the cornering > picture I posted. They are outlined in the FAQ item on Descending. > > Jobst Brandt   Date: 26 Oct 2007 06:51:16 From: joseph.santaniello@gmail.com Subject: Re: Gyroscopic forces revisited On Oct 26, 2:29 pm, dabac <dabac.2z2...@no- mx.forums.cyclingforums.com > wrote: > joseph.santanie...@gmail.com Wrote: > > > > > On Oct 26, 12:35 pm, dabac <dabac.2z2...@no- > > mx.forums.cyclingforums.com> wrote: > > > Joseph Santaniello Wrote: > > > > > > I have witnessed descending in a similar fashion, but with both > > > > > hands behind the rider's butt. This position is significantly > > more > > > > > aerodynamic, as evidenced by the riders marked speed increase > > upon > > > > > moving the arms back like that. > > > > jobst.bra...@stanfordalumni.org Wrote: > > > > > I doubt that this does what you say it does because frontal area > > is > > > > what causes drag, the closing area behind the rider being a bluff > > body > > > > and has no streamlining (as a long pointed tail). Frontal area is > > > > smaller with the shoulders pulled in in front of the rider with > > arms > > > > in > > > > chest, elbows on knees. > > > > Then why does the tuck used by ski jumpers and speed skaters look > > the > > > way it does? According to your line of reasoning they should be > > faster > > > if they kept their arms against the chest. How strange that none of > > > those guys have discovered such an easy way to gain an edge... > > > > -- > > > dabac > > > Ski jumpers want lift, speed-skaters need to keep their CG from being > > too far forward. I'm just playing Devil's advocate, as I agree with > > you. > > > Joseph > > Ski jumpers want lift while in-the-air, while on-the-piste(runway, > ramp?) speed is probably more desirable. If arms to the chest made them > faster down the ramp the jump-off would be an excellent time to > reposition their arms as they're already moving the rest of their body > around. > I thought for a moment that speed skaters might suffer interference > between their knees and elbows if they tucked their arms to their chest, > but some quick posing on the inlines seems to show that as not > particularly probable. > > I suppose Icould do a roll-out test with arms-to-the-sides and > arms-to-the-chest on my inlines (at least no possible changes in tire > pressure..) but I can already see the protests about "anecdotal" and > "too many uncontrolled variables" so it doesn't seem to be worth the > effort. > > -- > dabac It wasn't long ago that the V-style revolutionized ski jumping. Maybe a super-man position on the ramp would increase launch speed. Feel like testing THAT one out? ;-) As for skates, my elbows would interfere with my legs if I had my arms jammed up under my belly. But maybe... Joseph     Date: 27 Oct 2007 00:13:25 From: dabac Subject: Re: Gyroscopic forces revisited joseph.santaniello@gmail.com Wrote: > > It wasn't long ago that the V-style revolutionized ski jumping. Maybe > a super-man position on the ramp would increase launch speed. Feel > like testing THAT one out? ;-) Nah, I'll pass. Let's leave THAT to '*Eddie the eagle*' (http://tinyurl.com/63tpg) instead. joseph.santaniello@gmail.com Wrote: > > As for skates, my elbows would interfere with my legs if I had my arms > jammed up under my belly. But maybe... Well the amount of floor I had available didn't really lend itself to an all-out test, that'll have to wait until either the leaves have cleared away from the roads or until the ice has settled. And of course I don't know how representative MY tuck is compared to a "real" skater. But apart from that it appears like an average anatomy very well would offer some options other than boring old arms-at-the-sides... -- dabac   Date: 26 Oct 2007 06:48:02 From: joseph.santaniello@gmail.com Subject: Re: Gyroscopic forces revisited On Oct 26, 3:18 pm, Jon_C <JonCCro...@gmail.com > wrote: > On Oct 24, 11:04 am, "joseph.santanie...@gmail.com" > > > > <joseph.santanie...@gmail.com> wrote: > > On Oct 24, 2:40 pm, Jon_C <JonCCro...@gmail.com> wrote: > > > > On Oct 23, 10:51 pm, jobst.bra...@stanfordalumni.org wrote: > > > > > Jon Crouch writes: > > > > > How about a wheel on it's own? If you roll a wheel and it leans o= ver > > > > > it turns in that direction. The reason is that the contact patch = of > > > > > a leaning tyre is parallel to the ground and so at an angle to the > > > > > wheel's axle so the wheel acts as a rolling cone and turns into t= he > > > > > corner (which maintains it's balance). I'm pretty sure that's the > > > > > main steering force involved in no-handed riding. > > > > > (I seem to remember reading about someone testing this on a > > > > > motorbike and finding that on a prolonged corners the handlebars = are > > > > > actually turned slightly outward instead of inward as you;d > > > > > expect. I think that was due to rake/trail but I don;t have time = to > > > > > think about that right now :) > > > > > I think you ought to go to a velodrome. These tracks have sloped > > > > straights and more steeply banked circular curves at each end. Rid= ers > > > > have no problem riding no-hands on these tracks, either on the > > > > straight sections or in the banked curve (if they are going fast > > > > enough to not strike a pedal). > > > > > If you don't have a track, try a crowned paved street and notice how > > > > the angle with the pavement has essentially no effect on steering, = the > > > > centerline of the contact patch moving less than 1/8" off center fo= r a > > > > 10=B0 side slope. > > > > > Jobst Brandt > > > > To go in a straight line on an angled surface you'd have to keep the > > > steering pointed slightly down the slope (ie. left on a velodrome > > > straight) to counter-act the steering force from the slope. Maybe a > > > slight weave-like cycle occurs where the rightward steering force from > > > the slope causes the rider to 'fall' left (centrifugal force) which > > > causes the bike to steer back to the left and so on. Aslo, shifting > > > the rider's center of gravity to the right will lean the bike frame to > > > the left slightly, causing the front wheel to 'flop' (and steer) left > > > countering camber thrust from the slope. > > > > I don't think the 1/8" offset of the contact patch is too relevant > > > btw. Camber thrust is due the angle between the wheel's axle and the > > > ground (cone effect) not contact patch offset. > > > I don't think you do need have the steering anything but straight on a > > sloped surface. Not having a velodrome handy, that's why I broke out > > my trusty plywood board. I too thought the angle of the wheels on the > > surface had something to do with it, but it doesn't. I put my bike on > > the board and fiddled with it quite a bit, and the angle of contact > > alone does not appear to matter. What does matter is where the center > > of mass is relative to the contact patches and the combined gravity > > and turn forces. When the center of mass is along the centerline of > > the bike and these forces are acting through it to the patches, the > > steering satys straight no matter the angle of the wheels to the road. > > > Using a sloped board you can simulate the forces on a riderless bike > > during a turn. (On Mars...) > > > Perhaps some minimum speed is necessary to not fall over riding on a > > sloped velodrome, but I suspect this is from requiring a certain > > amount of momentum to be able to "climb" the banking everytime you > > make a steering correction up-track. > > > Joseph > > Joseph, your experiment fails because your bike is stationary on the > board. Camber thrust is a turning force caused by a wheel _rolling_ on > a surface it's not perpendicular to (rolling cone effect). It's > impossible to create camber thrust with a stationary wheel.http://picasaw= eb.google.com/JonCCrouch/UntitledAlbum/photo#5125629114... > > Also by being stationary you're eliminating the the effect of wheel > drag which, combined with trail (shopping cart wheel effect) gives the > wheel a tendency to point straight ahead and combined with an offset > contact patch (when leaning) give the wheel a tendency to turn inwards. Very interesting. Do you think these things contribute significantly to turning a leaned bike? Or are they just what accounts for the scrubbing sound? Joseph   Date: 26 Oct 2007 13:18:54 From: Jon_C Subject: Re: Gyroscopic forces revisited On Oct 24, 11:04 am, "joseph.santanie...@gmail.com" <joseph.santanie...@gmail.com > wrote: > On Oct 24, 2:40 pm, Jon_C <JonCCro...@gmail.com> wrote: > > > > > On Oct 23, 10:51 pm, jobst.bra...@stanfordalumni.org wrote: > > > > Jon Crouch writes: > > > > How about a wheel on it's own? If you roll a wheel and it leans over > > > > it turns in that direction. The reason is that the contact patch of > > > > a leaning tyre is parallel to the ground and so at an angle to the > > > > wheel's axle so the wheel acts as a rolling cone and turns into the > > > > corner (which maintains it's balance). I'm pretty sure that's the > > > > main steering force involved in no-handed riding. > > > > (I seem to remember reading about someone testing this on a > > > > motorbike and finding that on a prolonged corners the handlebars are > > > > actually turned slightly outward instead of inward as you;d > > > > expect. I think that was due to rake/trail but I don;t have time to > > > > think about that right now :) > > > > I think you ought to go to a velodrome. These tracks have sloped > > > straights and more steeply banked circular curves at each end. Riders > > > have no problem riding no-hands on these tracks, either on the > > > straight sections or in the banked curve (if they are going fast > > > enough to not strike a pedal). > > > > If you don't have a track, try a crowned paved street and notice how > > > the angle with the pavement has essentially no effect on steering, the > > > centerline of the contact patch moving less than 1/8" off center for a > > > 10=B0 side slope. > > > > Jobst Brandt > > > To go in a straight line on an angled surface you'd have to keep the > > steering pointed slightly down the slope (ie. left on a velodrome > > straight) to counter-act the steering force from the slope. Maybe a > > slight weave-like cycle occurs where the rightward steering force from > > the slope causes the rider to 'fall' left (centrifugal force) which > > causes the bike to steer back to the left and so on. Aslo, shifting > > the rider's center of gravity to the right will lean the bike frame to > > the left slightly, causing the front wheel to 'flop' (and steer) left > > countering camber thrust from the slope. > > > I don't think the 1/8" offset of the contact patch is too relevant > > btw. Camber thrust is due the angle between the wheel's axle and the > > ground (cone effect) not contact patch offset. > > I don't think you do need have the steering anything but straight on a > sloped surface. Not having a velodrome handy, that's why I broke out > my trusty plywood board. I too thought the angle of the wheels on the > surface had something to do with it, but it doesn't. I put my bike on > the board and fiddled with it quite a bit, and the angle of contact > alone does not appear to matter. What does matter is where the center > of mass is relative to the contact patches and the combined gravity > and turn forces. When the center of mass is along the centerline of > the bike and these forces are acting through it to the patches, the > steering satys straight no matter the angle of the wheels to the road. > > Using a sloped board you can simulate the forces on a riderless bike > during a turn. (On Mars...) > > Perhaps some minimum speed is necessary to not fall over riding on a > sloped velodrome, but I suspect this is from requiring a certain > amount of momentum to be able to "climb" the banking everytime you > make a steering correction up-track. > > Joseph Joseph, your experiment fails because your bike is stationary on the board. Camber thrust is a turning force caused by a wheel _rolling_ on a surface it's not perpendicular to (rolling cone effect). It's impossible to create camber thrust with a stationary wheel. http://picasaweb.google.com/JonCCrouch/UntitledAlbum/photo#5125629114602972= 546 Also by being stationary you're eliminating the the effect of wheel drag which, combined with trail (shopping cart wheel effect) gives the wheel a tendency to point straight ahead and combined with an offset contact patch (when leaning) give the wheel a tendency to turn inwards.     Date: 30 Oct 2007 20:38:21 From: bjw@mambo.ucolick.org Subject: Re: Gyroscopic forces revisited On Oct 30, 10:23 am, "Kerry Montgomery" <kamon...@teleport.com > wrote: > <b...@mambo.ucolick.org> wrote in message > > > Now consider the coin that is tilted and rolling. The > > forces on the coin are gravity acting straight down, > > a small force of rolling friction acting against the > > direction of roll, a normal force pushing up, and a > > centripetal friction force pushing inward. The > > centripetal friction force is what keeps the coin > > from low-siding. Its effect is to make the coin > > travel in a curve. > > Ben, > I agree with your description of a coin rolling straight on a hard floor, > and think it's a point of clarity in this discussion. Am not so sure of your > description of a tilted and rolling coin. If there's "a centripetal friction > force pushing inward", that would seem to make the coin low-side even > faster. I'd think an "inward" force would be one pushing from the center of > the coin in the same direction as the top of the coin is tilted. If the coin > is tilted and rolling, gravity tries to pull it down in the direction of > tilt, but angular momentum is what prevents it from immediately falling. > Kerry Centripetal (not centrifugal) means a force that is pointed inward toward the center of an arc. In this case, I mean a force parallel to the floor, pointing from the contact patch inward to the arc path that the rolling coin takes. A frictional force has to be exerted at the point of contact. You can't have a (frictional) force "pushing from the center of the coin" because nothing is touching it there. Gravity, however, can be described as a force acting at the center of mass. In this case, gravity is trying to make the coin fall over, which would kick the low side out, and the frictional force is resisting that. Ben   Date: 26 Oct 2007 04:21:03 From: joseph.santaniello@gmail.com Subject: Re: Gyroscopic forces revisited On Oct 26, 12:35 pm, dabac <dabac.2z2...@no- mx.forums.cyclingforums.com > wrote: > Joseph Santaniello Wrote: > > > > > > I have witnessed descending in a similar fashion, but with both > > > hands behind the rider's butt. This position is significantly more > > > aerodynamic, as evidenced by the riders marked speed increase upon > > > moving the arms back like that. > > jobst.bra...@stanfordalumni.org Wrote: > > > > > I doubt that this does what you say it does because frontal area is > > what causes drag, the closing area behind the rider being a bluff body > > and has no streamlining (as a long pointed tail). Frontal area is > > smaller with the shoulders pulled in in front of the rider with arms > > in > > chest, elbows on knees. > > Then why does the tuck used by ski jumpers and speed skaters look the > way it does? According to your line of reasoning they should be faster > if they kept their arms against the chest. How strange that none of > those guys have discovered such an easy way to gain an edge... > > -- > dabac Ski jumpers want lift, speed-skaters need to keep their CG from being too far forward. I'm just playing Devil's advocate, as I agree with you. Joseph     Date: 26 Oct 2007 22:29:24 From: dabac Subject: Re: Gyroscopic forces revisited joseph.santaniello@gmail.com Wrote: > On Oct 26, 12:35 pm, dabac <dabac.2z2...@no- > mx.forums.cyclingforums.com> wrote: > > Joseph Santaniello Wrote: > > > > > > > > > > I have witnessed descending in a similar fashion, but with both > > > > hands behind the rider's butt. This position is significantl > more > > > > aerodynamic, as evidenced by the riders marked speed increas > upon > > > > moving the arms back like that. > > > > jobst.bra...@stanfordalumni.org Wrote: > > > > > > > > > I doubt that this does what you say it does because frontal are > is > > > what causes drag, the closing area behind the rider being a bluf > body > > > and has no streamlining (as a long pointed tail). Frontal area is > > > smaller with the shoulders pulled in in front of the rider wit > arms > > > in > > > chest, elbows on knees. > > > > Then why does the tuck used by ski jumpers and speed skaters loo > the > > way it does? According to your line of reasoning they should b > faster > > if they kept their arms against the chest. How strange that none of > > those guys have discovered such an easy way to gain an edge... > > > > -- > > dabac > > Ski jumpers want lift, speed-skaters need to keep their CG from being > too far forward. I'm just playing Devil's advocate, as I agree with > you. > > Joseph Ski jumpers want lift while in-the-air, while on-the-piste(runway ramp?) speed is probably more desirable. If arms to the chest made the faster down the ramp the jump-off would be an excellent time t reposition their arms as they're already moving the rest of their bod around. I thought for a moment that speed skaters might suffer interferenc between their knees and elbows if they tucked their arms to their chest but some quick posing on the inlines seems to show that as no particularly probable. I suppose Icould do a roll-out test with arms-to-the-sides an arms-to-the-chest on my inlines (at least no possible changes in tir pressure..) but I can already see the protests about "anecdotal" an "too many uncontrolled variables" so it doesn't seem to be worth th effort -- dabac   Date: 26 Oct 2007 02:29:05 From: joseph.santaniello@gmail.com Subject: Re: Gyroscopic forces revisited On Oct 26, 5:30 am, velodancer <commerc...@yahoo.com > wrote: > On Oct 25, 10:41 am, jobst.bra...@stanfordalumni.org wrote: > > > I don't understand what you mean by steering angle. All turns are > > made with countersteer whether running, ice skating, snowboarding or > > bicycling. These are all balanced modes of motion and follow the same > > principles. > > Far be it from me to debate an engineer, but I'm not understanding > this. I would have expected the term countersteer to involve > mechanical objects with a hinged steering portion, and only those with > two wheels or skis inline. I have done all the activities listed above > and can only picture countersteering on a bicycle. I may bend the > lateral knee a bit deeper before turning while running or skating > which would cause my shoulder to drop to the other side, but only to > push off of that foot. Countersteering probably is in effect while on > one skate though to initiate a turn. Snowboard turns involve falling > to one side to facilitate an edge. Snowboarding is a good example. What do you need to do to precipitate falling to one side when you are in a steady upright balanced state? You need to countersteer to get your feet swept out from under you, and then you catch it before you kiss the ground and use the resulting lean to turn. I agree however that running or skiing is different because to lean you can just have one leg allow the body to fall in it's direction. Joseph > Seems like if you are saying that a vehicle that has parallel wheels > or devices such as a car would also depend on countersteering? I > wouldn't think so, any more for devices that effectively have some > other sort of parallelism such as people running. > > I read the article you wrote about steering on sheldonbrown and would > have to disagree that all bicycle manuevering requires > countersteering. I played with this this afternoon, and could detect > no countersteering below five miles an hour. I turned the wheel in the > direction I wanted to go (careful not to lean or apply an initial > force away from the direction of the turn on the bars), and that is > the direction I went. The countersteering effect did not feel strong > below 10 miles an hour on the small wheeled bike I was using. > > It strikes me that these are also the speeds that have been discussed > in this thread related to no hands riding. Are they actually related?   Date: 26 Oct 2007 02:23:30 From: joseph.santaniello@gmail.com Subject: Re: Gyroscopic forces revisited On Oct 25, 11:52 pm, jobst.bra...@stanfordalumni.org wrote: > Joseph Santaniello writes: > >> With both hands on the stem my elbows meet nicely in my gut, just > >> above my knees. From what you say, your bicycle seems to be too > >> short if you cannot bring your elbows in. > > Hands on the stem? Or the bars? With both hands in fists close > > together as though I were holding the bars with my thumbs wrapped > > under the stem, there is no angle I can hold my arms to get my > > elbows to meet. It wouldn't matter how long or short my bike is, my > > arms can't do it. With forearms rotated I can do it, but I wouldn't > > be able to hold the bar. > > I don't understand. I can bring my elbows together clasping hands as > in prayer or both over one another on the stem. Leave the bars out of > this. I see. That's different. I'll try that. > >>> But it is steering angle, not lean angle I was thinking about. > >>> But I suppose this just leads us to the drifting idea, which I > >>> agree is pretty much not do-able. > >> You have no option with steering angle, the road prescribes that > >> because You must get around the curve. That can be modified only a > >> little by choosing an ideal arc that doesn't infringe on oncoming > >> traffic or the edge of the road. > > I guess I was imagining it might be possible to force the bike into > > some combination of lean and steering angle that would not be the > > same angles the bike would find on it's own ridden the same line and > > speed no handed. But since wheel slip is out of the question, the > > only thing that would accomplish is either side loads, or > > unnecessarily steep bike lean. > > Those two parameters are not independent and are given by the curve > radius and speed. As I mentioned in the article, leaning off the > bicycle doesn't alter the plumb line from CG to ground contact line. No, but it does alter the angle the bike alone is leaning and subsequently the steering angle (by which I mean how far off center the stem is) and the relative location of the patches. Joseph     Date: 26 Oct 2007 16:02:54 From: Subject: Re: Gyroscopic forces revisited Joseph Santaniello writes: >>>> With both hands on the stem my elbows meet nicely in my gut, just >>>> above my knees. From what you say, your bicycle seems to be too >>>> short if you cannot bring your elbows in. >>> Hands on the stem? Or the bars? With both hands in fists close >>> together as though I were holding the bars with my thumbs wrapped >>> under the stem, there is no angle I can hold my arms to get my >>> elbows to meet. It wouldn't matter how long or short my bike is, >>> my arms can't do it. With forearms rotated I can do it, but I >>> wouldn't be able to hold the bar. >> I don't understand. I can bring my elbows together clasping hands >> as in prayer or both over one another on the stem. Leave the bars >> out of this. > I see. That's different. I'll try that. >>>>> But it is steering angle, not lean angle I was thinking about. >>>>> But I suppose this just leads us to the drifting idea, which I >>>>> agree is pretty much not do-able. >>>> You have no option with steering angle, the road prescribes that >>>> because You must get around the curve. That can be modified only >>>> a little by choosing an ideal arc that doesn't infringe on >>>> oncoming traffic or the edge of the road. >>> I guess I was imagining it might be possible to force the bike >>> into some combination of lean and steering angle that would not be >>> the same angles the bike would find on it's own ridden the same >>> line and speed no handed. But since wheel slip is out of the >>> question, the only thing that would accomplish is either side >>> loads, or unnecessarily steep bike lean. >> Those two parameters are not independent and are given by the curve >> radius and speed. As I mentioned in the article, leaning off the >> bicycle doesn't alter the plumb line from CG to ground contact >> line. > No, but it does alter the angle the bike alone is leaning and > subsequently the steering angle (by which I mean how far off center > the stem is) and the relative location of the patches. The steering angle does not change. The track around the curve remains unchanged, only the rider and bicycle are no longer in the same plane. Following the same track requires the same steering angle. You can't fool geometry, or gravity for that matter. Jobst Brandt   Date: 25 Oct 2007 20:30:30 From: velodancer Subject: Re: Gyroscopic forces revisited On Oct 25, 10:41 am, jobst.bra...@stanfordalumni.org wrote: > I don't understand what you mean by steering angle. All turns are > made with countersteer whether running, ice skating, snowboarding or > bicycling. These are all balanced modes of motion and follow the same > principles. Far be it from me to debate an engineer, but I'm not understanding this. I would have expected the term countersteer to involve mechanical objects with a hinged steering portion, and only those with two wheels or skis inline. I have done all the activities listed above and can only picture countersteering on a bicycle. I may bend the lateral knee a bit deeper before turning while running or skating which would cause my shoulder to drop to the other side, but only to push off of that foot. Countersteering probably is in effect while on one skate though to initiate a turn. Snowboard turns involve falling to one side to facilitate an edge. Seems like if you are saying that a vehicle that has parallel wheels or devices such as a car would also depend on countersteering? I wouldn't think so, any more for devices that effectively have some other sort of parallelism such as people running. I read the article you wrote about steering on sheldonbrown and would have to disagree that all bicycle manuevering requires countersteering. I played with this this afternoon, and could detect no countersteering below five miles an hour. I turned the wheel in the direction I wanted to go (careful not to lean or apply an initial force away from the direction of the turn on the bars), and that is the direction I went. The countersteering effect did not feel strong below 10 miles an hour on the small wheeled bike I was using. It strikes me that these are also the speeds that have been discussed in this thread related to no hands riding. Are they actually related?     Date: 26 Oct 2007 15:58:39 From: Subject: Re: Gyroscopic forces revisited Someone writes: >> I don't understand what you mean by steering angle. All turns are >> made with countersteer whether running, ice skating, snowboarding or >> bicycling. These are all balanced modes of motion and follow the same >> principles. > Far be it from me to debate an engineer, but I'm not understanding > this. I would have expected the term countersteer to involve > mechanical objects with a hinged steering portion, and only those > with two wheels or skis in-line. I have done all the activities > listed above and can only picture countersteering on a bicycle. I > may bend the lateral knee a bit deeper before turning while running > or skating which would cause my shoulder to drop to the other side, > but only to push off of that foot. Countersteering probably is in > effect while on one skate though to initiate a turn. Snowboard > turns involve falling to one side to facilitate an edge. Countersteer is the means by which one induces lean into a turn. Running is the easiest way of experiencing how it works. Before taking a sharp turn, the feet must take a course opposite to the turn to induce the lean one needs to achieve a change in direction. > Seems like if you are saying that a vehicle that has parallel wheels > or devices such as a car would also depend on countersteering? I > wouldn't think so, any more for devices that effectively have some > other sort of parallelism such as people running. Try it. > I read the article you wrote about steering on sheldonbrown and > would have to disagree that all bicycle maneuvering requires > countersteering. I played with this this afternoon, and could > detect no countersteering below five miles an hour. I turned the > wheel in the direction I wanted to go (careful not to lean or apply > an initial force away from the direction of the turn on the bars), > and that is the direction I went. The countersteering effect did > not feel strong below 10 miles an hour on the small wheeled bike I > was using. I think you are deceiving your senses. How do you induce lean into the turn you are making? > It strikes me that these are also the speeds that have been > discussed in this thread related to no hands riding. Are they > actually related? Try it. Jobst Brandt   Date: 25 Oct 2007 21:42:51 From: joseph.santaniello@gmail.com Subject: Re: Gyroscopic forces revisited On Oct 25, 11:27 pm, jobst.bra...@stanfordalumni.org wrote: > With both hands on the stem my elbows meet nicely in my gut, just > above my knees. From what you say, your bicycle seems to be too short > if you cannot bring your elbows in. Hands on the stem? Or the bars? With both hands in fists close together as though I were holding the bars with my thumbs wrapped under the stem, there is no angle I can hold my arms to get my elbows to meet. It wouldn't matter how long or short my bike is, my arms can't do it. With forearms rotated I can do it, but I wouldn't be able to hold the bar. > > But it is steering angle, not lean angle I was thinking about. But I > > suppose this just leads us to the drifting idea, which I agree is > > pretty much not do-able. > > You have no option with steering angle, the road prescribes that > because You must get around the curve. That can be modified only a > little by choosing an ideal arc that doesn't infringe on oncoming > traffic or the edge of the road. I guess I was imagining it might be possible to force the bike into some combination of lean and steering angle that would not be the same angles the bike would find on it's own ridden the same line and speed no handed. But since wheel slip is out of the question, the only thing that would accomplish is either side loads, or uneccesarily steep bike lean. Joseph     Date: 25 Oct 2007 21:52:43 From: Subject: Re: Gyroscopic forces revisited Joseph Santaniello writes: >> With both hands on the stem my elbows meet nicely in my gut, just >> above my knees. From what you say, your bicycle seems to be too >> short if you cannot bring your elbows in. > Hands on the stem? Or the bars? With both hands in fists close > together as though I were holding the bars with my thumbs wrapped > under the stem, there is no angle I can hold my arms to get my > elbows to meet. It wouldn't matter how long or short my bike is, my > arms can't do it. With forearms rotated I can do it, but I wouldn't > be able to hold the bar. I don't understand. I can bring my elbows together clasping hands as in prayer or both over one another on the stem. Leave the bars out of this. >>> But it is steering angle, not lean angle I was thinking about. >>> But I suppose this just leads us to the drifting idea, which I >>> agree is pretty much not do-able. >> You have no option with steering angle, the road prescribes that >> because You must get around the curve. That can be modified only a >> little by choosing an ideal arc that doesn't infringe on oncoming >> traffic or the edge of the road. > I guess I was imagining it might be possible to force the bike into > some combination of lean and steering angle that would not be the > same angles the bike would find on it's own ridden the same line and > speed no handed. But since wheel slip is out of the question, the > only thing that would accomplish is either side loads, or > unnecessarily steep bike lean. Those two parameters are not independent and are given by the curve radius and speed. As I mentioned in the article, leaning off the bicycle doesn't alter the plumb line from CG to ground contact line. Jobst Brandt       Date: 31 Oct 2007 13:42:35 From: Jon_C Subject: Re: Gyroscopic forces revisited On Oct 30, 3:15 pm, jobst.bra...@stanfordalumni.org wrote: > > Let's look at this in another situation. Merry-go-rounds of old were > larger in diameter and went fast enough that it was hard to catch the > golden ring on the ring dispenser as the "horseback" rider reached for > it, especially if the horse was at the top of its stride or bottom. > > These horses were tilted so that at running speed of the ride, the > connecting rod (brass covered pole) was tilted so that its projection > to the deck was in the plum line, about 10 degrees from vertical. > This way it would not have as much friction with the tube into which > it projected as it went up and down. Let's stand on that deck astride > a bicycle with foot on a support so that the bicycle can be balanced > leaning at the appropriate 10 degrees as the ride rotates. The net > load of the bicycle and rider is in the same position it would be if > this were a road and the rider were progressing at about 15mph (about > the speed of the merry-go-round). > > I see no scrub or torque at either wheel although all other conditions > of riding a bicycle in such a curve were occurring. If that isn't > good enough, lets say there are no obstacles on the surface of this > merry-go-round other than a painted broad stripe. If the rider were > to start riding on this stripe, I see no change other than the lean > angle, that would depend on whether riding with or opposite to the > rotation of the deck. > > Let's make that stipe with matte white paint so we can see "scrub" and > whatever kind of slip the tires are doing. > > Jobst Brandt You've picked an odd scenario. Riding at the same speed and direction as the merry-go-round, your wheels wouldn't turn. Riding at the same speed and opposite direction, your wheels would turn but you wouldn't move so you wouldn't have to lean. How about this scenario to illustrate scrub/slip: Draw a line straight line on the ground and stand a bike on the line with handlebars locked straight ahead. Now lean the bike at 45 degrees and push directly down through the bike to mimic the force during cornering. When you load the tyres like this they'll obviously deflect slightly and the wheel will move ever so slightly off of the line (even though the contact patch didn;t move). Now the tyres are only loaded/stressed around the contact patch while the rest of them are in their normal relaxed state. If you then roll the bike forward an inch or two the tyres move onto a relaxed section which deflect again moving the wheel even further off line. The further you roll the bike the further the tyres will 'walk' off of the line. The tyres aren't actually sliding but are still deviating from the direction that the bike is steering. This is one of the causes of 'slip' during cornering and is unavoidable with flexible tyres.         Date: 31 Oct 2007 19:00:41 From: Subject: Re: Gyroscopic forces revisited Jon Crouch writes: >> Let's look at this in another situation. Merry-go-rounds of old >> were larger in diameter and went fast enough that it was hard to >> catch the golden ring on the ring dispenser as the "horseback" >> rider reached for it, especially if the horse was at the top of its >> stride or bottom. >> These horses were tilted so that at running speed of the ride, the >> connecting rod (brass covered pole) was tilted so that its >> projection to the deck was in the plum line, about 10 degrees from >> vertical. This way it would not have as much friction with the >> tube into which it projected as it went up and down. Let's stand >> on that deck astride a bicycle with foot on a support so that the >> bicycle can be balanced leaning at the appropriate 10 degrees as >> the ride rotates. The net load of the bicycle and rider is in the >> same position it would be if this were a road and the rider were >> progressing at about 15mph (about the speed of the merry-go-round). >> I see no scrub or torque at either wheel although all other >> conditions of riding a bicycle in such a curve were occurring. If >> that isn't good enough, lets say there are no obstacles on the >> surface of this merry-go-round other than a painted broad stripe. >> If the rider were to start riding on this stripe, I see no change >> other than the lean angle, that would depend on whether riding with >> or opposite to the rotation of the deck. >> Let's make that stripe with matte white paint so we can see "scrub" >> and whatever kind of slip the tires are doing. > You've picked an odd scenario. Riding at the same speed and > direction as the merry-go-round, your wheels wouldn't turn. I think you missed the meaning: >> If the rider were to start riding on this stripe, I see no change >> other than the lean angle, that would depend on whether riding with >> or opposite to the rotation of the deck. This means riding relative to the rotation of the merry-go-round. > Riding at the same speed and opposite direction, your wheels would > turn but you wouldn't move so you wouldn't have to lean. > How about this scenario to illustrate scrub/slip: > Draw a line straight line on the ground and stand a bike on the line > with handlebars locked straight ahead. Now lean the bike at 45 > degrees and push directly down through the bike to mimic the force > during cornering. The force of cornering comes from centrifugal force from curvilinear motion. That force is missing in the experiment. If you are concerned about the scrub on the road (with little to no force) you are barking up the wrong tree. The scrub is so small that you cannot feel it when rolling the bicycle. Besides, what has this to do with control forces of cornering? As I mentioned a few times, if you try riding a flat tire full of Slime, the bicycle will walk off the road sideways from the crown of the road. I have done this and am aware what effect lateral tire deflection has on tracking and how it occurs. This is similar to cornering with knobby tires whose knobs walk sideways in curves. The difference is that the knobs have elastic resistance as well as limited deflection, in contrast to the slime filled tire that offerers practically no lateral force. > When you load the tyres like this they'll obviously deflect slightly > and the wheel will move ever so slightly off of the line (even though > the contact patch didn't move). Now the tyres are only loaded/stressed > around the contact patch while the rest of them are in their normal > relaxed state. If you then roll the bike forward an inch or two the > tyres move onto a relaxed section which deflect again moving the wheel > even further off line. The further you roll the bike the further the > tyres will 'walk' off of the line. The tyres aren't actually sliding > but are still deviating from the direction that the bike is steering. > This is one of the causes of 'slip' during cornering and is > unavoidable with flexible tyres. You ought to examine how far off center the center of pressure is in a fairly steeply leaning curve. This is not a major asymmetric distortion of the tire cross section. OK, now what? Jobst Brandt       Date: 30 Oct 2007 15:14:24 From: joseph.santaniello@gmail.com Subject: Re: Gyroscopic forces revisited On Oct 30, 8:15 pm, jobst.bra...@stanfordalumni.org wrote: > Tim McNamara writes: > >>>> If the steering angle of the front wheel is small, the bicycle is > >>>> not being steered by tire scrub, but by lean angle. This is > >>>> exactly analogous to the rolling tilted penny. This means that > >>>> the bike or penny will describe a circular segment of radius r, > >>>> where the centripetal acceleration a = v^2/r. > >>>> The acceleration points inward toward the center of the circle. > >>>> The acceleration is from a force exerted by the ground on the > >>>> bike. That force is exerted along the axis of the bike; it has a > >>>> vertical and horizontal component. The vertical component > >>>> balances the gravity force on the bike+rider, Fgrav=mg. The > >>>> horizontal component provides the centripetal acceleration. > >>>> Assuming flat ground and that the bike is tilted by an angle > >>>> theta from vertical, the centripetal force is Fcen = mg > >>>> tan(theta). You can get the radius of the turn from v^2/r = g > >>>> tan(theta), so r = v^2 / (g tan theta) The radius just depends on > >>>> speed and lean angle. For example, if you go round a turn at 10 > >>>> m/s with a lean angle of 25 deg, your turn radius is 21 m. > >>> So does this steering effect depend on the angle between the wheel > >>> and the ground, or on the angle between the line from the wheel's > >>> CG to the contact point and the ground? > >> (Speaking about rolling coins at least) The angle from contact > >> point to center-of-mass. Another way of looking at this is that > >> the gravity force, when the CG is not over the contact patch, > >> exerts a torque that changes the angular momentum direction of the > >> coin. > > Which seems to me would require little if any scrub and allows coins > > to act this way despite a very low coefficient of friction with the > > planar surface which limits "scrub" and lateral reaction forces. > > The reason I brought up my question is that it points out a problem > > with using the rolling coin metaphor to explain bicycle steering. > > The bicycle has two rolling hoops out of plane with each other, at > > an angle to the ground, and rolling through different turn radii. > > Not to mention a deformable high-friction contact patch. Maybe it'd > > be simpler to just talk about bikes. > > Let's look at this in another situation. Merry-go-rounds of old were > larger in diameter and went fast enough that it was hard to catch the > golden ring on the ring dispenser as the "horseback" rider reached for > it, especially if the horse was at the top of its stride or bottom. > > These horses were tilted so that at running speed of the ride, the > connecting rod (brass covered pole) was tilted so that its projection > to the deck was in the plum line, about 10 degrees from vertical. > This way it would not have as much friction with the tube into which > it projected as it went up and down. Let's stand on that deck astride > a bicycle with foot on a support so that the bicycle can be balanced > leaning at the appropriate 10 degrees as the ride rotates. The net > load of the bicycle and rider is in the same position it would be if > this were a road and the rider were progressing at about 15mph (about > the speed of the merry-go-round). > > I see no scrub or torque at either wheel although all other conditions > of riding a bicycle in such a curve were occurring. If that isn't > good enough, lets say there are no obstacles on the surface of this > merry-go-round other than a painted broad stripe. If the rider were > to start riding on this stripe, I see no change other than the lean > angle, that would depend on whether riding with or opposite to the > rotation of the deck. > > Let's make that stipe with matte white paint so we can see "scrub" and > whatever kind of slip the tires are doing. > > Jobst Brandt So you think narrow bike tires have no scrub, and no slip angle and thus the patches are always tangent to the arc of the curve being ridden? Joseph   Date: 25 Oct 2007 20:08:42 From: joseph.santaniello@gmail.com Subject: Re: Gyroscopic forces revisited On Oct 25, 9:45 pm, "Tom Nakashima" <t...@slac.stanford.edu > wrote: > That's usually the most common mistake of a newbie, that and weaving. > We had a few Cat 4/5 moving up to the Cat 3, their weaving for position was > dangerous. It's purely a hypothetical concern since I'm nowhere near a velodrome, but my main worry would be taking somebody out through some fault of mine. I'd stay clear of mass starts until I was good and ready. But man that would be fun. I think track racing might be my thing. I've only ridden on a velodrome a handful of times, so I never really got comfortable. That's why I don't do speed-skate racing. Other competitors would operate under the premise that I knew what I was doing, and give me margins to match, and if I came up short and somebody got hurt, I'd feel awful bad! Joseph     Date: 25 Oct 2007 13:24:04 From: Tom Nakashima Subject: Re: Gyroscopic forces revisited <joseph.santaniello@gmail.com > wrote in message news:1193342922.220886.106450@k79g2000hse.googlegroups.com... > On Oct 25, 9:45 pm, "Tom Nakashima" <t...@slac.stanford.edu> wrote: > >> That's usually the most common mistake of a newbie, that and weaving. >> We had a few Cat 4/5 moving up to the Cat 3, their weaving for position >> was >> dangerous. > > It's purely a hypothetical concern since I'm nowhere near a velodrome, > but my main worry would be taking somebody out through some fault of > mine. I'd stay clear of mass starts until I was good and ready. But > man that would be fun. I think track racing might be my thing. I've > only ridden on a velodrome a handful of times, so I never really got > comfortable. > Joseph > The funniest thing I've seen was this new kid, must have been age 15 or so, first time on a track bike and velodrome. Dressed in cycling shorts and a plain cotton white t-shirt. Helmet looked like it was two sizes too big. On a borrowed track bike, he blew away all the Cat 4/5 racers, so they put him in with the Cat 3. He was near the front of the pack, grinding his teeth and scared to death. He didn't win the Cat 3 race, but he sure gave them a run for their money. -tom   Date: 25 Oct 2007 19:50:58 From: joseph.santaniello@gmail.com Subject: Re: Gyroscopic forces revisited On Oct 25, 8:41 pm, jobst.bra...@stanfordalumni.org wrote: > Joseph Santaniello writes: > >>> I have witnessed descending in a similar fashion, but with both > >>> hands behind the rider's butt. This position is significantly > >>> more aerodynamic, as evidenced by the riders marked speed increase > >>> upon moving the arms back like that. > >> I doubt that this does what you say it does because frontal area is > >> what causes drag, the closing area behind the rider being a bluff > >> body and has no streamlining (as a long pointed tail). Frontal > >> area is smaller with the shoulders pulled in in front of the rider > >> with arms in chest, elbows on knees. > > This was in Italy. Since I am much larger than everyone else I rode > > with there, I was able to keep up on less demanding descents. What > > I lacked in balls/skill for the turns I was able to make up (to a > > degree) on the straight sections. When I would roll up on one of > > these guys when they were in a normal hands-on-the-tops tuck, and > > start to pass them, they would have none of it. They would then > > switch to the hand behind the butt position, and they would then > > start to roll away from me. > > I think you were also heavier while they not trying to crouch tightly > enough. Small changes make significant differences. There was a > large Swiss pro who had an ostrich mentality and leaned over the bars > to get his nose as close to the front tire as possible, all the while > with his rear end way up in the air. Not only was this an ungainly > position in which braking was barely possible but it was unstable. By larger I meant heavier. Maybe it was just me going slower from the additional drag from having my mouth gape open when I saw what they were pulling! > > The only thing I can think of is that with hands on the tops, the > > elbows splay a bit. I've seen other try that with an underhanded > > grip to allow the elbows to come in closer. > > The elbows are in my gut just above the knees. This goes to show that the ideal position can be a highly individual thing. When I put my hands together as though the were on the tops close together, near my chin as the would be were I to try you position, my elbows stick out wider than my shoulders. With an underhand grip I can get them to touch. And that is how I do my "super- tuck", which I hardly ever do anyway, as my weight usually more than makes up for any less than perfect aerodynamics. > > > It was definitely faster doing the speed-skater position. > > Maybe some brave soul wants to do some roll-downs to test it? > >>> When you are cornering tight turns like a hairpin, do you make > >>> conscious manual adjustments to your steering angle? > >> I don't understand what you mean by steering angle. All turns are > >> made with countersteer whether running, ice skating, snowboarding > >> or bicycling. These are all balanced modes of motion and follow > >> the same principles. > > http://www.sheldonbrown.com/brandt/descending.html>> > > > I've read the article before (and again just now), and it is quite > > informative. > > I wasn't thinking about countersteer, I was thinking about during > > the turn, well after it is initiated. Do you apply some conscious > > pressure one way or the other, or is it exactly the same as when you > > ride no hands, just that your hands are there to maintain control > > should you hit a pebble or something? > > I don't know where one would apply conscious pressure when cornering. > I think the picture at: > > http://www.sheldonbrown.com/brandt/ > > typifies the stance. You'll notice that the wheels are in a nearly > straight line, indicating this is a high speed turn. Hairpin turns > are tediously slow and need to be cut in such a way to make them less > tight. One of my favorite pictures. Ok, forget hairpins, it is really any high-G turn I was thinking about. > > It is clear from riding no-hands that the steering finds it's own > > angle, but I wonder if man-handling it to some other slightly > > different angle might be advantageous in some way. > > You must be thinking of the M/C pseudo racers, who shift their butts > from side to side when entering curves, leaning at about 30 degrees > from the perpendicular to the road. But it is steering angle, not lean angle I was thinking about. But I suppose this just leads us to the drifting idea, which I agree is pretty much not do-able. Joseph     Date: 26 Oct 2007 19:36:27 From: Tom Sherman Subject: Re: Gyroscopic forces revisited joseph.santaniello@gmail.com aka Joseph Santaniello wrote: > On Oct 25, 8:41 pm, jobst.bra...@stanfordalumni.org wrote: >> >> http://www.sheldonbrown.com/brandt/ >> >> typifies the stance. You'll notice that the wheels are in a nearly >> straight line, indicating this is a high speed turn. Hairpin turns >> are tediously slow and need to be cut in such a way to make them less >> tight. > > One of my favorite pictures... Bill Sornson has a framed, poster size version on his bedroom wall. ;) -- Tom Sherman - Holstein-Friesland Bovinia When did ignorance of biology become a "family value"?       Date: 26 Oct 2007 19:12:22 From: Bill Sornson Subject: Re: Gyroscopic forces revisited Tom Sherman wrote: > joseph.santaniello@gmail.com aka Joseph Santaniello wrote: >> On Oct 25, 8:41 pm, jobst.bra...@stanfordalumni.org wrote: >>> >>> http://www.sheldonbrown.com/brandt/ >>> >>> typifies the stance. You'll notice that the wheels are in a nearly >>> straight line, indicating this is a high speed turn. Hairpin turns >>> are tediously slow and need to be cut in such a way to make them >>> less tight. >> >> One of my favorite pictures... > > Bill Sornson has a framed, poster size version on his bedroom wall. ;) In some circles, competitive dart throwing is considered a sport. Bill "wins bets in taverns, too" S.     Date: 25 Oct 2007 21:27:39 From: Subject: Re: Gyroscopic forces revisited Joseph Santaniello writes: >>>>> I have witnessed descending in a similar fashion, but with both >>>>> hands behind the rider's butt. This position is significantly >>>>> more aerodynamic, as evidenced by the riders marked speed increase >>>>> upon moving the arms back like that. >>>> I doubt that this does what you say it does because frontal area is >>>> what causes drag, the closing area behind the rider being a bluff >>>> body and has no streamlining (as a long pointed tail). Frontal >>>> area is smaller with the shoulders pulled in in front of the rider >>>> with arms in chest, elbows on knees. >>> This was in Italy. Since I am much larger than everyone else I rode >>> with there, I was able to keep up on less demanding descents. What >>> I lacked in balls/skill for the turns I was able to make up (to a >>> degree) on the straight sections. When I would roll up on one of >>> these guys when they were in a normal hands-on-the-tops tuck, and >>> start to pass them, they would have none of it. They would then >>> switch to the hand behind the butt position, and they would then >>> start to roll away from me. >> I think you were also heavier while they not trying to crouch tightly >> enough. Small changes make significant differences. There was a >> large Swiss pro who had an ostrich mentality and leaned over the bars >> to get his nose as close to the front tire as possible, all the while >> with his rear end way up in the air. Not only was this an ungainly >> position in which braking was barely possible but it was unstable. > By larger I meant heavier. Maybe it was just me going slower from the > additional drag from having my mouth gape open when I saw what they > were pulling! >>> The only thing I can think of is that with hands on the tops, the >>> elbows splay a bit. I've seen other try that with an underhanded >>> grip to allow the elbows to come in closer. >> The elbows are in my gut just above the knees. > This goes to show that the ideal position can be a highly individual > thing. When I put my hands together as though the were on the tops > close together, near my chin as the would be were I to try you > position, my elbows stick out wider than my shoulders. With an > underhand grip I can get them to touch. And that is how I do my > "super- tuck", which I hardly ever do anyway, as my weight usually > more than makes up for any less than perfect aerodynamics. With both hands on the stem my elbows meet nicely in my gut, just above my knees. From what you say, your bicycle seems to be too short if you cannot bring your elbows in. >>> It was definitely faster doing the speed-skater position. Maybe >>> some brave soul wants to do some roll-downs to test it? >>>>> When you are cornering tight turns like a hairpin, do you make >>>>> conscious manual adjustments to your steering angle? >>>> I don't understand what you mean by steering angle. All turns >>>> are made with countersteer whether running, ice skating, >>>> snowboarding or bicycling. These are all balanced modes of >>>> motion and follow the same principles. http://www.sheldonbrown.com/brandt/descending.html >> >>> I've read the article before (and again just now), and it is quite >>> informative. >>> I wasn't thinking about countersteer, I was thinking about during >>> the turn, well after it is initiated. Do you apply some conscious >>> pressure one way or the other, or is it exactly the same as when >>> you ride no hands, just that your hands are there to maintain >>> control should you hit a pebble or something? >> I don't know where one would apply conscious pressure when >> cornering. I think the picture at: http://www.sheldonbrown.com/brandt/ >> typifies the stance. You'll notice that the wheels are in a nearly >> straight line, indicating this is a high speed turn. Hairpin turns >> are tediously slow and need to be cut in such a way to make them >> less tight. > One of my favorite pictures. Ok, forget hairpins, it is really any > high-G turn I was thinking about. >>> It is clear from riding no-hands that the steering finds it's own >>> angle, but I wonder if man-handling it to some other slightly >>> different angle might be advantageous in some way. >> You must be thinking of the M/C pseudo racers, who shift their >> butts from side to side when entering curves, leaning at about 30 >> degrees from the perpendicular to the road. > But it is steering angle, not lean angle I was thinking about. But I > suppose this just leads us to the drifting idea, which I agree is > pretty much not do-able. You have no option with steering angle, the road prescribes that because You must get around the curve. That can be modified only a little by choosing an ideal arc that doesn't infringe on oncoming traffic or the edge of the road. Jobst Brandt       Date: 30 Oct 2007 20:50:54 From: bjw@mambo.ucolick.org Subject: Re: Gyroscopic forces revisited On Oct 30, 1:53 am, Ben C <spams...@spam.eggs > wrote: > On 2007-10-30, b...@mambo.ucolick.org <b...@mambo.ucolick.org> wrote: > > > On Oct 29, 12:32 pm, Ben C <spams...@spam.eggs> wrote: > >> So does this steering effect depend on the angle between the wheel and > >> the ground, or on the angle between the line from the wheel's CG to the > >> contact point and the ground? > > > (Speaking about rolling coins at least) The angle > > from contact point to center-of-mass. Another > > way of looking at this is that the gravity force, > > when the CG is not over the contact patch, exerts > > a torque that changes the angular momentum > > direction of the coin. > > > This is easy to test experimentally with coins. > > Take a quarter and use scotch tape to tape two > > nickels to one side, creating an asymmetrically > > weighted wheel. (In the UK, you'll have to > > use different denominations and whatever the > > British word is for sticky tape.) Make sure the > > tape doesn't stick out so the quarter can roll > > smoothly. Now try rolling this thing straight > > on the floor, and try rolling an unadorned > > quarter for comparison. > > > I just tried this and while of course you can't > > roll either of them perfectly straight with > > your hand, it's clear that the asymmetrically > > weighted coin always deviates from straight more > > quickly, turning faster, in a tighter radius, > > with the weight inward. > > Yes but then the weight will cause it to lean, so it's hard to say > whether it's the lean that's causing it to steer, or the uneven weight. I was answering the question about whether the important angle is the angle of the wheel (I assume you're asking about the line of symmetry of the tread/ contact patch), or the angle of the center of mass. The experiment shows it's the center of mass that defines "lean." When I take my asymmetrical coin and roll it straight on, its rim is contacting the ground straight on, but the coin is already "leaning" because of the extra weight on one side. You can't separate steer caused by "lean" and by uneven weight - they mean the same thing. You could do the same experiment by rolling a spinning front wheel with weights securely attached to one side. I'll leave it to Carl Fogel to rig that up. If you haven't already, just try the coin experiment. Then go further and try (1) a quarter, (2) a quarter with 2 nickels on one side, (3) a quarter with a nickel on each side. It is really obvious that the asymmetric (2) turns more quickly, and the heavier symmetric (3) is much more stable and easy to roll straight than (1). This doesn't tell you that gyroscopic action governs how a bicycle turns, of course. It does show that you don't need to invoke tire scrub to describe the turning of a tilted hoop. To model how the bicycle turns, you need to consider how steering geometry initiates the lean. And I think that may take us back to square one. Ben         Date: 30 Oct 2007 18:26:33 From: Tim McNamara Subject: Re: Gyroscopic forces revisited In article <1193777454.424087.249760@e9g2000prf.googlegroups.com >, "bjw@mambo.ucolick.org" <bjw@mambo.ucolick.org > wrote: > On Oct 30, 1:53 am, Ben C <spams...@spam.eggs> wrote: > > On 2007-10-30, b...@mambo.ucolick.org <b...@mambo.ucolick.org> > > wrote: > > > > > On Oct 29, 12:32 pm, Ben C <spams...@spam.eggs> wrote: > > >> So does this steering effect depend on the angle between the > > >> wheel and the ground, or on the angle between the line from the > > >> wheel's CG to the contact point and the ground? > > > > > (Speaking about rolling coins at least) The angle from contact > > > point to center-of-mass. Another way of looking at this is that > > > the gravity force, when the CG is not over the contact patch, > > > exerts a torque that changes the angular momentum direction of > > > the coin. > > > > > This is easy to test experimentally with coins. Take a quarter > > > and use scotch tape to tape two nickels to one side, creating an > > > asymmetrically weighted wheel. (In the UK, you'll have to use > > > different denominations and whatever the British word is for > > > sticky tape.) Make sure the tape doesn't stick out so the > > > quarter can roll smoothly. Now try rolling this thing straight > > > on the floor, and try rolling an unadorned quarter for > > > comparison. > > > > > I just tried this and while of course you can't roll either of > > > them perfectly straight with your hand, it's clear that the > > > asymmetrically weighted coin always deviates from straight more > > > quickly, turning faster, in a tighter radius, with the weight > > > inward. > > > > Yes but then the weight will cause it to lean, so it's hard to say > > whether it's the lean that's causing it to steer, or the uneven > > weight. > > I was answering the question about whether the important angle is the > angle of the wheel (I assume you're asking about the line of symmetry > of the tread/ contact patch), or the angle of the center of mass. The > experiment shows it's the center of mass that defines "lean." When I > take my asymmetrical coin and roll it straight on, its rim is > contacting the ground straight on, but the coin is already "leaning" > because of the extra weight on one side. You can't separate steer > caused by "lean" and by uneven weight - they mean the same thing. If you lean to one side on a bike while (attempting to) keep the front wheel in plane with the rest of the bike, does it try to turn to the side you're leaning towards? It would if "center of mass equals lean" (by which I assume you mean the distance between the center of mass and the contact point normal to the plane of the coin/wheel). On a single wheel or coin, this would obviously be true, but because a bike's wheels can be out of plane and track different lines it might be less true in this application. > You could do the same experiment by rolling a spinning front wheel > with weights securely attached to one side. I'll leave it to Carl > Fogel to rig that up. Still remembering that a bike is two wheels slightly out of plane in a compound angle, with each wheel tracking a slightly different radius turn. That limits the utility if using a single wheel or a coin to analyze what's happening. > If you haven't already, just try the coin experiment. Then go > further and try (1) a quarter, (2) a quarter with 2 nickels on one > side, (3) a quarter with a nickel on each side. It is really obvious > that the asymmetric (2) turns more quickly, and the heavier symmetric > (3) is much more stable and easy to roll straight than (1). > > This doesn't tell you that gyroscopic action governs how a bicycle > turns, of course. It does show that you don't need to invoke tire > scrub to describe the turning of a tilted hoop. To model how the > bicycle turns, you need to consider how steering geometry initiates > the lean. And I think that may take us back to square one. The lean causes the center of mass to fall to the inside of the turn- isn't that basically what you are saying? The center of mass thus takes a shorter radius turn than the contact point(s) and would tend to pull the contact point in an arc because of gravity. If I understand this correctly, this would result in different behavior if the coin or the bike is leaning on a horizontal plane or vertical on a tilted plane. In the latter, the center of mass is still lined up above the contact points in the field of gravity and would have less tendency to describe an arc. Otherwise the bike would tend to climb the banking.           Date: 31 Oct 2007 02:57:00 From: Ben C Subject: Re: Gyroscopic forces revisited On 2007-10-30, Tim McNamara <timmcn@bitstream.net > wrote: > In article <1193777454.424087.249760@e9g2000prf.googlegroups.com>, > "bjw@mambo.ucolick.org" <bjw@mambo.ucolick.org> wrote: [...] >> I was answering the question about whether the important angle is the >> angle of the wheel (I assume you're asking about the line of symmetry >> of the tread/ contact patch), or the angle of the center of mass. The >> experiment shows it's the center of mass that defines "lean." When I >> take my asymmetrical coin and roll it straight on, its rim is >> contacting the ground straight on, but the coin is already "leaning" >> because of the extra weight on one side. You can't separate steer >> caused by "lean" and by uneven weight - they mean the same thing. > > If you lean to one side on a bike while (attempting to) keep the front > wheel in plane with the rest of the bike, does it try to turn to the > side you're leaning towards? It would if "center of mass equals lean" > (by which I assume you mean the distance between the center of mass and > the contact point normal to the plane of the coin/wheel). On a single > wheel or coin, this would obviously be true, but because a bike's wheels > can be out of plane and track different lines it might be less true in > this application. I think leaning to the left while holding the bicycle steering dead straight would effectively give you a slip-angle steering you back to the right. So you'd go straight. Shortly after that you'd fall off to the left.         Date: 30 Oct 2007 15:39:10 From: Subject: Re: Gyroscopic forces revisited On Tue, 30 Oct 2007 20:50:54 -0000, "bjw@mambo.ucolick.org" <bjw@mambo.ucolick.org > wrote: [snip] >I was answering the question about whether the >important angle is the angle of the wheel (I assume >you're asking about the line of symmetry of the tread/ >contact patch), or the angle of the center of mass. >The experiment shows it's the center of mass that defines >"lean." When I take my asymmetrical coin and roll it straight >on, its rim is contacting the ground straight on, but the >coin is already "leaning" because of the extra weight >on one side. You can't separate steer caused by >"lean" and by uneven weight - they mean the same thing. > >You could do the same experiment by rolling a spinning >front wheel with weights securely attached to one side. >I'll leave it to Carl Fogel to rig that up. [snip] Sir, To perform my late promise to you, I shall without further ceremony acquaint you, that at the end of October in the Year 2007, I procured me a circular bicycle wheel and a 15-pound weight, to try therewith the celebrated Phænomena of whatever is being discussed: http://i19.tinypic.com/4xwqbn4.jpg And in order thereto, having checked that my neighbors were not watching, I roped said weight to said wheel and rolled it down the matchlessly smooth driveway of Fogel Labs. The wheel curved sharply toward the weighted side, whether the wheel was upright or even tilted noticeably _away_ from the weighted side, tho' the preponderance of mass remained either centered or else still on the weighted side of the wheel. Further tests showed that the wheel was happy to curve either left or right, according to whichever side the weight was hung, thus eliminating any question of bias on the part of the worthy masons who originally laid the noble driveway at Fogel Labs. The curve was in fact so pronounced that the wheel invariably reversed its direction before crashing into the ground cover in the decorative garden bordering the spacious driveway at Fogel Labs. On the final trial, a plastic sprinkler head some two feet high was struck by the wheel as it headed back the way it came, whereupon the sprinkler broke off, utterly ruining $0.50 worth of costly white irrigation parts: http://i13.tinypic.com/4tghr49.jpg The shattered white connector lies near the dark threaded stump protruding from the ground cover, which leads thence to the very river of the Arkansas. The experimenter's faith that the wheel would fall weighted-side down was amply justified. Had the sprinkler system been turned on, the water would have flowed everywhere and thus convinced even the very senses of the onlookers that they had discovered an incomparably powerful force which had hitherto been entirely unknown in nature. Cheers, Isaac Newcomen         Date: 30 Oct 2007 16:22:47 From: Ben C Subject: Re: Gyroscopic forces revisited On 2007-10-30, bjw@mambo.ucolick.org <bjw@mambo.ucolick.org > wrote: > On Oct 30, 1:53 am, Ben C <spams...@spam.eggs> wrote: >> On 2007-10-30, b...@mambo.ucolick.org <b...@mambo.ucolick.org> wrote: [...] >> > I just tried this and while of course you can't >> > roll either of them perfectly straight with >> > your hand, it's clear that the asymmetrically >> > weighted coin always deviates from straight more >> > quickly, turning faster, in a tighter radius, >> > with the weight inward. >> >> Yes but then the weight will cause it to lean, so it's hard to say >> whether it's the lean that's causing it to steer, or the uneven weight. > > I was answering the question about whether the > important angle is the angle of the wheel (I assume > you're asking about the line of symmetry of the tread/ > contact patch), or the angle of the center of mass. Yes, that's the question. And I think it's the angle of the CG. If you start it off with the coin vertical, then in no time the coin falls to the left. It's then hard to say whether it's the tilt of the coin's plane with respect to the table or the fact that its CG projects to the left of its contact point that's making it steer. That's the point I was trying to make. > The experiment shows it's the center of mass that defines > "lean." When I take my asymmetrical coin and roll it straight > on, its rim is contacting the ground straight on, but the > coin is already "leaning" because of the extra weight > on one side. You can't separate steer caused by > "lean" and by uneven weight - they mean the same thing. If you could carefully stand up your asymmetrical coin so its CG was directly above its contact point, it would be tilting to the right (let's say the weights are taped onto the left). Obviously it's hard to balance it like that, but suppose you could. Then you roll it forwards. Does it steer to the right, or go straight on? I'm pretty sure it goes straight on. If the coin's weight is acting through its CG, then the only force between it and the floor is up/down. So no steering. The floor doesn't care in the least what the angle of the coin is, only where its weight is acting. If the coin ('s weight) is leaning to the right, and its contact has sufficient friction, then it's pushing the floor to the left. That means the floor is pushing it to the right. That's the force that's making it turn right. I think this is what you (and others) have been saying. So just to spell it out again: this means that when you turn a given corner at a given speed on a bicycle, there's not really any choice about how much steering angle you apply. It doesn't depend on rider position, except perhaps for some of the more subtle effects of steering geometry Joseph mentioned, and perhaps for some small amount of camber-thrust from the deformable tyres, assuming that that's a different effect from what we've been saying about the coin.       Date: 29 Oct 2007 01:57:27 From: joseph.santaniello@gmail.com Subject: Re: Gyroscopic forces revisited On Oct 29, 9:19 am, "b...@mambo.ucolick.org" <b...@mambo.ucolick.org > wrote: > > The acceleration points inward toward the center of > > the circle. The acceleration is from a force exerted > > by the ground on the bike. That force is exerted > > along the axis of the bike; it has a vertical and > > horizontal component. The vertical component > > balances the gravity force on the bike+rider, Fgrav=mg. > > The horizontal component provides the centripetal > > acceleration. Assuming flat ground and that the > > bike is tilted by an angle theta from vertical, > > the centripetal force is Fcen = mg tan(theta). > > You can get the radius of the turn from > > v^2/r = g tan(theta), so r = v^2 / (g tan theta) > > The radius just depends on speed and lean angle. > > For example, if you go round a turn at 10 m/s > > with a lean angle of 25 deg, your turn radius is 21 m. > > In case it's not clear, the horizontal component > of this force comes from friction between tires > and road. That friction has an upper limit of > Ffric = mu* mg, where mu is the coefficient of > friction. If the friction isn't large enough to > supply the centripetal force, then you slide out > and low-side the bike. Thus it's impossible > to lean a bike to arbitrarily high lean angle, > as most of us have probably discovered at one > time or another. > > Ben Does a bike tire offer the same amount of useful traction no matter what the slip angle? Or does some of the available friction get used up in non-useful way (ie in a direction that does not help keep you from crashing) if you have a greater slip angle? Joseph   Date: 25 Oct 2007 18:10:13 From: joseph.santaniello@gmail.com Subject: Re: Gyroscopic forces revisited On Oct 25, 7:41 pm, jobst.bra...@stanfordalumni.org wrote: > Joseph Santaniello writes: > >>>>> Two weeks ago I converted an old flatbar into an upright, higher > >>>>> stem, bars that come back, B-66 saddle. Now, miraculously, I can > >>>>> ride with no hands. When I tried with the flatbar, I had to shift > >>>>> my body up and back because I was leaning so far forward, but if I > >>>>> am already up and back, like I am now, I just let go and I am > >>>>> fine. > >>>>> No difference in wheel size, no difference in speed, but big > >>>>> difference in COG. COG is not just height, but longitudinal and > >>>>> lateral weight, too, or so I was told when I had to make such > >>>>> calculations. > >>>> This is more likely a skill issue in simultaneously letting go of > >>>> your hands while also repositioning your body. It is much easier > >>>> to initiate no hands if your fingertips are just resting lightly on > >>>> the handlebars rather than supporting weight. While you are > >>>> correct that the higher COG does make a difference, it is the same > >>>> COG riding no hands no matter where the bars are placed (after all, > >>>> you are not using them). > >>> The COG can move when riding no hands vs with the bars. On my bike > >>> my upper body is leaned forward at maybe 45deg and my arms are > >>> forward some amount when I ride holding the bars. If I want I can > >>> ride no hands by just letting go and keeping my body leaned over, > >>> but that isn't too comfortable. So I sit up with my upper body > >>> vertical and arms to the side. I'm sure this moves the COG at least > >>> 20cm or so. > >>> But I agree that the COG placement probably isn't the cause of his > >>> previous difficulties. > >> As I mentioned in the past, all my relatively straight top speed > >> descents are done essentially no-hands, at speeds at times exceeding > >> 60mph. At such times my hands are on the bar stem, elbows on knees > >> that are against the top tube and my back as low as I can get without > >> my chin bouncing on my hands. If there is no cross wind, this is a > >> highly stable position in which I can lean steeply into corners. > >> Steering is done by moving my upper body to either side of center, the > >> motion being slight and not apparent to an observer on a following > >> bicycle. It is in this position that "countersteer" is most apparent > >> because taking curves must be done by first leaning out of the turn > >> and then neutralizing that motion by returning to center. > > I have witnessed descending in a similar fashion, but with both > > hands behind the rider's butt. This position is significantly more > > aerodynamic, as evidenced by the riders marked speed increase upon > > moving the arms back like that. > > I doubt that this does what you say it does because frontal area is > what causes drag, the closing area behind the rider being a bluff body > and has no streamlining (as a long pointed tail). Frontal area is > smaller with the shoulders pulled in in front of the rider with arms in > chest, elbows on knees. This was in Italy. Since I am much larger than everyone else I rode with there, I was able to keep up on th eless demanding descents. What I lacked in balls/skill for the turns I was able to make up (to a degree) on the straight sections. When I would roll up on one of these guys when they were in a normal hands-on-the-tops tuck, and start to pass them, they would have none of it. They would then switch to the hand behind the butt position, and they would then start to roll away from me. The only thing I can think of is that with hands on the tops, the elbows splay a bit. I've seen other try that with an underhanded grip to allow the elbows to come in closer. It was definitely faster doing the speed-skater position. Maybe some brave soul wants to do some roll-downs to test it? > > When you are cornering tight turns like a hairpin, do you make > > conscious manual adjustments to your steering angle? > > I don't understand what you mean by steering angle. All turns are > made with countersteer whether running, ice skating, snowboarding or > bicycling. These are all balanced modes of motion and follow the same > principles. > > http://www.sheldonbrown.com/brandt/descending.html > I've read the article before (and again just now), and it is quite informative. I wasn't thinking about countersteer, I was thinking about during the turn, well after it is initiated. Do you apply some conscious pressure one way or the other, or is it exactly the same as when you ride no hands, just that your hands are there to maintain control should you hit a pebble or something? It is clear from riding not hands that the steering finds it's own angle, but I wonder if man-handling it to some other slightly different angle might be advantageous in some way. Joseph     Date: 25 Oct 2007 18:41:12 From: Subject: Re: Gyroscopic forces revisited Joseph Santaniello writes: >>> I have witnessed descending in a similar fashion, but with both >>> hands behind the rider's butt. This position is significantly >>> more aerodynamic, as evidenced by the riders marked speed increase >>> upon moving the arms back like that. >> I doubt that this does what you say it does because frontal area is >> what causes drag, the closing area behind the rider being a bluff >> body and has no streamlining (as a long pointed tail). Frontal >> area is smaller with the shoulders pulled in in front of the rider >> with arms in chest, elbows on knees. > This was in Italy. Since I am much larger than everyone else I rode > with there, I was able to keep up on less demanding descents. What > I lacked in balls/skill for the turns I was able to make up (to a > degree) on the straight sections. When I would roll up on one of > these guys when they were in a normal hands-on-the-tops tuck, and > start to pass them, they would have none of it. They would then > switch to the hand behind the butt position, and they would then > start to roll away from me. I think you were also heavier while they not trying to crouch tightly enough. Small changes make significant differences. There was a large Swiss pro who had an ostrich mentality and leaned over the bars to get his nose as close to the front tire as possible, all the while with his rear end way up in the air. Not only was this an ungainly position in which braking was barely possible but it was unstable. > The only thing I can think of is that with hands on the tops, the > elbows splay a bit. I've seen other try that with an underhanded > grip to allow the elbows to come in closer. The elbows are in my gut just above the knees. > It was definitely faster doing the speed-skater position. > Maybe some brave soul wants to do some roll-downs to test it? >>> When you are cornering tight turns like a hairpin, do you make >>> conscious manual adjustments to your steering angle? >> I don't understand what you mean by steering angle. All turns are >> made with countersteer whether running, ice skating, snowboarding >> or bicycling. These are all balanced modes of motion and follow >> the same principles. http://www.sheldonbrown.com/brandt/descending.html >> > I've read the article before (and again just now), and it is quite > informative. > I wasn't thinking about countersteer, I was thinking about during > the turn, well after it is initiated. Do you apply some conscious > pressure one way or the other, or is it exactly the same as when you > ride no hands, just that your hands are there to maintain control > should you hit a pebble or something? I don't know where one would apply conscious pressure when cornering. I think the picture at: http://www.sheldonbrown.com/brandt/ typifies the stance. You'll notice that the wheels are in a nearly straight line, indicating this is a high speed turn. Hairpin turns are tediously slow and need to be cut in such a way to make them less tight. > It is clear from riding no-hands that the steering finds it's own > angle, but I wonder if man-handling it to some other slightly > different angle might be advantageous in some way. You must be thinking of the M/C pseudo racers, who shift their butts from side to side when entering curves, leaning at about 30 degrees from the perpendicular to the road. Jobst Brandt       Date: 30 Oct 2007 08:42:33 From: bjw@mambo.ucolick.org Subject: Re: Gyroscopic forces revisited On Oct 29, 12:32 pm, Ben C <spams...@spam.eggs > wrote: > On 2007-10-29, b...@mambo.ucolick.org <b...@mambo.ucolick.org> wrote: > > > This discussion is going nowhere because you guys > > aren't stating this problem in simple generally > > accepted physics terms. > > I think there is some confusion (at least in my mind) about whether > camber-thrust is another name for the penny-steering effect, or > something else, that only applies when you've got deformable tyres and > is only significant if they're fairly wide. Yes. This is why I refused to use the phrase camber thrust. I think it's confusing the issue. I understand the penny-steering effect. When you add fat tires, or bicycle steering geometry with trail and wheel flop, things get a little more complicated. > > If the steering angle of the front wheel is small, > > the bicycle is not being steered by tire scrub, but > > by lean angle. This is exactly analogous to the > > rolling tilted penny. This means that the bike or > > penny will describe a circlular segment of radius r, > > where the centripetal acceleration a = v^2/r. > > > The acceleration points inward toward the center of > > the circle. The acceleration is from a force exerted > > by the ground on the bike. That force is exerted > > along the axis of the bike; it has a vertical and > > horizontal component. The vertical component > > balances the gravity force on the bike+rider, Fgrav=mg. > > The horizontal component provides the centripetal > > acceleration. Assuming flat ground and that the > > bike is tilted by an angle theta from vertical, > > the centripetal force is Fcen = mg tan(theta). > > You can get the radius of the turn from > > v^2/r = g tan(theta), so r = v^2 / (g tan theta) > > The radius just depends on speed and lean angle. > > For example, if you go round a turn at 10 m/s > > with a lean angle of 25 deg, your turn radius is 21 m. > > So does this steering effect depend on the angle between the wheel and > the ground, or on the angle between the line from the wheel's CG to the > contact point and the ground? (Speaking about rolling coins at least) The angle from contact point to center-of-mass. Another way of looking at this is that the gravity force, when the CG is not over the contact patch, exerts a torque that changes the angular momentum direction of the coin. This is easy to test experimentally with coins. Take a quarter and use scotch tape to tape two nickels to one side, creating an asymmetrically weighted wheel. (In the UK, you'll have to use different denominations and whatever the British word is for sticky tape.) Make sure the tape doesn't stick out so the quarter can roll smoothly. Now try rolling this thing straight on the floor, and try rolling an unadorned quarter for comparison. I just tried this and while of course you can't roll either of them perfectly straight with your hand, it's clear that the asymmetrically weighted coin always deviates from straight more quickly, turning faster, in a tighter radius, with the weight inward. Ben         Date: 30 Oct 2007 13:38:09 From: Tim McNamara Subject: Re: Gyroscopic forces revisited In article <1193733753.797944.288690@v29g2000prd.googlegroups.com >, "bjw@mambo.ucolick.org" <bjw@mambo.ucolick.org > wrote: > On Oct 29, 12:32 pm, Ben C <spams...@spam.eggs> wrote: > > On 2007-10-29, b...@mambo.ucolick.org <b...@mambo.ucolick.org> > > wrote: > > > > > This discussion is going nowhere because you guys aren't stating > > > this problem in simple generally accepted physics terms. > > > > I think there is some confusion (at least in my mind) about whether > > camber-thrust is another name for the penny-steering effect, or > > something else, that only applies when you've got deformable tyres > > and is only significant if they're fairly wide. > > Yes. This is why I refused to use the phrase camber thrust. I think > it's confusing the issue. I understand the penny-steering effect. > When you add fat tires, or bicycle steering geometry with trail and > wheel flop, things get a little more complicated. > > > > If the steering angle of the front wheel is small, the bicycle is > > > not being steered by tire scrub, but by lean angle. This is > > > exactly analogous to the rolling tilted penny. This means that > > > the bike or penny will describe a circlular segment of radius r, > > > where the centripetal acceleration a = v^2/r. > > > > > The acceleration points inward toward the center of the circle. > > > The acceleration is from a force exerted by the ground on the > > > bike. That force is exerted along the axis of the bike; it has a > > > vertical and horizontal component. The vertical component > > > balances the gravity force on the bike+rider, Fgrav=mg. The > > > horizontal component provides the centripetal acceleration. > > > Assuming flat ground and that the bike is tilted by an angle > > > theta from vertical, the centripetal force is Fcen = mg > > > tan(theta). You can get the radius of the turn from v^2/r = g > > > tan(theta), so r = v^2 / (g tan theta) The radius just depends on > > > speed and lean angle. For example, if you go round a turn at 10 > > > m/s with a lean angle of 25 deg, your turn radius is 21 m. > > > > So does this steering effect depend on the angle between the wheel > > and the ground, or on the angle between the line from the wheel's > > CG to the contact point and the ground? > > (Speaking about rolling coins at least) The angle from contact point > to center-of-mass. Another way of looking at this is that the > gravity force, when the CG is not over the contact patch, exerts a > torque that changes the angular momentum direction of the coin. Which seems to me would require little if any scrub and allows coins to act this way despite a very low coefficient of friction with the planar surface which limits "scrub" and lateral reaction forces. The reason I brought up my question is that it points out a problem with using the rolling coin metaphor to explain bicycle steering. The bicycle has two rolling hoops out of plane with each other, at an angle to the ground, and rolling through different turn radii. Not to mention a deformable high-friction contact patch. Maybe it'd be simpler to just talk about bikes.           Date: 30 Oct 2007 19:15:02 From: Subject: Re: Gyroscopic forces revisited Tim McNamara writes: >>>> If the steering angle of the front wheel is small, the bicycle is >>>> not being steered by tire scrub, but by lean angle. This is >>>> exactly analogous to the rolling tilted penny. This means that >>>> the bike or penny will describe a circular segment of radius r, >>>> where the centripetal acceleration a = v^2/r. >>>> The acceleration points inward toward the center of the circle. >>>> The acceleration is from a force exerted by the ground on the >>>> bike. That force is exerted along the axis of the bike; it has a >>>> vertical and horizontal component. The vertical component >>>> balances the gravity force on the bike+rider, Fgrav=mg. The >>>> horizontal component provides the centripetal acceleration. >>>> Assuming flat ground and that the bike is tilted by an angle >>>> theta from vertical, the centripetal force is Fcen = mg >>>> tan(theta). You can get the radius of the turn from v^2/r = g >>>> tan(theta), so r = v^2 / (g tan theta) The radius just depends on >>>> speed and lean angle. For example, if you go round a turn at 10 >>>> m/s with a lean angle of 25 deg, your turn radius is 21 m. >>> So does this steering effect depend on the angle between the wheel >>> and the ground, or on the angle between the line from the wheel's >>> CG to the contact point and the ground? >> (Speaking about rolling coins at least) The angle from contact >> point to center-of-mass. Another way of looking at this is that >> the gravity force, when the CG is not over the contact patch, >> exerts a torque that changes the angular momentum direction of the >> coin. > Which seems to me would require little if any scrub and allows coins > to act this way despite a very low coefficient of friction with the > planar surface which limits "scrub" and lateral reaction forces. > The reason I brought up my question is that it points out a problem > with using the rolling coin metaphor to explain bicycle steering. > The bicycle has two rolling hoops out of plane with each other, at > an angle to the ground, and rolling through different turn radii. > Not to mention a deformable high-friction contact patch. Maybe it'd > be simpler to just talk about bikes. Let's look at this in another situation. Merry-go-rounds of old were larger in diameter and went fast enough that it was hard to catch the golden ring on the ring dispenser as the "horseback" rider reached for it, especially if the horse was at the top of its stride or bottom. These horses were tilted so that at running speed of the ride, the connecting rod (brass covered pole) was tilted so that its projection to the deck was in the plum line, about 10 degrees from vertical. This way it would not have as much friction with the tube into which it projected as it went up and down. Let's stand on that deck astride a bicycle with foot on a support so that the bicycle can be balanced leaning at the appropriate 10 degrees as the ride rotates. The net load of the bicycle and rider is in the same position it would be if this were a road and the rider were progressing at about 15mph (about the speed of the merry-go-round). I see no scrub or torque at either wheel although all other conditions of riding a bicycle in such a curve were occurring. If that isn't good enough, lets say there are no obstacles on the surface of this merry-go-round other than a painted broad stripe. If the rider were to start riding on this stripe, I see no change other than the lean angle, that would depend on whether riding with or opposite to the rotation of the deck. Let's make that stipe with matte white paint so we can see "scrub" and whatever kind of slip the tires are doing. Jobst Brandt         Date: 30 Oct 2007 03:53:03 From: Ben C Subject: Re: Gyroscopic forces revisited On 2007-10-30, bjw@mambo.ucolick.org <bjw@mambo.ucolick.org > wrote: > On Oct 29, 12:32 pm, Ben C <spams...@spam.eggs> wrote: >> On 2007-10-29, b...@mambo.ucolick.org <b...@mambo.ucolick.org> wrote: >> >> > This discussion is going nowhere because you guys >> > aren't stating this problem in simple generally >> > accepted physics terms. >> >> I think there is some confusion (at least in my mind) about whether >> camber-thrust is another name for the penny-steering effect, or >> something else, that only applies when you've got deformable tyres and >> is only significant if they're fairly wide. > > Yes. This is why I refused to use the phrase > camber thrust. I think it's confusing the issue. > I understand the penny-steering effect. I think I'm starting to now too, thanks to your explanations. [...] >> So does this steering effect depend on the angle between the wheel and >> the ground, or on the angle between the line from the wheel's CG to the >> contact point and the ground? > > (Speaking about rolling coins at least) The angle > from contact point to center-of-mass. Another > way of looking at this is that the gravity force, > when the CG is not over the contact patch, exerts > a torque that changes the angular momentum > direction of the coin. > > This is easy to test experimentally with coins. > Take a quarter and use scotch tape to tape two > nickels to one side, creating an asymmetrically > weighted wheel. (In the UK, you'll have to > use different denominations and whatever the > British word is for sticky tape.) Make sure the > tape doesn't stick out so the quarter can roll > smoothly. Now try rolling this thing straight > on the floor, and try rolling an unadorned > quarter for comparison. > > I just tried this and while of course you can't > roll either of them perfectly straight with > your hand, it's clear that the asymmetrically > weighted coin always deviates from straight more > quickly, turning faster, in a tighter radius, > with the weight inward. Yes but then the weight will cause it to lean, so it's hard to say whether it's the lean that's causing it to steer, or the uneven weight.           Date: 31 Oct 2007 00:18:25 From: Michael Press Subject: Re: Gyroscopic forces revisited In article <slrnfids8v.t6q.spamspam@bowser.marioworld >, Ben C <spamspam@spam.eggs > wrote: > On 2007-10-30, bjw@mambo.ucolick.org <bjw@mambo.ucolick.org> wrote: > > On Oct 29, 12:32 pm, Ben C <spams...@spam.eggs> wrote: > >> On 2007-10-29, b...@mambo.ucolick.org <b...@mambo.ucolick.org> wrote: > >> > >> > This discussion is going nowhere because you guys > >> > aren't stating this problem in simple generally > >> > accepted physics terms. > >> > >> I think there is some confusion (at least in my mind) about whether > >> camber-thrust is another name for the penny-steering effect, or > >> something else, that only applies when you've got deformable tyres and > >> is only significant if they're fairly wide. > > > > Yes. This is why I refused to use the phrase > > camber thrust. I think it's confusing the issue. > > I understand the penny-steering effect. > > I think I'm starting to now too, thanks to your explanations. > > [...] > >> So does this steering effect depend on the angle between the wheel and > >> the ground, or on the angle between the line from the wheel's CG to the > >> contact point and the ground? > > > > (Speaking about rolling coins at least) The angle > > from contact point to center-of-mass. Another > > way of looking at this is that the gravity force, > > when the CG is not over the contact patch, exerts > > a torque that changes the angular momentum > > direction of the coin. > > > > This is easy to test experimentally with coins. > > Take a quarter and use scotch tape to tape two > > nickels to one side, creating an asymmetrically > > weighted wheel. (In the UK, you'll have to > > use different denominations and whatever the > > British word is for sticky tape.) Make sure the > > tape doesn't stick out so the quarter can roll > > smoothly. Now try rolling this thing straight > > on the floor, and try rolling an unadorned > > quarter for comparison. > > > > I just tried this and while of course you can't > > roll either of them perfectly straight with > > your hand, it's clear that the asymmetrically > > weighted coin always deviates from straight more > > quickly, turning faster, in a tighter radius, > > with the weight inward. > > Yes but then the weight will cause it to lean, so it's hard to say > whether it's the lean that's causing it to steer, or the uneven weight. Simplify. Think of a cone rolling on a flat surface. -- Michael Press       Date: 29 Oct 2007 01:19:32 From: bjw@mambo.ucolick.org Subject: Re: Gyroscopic forces revisited > The acceleration points inward toward the center of > the circle. The acceleration is from a force exerted > by the ground on the bike. That force is exerted > along the axis of the bike; it has a vertical and > horizontal component. The vertical component > balances the gravity force on the bike+rider, Fgrav=mg. > The horizontal component provides the centripetal > acceleration. Assuming flat ground and that the > bike is tilted by an angle theta from vertical, > the centripetal force is Fcen = mg tan(theta). > You can get the radius of the turn from > v^2/r = g tan(theta), so r = v^2 / (g tan theta) > The radius just depends on speed and lean angle. > For example, if you go round a turn at 10 m/s > with a lean angle of 25 deg, your turn radius is 21 m. In case it's not clear, the horizontal component of this force comes from friction between tires and road. That friction has an upper limit of Ffric = mu* mg, where mu is the coefficient of friction. If the friction isn't large enough to supply the centripetal force, then you slide out and low-side the bike. Thus it's impossible to lean a bike to arbitrarily high lean angle, as most of us have probably discovered at one time or another. Ben   Date: 25 Oct 2007 08:43:59 From: joseph.santaniello@gmail.com Subject: Re: Gyroscopic forces revisited On Oct 25, 9:15 am, jobst.bra...@stanfordalumni.org wrote: > Joseph Santaniello writes: > >>> Two weeks ago I converted an old flatbar into an upright, higher > >>> stem, bars that come back, B-66 saddle. Now, miraculously, I can > >>> ride with no hands. When I tried with the flatbar, I had to shift > >>> my body up and back because I was leaning so far forward, but if I > >>> am already up and back, like I am now, I just let go and I am > >>> fine. > >>> No difference in wheel size, no difference in speed, but big > >>> difference in COG. COG is not just height, but longitudinal and > >>> lateral weight, too, or so I was told when I had to make such > >>> calculations. > >> This is more likely a skill issue in simultaneously letting go of > >> your hands while also repositioning your body. It is much easier > >> to initiate no hands if your fingertips are just resting lightly on > >> the handlebars rather than supporting weight. While you are > >> correct that the higher COG does make a difference, it is the same > >> COG riding no hands no matter where the bars are placed (after all, > >> you are not using them). > > The COG can move when riding no hands vs with the bars. On my bike > > my upper body is leaned forward at maybe 45deg and my arms are > > forward some amount when I ride holding the bars. If I want I can > > ride no hands by just letting go and keeping my body leaned over, > > but that isn't too comfortable. So I sit up with my upper body > > vertical and arms to the side. I'm sure this moves the COG at least > > 20cm or so. > > But I agree that the COG placement probably isn't the cause of his > > previous difficulties. > > As I mentioned in the past, all my relatively straight top speed > descents are done essentially no-hands, at speeds at times exceeding > 60mph. At such times my hands are on the bar stem, elbows on knees > that are against the top tube and my back as low as I can get without > my chin bouncing on my hands. If there is no cross wind, this is a > highly stable position in which I can lean steeply into corners. > > Steering is done by moving my upper body to either side of center, the > motion being slight and not apparent to an observer on a following > bicycle. It is in this position that "countersteer" is most apparent > because taking curves must be done by first leaning out of the turn > and then neutralizing that motion by returning to center. > > Jobst Brandt I have witnessed descending in a similar fashion, but with both hands behind the rider's butt. This position is significantly more aerodynamic, as evidenced by the riders marked speed increase upon moving the arms back like that. When you are cornering tight turns like a hairpin, do you make conscious manual adjustments to your steering angle? Joseph     Date: 25 Oct 2007 17:41:25 From: Subject: Re: Gyroscopic forces revisited Joseph Santaniello writes: >>>>> Two weeks ago I converted an old flatbar into an upright, higher >>>>> stem, bars that come back, B-66 saddle. Now, miraculously, I can >>>>> ride with no hands. When I tried with the flatbar, I had to shift >>>>> my body up and back because I was leaning so far forward, but if I >>>>> am already up and back, like I am now, I just let go and I am >>>>> fine. >>>>> No difference in wheel size, no difference in speed, but big >>>>> difference in COG. COG is not just height, but longitudinal and >>>>> lateral weight, too, or so I was told when I had to make such >>>>> calculations. >>>> This is more likely a skill issue in simultaneously letting go of >>>> your hands while also repositioning your body. It is much easier >>>> to initiate no hands if your fingertips are just resting lightly on >>>> the handlebars rather than supporting weight. While you are >>>> correct that the higher COG does make a difference, it is the same >>>> COG riding no hands no matter where the bars are placed (after all, >>>> you are not using them). >>> The COG can move when riding no hands vs with the bars. On my bike >>> my upper body is leaned forward at maybe 45deg and my arms are >>> forward some amount when I ride holding the bars. If I want I can >>> ride no hands by just letting go and keeping my body leaned over, >>> but that isn't too comfortable. So I sit up with my upper body >>> vertical and arms to the side. I'm sure this moves the COG at least >>> 20cm or so. >>> But I agree that the COG placement probably isn't the cause of his >>> previous difficulties. >> As I mentioned in the past, all my relatively straight top speed >> descents are done essentially no-hands, at speeds at times exceeding >> 60mph. At such times my hands are on the bar stem, elbows on knees >> that are against the top tube and my back as low as I can get without >> my chin bouncing on my hands. If there is no cross wind, this is a >> highly stable position in which I can lean steeply into corners. >> Steering is done by moving my upper body to either side of center, the >> motion being slight and not apparent to an observer on a following >> bicycle. It is in this position that "countersteer" is most apparent >> because taking curves must be done by first leaning out of the turn >> and then neutralizing that motion by returning to center. > I have witnessed descending in a similar fashion, but with both > hands behind the rider's butt. This position is significantly more > aerodynamic, as evidenced by the riders marked speed increase upon > moving the arms back like that. I doubt that this does what you say it does because frontal area is what causes drag, the closing area behind the rider being a bluff body and has no streamlining (as a long pointed tail). Frontal area is smaller with the shoulders pulled in in front of the rider with arms in chest, elbows on knees. > When you are cornering tight turns like a hairpin, do you make > conscious manual adjustments to your steering angle? I don't understand what you mean by steering angle. All turns are made with countersteer whether running, ice skating, snowboarding or bicycling. These are all balanced modes of motion and follow the same principles. http://www.sheldonbrown.com/brandt/descending.html Jobst Brandt       Date: 29 Oct 2007 01:05:34 From: bjw@mambo.ucolick.org Subject: Re: Gyroscopic forces revisited On Oct 28, 3:30 pm, Ben C <spams...@spam.eggs > wrote: > On 2007-10-28, joseph.santanie...@gmail.com <joseph.santanie...@gmail.com> wrote: > > > Why? Isn't camber-thrust just dependent upon the wheel's lean angle? > > Good question. Thinking back to Jon_C's diagram and the idea that > camber-thrust is a consequence of the tyre deforming as if it were part > of a cone, it looks like you should be right. > > But I'm not sure I understand camber-thrust very well. A penny rolling > along and tilted to one side spins round in a tight circle, but it's not > deforming like a tyre. So how does that work? This discussion is going nowhere because you guys aren't stating this problem in simple generally accepted physics terms. If the steering angle of the front wheel is small, the bicycle is not being steered by tire scrub, but by lean angle. This is exactly analogous to the rolling tilted penny. This means that the bike or penny will describe a circlular segment of radius r, where the centripetal acceleration a = v^2/r. The acceleration points inward toward the center of the circle. The acceleration is from a force exerted by the ground on the bike. That force is exerted along the axis of the bike; it has a vertical and horizontal component. The vertical component balances the gravity force on the bike+rider, Fgrav=mg. The horizontal component provides the centripetal acceleration. Assuming flat ground and that the bike is tilted by an angle theta from vertical, the centripetal force is Fcen = mg tan(theta). You can get the radius of the turn from v^2/r = g tan(theta), so r = v^2 / (g tan theta) The radius just depends on speed and lean angle. For example, if you go round a turn at 10 m/s with a lean angle of 25 deg, your turn radius is 21 m. That's pretty much it. I think other factors like the size of your tire are negligible as long as you stick to relatively narrow bicycle tires. Once you start talking about either tires with a square profile; or vehicles that don't lean, such as cars; or walking-speed bicycles; then the vehicle can be turned by tire scrub forces and you have to think about ways of accounting for it. But for a bicycle at speed, those are not necessary. Ben         Date: 29 Oct 2007 14:32:17 From: Ben C Subject: Re: Gyroscopic forces revisited On 2007-10-29, bjw@mambo.ucolick.org <bjw@mambo.ucolick.org > wrote: > On Oct 28, 3:30 pm, Ben C <spams...@spam.eggs> wrote: >> On 2007-10-28, joseph.santanie...@gmail.com <joseph.santanie...@gmail.com> wrote: >> >> > Why? Isn't camber-thrust just dependent upon the wheel's lean angle? >> >> Good question. Thinking back to Jon_C's diagram and the idea that >> camber-thrust is a consequence of the tyre deforming as if it were part >> of a cone, it looks like you should be right. >> >> But I'm not sure I understand camber-thrust very well. A penny rolling >> along and tilted to one side spins round in a tight circle, but it's not >> deforming like a tyre. So how does that work? > > This discussion is going nowhere because you guys > aren't stating this problem in simple generally > accepted physics terms. I think there is some confusion (at least in my mind) about whether camber-thrust is another name for the penny-steering effect, or something else, that only applies when you've got deformable tyres and is only significant if they're fairly wide. > If the steering angle of the front wheel is small, > the bicycle is not being steered by tire scrub, but > by lean angle. This is exactly analogous to the > rolling tilted penny. This means that the bike or > penny will describe a circlular segment of radius r, > where the centripetal acceleration a = v^2/r. > > The acceleration points inward toward the center of > the circle. The acceleration is from a force exerted > by the ground on the bike. That force is exerted > along the axis of the bike; it has a vertical and > horizontal component. The vertical component > balances the gravity force on the bike+rider, Fgrav=mg. > The horizontal component provides the centripetal > acceleration. Assuming flat ground and that the > bike is tilted by an angle theta from vertical, > the centripetal force is Fcen = mg tan(theta). > You can get the radius of the turn from > v^2/r = g tan(theta), so r = v^2 / (g tan theta) > The radius just depends on speed and lean angle. > For example, if you go round a turn at 10 m/s > with a lean angle of 25 deg, your turn radius is 21 m. > > That's pretty much it. I think other factors like > the size of your tire are negligible as long as you > stick to relatively narrow bicycle tires. So does this steering effect depend on the angle between the wheel and the ground, or on the angle between the line from the wheel's CG to the contact point and the ground?       Date: 28 Oct 2007 13:02:36 From: joseph.santaniello@gmail.com Subject: Re: Gyroscopic forces revisited On Oct 28, 7:41 pm, Ben C <spams...@spam.eggs > wrote: > On 2007-10-28, joseph.santanie...@gmail.com <joseph.santanie...@gmail.com> wrote: > [...] > > > And this camber-thrust issue. How big is this camber-thrust force as a > > percent of the total force at the tire, and does this percent change > > based on turn/lean angle? And if it does, is there any reason to > > prefer one force over the other in terms of traction and how the tire > > interacts with the road? In other words, does leaning the bike less > > (and steering more) rely more on slip-angle, and leaning the bike more > > rely to a lesser degree on slip-angle and a greater degree on camber- > > thrust? > > I think so, that was what I gathered from the Tony Foale article. Duh. I guess I should have read that before... So, does that mean there is any reason to prefer camber-thrust to have a greater portion of the task? It seems it would, as slip angle generated forces have a larger drag element. How do you figure out what the camber thrust force values are for given lean angles? What sort of slip angles do bicycle tires have? It would be interesting to see a graph showing the relative importance of each component for different lean angles and speeds (and by extension turn radii). Joseph         Date: 28 Oct 2007 17:37:42 From: Ben C Subject: Re: Gyroscopic forces revisited On 2007-10-28, joseph.santaniello@gmail.com <joseph.santaniello@gmail.com > wrote: > On Oct 28, 7:41 pm, Ben C <spams...@spam.eggs> wrote: >> On 2007-10-28, joseph.santanie...@gmail.com <joseph.santanie...@gmail.com> wrote: >> [...] >> >> > And this camber-thrust issue. How big is this camber-thrust force as a >> > percent of the total force at the tire, and does this percent change >> > based on turn/lean angle? And if it does, is there any reason to >> > prefer one force over the other in terms of traction and how the tire >> > interacts with the road? In other words, does leaning the bike less >> > (and steering more) rely more on slip-angle, and leaning the bike more >> > rely to a lesser degree on slip-angle and a greater degree on camber- >> > thrust? >> >> I think so, that was what I gathered from the Tony Foale article. > > Duh. I guess I should have read that before... > > So, does that mean there is any reason to prefer camber-thrust to have > a greater portion of the task? > > It seems it would, as slip angle generated forces have a larger drag > element. Perhaps although on a motorbike or descending a hill on a bicycle, you're probably slowing down for the corners anyway. A bit of drag just saves the brakes. More important questions may be which is more grippy and which is more controllable. > How do you figure out what the camber thrust force values are for > given lean angles? I don't know. > What sort of slip angles do bicycle tires have? If camber-thrust is insignificant for bicycles (I think Jobst was saying it was) then it shouldn't be too hard to work out the slip angle for a given radius and speed.       Date: 26 Oct 2007 20:35:20 From: dabac Subject: Re: Gyroscopic forces revisited Joseph Santaniello Wrote: > > > > I have witnessed descending in a similar fashion, but with both > > hands behind the rider's butt. This position is significantly more > > aerodynamic, as evidenced by the riders marked speed increase upon > > moving the arms back like that. jobst.brandt@stanfordalumni.org Wrote: > > I doubt that this does what you say it does because frontal area is > what causes drag, the closing area behind the rider being a bluff body > and has no streamlining (as a long pointed tail). Frontal area is > smaller with the shoulders pulled in in front of the rider with arms > in > chest, elbows on knees. Then why does the tuck used by ski jumpers and speed skaters look the way it does? According to your line of reasoning they should be faster if they kept their arms against the chest. How strange that none of those guys have discovered such an easy way to gain an edge... -- dabac         Date: 26 Oct 2007 13:51:15 From: Tim McNamara Subject: Re: Gyroscopic forces revisited In article <dabac.2z247b@no-mx.forums.cyclingforums.com >, dabac <dabac.2z247b@no-mx.forums.cyclingforums.com > wrote: > Joseph Santaniello Wrote: > > > > > > > I have witnessed descending in a similar fashion, but with both > > > hands behind the rider's butt. This position is significantly > > > more aerodynamic, as evidenced by the riders marked speed > > > increase upon moving the arms back like that. > > jobst.brandt@stanfordalumni.org Wrote: > > > > I doubt that this does what you say it does because frontal area is > > what causes drag, the closing area behind the rider being a bluff > > body and has no streamlining (as a long pointed tail). Frontal > > area is smaller with the shoulders pulled in in front of the rider > > with arms in chest, elbows on knees. > > Then why does the tuck used by ski jumpers and speed skaters look the > way it does? According to your line of reasoning they should be > faster if they kept their arms against the chest. How strange that > none of those guys have discovered such an easy way to gain an > edge... Never having speed skated or ski jumped, I am speculating here. My understanding about ski jumping is that the position they adopt is one intended to allow small corrections in midair. My understanding about the speed skating position is that it's about ergonomics as well as aerodynamics. Aerodynamics is only part of the equation. I recall reading about Miguel Indurain's trip to the wind tunnel. They found a position that would be much faster than the one he usually used. He then pointed out that he couldn't actually pedal the bike in that position.         Date: 26 Oct 2007 16:10:00 From: Subject: Re: Gyroscopic forces revisited who? writes: >>> I have witnessed descending in a similar fashion, but with both >>> hands behind the rider's butt. This position is significantly more >>> aerodynamic, as evidenced by the riders marked speed increase upon >>> moving the arms back like that. >> I doubt that this does what you say it does because frontal area is >> what causes drag, the closing area behind the rider being a bluff >> body and has no streamlining (as a long pointed tail). Frontal >> area is smaller with the shoulders pulled in in front of the rider >> with arms in chest, elbows on knees. > Then why does the tuck used by ski jumpers and speed skaters look > the way it does? According to your line of reasoning they should be > faster if they kept their arms against the chest. How strange that > none of those guys have discovered such an easy way to gain an > edge... You cant ice skate with your arms tucked into you stomach and arms in front add to drag slightly, skaters wearing slick suits for that reason. Similarly ski jumping relies on a smooth frontal area and with arms to the rear, one can make corrections in flight in the event of a breeze or imbalance, something not reasonably possible with arms doubled up on ones chest. You might notice that the bicyclist does not have a smooth frontal face when descending in a crouch. Jobst Brandt           Date: 26 Oct 2007 09:15:13 From: Tom Nakashima Subject: Re: Gyroscopic forces revisited <jobst.brandt@stanfordalumni.org > wrote in message news:47221158$0$14068$742ec2ed@news.sonic.net... > Similarly ski jumping relies on a smooth frontal area and > with arms to the rear, one can make corrections in flight in the event > of a breeze or imbalance, something not reasonably possible with arms > doubled up on ones chest. > Jobst Brandt http://www.funnyclick.net/data/sumo-ski-jumping.jpg -tom       Date: 25 Oct 2007 11:02:55 From: Tom Nakashima Subject: Re: Gyroscopic forces revisited <jobst.brandt@stanfordalumni.org > wrote in message news:4720d545$0$14112$742ec2ed@news.sonic.net... > Joseph Santaniello writes: > >> I have witnessed descending in a similar fashion, but with both >> hands behind the rider's butt. This position is significantly more >> aerodynamic, as evidenced by the riders marked speed increase upon >> moving the arms back like that. > > I doubt that this does what you say it does because frontal area is > what causes drag, the closing area behind the rider being a bluff body > and has no streamlining (as a long pointed tail). Frontal area is > smaller with the shoulders pulled in in front of the rider with arms in > chest, elbows on knees. > Jobst Brandt The banned style of Graeme Obree. http://www.youtube.com/watch?v=ml6KT5MArC8 -tom   Date: 24 Oct 2007 23:53:55 From: joseph.santaniello@gmail.com Subject: Re: Gyroscopic forces revisited On Oct 25, 12:00 am, velodancer <commerc...@yahoo.com > wrote: > On Oct 24, 2:26 pm, vey <jun...@ericvey.com> wrote: > > > Two weeks ago I converted an old flatbar into an upright, higher stem, > > bars that come back, B-66 saddle. Now, miraculously, I can ride with no > > hands. When I tried with the flatbar, I had to shift my body up and back > > because I was leaning so far forward, but if I am already up and back, > > like I am now, I just let go and I am fine. > > > No difference in wheel size, no difference in speed, but big difference > > in COG. COG is not just height, but longitudinal and lateral weight, > > too, or so I was told when I had to make such calculations. > > This is more likely a skill issue in simultaneously letting go of your > hands while also repositioning your body. It is much easier to > initiate no hands if your fingertips are just resting lightly on the > handlebars rather than supporting weight. While you are correct that > the higher COG does make a difference, it is the same COG riding no > hands no matter where the bars are placed (after all, you are not > using them). The COG can move when riding no hands vs with the bars. On my bike my upper body is leaned forward at maybe 45deg and my arms are forward some amount when I ride holding the bars. If I want I can ride no hands by just letting go and keeping my body leaned over, but that isn't too comfortable. So I sit up with my upper body vertical and arms to the side. I'm sure this moves the COG at least 20cm or so. But I agree that the COG placement probably isn't the cause of his previous difficulties. Joseph     Date: 25 Oct 2007 07:15:52 From: Subject: Re: Gyroscopic forces revisited Joseph Santaniello writes: >>> Two weeks ago I converted an old flatbar into an upright, higher >>> stem, bars that come back, B-66 saddle. Now, miraculously, I can >>> ride with no hands. When I tried with the flatbar, I had to shift >>> my body up and back because I was leaning so far forward, but if I >>> am already up and back, like I am now, I just let go and I am >>> fine. >>> No difference in wheel size, no difference in speed, but big >>> difference in COG. COG is not just height, but longitudinal and >>> lateral weight, too, or so I was told when I had to make such >>> calculations. >> This is more likely a skill issue in simultaneously letting go of >> your hands while also repositioning your body. It is much easier >> to initiate no hands if your fingertips are just resting lightly on >> the handlebars rather than supporting weight. While you are >> correct that the higher COG does make a difference, it is the same >> COG riding no hands no matter where the bars are placed (after all, >> you are not using them). > The COG can move when riding no hands vs with the bars. On my bike > my upper body is leaned forward at maybe 45deg and my arms are > forward some amount when I ride holding the bars. If I want I can > ride no hands by just letting go and keeping my body leaned over, > but that isn't too comfortable. So I sit up with my upper body > vertical and arms to the side. I'm sure this moves the COG at least > 20cm or so. > But I agree that the COG placement probably isn't the cause of his > previous difficulties. As I mentioned in the past, all my relatively straight top speed descents are done essentially no-hands, at speeds at times exceeding 60mph. At such times my hands are on the bar stem, elbows on knees that are against the top tube and my back as low as I can get without my chin bouncing on my hands. If there is no cross wind, this is a highly stable position in which I can lean steeply into corners. Steering is done by moving my upper body to either side of center, the motion being slight and not apparent to an observer on a following bicycle. It is in this position that "countersteer" is most apparent because taking curves must be done by first leaning out of the turn and then neutralizing that motion by returning to center. Jobst Brandt       Date: 28 Oct 2007 13:10:57 From: joseph.santaniello@gmail.com Subject: Re: Gyroscopic forces revisited On Oct 28, 8:59 pm, Ben C <spams...@spam.eggs > wrote: > On 2007-10-28, Ben C <spams...@spam.eggs> wrote: > > > On 2007-10-28, joseph.santanie...@gmail.com <joseph.santanie...@gmail.com> wrote: > [...] > >> I think it is true that a lean angle either side of "natural" requires > >> some manual steering torque, but the steering angle I still think is > >> different. For a given lean angle and steering angle, the front patch > >> is at particular place on the wheel. AS the lean increases, keeping > >> the stering angle the same, the patch moves forward on the wheel. > >> I think this means the arc described by the two patches changes, > >> meaning that if the arc is to remain the same, the steering angle has > >> to change. > > > Furthermore, if you read that link posted by Jon_C, the more you lean > > the bike the more camber thrust you get, and therefore the less slip > > angle you need, to the point where you might even have to steer slightly > > the wrong way to get around the corner. > > Sorry, I got that wrong-- I think that only applies to "real" CG lean as > opposed to keeping the CG the same and making the bike more or less > upright. Why? Isn't camber-thrust just dependent upon the wheel's lean angle? So once camber-thrust is used up, slip angle needs to supply the rest? CG doesn't have anything to do with it, other than whether you crash or not! > In other words, apart from what you've said about steering geometry, the > amount of camber thrust and slip angle you get/need depends on curve and > speed, not on rider position. Not at all. Think about those BMX guys doing tricks. They can lean the bike over to some crazy angle (60deg+?) while rolling at a wlking pace (easy with such a low top-tube, they can just stand) and do turns with a very tight radius, with almost no steering turn at all. The same turn upright requires lots of turn and thus slip angle. In both cases the CG is in the same place, but the rider position and thus the lean angle is quite different. Joseph         Date: 28 Oct 2007 17:30:43 From: Ben C Subject: Re: Gyroscopic forces revisited On 2007-10-28, joseph.santaniello@gmail.com <joseph.santaniello@gmail.com > wrote: > On Oct 28, 8:59 pm, Ben C <spams...@spam.eggs> wrote: >> On 2007-10-28, Ben C <spams...@spam.eggs> wrote: [...] >> > Furthermore, if you read that link posted by Jon_C, the more you lean >> > the bike the more camber thrust you get, and therefore the less slip >> > angle you need, to the point where you might even have to steer slightly >> > the wrong way to get around the corner. >> >> Sorry, I got that wrong-- I think that only applies to "real" CG lean as >> opposed to keeping the CG the same and making the bike more or less >> upright. > > Why? Isn't camber-thrust just dependent upon the wheel's lean angle? Good question. Thinking back to Jon_C's diagram and the idea that camber-thrust is a consequence of the tyre deforming as if it were part of a cone, it looks like you should be right. But I'm not sure I understand camber-thrust very well. A penny rolling along and tilted to one side spins round in a tight circle, but it's not deforming like a tyre. So how does that work?       Date: 28 Oct 2007 10:27:32 From: joseph.santaniello@gmail.com Subject: Re: Gyroscopic forces revisited On Oct 28, 4:38 pm, Ben C <spams...@spam.eggs > wrote: > On 2007-10-28, joseph.santanie...@gmail.com <joseph.santanie...@gmail.com> wrote: > > > > > On Oct 28, 10:29 am, Ben C <spams...@spam.eggs> wrote: > >> On 2007-10-26, joseph.santanie...@gmail.com <joseph.santanie...@gmail.com> wrote: > [...] > >> > I'm not talking about the CG to patch angle, that is of course defined > >> > by the arc of the turn and the speed. I'm talking about the lean angle > >> > of the center-line of the bike. In other words the lateral placement > >> > of the CG as happens when you stick a knee out one way or the other. > >> > Sticking your knee out does not change the ground-patch-CG angle, but > >> > it does change the ground-patch-bike angle. > > >> Don't you mean the other way round? Sticking your knee out does change > >> ground-patch-CG angle, since it moves CG, but doesn't change > >> ground-patch-bike angle. > > > The only way the CG "moves" is relative to the bike. The lean (ground- > > patch-CG) stays the same for a given speed and radius. If you change > > the angle, the radius changes. > > Yes, that's all I was trying to say. I misunderstood you originally. You > were talking about sticking your knee out while at the same time > adjusting the bicycle's angle of lean to compensate, so keeping the CG > angle the same. > > I read it as sticking the knee out without compensating, which keeps the > bike angle the same, but changes the CG angle, tightening the turn if > you lean the knee into the corner. > > [...] > > >> I would have thought the steering angle was a given, just like angle > >> between patches and CG, for a given turn-- you've got to go round the > >> same arc without falling off. > > > I think it is true that a lean angle either side of "natural" requires > > some manual steering torque, but the steering angle I still think is > > different. For a given lean angle and steering angle, the front patch > > is at particular place on the wheel. AS the lean increases, keeping > > the stering angle the same, the patch moves forward on the wheel. > > I think this means the arc described by the two patches changes, > > meaning that if the arc is to remain the same, the steering angle has > > to change. > > Furthermore, if you read that link posted by Jon_C, the more you lean > the bike the more camber thrust you get, and therefore the less slip > angle you need, to the point where you might even have to steer slightly > the wrong way to get around the corner. > > So I think you're right, lean angle does definitely alter the amount of > steering lock required. > > > If you hold a bike upright and the steering off to one side at say > > 10deg, the patch is still more or less rigt at the bottom (6 o-clock) > > of the wheel. If you lean the bike keeping the steering the same, the > > patch moves closer to 3 or 9 (depending on which side you look from). > > This at the least changes the steering torque that comes from trail, > > and I think changes the relative angles of the patches (seen from > > above) and thus the arc of the turn. > > That sounds plausible. > > > So the question remains whether there is some optimal angle where > > traction is optimized? > > I have no idea of course... but at the end of the day the tyre has to > provide enough steering force to get you around the corner. Whether it's > doing that with more or less steering lock, perhaps the limit of the > force it can apply is the same anyway. Yes, and as Jobst points out, the force vectors at the patch are the same anyway. Different lean angles just put the patches in different places. That means that in terms of traction, it doesn't really matter what lean angle the bike has, but in practical terms, the "natural" steering angle seems best because as he also points out, it is less susceptible to being upset by bumps. But all of this is assuming that traction is fixed, in a way. Sort of like ice skates, and ignores what if any slip angle plays into this. And this camber-thrust issue. How big is this camber-thrust force as a percent of the total force at the tire, and does this percent change based on turn/lean angle? And if it does, is there any reason to prefer one force over the other in terms of traction and how the tire interacts with the road? In other words, does leaning the bike less (and steering more) rely more on slip-angle, and leaning the bike more rely to a lesser degree on slip-angle and a greater degree on camber- thrust? Joseph         Date: 28 Oct 2007 13:41:17 From: Ben C Subject: Re: Gyroscopic forces revisited On 2007-10-28, joseph.santaniello@gmail.com <joseph.santaniello@gmail.com > wrote: [...] > And this camber-thrust issue. How big is this camber-thrust force as a > percent of the total force at the tire, and does this percent change > based on turn/lean angle? And if it does, is there any reason to > prefer one force over the other in terms of traction and how the tire > interacts with the road? In other words, does leaning the bike less > (and steering more) rely more on slip-angle, and leaning the bike more > rely to a lesser degree on slip-angle and a greater degree on camber- > thrust? I think so, that was what I gathered from the Tony Foale article.   Date: 24 Oct 2007 23:47:56 From: joseph.santaniello@gmail.com Subject: Re: Gyroscopic forces revisited On Oct 24, 11:26 pm, vey <jun...@ericvey.com > wrote: > joseph.santanie...@gmail.com wrote: > > Using a sloped board you can simulate the forces on a riderless bike > > during a turn. (On Mars...) > > I tried an experiment today. On a flat surface, I haven't been able to > ride without hands since I retired my old three speed years ago. Since > then, I've been riding flatbars and if I let go I would just about crash > right then and there. Today, I was able to do it easily. > > Two weeks ago I converted an old flatbar into an upright, higher stem, > bars that come back, B-66 saddle. Now, miraculously, I can ride with no > hands. When I tried with the flatbar, I had to shift my body up and back > because I was leaning so far forward, but if I am already up and back, > like I am now, I just let go and I am fine. > > No difference in wheel size, no difference in speed, but big difference > in COG. COG is not just height, but longitudinal and lateral weight, > too, or so I was told when I had to make such calculations. I think the difference is practice. As you get more comfortable riding no hands you will be able to do the transition easier. I'm sure it was the upper body movement during the transfer that was the culprit, not the placement of the COG. Joseph   Date: 24 Oct 2007 15:00:07 From: velodancer Subject: Re: Gyroscopic forces revisited On Oct 24, 2:26 pm, vey <jun...@ericvey.com > wrote: > Two weeks ago I converted an old flatbar into an upright, higher stem, > bars that come back, B-66 saddle. Now, miraculously, I can ride with no > hands. When I tried with the flatbar, I had to shift my body up and back > because I was leaning so far forward, but if I am already up and back, > like I am now, I just let go and I am fine. > > No difference in wheel size, no difference in speed, but big difference > in COG. COG is not just height, but longitudinal and lateral weight, > too, or so I was told when I had to make such calculations. This is more likely a skill issue in simultaneously letting go of your hands while also repositioning your body. It is much easier to initiate no hands if your fingertips are just resting lightly on the handlebars rather than supporting weight. While you are correct that the higher COG does make a difference, it is the same COG riding no hands no matter where the bars are placed (after all, you are not using them).     Date: 28 Oct 2007 22:07:22 From: Joe Riel Subject: Re: Gyroscopic forces revisited jobst.brandt@stanfordalumni.org writes: > Let's leave "camber thrust" out of this until one can establish that > it has any effect on bicycle cornering. As far as I can find, it is > an effect found mainly in tires with cylindrical tread (wide car tires) > where it has little effect on traction, having insignificant angles. Check out Milliken's camber car: http://www.autoblog.com/2007/06/27/2007-goodwood-festival-of-speed-open-wheel-racers/ It was a test vehicle, not a real race car. But interesting none-the-less. It's cornering ability was supposedly exceptional. -- Joe Riel       Date: 29 Oct 2007 14:01:15 From: Subject: Re: Gyroscopic forces revisited Joe Riel writes: >> Let's leave "camber thrust" out of this until one can establish that >> it has any effect on bicycle cornering. As far as I can find, it is >> an effect found mainly in tires with cylindrical tread (wide car tires) >> where it has little effect on traction, having insignificant angles. > Check out Milliken's camber car: > http://www.autoblog.com/2007/06/27/2007-goodwood-festival-of-speed-open-wheel-racers/ > It was a test vehicle, not a real race car. But interesting > none-the-less. It's cornering ability was supposedly exceptional. I don't see any such vehicles in use in competition. What was the fatal flaw? I worked in race car design and recall that we kept camber as close to zero as practical. Slight negative camber was used to assure the outside curve wheels would go to full width road contact under lateral cornering force. Today that is done by belted tires specifically designed to not tilt the contact patch. Jobst Brandt     Date: 28 Oct 2007 06:02:59 From: joseph.santaniello@gmail.com Subject: Re: Gyroscopic forces revisited On Oct 28, 10:29 am, Ben C <spams...@spam.eggs > wrote: > On 2007-10-26, joseph.santanie...@gmail.com <joseph.santanie...@gmail.com> wrote: > > > On Oct 26, 6:46 pm, jobst.bra...@stanfordalumni.org wrote: > [...] > >> You can't have different angles of lean. As I said you can't fool > >> geometry and gravity. It may appear as a different angle of lean but > >> it isn't. The plumb line from the CG to the road is the same at the > >> same speed around the same curve no matter what the rider does. > > > I'm not talking about the CG to patch angle, that is of course defined > > by the arc of the turn and the speed. I'm talking about the lean angle > > of the center-line of the bike. In other words the lateral placement > > of the CG as happens when you stick a knee out one way or the other. > > Sticking your knee out does not change the ground-patch-CG angle, but > > it does change the ground-patch-bike angle. > > Don't you mean the other way round? Sticking your knee out does change > ground-patch-CG angle, since it moves CG, but doesn't change > ground-patch-bike angle. The only way the CG "moves" is relative to the bike. The lean (ground- patch-CG) stays the same for a given speed and radius. If you change the angle, the radius changes. > > > > When the CG is in the centerline of the bike and the combined forces > > of gravity and turn accleration are working along this centerline to > > the patches, the steering remains straight. Such as when riding > > straight. When the CG is out of the centerline, the steering moves to > > that side. > > > When riding no hands in a constant turn, the steering is NOT straight, > > so necesarily the CG is not on the center-line of the bike. If the CG > > were in the centerline of the bike, the combined forces would be > > acting along the centerline through the patches just as if the bike > > were on a velodrome straight, and the steering angle would be straight > > too. But the steering cannot be straight because we are in the middle > > of a turn! > > > Riding in a turn with hands on the bars it is possible to have the CG > > on the centerline of the bike, but to do so the steering would need to > > be manually turned. > > >> > I think because of the angle of the steering axis and the resulting > >> > trail, the position and orientation of the patches does change if you > >> > hold the steering angle constant. Perhaps only slightly. > > >> That has no effect on the path taken. Besides, when leaning off the > >> plane of the bicycle, corrective force is required on the handlebar to > >> make up for the odd geometry. This is less apparent on an M/C because > >> it is far heavier but still requires steering force to hold the line. > >> You can't, for instance, ride no-hands when leaning off to one side. > > > That corrective force is what I am interested in. Is there some > > optimal (in terms of traction, or max possible cornering speed) bike > > to ground angle that is NOT the "natural" angle the bike would take > > were it ridden no hands, that of necessity requires some corrective > > force. Or is any lean angle that requires corrective force worse in > > terms of traction than the natural angle? > > If I understand what you're saying, the idea is that to hold a bike in a > turn while keeping CG through bike-centreline requires some steering > torque on the handlebars, and therefore a bit more traction from the > front tyre? I think you see what I am saying, but I don't think it means more traction is required, just that it is different. Whether it needs more or less traction is sort of what I am trying to figure out. > I don't know if that's true. There must still be a steering torque even > at the natural angle, it's just that the rider isn't applying it with > his hands. This is true, the steering torque is applied by the combined cornering force and gravity making the steering flop to the side. > > The best thing for fast cornering is smoothness. The cornering force > dictated by the speed and radius of the turn is an average. Constant > overcorrections will create peaks either side of the average. When a > peak exceeds tyre grip then you're on your ear. > > So the question is what rider position allows the finest control. I > should think for that you do want a position that does require some > manual steering torque (i.e. away from the "natural angle"), since > that's easier to modulate. That's a good point. If you are right on the center in a way, you have to make corrections both ways, but if you have constant pressure, you only have to modulate how much pressure. > > The other issue (which you may have covered) is keeping a good angle > between tyres and road. If you watch Moto GP, they sort of hang off the > inside of the bike in turns. That keeps the bike itself more upright > than it would otherwise be, so it's not cornering on the rim instead of > the tyre. I don't think anyone ever gets such extreme angles of lean on > a bicycle though. > > But leaning across the bicycle to the outside of the turn is probably a > bad idea because it makes the angle between the wheels and the road more > acute. > > >> > Suppose you walk a bike perfectly upright along some arc. The > >> > steering will need to be turned to some angle to achieve this. If > >> > you then walk the bike along that same arc, but with some arbitrary > >> > angle of lean, I contend that the steering angle (in terms of stem > >> > to center- centerline) will be different. Maybe the patches will > >> > describe the same arc, but the steering will be at a different > >> > angle, and the front patch will be at a different place on the > >> > wheel. > > Maybe the steering angle is the same, it's just that at lean angles > either side of "natural", a manual steering torque is required to hold > the correct steering angle. > > I would have thought the steering angle was a given, just like angle > between patches and CG, for a given turn-- you've got to go round the > same arc without falling off. I think it is true that a lean angle either side of "natural" requires some manual steering torque, but the steering angle I still think is different. For a given lean angle and steering angle, the front patch is at particular place on the wheel. AS the lean increases, keeping the stering angle the same, the patch moves forward on the wheel. I think this means the arc described by the two patches changes, meaning that if the arc is to remain the same, the steering angle has to change. If you hold a bike upright and the steering off to one side at say 10deg, the patch is still more or less rigt at the bottom (6 o-clock) of the wheel. If you lean the bike keeping the steering the same, the patch moves closer to 3 or 9 (depending on which side you look from). This at the least changes the steering torque that comes from trail, and I think changes the relative angles of the patches (seen from above) and thus the arc of the turn. So the question remains whether there is some optimal angle where traction is optimized? Joseph       Date: 28 Oct 2007 10:38:25 From: Ben C Subject: Re: Gyroscopic forces revisited On 2007-10-28, joseph.santaniello@gmail.com <joseph.santaniello@gmail.com > wrote: > On Oct 28, 10:29 am, Ben C <spams...@spam.eggs> wrote: >> On 2007-10-26, joseph.santanie...@gmail.com <joseph.santanie...@gmail.com> wrote: [...] >> > I'm not talking about the CG to patch angle, that is of course defined >> > by the arc of the turn and the speed. I'm talking about the lean angle >> > of the center-line of the bike. In other words the lateral placement >> > of the CG as happens when you stick a knee out one way or the other. >> > Sticking your knee out does not change the ground-patch-CG angle, but >> > it does change the ground-patch-bike angle. >> >> Don't you mean the other way round? Sticking your knee out does change >> ground-patch-CG angle, since it moves CG, but doesn't change >> ground-patch-bike angle. > > The only way the CG "moves" is relative to the bike. The lean (ground- > patch-CG) stays the same for a given speed and radius. If you change > the angle, the radius changes. Yes, that's all I was trying to say. I misunderstood you originally. You were talking about sticking your knee out while at the same time adjusting the bicycle's angle of lean to compensate, so keeping the CG angle the same. I read it as sticking the knee out without compensating, which keeps the bike angle the same, but changes the CG angle, tightening the turn if you lean the knee into the corner. [...] >> I would have thought the steering angle was a given, just like angle >> between patches and CG, for a given turn-- you've got to go round the >> same arc without falling off. > > I think it is true that a lean angle either side of "natural" requires > some manual steering torque, but the steering angle I still think is > different. For a given lean angle and steering angle, the front patch > is at particular place on the wheel. AS the lean increases, keeping > the stering angle the same, the patch moves forward on the wheel. > I think this means the arc described by the two patches changes, > meaning that if the arc is to remain the same, the steering angle has > to change. Furthermore, if you read that link posted by Jon_C, the more you lean the bike the more camber thrust you get, and therefore the less slip angle you need, to the point where you might even have to steer slightly the wrong way to get around the corner. So I think you're right, lean angle does definitely alter the amount of steering lock required. > If you hold a bike upright and the steering off to one side at say > 10deg, the patch is still more or less rigt at the bottom (6 o-clock) > of the wheel. If you lean the bike keeping the steering the same, the > patch moves closer to 3 or 9 (depending on which side you look from). > This at the least changes the steering torque that comes from trail, > and I think changes the relative angles of the patches (seen from > above) and thus the arc of the turn. That sounds plausible. > So the question remains whether there is some optimal angle where > traction is optimized? I have no idea of course... but at the end of the day the tyre has to provide enough steering force to get you around the corner. Whether it's doing that with more or less steering lock, perhaps the limit of the force it can apply is the same anyway.         Date: 28 Oct 2007 23:50:29 From: Subject: Re: Gyroscopic forces revisited Ben C? writes: >>>> I'm not talking about the CG to patch angle, that is of course >>>> defined by the arc of the turn and the speed. I'm talking about >>>> the lean angle of the center-line of the bike. In other words >>>> the lateral placement of the CG as happens when you stick a knee >>>> out one way or the other. Sticking your knee out does not change >>>> the ground-patch-CG angle, but it does change the >>>> ground-patch-bike angle. >>> Don't you mean the other way round? Sticking your knee out does >>> change ground-patch-CG angle, since it moves CG, but doesn't >>> change ground-patch-bike angle. >> The only way the CG "moves" is relative to the bike. The lean >> (ground- patch-CG) stays the same for a given speed and radius. If >> you change the angle, the radius changes. > Yes, that's all I was trying to say. I misunderstood you > originally. You were talking about sticking your knee out while at > the same time adjusting the bicycle's angle of lean to compensate, > so keeping the CG angle the same. > I read it as sticking the knee out without compensating, which keeps > the bike angle the same, but changes the CG angle, tightening the > turn if you lean the knee into the corner. >>> I would have thought the steering angle was a given, just like >>> angle between patches and CG, for a given turn-- you've got to go >>> round the same arc without falling off. >> I think it is true that a lean angle either side of "natural" >> requires some manual steering torque, but the steering angle I >> still think is different. For a given lean angle and steering >> angle, the front patch is at particular place on the wheel. AS the >> lean increases, keeping the steering angle the same, the patch moves >> forward on the wheel. I think this means the arc described by the >> two patches changes, meaning that if the arc is to remain the same, >> the steering angle has to change. > Furthermore, if you read that link posted by Jon_C, the more you > lean the bike the more camber thrust you get, and therefore the less > slip angle you need, to the point where you might even have to steer > slightly the wrong way to get around the corner. Let's leave "camber thrust" out of this until one can establish that it has any effect on bicycle cornering. As far as I can find, it is an effect found mainly in tires with cylindrical tread (wide car tires) where it has little effect on traction, having insignificant angles. > So I think you're right, lean angle does definitely alter the amount > of steering lock required. Where do these pseudo technical terms arise? Steering angle is adequate. Nothing is locked or even close to lock (the mechanical limit of steering angle). This has nothing to do with "camber torque" but rather geometry, as I mentioned in another reply. >> If you hold a bike upright and the steering off to one side at say >> 10deg, the patch is still more or less right at the bottom (6 >> o-clock) of the wheel. If you lean the bike keeping the steering >> the same, the patch moves closer to 3 or 9 (depending on which side >> you look from). This at the least changes the steering torque that >> comes from trail, and I think changes the relative angles of the >> patches (seen from above) and thus the arc of the turn. This is getting too complicated to follow. Go out and ride bike! > That sounds plausible. Not to me, but I can't prove it, not understanding what is being proposed. I do much high speed cornering and find no such complicated concepts that seem to find fertile ground mainly with people who don't corner at high speed. >> So the question remains whether there is some optimal angle where >> traction is optimized? WHAT!!! There is a limit that is defined by traction. > I have no idea of course... but at the end of the day the tyre has > to provide enough steering force to get you around the corner. > Whether it's doing that with more or less steering lock, perhaps the > limit of the force it can apply is the same anyway. Jobst Brandt           Date: 29 Oct 2007 03:04:48 From: Ben C Subject: Re: Gyroscopic forces revisited On 2007-10-28, jobst.brandt@stanfordalumni.org <jobst.brandt@stanfordalumni.org > wrote: > Ben C? writes: [...] >> So I think you're right, lean angle does definitely alter the amount >> of steering lock required. > > Where do these pseudo technical terms arise? Steering angle is > adequate. The reason I say "steering lock" rather than "steering angle" is we're already discussing angle of the bicycle to the road and angle of the rider's CG to the road. The word "lock" was just intended to make it clear that we're talking about twisting the bars round.             Date: 29 Oct 2007 14:07:46 From: Subject: Re: Gyroscopic forces revisited Ben C? writes: >>> So I think you're right, lean angle does definitely alter the >>> amount of steering lock required. >> Where do these pseudo technical terms arise? Steering angle is >> adequate. > The reason I say "steering lock" rather than "steering angle" is > we're already discussing angle of the bicycle to the road and angle > of the rider's CG to the road. The word "lock" was just intended to > make it clear that we're talking about twisting the bars round. How does "lock" do that and how is steering angle in opposition to that? Jobst Brandt               Date: 29 Oct 2007 09:36:17 From: Ben C Subject: Re: Gyroscopic forces revisited On 2007-10-29, jobst.brandt@stanfordalumni.org <jobst.brandt@stanfordalumni.org > wrote: > Ben C? writes: > >>>> So I think you're right, lean angle does definitely alter the >>>> amount of steering lock required. > >>> Where do these pseudo technical terms arise? Steering angle is >>> adequate. > >> The reason I say "steering lock" rather than "steering angle" is >> we're already discussing angle of the bicycle to the road and angle >> of the rider's CG to the road. The word "lock" was just intended to >> make it clear that we're talking about twisting the bars round. > > How does "lock" do that and how is steering angle in opposition to > that? I was just trying to find a synonym rather than overload the word "angle" some more. That's all. When driving cars, people often say things like "apply some more lock". Literally this is perhaps not very logical but I hoped people would know what I meant. Perhaps it's a British-English expression.                 Date: 29 Oct 2007 18:37:54 From: Subject: Re: Gyroscopic forces revisited Ben C? writes: >>>>> So I think you're right, lean angle does definitely alter the >>>>> amount of steering lock required. >>>> Where do these pseudo technical terms arise? Steering angle is >>>> adequate. >>> The reason I say "steering lock" rather than "steering angle" is >>> we're already discussing angle of the bicycle to the road and >>> angle of the rider's CG to the road. The word "lock" was just >>> intended to make it clear that we're talking about twisting the >>> bars round. >> How does "lock" do that and how is steering angle in opposition to >> that? > I was just trying to find a synonym rather than overload the word > "angle" some more. That's all. > When driving cars, people often say things like "apply some more > lock". Literally this is perhaps not very logical but I hoped > people would know what I meant. Perhaps it's a British-English > expression. Lock is bad jargon from olden days when cars could shimmy enough for the steering wheel to spin from limit to limit of rotation, the limit being locked against an end stop. There's nothing wrong with "angle". Jobst Brandt                   Date: 29 Oct 2007 14:35:11 From: Ben C Subject: Re: Gyroscopic forces revisited On 2007-10-29, jobst.brandt@stanfordalumni.org <jobst.brandt@stanfordalumni.org > wrote: > Ben C? writes: [...] >> I was just trying to find a synonym rather than overload the word >> "angle" some more. That's all. > >> When driving cars, people often say things like "apply some more >> lock". Literally this is perhaps not very logical but I hoped >> people would know what I meant. Perhaps it's a British-English >> expression. > > Lock is bad jargon from olden days when cars could shimmy enough for > the steering wheel to spin from limit to limit of rotation, the limit > being locked against an end stop. There's nothing wrong with "angle". Quibble not that you be not quibbled. I've lost count of the number of times I've let you get away with the word "hydroplane". Why am I so generous?                     Date: 31 Oct 2007 21:15:11 From: Tom Sherman Subject: Re: Gyroscopic forces revisited Ben C? wrote: > On 2007-10-29, jobst.brandt@stanfordalumni.org <jobst.brandt@stanfordalumni.org> wrote: >> Ben C? writes: > [...] >>> I was just trying to find a synonym rather than overload the word >>> "angle" some more. That's all. >>> When driving cars, people often say things like "apply some more >>> lock". Literally this is perhaps not very logical but I hoped >>> people would know what I meant. Perhaps it's a British-English >>> expression. >> Lock is bad jargon from olden days when cars could shimmy enough for >> the steering wheel to spin from limit to limit of rotation, the limit >> being locked against an end stop. There's nothing wrong with "angle". > > Quibble not that you be not quibbled. > > I've lost count of the number of times I've let you get away with the > word "hydroplane". Why am I so generous? Hydroplane is the correct term in American English. Google "Treaty of Paris". -- Tom Sherman - Holstein-Friesland Bovinia When did ignorance of biology become a "family value"?         Date: 28 Oct 2007 14:59:31 From: Ben C Subject: Re: Gyroscopic forces revisited On 2007-10-28, Ben C <spamspam@spam.eggs > wrote: > On 2007-10-28, joseph.santaniello@gmail.com <joseph.santaniello@gmail.com> wrote: [...] >> I think it is true that a lean angle either side of "natural" requires >> some manual steering torque, but the steering angle I still think is >> different. For a given lean angle and steering angle, the front patch >> is at particular place on the wheel. AS the lean increases, keeping >> the stering angle the same, the patch moves forward on the wheel. >> I think this means the arc described by the two patches changes, >> meaning that if the arc is to remain the same, the steering angle has >> to change. > > Furthermore, if you read that link posted by Jon_C, the more you lean > the bike the more camber thrust you get, and therefore the less slip > angle you need, to the point where you might even have to steer slightly > the wrong way to get around the corner. Sorry, I got that wrong-- I think that only applies to "real" CG lean as opposed to keeping the CG the same and making the bike more or less upright. In other words, apart from what you've said about steering geometry, the amount of camber thrust and slip angle you get/need depends on curve and speed, not on rider position.   Date: 24 Oct 2007 09:57:12 From: velodancer Subject: Re: Gyroscopic forces revisited > I think we need to define "clown-bike" here. Without defining a clown bike, their use is walking speed or below, the most difficult speed to ride any bike no hands. At speeds higher than walking, these bikes may be very close to regular bikes in the ability to balance no hands. While it has been a while since I've been in college, I would expect gyroscopic forces to have a direct relationship to the linear speed of the wheel (at the tire contact) and the effective rotating mass at the tire contact. Thus a small wheel with the same effective mass as a larger wheel, will exhibit the exact same gyroscopic effect as the larger wheel. Small wheel bikes generally have faster handling than larger wheeled bikes. I believe that much of it is due to the shorter contact patch along the line of direction, but the smaller radius also means that the wheel will fall into a turn with greater acceleration. This lowered forward stability for me seems to make a small wheeled bike easier to ride no hands at a significantly lower speed than a larger wheeled bike (no experience below 16"). I've noticed this surprising experience a number of times over the last 20 years. I will say that most find it harder to ride small wheel bikes no hands at any speed, but I conjecture that this is because they are not used to the faster handling. I have the same experience on large wheeled bikes with faster handling, I can ride them slower no hands. Recumbents, due their typically low COG, are significantly harder to ride no hands. I've never been able to do it at speeds under 10 miles an hour and even then it can be very difficult to impossible depending on the model.   Date: 24 Oct 2007 08:04:25 From: joseph.santaniello@gmail.com Subject: Re: Gyroscopic forces revisited On Oct 24, 2:40 pm, Jon_C <JonCCro...@gmail.com > wrote: > On Oct 23, 10:51 pm, jobst.bra...@stanfordalumni.org wrote: > > > > > Jon Crouch writes: > > > How about a wheel on it's own? If you roll a wheel and it leans over > > > it turns in that direction. The reason is that the contact patch of > > > a leaning tyre is parallel to the ground and so at an angle to the > > > wheel's axle so the wheel acts as a rolling cone and turns into the > > > corner (which maintains it's balance). I'm pretty sure that's the > > > main steering force involved in no-handed riding. > > > (I seem to remember reading about someone testing this on a > > > motorbike and finding that on a prolonged corners the handlebars are > > > actually turned slightly outward instead of inward as you;d > > > expect. I think that was due to rake/trail but I don;t have time to > > > think about that right now :) > > > I think you ought to go to a velodrome. These tracks have sloped > > straights and more steeply banked circular curves at each end. Riders > > have no problem riding no-hands on these tracks, either on the > > straight sections or in the banked curve (if they are going fast > > enough to not strike a pedal). > > > If you don't have a track, try a crowned paved street and notice how > > the angle with the pavement has essentially no effect on steering, the > > centerline of the contact patch moving less than 1/8" off center for a > > 10=B0 side slope. > > > Jobst Brandt > > To go in a straight line on an angled surface you'd have to keep the > steering pointed slightly down the slope (ie. left on a velodrome > straight) to counter-act the steering force from the slope. Maybe a > slight weave-like cycle occurs where the rightward steering force from > the slope causes the rider to 'fall' left (centrifugal force) which > causes the bike to steer back to the left and so on. Aslo, shifting > the rider's center of gravity to the right will lean the bike frame to > the left slightly, causing the front wheel to 'flop' (and steer) left > countering camber thrust from the slope. > > I don't think the 1/8" offset of the contact patch is too relevant > btw. Camber thrust is due the angle between the wheel's axle and the > ground (cone effect) not contact patch offset. I don't think you do need have the steering anything but straight on a sloped surface. Not having a velodrome handy, that's why I broke out my trusty plywood board. I too thought the angle of the wheels on the surface had something to do with it, but it doesn't. I put my bike on the board and fiddled with it quite a bit, and the angle of contact alone does not appear to matter. What does matter is where the center of mass is relative to the contact patches and the combined gravity and turn forces. When the center of mass is along the centerline of the bike and these forces are acting through it to the patches, the steering satys straight no matter the angle of the wheels to the road. Using a sloped board you can simulate the forces on a riderless bike during a turn. (On Mars...) Perhaps some minimum speed is necessary to not fall over riding on a sloped velodrome, but I suspect this is from requiring a certain amount of momentum to be able to "climb" the banking everytime you make a steering correction up-track. Joseph     Date: 29 Oct 2007 20:15:01 From: datakoll Subject: Re: Gyroscopic forces revisited http://news.yahoo.com/comics/bc;_ylt=At_bHmDn6UcAvY48QAkYFpsO_b4F is the brass disc system consumerabble? will I Damon Hill Clingman's Dome?     Date: 26 Oct 2007 11:31:42 From: joseph.santaniello@gmail.com Subject: Re: Gyroscopic forces revisited On Oct 26, 6:46 pm, jobst.bra...@stanfordalumni.org wrote: > Joseph Santaniello writes: > >>>>>> With both hands on the stem my elbows meet nicely in my gut, just > >>>>>> above my knees. From what you say, your bicycle seems to be too > >>>>>> short if you cannot bring your elbows in. > >>>>> Hands on the stem? Or the bars? With both hands in fists close > >>>>> together as though I were holding the bars with my thumbs wrapped > >>>>> under the stem, there is no angle I can hold my arms to get my > >>>>> elbows to meet. It wouldn't matter how long or short my bike is, > >>>>> my arms can't do it. With forearms rotated I can do it, but I > >>>>> wouldn't be able to hold the bar. > >>>> I don't understand. I can bring my elbows together clasping hands > >>>> as in prayer or both over one another on the stem. Leave the bars > >>>> out of this. > >>> I see. That's different. I'll try that. > >>>>>>> But it is steering angle, not lean angle I was thinking about. > >>>>>>> But I suppose this just leads us to the drifting idea, which I > >>>>>>> agree is pretty much not do-able. > >>>>>> You have no option with steering angle, the road prescribes that > >>>>>> because You must get around the curve. That can be modified only > >>>>>> a little by choosing an ideal arc that doesn't infringe on > >>>>>> oncoming traffic or the edge of the road. > >>>>> I guess I was imagining it might be possible to force the bike > >>>>> into some combination of lean and steering angle that would not be > >>>>> the same angles the bike would find on it's own ridden the same > >>>>> line and speed no handed. But since wheel slip is out of the > >>>>> question, the only thing that would accomplish is either side > >>>>> loads, or unnecessarily steep bike lean. > >>>> Those two parameters are not independent and are given by the curve > >>>> radius and speed. As I mentioned in the article, leaning off the > >>>> bicycle doesn't alter the plumb line from CG to ground contact > >>>> line. > >>> No, but it does alter the angle the bike alone is leaning and > >>> subsequently the steering angle (by which I mean how far off center > >>> the stem is) and the relative location of the patches. > >> The steering angle does not change. The track around the curve > >> remains unchanged, only the rider and bicycle are no longer in the > >> same plane. Following the same track requires the same steering > >> angle. You can't fool geometry, or gravity for that matter. > > For a given angle of lean on a given curve, the steering angle will be > > the same, but for the same curve at different angles of lean, I think > > the steering angle will be different. > > You can't have different angles of lean. As I said you can't fool > geometry and gravity. It may appear as a different angle of lean but > it isn't. The plumb line from the CG to the road is the same at the > same speed around the same curve no matter what the rider does. I'm not talking about the CG to patch angle, that is of course defined by the arc of the turn and the speed. I'm talking about the lean angle of the center-line of the bike. In other words the lateral placement of the CG as happens when you stick a knee out one way or the other. Sticking your knee out does not change the ground-patch-CG angle, but it does change the ground-patch-bike angle. When the CG is in the centerline of the bike and the combined forces of gravity and turn accleration are working along this centerline to the patches, the steering remains straight. Such as when riding straight. When the CG is out of the centerline, the steering moves to that side. When riding no hands in a constant turn, the steering is NOT straight, so necesarily the CG is not on the center-line of the bike. If the CG were in the centerline of the bike, the combined forces would be acting along the centerline through the patches just as if the bike were on a velodrome straight, and the steering angle would be straight too. But the steering cannot be straight because we are in the middle of a turn! Riding in a turn with hands on the bars it is possible to have the CG on the centerline of the bike, but to do so the steering would need to be manually turned. > > I think because of the angle of the steering axis and the resulting > > trail, the position and orientation of the patches does change if you > > hold the steering angle constant. Perhaps only slightly. > > That has no effect on the path taken. Besides, when leaning off the > plane of the bicycle, corrective force is required on the handlebar to > make up for the odd geometry. This is less apparent on an M/C because > it is far heavier but still requires steering force to hold the line. > You can't, for instance, ride no-hands when leaning off to one side. That corrective force is what I am interested in. Is there some optimal (in terms of traction, or max possible cornering speed) bike to ground angle that is NOT the "natural" angle the bike would take were it ridden no hands, that of necessity requires some corrective force. Or is any lean angle that requires corrective force worse in terms of traction than the natural angle? > > Suppose you walk a bike perfectly upright along some arc. The > > steering will need to be turned to some angle to achieve this. If > > you then walk the bike along that same arc, but with some arbitrary > > angle of lean, I contend that the steering angle (in terms of stem > > to center- centerline) will be different. Maybe the patches will > > describe the same arc, but the steering will be at a different > > angle, and the front patch will be at a different place on the > > wheel. > > You are holding the bicycle. It is not balanced in the curve. This > is not a two wheeled experiment but one with a trainer (training > wheels). The experiment could be done riding, but it would be harder to hold the arc. Ride once keeping the bike as upright as possible, ride again leaning more. You'd need to use different speeds and lean your body out of the plane of the bike maybe, but the point would be to see if a bike following the same arc, but at different angles of lean has the same or different steering angle. Joseph       Date: 28 Oct 2007 04:29:27 From: Ben C Subject: Re: Gyroscopic forces revisited On 2007-10-26, joseph.santaniello@gmail.com <joseph.santaniello@gmail.com > wrote: > On Oct 26, 6:46 pm, jobst.bra...@stanfordalumni.org wrote: [...] >> You can't have different angles of lean. As I said you can't fool >> geometry and gravity. It may appear as a different angle of lean but >> it isn't. The plumb line from the CG to the road is the same at the >> same speed around the same curve no matter what the rider does. > > I'm not talking about the CG to patch angle, that is of course defined > by the arc of the turn and the speed. I'm talking about the lean angle > of the center-line of the bike. In other words the lateral placement > of the CG as happens when you stick a knee out one way or the other. > Sticking your knee out does not change the ground-patch-CG angle, but > it does change the ground-patch-bike angle. Don't you mean the other way round? Sticking your knee out does change ground-patch-CG angle, since it moves CG, but doesn't change ground-patch-bike angle. > When the CG is in the centerline of the bike and the combined forces > of gravity and turn accleration are working along this centerline to > the patches, the steering remains straight. Such as when riding > straight. When the CG is out of the centerline, the steering moves to > that side. > > When riding no hands in a constant turn, the steering is NOT straight, > so necesarily the CG is not on the center-line of the bike. If the CG > were in the centerline of the bike, the combined forces would be > acting along the centerline through the patches just as if the bike > were on a velodrome straight, and the steering angle would be straight > too. But the steering cannot be straight because we are in the middle > of a turn! > > Riding in a turn with hands on the bars it is possible to have the CG > on the centerline of the bike, but to do so the steering would need to > be manually turned. > >> > I think because of the angle of the steering axis and the resulting >> > trail, the position and orientation of the patches does change if you >> > hold the steering angle constant. Perhaps only slightly. >> >> That has no effect on the path taken. Besides, when leaning off the >> plane of the bicycle, corrective force is required on the handlebar to >> make up for the odd geometry. This is less apparent on an M/C because >> it is far heavier but still requires steering force to hold the line. >> You can't, for instance, ride no-hands when leaning off to one side. > > That corrective force is what I am interested in. Is there some > optimal (in terms of traction, or max possible cornering speed) bike > to ground angle that is NOT the "natural" angle the bike would take > were it ridden no hands, that of necessity requires some corrective > force. Or is any lean angle that requires corrective force worse in > terms of traction than the natural angle? If I understand what you're saying, the idea is that to hold a bike in a turn while keeping CG through bike-centreline requires some steering torque on the handlebars, and therefore a bit more traction from the front tyre? I don't know if that's true. There must still be a steering torque even at the natural angle, it's just that the rider isn't applying it with his hands. The best thing for fast cornering is smoothness. The cornering force dictated by the speed and radius of the turn is an average. Constant overcorrections will create peaks either side of the average. When a peak exceeds tyre grip then you're on your ear. So the question is what rider position allows the finest control. I should think for that you do want a position that does require some manual steering torque (i.e. away from the "natural angle"), since that's easier to modulate. The other issue (which you may have covered) is keeping a good angle between tyres and road. If you watch Moto GP, they sort of hang off the inside of the bike in turns. That keeps the bike itself more upright than it would otherwise be, so it's not cornering on the rim instead of the tyre. I don't think anyone ever gets such extreme angles of lean on a bicycle though. But leaning across the bicycle to the outside of the turn is probably a bad idea because it makes the angle between the wheels and the road more acute. >> > Suppose you walk a bike perfectly upright along some arc. The >> > steering will need to be turned to some angle to achieve this. If >> > you then walk the bike along that same arc, but with some arbitrary >> > angle of lean, I contend that the steering angle (in terms of stem >> > to center- centerline) will be different. Maybe the patches will >> > describe the same arc, but the steering will be at a different >> > angle, and the front patch will be at a different place on the >> > wheel. Maybe the steering angle is the same, it's just that at lean angles either side of "natural", a manual steering torque is required to hold the correct steering angle. I would have thought the steering angle was a given, just like angle between patches and CG, for a given turn-- you've got to go round the same arc without falling off.         Date: 28 Oct 2007 13:47:13 From: Subject: Re: Gyroscopic forces revisited Ben C? writes: >>> You can't have different angles of lean. As I said you can't fool >>> geometry and gravity. It may appear as a different angle of lean >>> but it isn't. The plumb line from the CG to the road is the same >>> at the same speed around the same curve no matter what the rider >>> does. >> I'm not talking about the CG to patch angle, that is of course >> defined by the arc of the turn and the speed. I'm talking about the >> lean angle of the center-line of the bike. In other words the >> lateral placement of the CG as happens when you stick a knee out >> one way or the other. Sticking your knee out does not change the >> ground-patch-CG angle, but it does change the ground-patch-bike >> angle. > Don't you mean the other way round? Sticking your knee out does > change ground-patch-CG angle, since it moves CG, but doesn't change > ground-patch-bike angle. No, the CG plumb line remains constant, only the configuration of the body changes, its CG remaining unchanged. If you stick your arm out for a hand signal, the rest of your body must change position so the CG of extended arm and the rest of the body remains constant. If you don't do that, you must alter the path through the curve through which you travel. When making such changes in body configuration, the angle of the bicycle wheel to the ground will change but the contact pressure and force vector remains unchanged. Therefore, nothing is gained by sticking one's knee out except a false feeling of security from not leaning the body so far into the curve... or the feeling of sticking the foot out in the event of a slip. Of course this doesn't help either at any reasonable speed. >> When the CG is in the centerline of the bike and the combined >> forces of gravity and turn acceleration are working along this >> centerline to the patches, the steering remains straight. Such as >> when riding straight. When the CG is out of the centerline, the >> steering moves to that side. >> When riding no hands in a constant turn, the steering is NOT >> straight, so necessarily the CG is not on the center-line of the >> bike. If the CG were in the centerline of the bike, the combined >> forces would be acting along the centerline through the patches >> just as if the bike were on a velodrome straight, and the steering >> angle would be straight too. But the steering cannot be straight >> because we are in the middle of a turn! >> Riding in a turn with hands on the bars it is possible to have the >> CG on the centerline of the bike, but to do so the steering would >> need to be manually turned. >>>> I think because of the angle of the steering axis and the >>>> resulting trail, the position and orientation of the patches does >>>> change if you hold the steering angle constant. Perhaps only >>>> slightly. >>> That has no effect on the path taken. Besides, when leaning off >>> the plane of the bicycle, corrective force is required on the >>> handlebar to make up for the odd geometry. This is less apparent >>> on an M/C because it is far heavier but still requires steering >>> force to hold the line. You can't, for instance, ride no-hands >>> when leaning off to one side. >> That corrective force is what I am interested in. Is there some >> optimal (in terms of traction, or max possible cornering speed) >> bike to ground angle that is NOT the "natural" angle the bike would >> take were it ridden no hands, that of necessity requires some >> corrective force. Or is any lean angle that requires corrective >> force worse in terms of traction than the natural angle? > If I understand what you're saying, the idea is that to hold a bike > in a turn while keeping CG through bike-centreline requires some > steering torque on the handlebars, and therefore a bit more traction > from the front tyre? As I said above, this does not change anything except the psyche of the rider. If you could hold the bicycle vertically and lean the body out sufficiently to ride the same path, the force on the tire would be no different. Only the portion of the tire tread contacting the road would be farther around the cross section but with the same forces. > I don't know if that's true. There must still be a steering torque > even at the natural angle, it's just that the rider isn't applying > it with his hands. There is on the handlebar but not on the road. That's why one should not lean off the bicycle for smooth fast cornering. Any bump in the road will cause a change in steering torque caused by trail. It is similar to statically standing on one pedal while leaning the bicycle. That takes torque on the steertube (aka handlebar). > The best thing for fast cornering is smoothness. The cornering > force dictated by the speed and radius of the turn is an average. > Constant over-corrections will create peaks either side of the > average. When a peak exceeds tyre grip then you're on your ear. > So the question is what rider position allows the finest control. I > should think for that you do want a position that does require some > manual steering torque (i.e. away from the "natural angle"), since > that's easier to modulate. I think I covered that in the FAQ... and the adjoining picture of riding around a fast curve. > The other issue (which you may have covered) is keeping a good angle > between tyres and road. If you watch Moto GP, they sort of hang off > the inside of the bike in turns. That keeps the bike itself more > upright than it would otherwise be, so it's not cornering on the rim > instead of the tyre. I don't think anyone ever gets such extreme > angles of lean on a bicycle though. You have no control over the angle of the tire to the road. That is given by speed and curvature of the path. I think you still have the concept of effective lean angle and speed confused. You cannot alter it. It is a pendulum effect. The pendulum knows where the plumb line is. > But leaning across the bicycle to the outside of the turn is > probably a bad idea because it makes the angle between the wheels > and the road more acute. That angle has nothing to do with traction forces or the line between CG and contact line on the road. >>>> Suppose you walk a bike perfectly upright along some arc. The >>>> steering will need to be turned to some angle to achieve this. >>>> If you then walk the bike along that same arc, but with some >>>> arbitrary angle of lean, I contend that the steering angle (in >>>> terms of stem to center- centerline) will be different. Maybe >>>> the patches will describe the same arc, but the steering will be >>>> at a different angle, and the front patch will be at a different >>>> place on the wheel. > Maybe the steering angle is the same, it's just that at lean angles > either side of "natural", a manual steering torque is required to > hold the correct steering angle. ... to hold the desired track through the curve. The steering angle is essentially straight ahead. See picture in FAQ. > I would have thought the steering angle was a given, just like angle > between patches and CG, for a given turn-- you've got to go round > the same arc without falling off. I can't decipher what you mean by this. Jobst Brandt           Date: 28 Oct 2007 10:24:44 From: Ben C Subject: Re: Gyroscopic forces revisited On 2007-10-28, jobst.brandt@stanfordalumni.org <jobst.brandt@stanfordalumni.org > wrote: > Ben C? writes: [...] >> The other issue (which you may have covered) is keeping a good angle >> between tyres and road. If you watch Moto GP, they sort of hang off >> the inside of the bike in turns. That keeps the bike itself more >> upright than it would otherwise be, so it's not cornering on the rim >> instead of the tyre. I don't think anyone ever gets such extreme >> angles of lean on a bicycle though. > > You have no control over the angle of the tire to the road. That is > given by speed and curvature of the path. The CG and the two contact points at the tyres define a plane. The angle of that plane to the road surface is what's given by speed and curvature of the path isn't it? But by shifting the rider position, the angle of the tyre to the road can be altered while keeping the angle of that plane described above the same. [...] >> I would have thought the steering angle was a given, just like angle >> between patches and CG, for a given turn-- you've got to go round >> the same arc without falling off. > > I can't decipher what you mean by this. Suppose you attach a little protractor to the steerer and measure the amount of steering lock applied. I'm suggesting that for a given curve and speed, the amount of lock will always be the same, no matter what riding position is adopted. I think Joseph is saying it won't be the same.             Date: 28 Oct 2007 23:37:36 From: Subject: Re: Gyroscopic forces revisited Ben C? writes: >>> The other issue (which you may have covered) is keeping a good >>> angle between tyres and road. If you watch Moto GP, they sort of >>> hang off the inside of the bike in turns. That keeps the bike >>> itself more upright than it would otherwise be, so it's not >>> cornering on the rim instead of the tyre. I don't think anyone >>> ever gets such extreme angles of lean on a bicycle though. >> You have no control over the angle of the tire to the road. That >> is given by speed and curvature of the path. > The CG and the two contact points at the tyres define a plane. The > angle of that plane to the road surface is what's given by speed and > curvature of the path isn't it? Altering the contact of the tire cross section does not change the path, only which portion of the tire tread makes contact with the road, and that is only a matter of a few degrees at that. > But by shifting the rider position, the angle of the tyre to the > road can be altered while keeping the angle of that plane described > above the same. I think you need to view this as a free body diagram with vertical and horizontal force vectors that add to lie in the effective plane of the bicycle frame and rider CG. >>> I would have thought the steering angle was a given, just like >>> angle between patches and CG, for a given turn-- you've got to go >>> round the same arc without falling off. >> I can't decipher what you mean by this. > Suppose you attach a little protractor to the steerer and measure > the amount of steering lock applied. I'm suggesting that for a > given curve and speed, the amount of lock will always be the same, > no matter what riding position is adopted. Yes, roughly so but assuming that with rider and bicycle in the same plane for a 45 degree lean is perverted by standing way off to the side sufficiently to make the bicycle be vertical, a different steering angle will result for the same curvilinear path. You can try this by manually walking the bicycle around a circular arc when vertically and when leaned at 45 degrees. Typically a stunt rider in a vertical barrel rides with practically zero steering angle as he rides horizontal circles on the wall. This is an example of the geometric circumstances. > I think Joseph is saying it won't be the same. Let him speak for himself. # According to Longfellow's epic, The Courtship of Miles Standish, # John Alden proposed to Priscilla Mullins on behalf of Standish and # she replied, "Why don't you speak for yourself, John?" Priscilla did # in fact marry John Alden at Plymouth. The records do not mention # Standish ever courting Priscilla. Jobst Brandt             Date: 28 Oct 2007 11:45:41 From: Kerry Montgomery Subject: Re: Gyroscopic forces revisited "Ben C" <spamspam@spam.eggs > wrote in message news:slrnfi9afd.8vo.spamspam@bowser.marioworld... > On 2007-10-28, jobst.brandt@stanfordalumni.org > <jobst.brandt@stanfordalumni.org> wrote: >> Ben C? writes: > [...] >>> The other issue (which you may have covered) is keeping a good angle >>> between tyres and road. If you watch Moto GP, they sort of hang off >>> the inside of the bike in turns. That keeps the bike itself more >>> upright than it would otherwise be, so it's not cornering on the rim >>> instead of the tyre. I don't think anyone ever gets such extreme >>> angles of lean on a bicycle though. >> >> You have no control over the angle of the tire to the road. That is >> given by speed and curvature of the path. > > The CG and the two contact points at the tyres define a plane. The angle > of that plane to the road surface is what's given by speed and curvature > of the path isn't it? > > But by shifting the rider position, the angle of the tyre to the road > can be altered while keeping the angle of that plane described above > the same. > > [...] >>> I would have thought the steering angle was a given, just like angle >>> between patches and CG, for a given turn-- you've got to go round >>> the same arc without falling off. >> >> I can't decipher what you mean by this. > > Suppose you attach a little protractor to the steerer and measure the > amount of steering lock applied. I'm suggesting that for a given curve > and speed, the amount of lock will always be the same, no matter what > riding position is adopted. > > I think Joseph is saying it won't be the same. Ben, I don't think constant lock + constant speed + any riding position = constant curve. Imagine a bicycle with tires wider than any other part of the bicycle. If you apply a little steering lock to the right and travel at a given speed, and assume a riding position that keeps the bike vertical, it will result in a curve to the right with a relatively large radius. If you then keep the steering lock the same, and assume a riding position that leans the bike over to the right until it is almost horizontal, the curve will have a radius of about the distance from the rear of the back tire to the front of the front tire. It seems that, with an intermediate amount of lean, the curve radius becomes smaller than it was when the bike was vertical. Kerry               Date: 28 Oct 2007 17:22:12 From: Ben C Subject: Re: Gyroscopic forces revisited On 2007-10-28, Kerry Montgomery <kamontgo@teleport.com > wrote: > > "Ben C" <spamspam@spam.eggs> wrote in message [...] >> Suppose you attach a little protractor to the steerer and measure the >> amount of steering lock applied. I'm suggesting that for a given curve >> and speed, the amount of lock will always be the same, no matter what >> riding position is adopted. >> >> I think Joseph is saying it won't be the same. > > Ben, > I don't think constant lock + constant speed + any riding position = > constant curve. Imagine a bicycle with tires wider than any other part of > the bicycle. If you apply a little steering lock to the right and travel at > a given speed, and assume a riding position that keeps the bike vertical, it > will result in a curve to the right with a relatively large radius. If you > then keep the steering lock the same, and assume a riding position that > leans the bike over to the right until it is almost horizontal, the curve > will have a radius of about the distance from the rear of the back tire to > the front of the front tire. It seems that, with an intermediate amount of > lean, the curve radius becomes smaller than it was when the bike was > vertical. I think that's right. Just to clarify what I meant: the speed and curve are given. Suppose we're taking a right turn with a radius of 10m at 15mph. Those conditions dictate a certain angle of lean. Now suppose I take the turn again but tip more of my weight to the right this time. The only way this is possible (same radius, same speed0, I will have to make the bicycle a bit more upright-- so that the angle the CG and contact points make with the ground stays exactly the same as it was before. Now that the bicycle is a bit more upright, does it need a tiny bit less steering lock to make that same turn at that same speed? That's the question we're asking.     Date: 24 Oct 2007 17:26:56 From: vey Subject: Re: Gyroscopic forces revisited joseph.santaniello@gmail.com wrote: > Using a sloped board you can simulate the forces on a riderless bike > during a turn. (On Mars...) I tried an experiment today. On a flat surface, I haven't been able to ride without hands since I retired my old three speed years ago. Since then, I've been riding flatbars and if I let go I would just about crash right then and there. Today, I was able to do it easily. Two weeks ago I converted an old flatbar into an upright, higher stem, bars that come back, B-66 saddle. Now, miraculously, I can ride with no hands. When I tried with the flatbar, I had to shift my body up and back because I was leaning so far forward, but if I am already up and back, like I am now, I just let go and I am fine. No difference in wheel size, no difference in speed, but big difference in COG. COG is not just height, but longitudinal and lateral weight, too, or so I was told when I had to make such calculations.   Date: 24 Oct 2007 12:40:08 From: Jon_C Subject: Re: Gyroscopic forces revisited On Oct 23, 10:51 pm, jobst.bra...@stanfordalumni.org wrote: > Jon Crouch writes: > > How about a wheel on it's own? If you roll a wheel and it leans over > > it turns in that direction. The reason is that the contact patch of > > a leaning tyre is parallel to the ground and so at an angle to the > > wheel's axle so the wheel acts as a rolling cone and turns into the > > corner (which maintains it's balance). I'm pretty sure that's the > > main steering force involved in no-handed riding. > > (I seem to remember reading about someone testing this on a > > motorbike and finding that on a prolonged corners the handlebars are > > actually turned slightly outward instead of inward as you;d > > expect. I think that was due to rake/trail but I don;t have time to > > think about that right now :) > > I think you ought to go to a velodrome. These tracks have sloped > straights and more steeply banked circular curves at each end. Riders > have no problem riding no-hands on these tracks, either on the > straight sections or in the banked curve (if they are going fast > enough to not strike a pedal). > > If you don't have a track, try a crowned paved street and notice how > the angle with the pavement has essentially no effect on steering, the > centerline of the contact patch moving less than 1/8" off center for a > 10=B0 side slope. > > Jobst Brandt To go in a straight line on an angled surface you'd have to keep the steering pointed slightly down the slope (ie. left on a velodrome straight) to counter-act the steering force from the slope. Maybe a slight weave-like cycle occurs where the rightward steering force from the slope causes the rider to 'fall' left (centrifugal force) which causes the bike to steer back to the left and so on. Aslo, shifting the rider's center of gravity to the right will lean the bike frame to the left slightly, causing the front wheel to 'flop' (and steer) left countering camber thrust from the slope. I don't think the 1/8" offset of the contact patch is too relevant btw. Camber thrust is due the angle between the wheel's axle and the ground (cone effect) not contact patch offset.     Date: 30 Oct 2007 01:07:09 From: bjw@mambo.ucolick.org Subject: Re: Gyroscopic forces revisited On Oct 29, 1:56 pm, Tim McNamara <tim...@bitstream.net > wrote: > In article <rubrum-E98E0B.12164429102...@newsclstr03.news.prodigy.net>, > Michael Press <rub...@pacbell.net> wrote: > > Tim McNamara <tim...@bitstream.net> wrote: > > > > A hoop rolling with its axis oblique to the roadway will turn in > > > > an arc. To turn there must be a reaction force at the contact > > > > patch. > > > > How about gravity? > > > Gravity acts perpendicular to a flat road surface. The reaction force > > to make the hoop roll on an arc of a circle acts parallel to the road > > surface. > > But hoops with extremely low coefficients of friction, and hence little > reaction force (e.g., "scrub"), display this behavior. It looks to me > that it's due to gravity not a reaction force against the ground. If > the hoop can't tilt, as in a two track vehicle such as a car, then a > reaction force is obviously necessary. But for a hoop such as a coin or > a tire, it's not clear to me that a scrub effect or reaction force > against the ground is necessary to cause a curving trajectory when > rolling across a flat surface. Consider a coin rolling straight on a hard floor. There is relatively little friction. But there is still friction, otherwise the coin would slide rather than roll. This is not "scrub" at the macroscopic level - it occurs even when the surfaces are hard and the contact patch is small. I think scrub is only a useful concept when you have a contact patch large enough that different pieces of the object in contact with the ground want to move on different paths. Now consider the coin that is tilted and rolling. The forces on the coin are gravity acting straight down, a small force of rolling friction acting against the direction of roll, a normal force pushing up, and a centripetal friction force pushing inward. The centripetal friction force is what keeps the coin from low-siding. Its effect is to make the coin travel in a curve. Ben       Date: 30 Oct 2007 10:23:58 From: Kerry Montgomery Subject: Re: Gyroscopic forces revisited <bjw@mambo.ucolick.org > wrote in message news:1193707935.390301.162090@z24g2000prh.googlegroups.com... > On Oct 29, 1:56 pm, Tim McNamara <tim...@bitstream.net> wrote: >> In article <rubrum-E98E0B.12164429102...@newsclstr03.news.prodigy.net>, >> Michael Press <rub...@pacbell.net> wrote: >> > Tim McNamara <tim...@bitstream.net> wrote: >> > > > A hoop rolling with its axis oblique to the roadway will turn in >> > > > an arc. To turn there must be a reaction force at the contact >> > > > patch. >> >> > > How about gravity? >> >> > Gravity acts perpendicular to a flat road surface. The reaction force >> > to make the hoop roll on an arc of a circle acts parallel to the road >> > surface. >> >> But hoops with extremely low coefficients of friction, and hence little >> reaction force (e.g., "scrub"), display this behavior. It looks to me >> that it's due to gravity not a reaction force against the ground. If >> the hoop can't tilt, as in a two track vehicle such as a car, then a >> reaction force is obviously necessary. But for a hoop such as a coin or >> a tire, it's not clear to me that a scrub effect or reaction force >> against the ground is necessary to cause a curving trajectory when >> rolling across a flat surface. > > Consider a coin rolling straight on a hard floor. > There is relatively little friction. But there is still friction, > otherwise the coin would slide rather than roll. > This is not "scrub" at the macroscopic level - it > occurs even when the surfaces are hard and the contact > patch is small. I think scrub is only a useful concept > when you have a contact patch large enough that > different pieces of the object in contact with the > ground want to move on different paths. > > Now consider the coin that is tilted and rolling. The > forces on the coin are gravity acting straight down, > a small force of rolling friction acting against the > direction of roll, a normal force pushing up, and a > centripetal friction force pushing inward. The > centripetal friction force is what keeps the coin > from low-siding. Its effect is to make the coin > travel in a curve. > Ben, I agree with your description of a coin rolling straight on a hard floor, and think it's a point of clarity in this discussion. Am not so sure of your description of a tilted and rolling coin. If there's "a centripetal friction force pushing inward", that would seem to make the coin low-side even faster. I'd think an "inward" force would be one pushing from the center of the coin in the same direction as the top of the coin is tilted. If the coin is tilted and rolling, gravity tries to pull it down in the direction of tilt, but angular momentum is what prevents it from immediately falling. Kerry   Date: 24 Oct 2007 04:58:51 From: joseph.santaniello@gmail.com Subject: Re: Gyroscopic forces revisited On Oct 24, 1:49 pm, Doc O'Leary <droleary.use...@4q2007.subsume.com > wrote: > In article <fflnm0$ko...@news.Stanford.EDU>, > "Tom Nakashima" <t...@slac.stanford.edu> wrote: > > > Someone tried to ride a clown's bike with mini wheels and published an > > article in one of the bicycle magazines. Said they had a hard time keeping > > their balance. I believe it was a contest to see if anyone could ride it a > > 100 ft. without putting their foot down. Nobody was able to do it. > > I'm not sure what that has to do with gyroscopic forces. Something like > the A-bike seems quite able to be balanced with mini wheels, likely for > the same reason (higher center of gravity) that it's easy to balance a > broom on your fingertip but not the dustpan. > I think we need to define "clown-bike" here. For me a clown bike is one with tiny wheels and short wheelbase, but reasonable normal seat and bar height. Picture a 12" kids' BMX with a huge seatpost and riser bars. Is that what you guys think of? Joseph     Date: 24 Oct 2007 06:47:50 From: Tom Nakashima Subject: Re: Gyroscopic forces revisited <joseph.santaniello@gmail.com > wrote in message news:1193227131.128703.71040@t8g2000prg.googlegroups.com... > I think we need to define "clown-bike" here. > Joseph http://www.pranks4u.com/media/brian02bike.72.jpg -tom       Date: 24 Oct 2007 16:49:59 From: Michael Press Subject: Re: Gyroscopic forces revisited In article <ffnie6$1dj$1@news.Stanford.EDU >, "Tom Nakashima" <tom@slac.stanford.edu > wrote: > <joseph.santaniello@gmail.com> wrote in message > news:1193227131.128703.71040@t8g2000prg.googlegroups.com... > > > I think we need to define "clown-bike" here. > > Joseph > > http://www.pranks4u.com/media/brian02bike.72.jpg Front wheel fork offset is rearward as on a grocery cart. -- Michael Press   Date: 24 Oct 2007 04:23:49 From: joseph.santaniello@gmail.com Subject: Re: Gyroscopic forces revisited On Oct 24, 4:59 am, jobst.bra...@stanfordalumni.org wrote: > Well, try riding no-hands at 2mph (less than walking speed). To what > do you attribute the inability to steer the bicycle at such a speed? > > Just try moving one knee side-to-side and see what steering you get at > those speeds. I just did that (took video too!) and it was as expected easy to ride at a fast pace, difficult but not impossible at a walking pace, and virtually impossible slower. Side to side knee movement at speed produces little effect, at walking has a very large effect, and at sub walking speed is not wise with double toe straps! So gyroscopic forces provide stability to the bike by resisting steering changes induced by lean. And of course gyroscipic forces also contibute to steering changes when the axis of rotation is tilted by leaning quickly. But a bike stays in a turn when the center of mass is outside of the center-line of the bike. This is why a riderless bike goes straight (more or less). If it doesn't go dead straight it is probably alignment or maybe even the off-center center of mass due to the drivetrain. My son helped me with some "ghost-rider" tests on some of our too-small kids bikes. Inconclusive as to whether they always end up going to the same side. If you tape a brick to the side of the seatpost of a bike and send it off on a riderless mission, it turns as the COM is off center and this combined with the steering geometry means the steering flops to the side. Gyroscopic forces keep it stable in this turn, just as they keep it stable in a straight line with no brick. So gyroscopic forces are necessary to keep a bike in a stable state, whether a straight ahead, or a constant arc turn. They keep the steering from making too twitchy changes, but they are not exclusively what makes the turn happen, nor what keep the steering at the proper angle during a turn. Joseph     Date: 24 Oct 2007 16:03:59 From: Subject: Re: Gyroscopic forces revisited Joseph Santaniello writes: >> Well, try riding no-hands at 2mph (less than walking speed). To >> what do you attribute the inability to steer the bicycle at such a >> speed? >> Just try moving one knee side-to-side and see what steering you get >> at those speeds. > I just did that (took video too!) and it was as expected easy to > ride at a fast pace, difficult but not impossible at a walking pace, > and virtually impossible slower. Side to side knee movement at > speed produces little effect, at walking has a very large effect, > and at sub walking speed is not wise with double toe straps! > So gyroscopic forces provide stability to the bike by resisting > steering changes induced by lean. And of course gyroscopic forces > also contribute to steering changes when the axis of rotation is > tilted by leaning quickly. You can try that by holding a front wheel in your hands and not that the steer reaction to tilting a spinning wheel is faster the slower the rotation rate but the force to tilt the axis is also smaller. When coasting downhill no-hands steering works well the greater the speed. It just doesn't react as fast, but then that is what you want anyway or you would crash in a hurry. > But a bike stays in a turn when the center of mass is outside of the > center-line of the bike. This is why a riderless bike goes straight > (more or less). If it doesn't go dead straight it is probably > alignment or maybe even the off-center center of mass due to the > drivetrain. My son helped me with some "ghost-rider" tests on some > of our too-small kids bikes. Inconclusive as to whether they always > end up going to the same side. I think that experiment doesn't demonstrate the relationship between gyroscopic forces and rider control by moving the hips to steer. That movement is, in essence, one that doesn't occur on a riderless bicycle. > If you tape a brick to the side of the seatpost of a bike and send > it off on a riderless mission, it turns as the COM is off center and > this combined with the steering geometry means the steering flops to > the side. Gyroscopic forces keep it stable in this turn, just as > they keep it stable in a straight line with no brick. > So gyroscopic forces are necessary to keep a bike in a stable state, > whether a straight ahead, or a constant arc turn. They keep the > steering from making too twitchy changes, but they are not > exclusively what makes the turn happen, nor what keep the steering > at the proper angle during a turn. As I said, the faster it turns the slower the response. Jobst Brandt   Date: 23 Oct 2007 23:48:51 From: joseph.santaniello@gmail.com Subject: Re: Gyroscopic forces revisited On Oct 23, 11:05 pm, "Tom Nakashima" <t...@slac.stanford.edu > wrote: > <joseph.santanie...@gmail.com> wrote in message > > news:1193172557.232179.68820@v23g2000prn.googlegroups.com... > > > > > On Oct 23, 9:41 pm, "Tom Nakashima" <t...@slac.stanford.edu> wrote: > >> <joseph.santanie...@gmail.com> wrote in message > > >>news:1193163959.365061.86370@q3g2000prf.googlegroups.com... > > >> > It was postulated that a clown-bike with tiny wheel would be difficult > >> > to ride no hands due to the small gyroscopic forces. > Joseph > > >>http://www.youtube.com/watch?v=uRvwRgs_ZXE > >> -tom > > > Do you think this robot being able to ride a bike at super slow speeds > > with obviously entirely manual steering correction is evidence that a > > clown bike is difficult to ride no hands, or not so difficult? > > > Joseph > > The hardest part the designers had to deal with robot riding the bike is > balance. The engineers installed a gyro sensor that detects angular > velocity and inclination, then transmit the data to a computer that adjust > the robot's balance. > > Someone tried to ride a clown's bike with mini wheels and published an > article in one of the bicycle magazines. Said they had a hard time keeping > their balance. I believe it was a contest to see if anyone could ride it a > 100 ft. without putting their foot down. Nobody was able to do it. > -tom That may have been some problem with the geometry of the clown bike. I've seen 2 people riding those crazy fold-up bikes with what looks like roller-blade wheels and a belt drive, and they seemed to be doing just fine. I was in a car both times, so I didn't have the oppurtunity to investigate further. Joseph   Date: 23 Oct 2007 15:59:57 From: vey Subject: Re: Gyroscopic forces revisited joseph.santaniello@gmail.com wrote: > Hi All, > > A while ago there was a thread discussing no-handed riding and the > role gyroscopic forces play in bike turning. I think it is a COG problem. The bikes I rode that I could ride with no hands relied less on the speed and more on how high the COG was. These MTB style bikes have a much higher COG than say, a Dutch commuter bike.   Date: 23 Oct 2007 13:49:17 From: joseph.santaniello@gmail.com Subject: Re: Gyroscopic forces revisited On Oct 23, 9:41 pm, "Tom Nakashima" <t...@slac.stanford.edu > wrote: > <joseph.santanie...@gmail.com> wrote in message > > news:1193163959.365061.86370@q3g2000prf.googlegroups.com... > > > > > It was postulated that a clown-bike with tiny wheel would be difficult > > to ride no hands due to the small gyroscopic forces. > Joseph > > http://www.youtube.com/watch?v=uRvwRgs_ZXE > -tom Do you think this robot being able to ride a bike at super slow speeds with obviously entirely manual steering correction is evidence that a clown bike is difficult to ride no hands, or not so difficult? Joseph     Date: 23 Oct 2007 21:50:27 From: Subject: Re: Gyroscopic forces revisited On Tue, 23 Oct 2007 13:49:17 -0700, "joseph.santaniello@gmail.com" <joseph.santaniello@gmail.com > wrote: >On Oct 23, 9:41 pm, "Tom Nakashima" <t...@slac.stanford.edu> wrote: >> <joseph.santanie...@gmail.com> wrote in message >> >> news:1193163959.365061.86370@q3g2000prf.googlegroups.com... >> >> >> >> > It was postulated that a clown-bike with tiny wheel would be difficult >> > to ride no hands due to the small gyroscopic forces. > Joseph >> >> http://www.youtube.com/watch?v=uRvwRgs_ZXE >> -tom > >Do you think this robot being able to ride a bike at super slow speeds >with obviously entirely manual steering correction is evidence that a >clown bike is difficult to ride no hands, or not so difficult? > >Joseph What was in the robot's backpack? I'm betting there was a gyro somewhere. -- Posted via a free Usenet account from http://www.teranews.com       Date: 24 Oct 2007 06:38:38 From: Tom Nakashima Subject: Re: Gyroscopic forces revisited <clare at snyder.on.ca > wrote in message news:469th3hkg09vi47ss3jqjv67r3nle42678@4ax.com... > On Tue, 23 Oct 2007 13:49:17 -0700, "joseph.santaniello@gmail.com" > <joseph.santaniello@gmail.com> wrote: > >>On Oct 23, 9:41 pm, "Tom Nakashima" <t...@slac.stanford.edu> wrote: >>> <joseph.santanie...@gmail.com> wrote in message >>> >>> news:1193163959.365061.86370@q3g2000prf.googlegroups.com... >>> >>> >>> >>> > It was postulated that a clown-bike with tiny wheel would be difficult >>> > to ride no hands due to the small gyroscopic forces. > Joseph >>> >>> http://www.youtube.com/watch?v=uRvwRgs_ZXE >>> -tom >> >>Do you think this robot being able to ride a bike at super slow speeds >>with obviously entirely manual steering correction is evidence that a >>clown bike is difficult to ride no hands, or not so difficult? >> >>Joseph > What was in the robot's backpack? I'm betting there was a gyro > somewhere. Not sure what's in the robot's backpack, the computer? The gyroscopic sensor is in robot's stomach, makes sense to balance the CG. Another video showing the robot riding the bike in circles, you can clearly seen the gyroscopic sensors. http://www.youtube.com/watch?v=Srwk-i5aXRQ enjoy, -tom     Date: 23 Oct 2007 14:05:03 From: Tom Nakashima Subject: Re: Gyroscopic forces revisited <joseph.santaniello@gmail.com > wrote in message news:1193172557.232179.68820@v23g2000prn.googlegroups.com... > On Oct 23, 9:41 pm, "Tom Nakashima" <t...@slac.stanford.edu> wrote: >> <joseph.santanie...@gmail.com> wrote in message >> >> news:1193163959.365061.86370@q3g2000prf.googlegroups.com... >> >> >> >> > It was postulated that a clown-bike with tiny wheel would be difficult >> > to ride no hands due to the small gyroscopic forces. > Joseph >> >> http://www.youtube.com/watch?v=uRvwRgs_ZXE >> -tom > > Do you think this robot being able to ride a bike at super slow speeds > with obviously entirely manual steering correction is evidence that a > clown bike is difficult to ride no hands, or not so difficult? > > Joseph > The hardest part the designers had to deal with robot riding the bike is balance. The engineers installed a gyro sensor that detects angular velocity and inclination, then transmit the data to a computer that adjust the robot's balance. Someone tried to ride a clown's bike with mini wheels and published an article in one of the bicycle magazines. Said they had a hard time keeping their balance. I believe it was a contest to see if anyone could ride it a 100 ft. without putting their foot down. Nobody was able to do it. -tom       Date: 24 Oct 2007 06:49:54 From: Doc O'Leary Subject: Re: Gyroscopic forces revisited In article <fflnm0$ko5$1@news.Stanford.EDU >, "Tom Nakashima" <tom@slac.stanford.edu > wrote: > Someone tried to ride a clown's bike with mini wheels and published an > article in one of the bicycle magazines. Said they had a hard time keeping > their balance. I believe it was a contest to see if anyone could ride it a > 100 ft. without putting their foot down. Nobody was able to do it. I'm not sure what that has to do with gyroscopic forces. Something like the A-bike seems quite able to be balanced with mini wheels, likely for the same reason (higher center of gravity) that it's easy to balance a broom on your fingertip but not the dustpan. -- My personal UDP list: 127.0.0.1, 4ax.com, buzzardnews.com, googlegroups.com, heapnode.com, localhost, ntli.net, teranews.com, vif.com, x-privat.org       Date: 24 Oct 2007 03:11:12 From: Subject: Re: Gyroscopic forces revisited Tom Nakashima writes: >>>> It was postulated that a clown-bike with tiny wheel would be >>>> difficult to ride no hands due to the small gyroscopic forces. > >>>> Joseph http://www.youtube.com/watch?v=uRvwRgs_ZXE >> Do you think this robot being able to ride a bike at super slow >> speeds with obviously entirely manual steering correction is >> evidence that a clown bike is difficult to ride no hands, or not so >> difficult? > The hardest part the designers had to deal with robot riding the > bike is balance. The engineers installed a gyro sensor that detects > angular velocity and inclination, then transmit the data to a > computer that adjust the robot's balance. > Someone tried to ride a clown's bike with mini wheels and published > an article in one of the bicycle magazines. Said they had a hard > time keeping their balance. I believe it was a contest to see if > anyone could ride it a 100 ft. without putting their foot down. > Nobody was able to do it. I think this is all over the hill stuff because Segway has a product and it is being used in some odd places, such as for flatfoots cruising the downtown sidewalks. Balance there uses gyroscopes in two directions. http://ai-depot.com/Robotics/112.html http://www.segway.com/ Jobst Brandt   Date: 23 Oct 2007 20:49:14 From: bjw@mambo.ucolick.org Subject: Re: Gyroscopic forces revisited On Oct 23, 11:25 am, "joseph.santanie...@gmail.com" <joseph.santanie...@gmail.com > wrote: > Hi All, > > A while ago there was a thread discussing no-handed riding and the > role gyroscopic forces play in bike turning. > > I hypothesized that gyroscopic forces were not that important, and > several things were pointed out to me that almost convinced me. A good > dose of insomnia allowed me to think about this for several hours and > I am now convinced that gyroscopic forces are virtually irrelevant for > riding bikes, hands or no-hands. > .... > So what does all that mean? It means that "flop" from an off-center > center of mass is what makes a bike turn, and thus while gyroscopic > forces make help the initial turn of the steering due to induced lean, > it is an unnecessary component that is ultimately irrelevant to turing > a bike. I don't think your experiment tested what you think it tested. First, gyro inertia experiments were done in a controlled way by David Jones many years ago. Andrew Muzi alluded to this. You must read the article at http://www.phys.lsu.edu/faculty/gonzalez/Teaching/Phys7221/vol59no9p51_56.pdf He found that the zero-gyro-inertia front wheel (URB 1) was very difficult to ride no-hands. Leaning the bike is how you turn at high speed. Go through a turn at high speed and you'll see that the front wheel is nearly straight. A bit of countersteer initiates the lean and the turn. I'm not sure I'd call this a gyroscopic force, because the flop comes from the steering geometry (fork trail), not from the gyro force on the wheel. As evidence, I don't think you lean the bike at a different angle with a heavy tire on the front wheel, even though that increases the moment of inertia a lot. Of course, you still lean the bike at low speed, but the allowable turn of the front wheel is much greater. When riding no hands, I believe the gyro inertia of the front wheel helps stabilize the front against excessive flop, and this is why it is more difficult to ride no-hands at low speed. So the gyro inertia doesn't help you turn. It helps you not turn. Ben   Date: 23 Oct 2007 12:41:55 From: Tom Nakashima Subject: Re: Gyroscopic forces revisited <joseph.santaniello@gmail.com > wrote in message news:1193163959.365061.86370@q3g2000prf.googlegroups.com... > > It was postulated that a clown-bike with tiny wheel would be difficult > to ride no hands due to the small gyroscopic forces. > Joseph > http://www.youtube.com/watch?v=uRvwRgs_ZXE -tom   Date: 23 Oct 2007 12:27:33 From: joseph.santaniello@gmail.com Subject: Re: Gyroscopic forces revisited On Oct 23, 8:54 pm, jobst.bra...@stanfordalumni.org wrote: > Joseph Santaniello writes: > > A while ago there was a thread discussing no-handed riding and the > > role gyroscopic forces play in bike turning. > > I hypothesized that gyroscopic forces were not that important, and > > several things were pointed out to me that almost convinced me. A > > good dose of insomnia allowed me to think about this for several > > hours and I am now convinced that gyroscopic forces are virtually > > irrelevant for riding bikes, hands or no-hands. > > It was postulated that a clown-bike with tiny wheel would be > > difficult to ride no hands due to the small gyroscopic forces. > > Since I don't have a clown-bike to play with, this remains a > > mystery. But I have observed little kids riding small wheel (10") > > bikes at below walking pace. I have also spun some 12" wheels in my > > hand a various speeds to feel what sort of gyroscopic forces are > > there. Not much. A little kid with a sense of balance not as > > developed as an adult can ride one of these bikes. I do not believe > > a kid could keep one of these bikes upright by manual correction > > alone. These small bikes are stable by themselves, and since the > > gyroscopic forces are so low, there must be something else at work > > here. This isn't proof or anything, this is just what got me > > thinking. > > You should have seen the demo at InterBike where an engineer built a > front wheel with a forward rotating flywheel (brass disk) between > the spokes driven by a small motor at about the speed you expect from > a 27" wheel at 10-15mph. He rode this bicycle no-hands at below 2mph, > steady as a rock, up and down the isles. Without the flywheel > turning independently, the bicycle was as difficult to ride no-hands > as any other bicycle with wheels that size. That sounds like fun! That seems to indicate that the gyroscopic forces counter the "flop" tendency and keep a bike from turing too much. I wonder if his gizmo would make a chopper bike rideable no hands... > > > I finally pulled out a big plywood board (much to the chagrin of my > > wife who imagines there are myriad things I could be better spending > > my time on) and propped it up at an angle and put a bike on it to > > simulate what happens when a bike is in a turn. It was suggested > > that in a turn a pendulum hung from the top-tube would hang parallel > > to the seat tube. When riding no hands in a constant radius turn, > > this cannot be the case, and I suspect that it is not the case with > > hands on the bars either. Experiments with my plywood board show > > that when force is applied straight down through the bike the > > steering remains straight no matter what the angle of the board > > (simulated angle of lean). In a turn, the steering cannot be > > straight, otherwise it wouldn't be called a turn, it would be called > > a crash. So in a turn (a no-handed one in particular) something has > > to be holding the steering at a non-straight angle. The only thing > > it can be is that the center of mass is moved to the side of the > > plane that is the centerline of the bike. This makes the steering > > flop into the turn. The combined force of gravity and the > > acceleration of the turn act from the center of mass through the > > contact patches of the tires. Since the center of mass is not in > > the plane of the bike's centerline, this means that the combined > > force is not parallel to the seat tube and thus a pendulum hanging > > from the top- tube could not be parallel with the seat tube. Riding > > with hands on the bars I suspect is the same. But a rider could > > force the bike to be in the same plane, but then they would need to > > hold the steering at the proper angle manually. This would no doubt > > require quite a bit of skill, and I believe in practice to be > > virtually impossible. But perhaps it is just this skill which > > separates the good from the great. > > Please discover why one should use paragraphs. You must have come > across this in school. Who says I went to school? Ok, I'll try. ;-) > > So what does all that mean? It means that "flop" from an off-center > > center of mass is what makes a bike turn, and thus while gyroscopic > > forces make help the initial turn of the steering due to induced > > lean, it is an unnecessary component that is ultimately irrelevant > > to turning a bike. > > The whole COM argument was brought about by thinking about how a > > radio controlled motorcycle I used to have worked. > > I think your research came up with the wrong result. > > An easily repeatable exercise of coasting down a smooth road at more > than 20mph riding no-hands, is to shake one knee from side to side > while resting the other one against the top tube for stability. I > think there is where you will see the effect the best. In addition, > shimmy on a bicycle cannot occur without gyroscopic forces of the > front wheel. I did that very same exercise today on my ride while I was polishing my theory. I don't argue that gyroscopic forces do not turn the steering from leaning the bike. I argue that the steering would turn anyway even if the gyroscopic forces were not there, and that the same force that would do this turning is the force that holds a bike with no hands in a constant arc turn. Joseph     Date: 24 Oct 2007 02:59:04 From: Subject: Re: Gyroscopic forces revisited Joseph Santaniello writes: >>> A while ago there was a thread discussing no-handed riding and the >>> role gyroscopic forces play in bike turning. I hypothesized that >>> gyroscopic forces were not that important, and several things were >>> pointed out to me that almost convinced me. A good dose of >>> insomnia allowed me to think about this for several hours and I am >>> now convinced that gyroscopic forces are virtually irrelevant for >>> riding bikes, hands or no-hands. It was postulated that a >>> clown-bike with tiny wheel would be difficult to ride no hands due >>> to the small gyroscopic forces. Since I don't have a clown-bike >>> to play with, this remains a mystery. But I have observed little >>> kids riding small wheel (10") bikes at below walking pace. I have >>> also spun some 12" wheels in my hand a various speeds to feel what >>> sort of gyroscopic forces are there. Not much. A little kid with >>> a sense of balance not as developed as an adult can ride one of >>> these bikes. I do not believe a kid could keep one of these bikes >>> upright by manual correction alone. These small bikes are stable >>> by themselves, and since the gyroscopic forces are so low, there >>> must be something else at work here. This isn't proof or >>> anything, this is just what got me thinking. >> You should have seen the demo at InterBike where an engineer built >> a front wheel with a forward rotating flywheel (brass disk) between >> the spokes driven by a small motor at about the speed you expect >> from a 27" wheel at 10-15mph. He rode this bicycle no-hands at >> below 2mph, steady as a rock, up and down the isles. Without the >> flywheel turning independently, the bicycle was as difficult to >> ride no-hands as any other bicycle with wheels that size. > That sounds like fun! That seems to indicate that the gyroscopic > forces counter the "flop" tendency and keep a bike from turning too > much. I wonder if his gizmo would make a chopper bike ridable no > hands... You are grasping at straws. The gyroscopic effect of the flywheel made the bicycle handle as though it were traveling above 10mph, a speed at which riding no hands is easy because there are gyroscopic steering forces available to the rider. >>> I finally pulled out a big plywood board (much to the chagrin of >>> my wife who imagines there are myriad things I could be better >>> spending my time on) and propped it up at an angle and put a bike >>> on it to simulate what happens when a bike is in a turn. It was >>> suggested that in a turn a pendulum hung from the top-tube would >>> hang parallel to the seat tube. When riding no hands in a >>> constant radius turn, this cannot be the case, and I suspect that >>> it is not the case with hands on the bars either. Experiments >>> with my plywood board show that when force is applied straight >>> down through the bike the steering remains straight no matter what >>> the angle of the board (simulated angle of lean). In a turn, the >>> steering cannot be straight, otherwise it wouldn't be called a >>> turn, it would be called a crash. So in a turn (a no-handed one >>> in particular) something has to be holding the steering at a >>> non-straight angle. The only thing it can be is that the center >>> of mass is moved to the side of the plane that is the centerline >>> of the bike. This makes the steering flop into the turn. The >>> combined force of gravity and the acceleration of the turn act >>> from the center of mass through the contact patches of the tires. >>> Since the center of mass is not in the plane of the bike's >>> centerline, this means that the combined force is not parallel to >>> the seat tube and thus a pendulum hanging from the top- tube could >>> not be parallel with the seat tube. Riding with hands on the bars >>> I suspect is the same. But a rider could force the bike to be in >>> the same plane, but then they would need to hold the steering at >>> the proper angle manually. This would no doubt require quite a >>> bit of skill, and I believe in practice to be virtually >>> impossible. But perhaps it is just this skill which separates the >>> good from the great. >> Please discover why one should use paragraphs. You must have come >> across this in school. > Who says I went to school? OK, I'll try. ;-) >>> So what does all that mean? It means that "flop" from an >>> off-center center of mass is what makes a bike turn, and thus >>> while gyroscopic forces make help the initial turn of the steering >>> due to induced lean, it is an unnecessary component that is >>> ultimately irrelevant to turning a bike. The whole COM argument >>> was brought about by thinking about how a radio controlled >>> motorcycle I used to have worked. >> I think your research came up with the wrong result. >> An easily repeatable exercise of coasting down a smooth road at >> more than 20mph riding no-hands, is to shake one knee from side to >> side while resting the other one against the top tube for >> stability. I think there is where you will see the effect the >> best. In addition, shimmy on a bicycle cannot occur without >> gyroscopic forces of the front wheel. > I did that very same exercise today on my ride while I was polishing > my theory. I don't argue that gyroscopic forces do not turn the > steering from leaning the bike. I argue that the steering would > turn anyway even if the gyroscopic forces were not there, and that > the same force that would do this turning is the force that holds a > bike with no hands in a constant arc turn. Well, try riding no-hands at 2mph (less than walking speed). To what do you attribute the inability to steer the bicycle at such a speed? Just try moving one knee side-to-side and see what steering you get at those speeds. Jobst Brandt   Date: 23 Oct 2007 19:08:54 From: Jon_C Subject: Re: Gyroscopic forces revisited How about a wheel on it's own? If you roll a wheel and it leans over it turns in that direction. The reason is that the contact patch of a leaning tyre is parallel to the ground and so at an angle to the wheel's axle so the wheel acts as a rolling cone and turns into the corner (which maintains it's balance). I'm pretty sure that's the main steering force involved in no-handed riding. (I seem to remember reading about someone testing this on a motorbike and finding that on a prolonged corners the handlebars are actually turned slightly outward instead of inward as you;d expect. I think that was due to rake/trail but I don;t have time to think about that right now :)     Date: 24 Oct 2007 02:51:02 From: Subject: Re: Gyroscopic forces revisited Jon Crouch writes: > How about a wheel on it's own? If you roll a wheel and it leans over > it turns in that direction. The reason is that the contact patch of > a leaning tyre is parallel to the ground and so at an angle to the > wheel's axle so the wheel acts as a rolling cone and turns into the > corner (which maintains it's balance). I'm pretty sure that's the > main steering force involved in no-handed riding. > (I seem to remember reading about someone testing this on a > motorbike and finding that on a prolonged corners the handlebars are > actually turned slightly outward instead of inward as you;d > expect. I think that was due to rake/trail but I don;t have time to > think about that right now :) I think you ought to go to a velodrome. These tracks have sloped straights and more steeply banked circular curves at each end. Riders have no problem riding no-hands on these tracks, either on the straight sections or in the banked curve (if they are going fast enough to not strike a pedal). If you don't have a track, try a crowned paved street and notice how the angle with the pavement has essentially no effect on steering, the centerline of the contact patch moving less than 1/8" off center for a 10=C2=B0 side slope. Jobst Brandt       Date: 29 Oct 2007 17:32:35 From: Jon_C Subject: Re: Gyroscopic forces revisited On Oct 29, 10:34 am, jobst.bra...@stanfordalumni.org wrote: > Jon Crouch writes: > > How would deflating the tyre increase camber thrust? You don't seem > > to have grasped the concept and yet you're going to great lengths to > > deny it's validity. > > It increases the width of the contact patch from which some torque > about the vertical axis of the wheel is supposedly acting. > > Jobst Brandt If you'd taken the time to read the articles you dismissed on camber force, you'd understand that the wheel acts as a rolling partial cone. (the red section in this pic http://picasaweb.google.com/JonCCrouch/UntitledAlbum/photo#5125629114602972546) Increasing the width of the cone section won't alter the turn radius.         Date: 29 Oct 2007 19:13:59 From: Subject: Re: Gyroscopic forces revisited Jon Crouch writes: >>> How would deflating the tyre increase camber thrust? You don't >>> seem to have grasped the concept and yet you're going to great >>> lengths to deny it's validity. >> It increases the width of the contact patch from which some torque >> about the vertical axis of the wheel is supposedly acting. > If you'd taken the time to read the articles you dismissed on camber > force, you'd understand that the wheel acts as a rolling partial cone. It isn't a part of a cone if it is knife-edged thin. If the tire has a broad contact patch and is tilted, then it is a partial cone. > the red section in this pic: http://picasaweb.google.com/JonCCrouch/UntitledAlbum/photo#5125629114602972546 > Increasing the width of the cone section won't alter the turn radius. This is a nice picture but it doesn't address the flexible nature of a tire that has a non-uniform pressure in the contact patch and makes contact with the road in a canoe shaped area, the point toughing down first and the sides later. It also does not include any analysis of the force vectors. If I ride around a ten foot circle with a lean of five degrees, and then ride in a 40 foot circle with a lean of five degrees, how does the "camber thrust" bear on this? I don't see the cone coming into play here. To think about tire scrub one might look at how tires wear. It is my experience that front tires wear through in the center even though their greatest traction demand is from cornering, and from the round cross section that makes the edge of the contact patch scrub forward, it being at a smaller radius than the center and more lightly loaded. This is best observed on smooth treaded tires. Jobst Brandt       Date: 26 Oct 2007 09:33:30 From: joseph.santaniello@gmail.com Subject: Re: Gyroscopic forces revisited On Oct 26, 6:02 pm, jobst.bra...@stanfordalumni.org wrote: > Joseph Santaniello writes: > >>>> With both hands on the stem my elbows meet nicely in my gut, just > >>>> above my knees. From what you say, your bicycle seems to be too > >>>> short if you cannot bring your elbows in. > >>> Hands on the stem? Or the bars? With both hands in fists close > >>> together as though I were holding the bars with my thumbs wrapped > >>> under the stem, there is no angle I can hold my arms to get my > >>> elbows to meet. It wouldn't matter how long or short my bike is, > >>> my arms can't do it. With forearms rotated I can do it, but I > >>> wouldn't be able to hold the bar. > >> I don't understand. I can bring my elbows together clasping hands > >> as in prayer or both over one another on the stem. Leave the bars > >> out of this. > > I see. That's different. I'll try that. > >>>>> But it is steering angle, not lean angle I was thinking about. > >>>>> But I suppose this just leads us to the drifting idea, which I > >>>>> agree is pretty much not do-able. > >>>> You have no option with steering angle, the road prescribes that > >>>> because You must get around the curve. That can be modified only > >>>> a little by choosing an ideal arc that doesn't infringe on > >>>> oncoming traffic or the edge of the road. > >>> I guess I was imagining it might be possible to force the bike > >>> into some combination of lean and steering angle that would not be > >>> the same angles the bike would find on it's own ridden the same > >>> line and speed no handed. But since wheel slip is out of the > >>> question, the only thing that would accomplish is either side > >>> loads, or unnecessarily steep bike lean. > >> Those two parameters are not independent and are given by the curve > >> radius and speed. As I mentioned in the article, leaning off the > >> bicycle doesn't alter the plumb line from CG to ground contact > >> line. > > No, but it does alter the angle the bike alone is leaning and > > subsequently the steering angle (by which I mean how far off center > > the stem is) and the relative location of the patches. > > The steering angle does not change. The track around the curve > remains unchanged, only the rider and bicycle are no longer in the > same plane. Following the same track requires the same steering > angle. You can't fool geometry, or gravity for that matter. > For a given angle of lean on a given curve, the steering angle will be the same, but for the same curve at different angles of lean, I think the steering angle will be different. I think because of the angle of the steering axis and the resulting trail, the poition and orientation of the patches does change if you hold the steering angle constant. Perhaps only slightly. Suppose you walk a bike perfectly upright along some arc. The steering will need to be turned to some angle to acheive this. If you then walk the bike along that same arc, but with some arbitrary angle of lean, I contend that the steering angle (in terms of stem to center- centerline) will be different. Maybe the patches will describe the same arc, but the steering will be at a different angle, and the front patch will be at a different place on the wheel. Joseph         Date: 26 Oct 2007 16:46:12 From: Subject: Re: Gyroscopic forces revisited Joseph Santaniello writes: >>>>>> With both hands on the stem my elbows meet nicely in my gut, just >>>>>> above my knees. From what you say, your bicycle seems to be too >>>>>> short if you cannot bring your elbows in. >>>>> Hands on the stem? Or the bars? With both hands in fists close >>>>> together as though I were holding the bars with my thumbs wrapped >>>>> under the stem, there is no angle I can hold my arms to get my >>>>> elbows to meet. It wouldn't matter how long or short my bike is, >>>>> my arms can't do it. With forearms rotated I can do it, but I >>>>> wouldn't be able to hold the bar. >>>> I don't understand. I can bring my elbows together clasping hands >>>> as in prayer or both over one another on the stem. Leave the bars >>>> out of this. >>> I see. That's different. I'll try that. >>>>>>> But it is steering angle, not lean angle I was thinking about. >>>>>>> But I suppose this just leads us to the drifting idea, which I >>>>>>> agree is pretty much not do-able. >>>>>> You have no option with steering angle, the road prescribes that >>>>>> because You must get around the curve. That can be modified only >>>>>> a little by choosing an ideal arc that doesn't infringe on >>>>>> oncoming traffic or the edge of the road. >>>>> I guess I was imagining it might be possible to force the bike >>>>> into some combination of lean and steering angle that would not be >>>>> the same angles the bike would find on it's own ridden the same >>>>> line and speed no handed. But since wheel slip is out of the >>>>> question, the only thing that would accomplish is either side >>>>> loads, or unnecessarily steep bike lean. >>>> Those two parameters are not independent and are given by the curve >>>> radius and speed. As I mentioned in the article, leaning off the >>>> bicycle doesn't alter the plumb line from CG to ground contact >>>> line. >>> No, but it does alter the angle the bike alone is leaning and >>> subsequently the steering angle (by which I mean how far off center >>> the stem is) and the relative location of the patches. >> The steering angle does not change. The track around the curve >> remains unchanged, only the rider and bicycle are no longer in the >> same plane. Following the same track requires the same steering >> angle. You can't fool geometry, or gravity for that matter. > For a given angle of lean on a given curve, the steering angle will be > the same, but for the same curve at different angles of lean, I think > the steering angle will be different. You can't have different angles of lean. As I said you can't fool geometry and gravity. It may appear as a different angle of lean but it isn't. The plumb line from the CG to the road is the same at the same speed around the same curve no matter what the rider does. > I think because of the angle of the steering axis and the resulting > trail, the position and orientation of the patches does change if you > hold the steering angle constant. Perhaps only slightly. That has no effect on the path taken. Besides, when leaning off the plane of the bicycle, corrective force is required on the handlebar to make up for the odd geometry. This is less apparent on an M/C because it is far heavier but still requires steering force to hold the line. You can't, for instance, ride no-hands when leaning off to one side. > Suppose you walk a bike perfectly upright along some arc. The > steering will need to be turned to some angle to achieve this. If > you then walk the bike along that same arc, but with some arbitrary > angle of lean, I contend that the steering angle (in terms of stem > to center- centerline) will be different. Maybe the patches will > describe the same arc, but the steering will be at a different > angle, and the front patch will be at a different place on the > wheel. You are holding the bicycle. It is not balanced in the curve. This is not a two wheeled experiment but one with a trainer (training wheels). Jobst Brandt       Date: 24 Oct 2007 06:31:45 From: Tom Nakashima Subject: Re: Gyroscopic forces revisited <jobst.brandt@stanfordalumni.org > wrote in message news:471eb316$0$14121$742ec2ed@news.sonic.net... >I think you ought to go to a velodrome. These tracks have sloped >straights and more steeply banked circular curves at each end. Riders >have no problem riding no-hands on these tracks, either on the >straight sections or in the banked curve (if they are going fast >enough to not strike a pedal). >If you don't have a track, try a crowned paved street and notice how >the angle with the pavement has essentially no effect on steering, the centerline of the contact patch moving less than 1/8" off center for a >10° side slope. >Jobst Brandt I think I may be able to answer this one. I was talking with co-worker Timothy Montagne who just got back from the track Nationals in Trexlertown, Pennsylvania. I forget the exact slope on the straights in degrees, but he told me it was so steep that you have to keep a minimum speed of 20 mph. or else you would fall over. So I can imagine your speed has to be greater than 20 mph to be able to ride with no hands, of course which is no problem with a track rider. When I ride the San Jose velodrome, I have no problem riding with no hands, but the slope is mild. I'm sure a min. speed is needed, just never crossed my mind. I'm riding with Tim today after work, so I can pick his brains more on the banked track. -tom         Date: 24 Oct 2007 23:47:45 From: sl Subject: Re: Gyroscopic forces revisited In article <ffnhg2$n2$1@news.Stanford.EDU >, Tom Nakashima <tom@slac.stanford.edu > wrote: > ><jobst.brandt@stanfordalumni.org> wrote in message >news:471eb316$0$14121$742ec2ed@news.sonic.net... > >>I think you ought to go to a velodrome. These tracks have sloped >>straights and more steeply banked circular curves at each end. Riders >>have no problem riding no-hands on these tracks, either on the >>straight sections or in the banked curve (if they are going fast >>enough to not strike a pedal). > >>If you don't have a track, try a crowned paved street and notice how >>the angle with the pavement has essentially no effect on steering, the >centerline of the contact patch moving less than 1/8" off center for a >>10° side slope. >>Jobst Brandt > >I think I may be able to answer this one. >I was talking with co-worker Timothy Montagne who just got back from the >track Nationals in Trexlertown, Pennsylvania. I forget the exact slope on >the straights in degrees, but he told me it was so steep that you have to >keep a minimum speed of 20 mph. or else you would fall over. So I can >imagine your speed has to be greater than 20 mph to be able to ride with no >hands, of course which is no problem with a track rider. I ride at Burnaby Velodrome, 200m wooden track, 47 degrees at the steepest part of the curve. It is difficult to stay on in the curve at anything under 30kph. We routinely see new riders slide off from going to slow. As recently as yesterday morning.           Date: 25 Oct 2007 07:20:23 From: Tom Nakashima Subject: Re: Gyroscopic forces revisited "sl" <sl@whiskey.enposte.net > wrote in message news:BQQTi.140654$1y4.11487@pd7urf2no... > In article <ffnhg2$n2$1@news.Stanford.EDU>, > Tom Nakashima <tom@slac.stanford.edu> wrote: >> >><jobst.brandt@stanfordalumni.org> wrote in message >>news:471eb316$0$14121$742ec2ed@news.sonic.net... >> >>>I think you ought to go to a velodrome. These tracks have sloped >>>straights and more steeply banked circular curves at each end. Riders >>>have no problem riding no-hands on these tracks, either on the >>>straight sections or in the banked curve (if they are going fast >>>enough to not strike a pedal). >> >>>If you don't have a track, try a crowned paved street and notice how >>>the angle with the pavement has essentially no effect on steering, the >>centerline of the contact patch moving less than 1/8" off center for a >>>10° side slope. >>>Jobst Brandt >> >>I think I may be able to answer this one. >>I was talking with co-worker Timothy Montagne who just got back from the >>track Nationals in Trexlertown, Pennsylvania. I forget the exact slope on >>the straights in degrees, but he told me it was so steep that you have to >>keep a minimum speed of 20 mph. or else you would fall over. So I can >>imagine your speed has to be greater than 20 mph to be able to ride with >>no >>hands, of course which is no problem with a track rider. > > I ride at Burnaby Velodrome, 200m wooden track, 47 degrees at the > steepest part of the curve. > > It is difficult to stay on in the curve at anything under 30kph. > > We routinely see new riders slide off from going to slow. As recently > as yesterday morning. > That sounds about right, 30kph would be 18.64114 mph. -tom           Date: 24 Oct 2007 20:42:33 From: Tim McNamara Subject: Re: Gyroscopic forces revisited In article <BQQTi.140654$1y4.11487@pd7urf2no >, sl@whiskey.enposte.net (sl) wrote: > In article <ffnhg2$n2$1@news.Stanford.EDU>, Tom Nakashima > <tom@slac.stanford.edu> wrote: > > > ><jobst.brandt@stanfordalumni.org> wrote in message > >news:471eb316$0$14121$742ec2ed@news.sonic.net... > > > >>I think you ought to go to a velodrome. These tracks have sloped > >>straights and more steeply banked circular curves at each end. > >>Riders have no problem riding no-hands on these tracks, either on > >>the straight sections or in the banked curve (if they are going > >>fast enough to not strike a pedal). > > > >>If you don't have a track, try a crowned paved street and notice > >>how the angle with the pavement has essentially no effect on > >>steering, the > >centerline of the contact patch moving less than 1/8" off center for > >a > >>10° side slope. Jobst Brandt > > > >I think I may be able to answer this one. I was talking with > >co-worker Timothy Montagne who just got back from the track > >Nationals in Trexlertown, Pennsylvania. I forget the exact slope on > >the straights in degrees, but he told me it was so steep that you > >have to keep a minimum speed of 20 mph. or else you would fall over. > >So I can imagine your speed has to be greater than 20 mph to be able > >to ride with no hands, of course which is no problem with a track > >rider. > > I ride at Burnaby Velodrome, 200m wooden track, 47 degrees at the > steepest part of the curve. > > It is difficult to stay on in the curve at anything under 30kph. > > We routinely see new riders slide off from going to slow. As recently > as yesterday morning. And yet sprinters can ride the entire lap, at any lateral position, at literally a walking pace. So that disproves the claim that you have to go at least 30 kph. I've ridden many times at the local wooden velodrome (Blaine MN, 250 m, 43 degree banks in corners) as slow as 10 mph. It is necessary to tilt the bike deliberately to the inside to avoid striking a pedal on the up-track side- failure to do so is probably what is causing your new riders to fall off. I have seen riders come to a complete stop in mid-corner without falling off. Perhaps proper slow cornering technique ought to be added to the introductory class. You do have an introductory class for new riders, right? It would reduce the number of crashing newbies.             Date: 25 Oct 2007 06:57:10 From: Tom Nakashima Subject: Re: Gyroscopic forces revisited "Tim McNamara" <timmcn@bitstream.net > wrote in message news:timmcn-9FFEAC.20423324102007@news.iphouse.com... > > Perhaps proper slow cornering technique ought to be added to the > introductory class. You do have an introductory class for new riders, > right? It would reduce the number of crashing newbies. As a track rider I'm surprised you don't know that most crashes on the track are due to the inexperience of riding in a tight pack. I've seen riders panic, stop pedaling, touch the rear wheel of the bike in front of them. It isn't until one becomes experienced in track riding that the newbie's become at ease. When I get back on my roadbike after riding the track, everything seems like slow motion. Try riding with no hands at 5 mph on the straights of a banked track next time you're out there. -tom               Date: 25 Oct 2007 13:59:37 From: Tim McNamara Subject: Re: Gyroscopic forces revisited In article <ffq7bn$3mm$1@news.Stanford.EDU >, "Tom Nakashima" <tom@slac.stanford.edu > wrote: > "Tim McNamara" <timmcn@bitstream.net> wrote in message > news:timmcn-9FFEAC.20423324102007@news.iphouse.com... > > > > Perhaps proper slow cornering technique ought to be added to the > > introductory class. You do have an introductory class for new > > riders, right? It would reduce the number of crashing newbies. > > As a track rider I'm surprised you don't know that most crashes on > the track are due to the inexperience of riding in a tight pack. I've > seen riders panic, stop pedaling, touch the rear wheel of the bike in > front of them. It isn't until one becomes experienced in track riding > that the newbie's become at ease. When I get back on my roadbike > after riding the track, everything seems like slow motion. The crashes I have seen on the velodrome have come from mostly from grounding a pedal (probably about 5 of those, two of which took me down when the riders slid down the track into me), a couple from touching wheels and once from a blowout the cause of which I don't know. All the pedal strikes occurred on the roll-out before the start of an event. Oddly enough, the wheel touches all occurred in Cat 1/2/3 races, but that was just luck of the draw (and smaller fields in those days- there are twice as many people racing track now than 10 years ago). There were way more crashes in the Cat 4/5 races. I once did stop pedaling after a sprint and got a reminder from the bike to not do *that* again. Fortunately we were all lined up across the track and there wasn't anyone behind me to get drop-kicked. The official had a little chat with me after the race, as you might expect. > Try riding with no hands at 5 mph on the straights of a banked track > next time you're out there. -tom I haven't ridden on the velodrome since 1998 or 99 (separated my shoulder in one of the above-mentioned crashes. Never got back on the track after that, realizing that at 40 years old I wasn't going to heal as well as in my youth and that I couldn't afford to lose time from work over a hobby). I don't even have my track bike any more.                 Date: 25 Oct 2007 12:45:27 From: Tom Nakashima Subject: Re: Gyroscopic forces revisited "Tim McNamara" <timmcn@bitstream.net > wrote in message news:timmcn-4006AA.13593725102007@news.iphouse.com... > In article <ffq7bn$3mm$1@news.Stanford.EDU>, > "Tom Nakashima" <tom@slac.stanford.edu> wrote: > >> "Tim McNamara" <timmcn@bitstream.net> wrote in message >> news:timmcn-9FFEAC.20423324102007@news.iphouse.com... >> > >> > Perhaps proper slow cornering technique ought to be added to the >> > introductory class. You do have an introductory class for new >> > riders, right? It would reduce the number of crashing newbies. >> >> As a track rider I'm surprised you don't know that most crashes on >> the track are due to the inexperience of riding in a tight pack. I've >> seen riders panic, stop pedaling, touch the rear wheel of the bike in >> front of them. It isn't until one becomes experienced in track riding >> that the newbie's become at ease. When I get back on my roadbike >> after riding the track, everything seems like slow motion. > > The crashes I have seen on the velodrome have come from mostly from > grounding a pedal (probably about 5 of those, two of which took me down > when the riders slid down the track into me), a couple from touching > wheels and once from a blowout the cause of which I don't know. All the > pedal strikes occurred on the roll-out before the start of an event. > > Oddly enough, the wheel touches all occurred in Cat 1/2/3 races, but > that was just luck of the draw (and smaller fields in those days- there > are twice as many people racing track now than 10 years ago). There > were way more crashes in the Cat 4/5 races. Agree, there seems to be more crashes in the Cat 4/5. The Cat 1-3 are more experienced, and when they see a newbie, most of the time they try to keep you in the back of the pack. > I once did stop pedaling > after a sprint and got a reminder from the bike to not do *that* again. > Fortunately we were all lined up across the track and there wasn't > anyone behind me to get drop-kicked. The official had a little chat > with me after the race, as you might expect. That's usually the most common mistake of a newbie, that and weaving. We had a few Cat 4/5 moving up to the Cat 3, their weaving for position was dangerous. > >> Try riding with no hands at 5 mph on the straights of a banked track >> next time you're out there. -tom > > I haven't ridden on the velodrome since 1998 or 99 (separated my > shoulder in one of the above-mentioned crashes. Never got back on the > track after that, realizing that at 40 years old I wasn't going to heal > as well as in my youth and that I couldn't afford to lose time from work > over a hobby). I don't even have my track bike any more. Got that right, I'm still have pain from my broken clavicle and that was two months ago. They said 6-weeks for a recovery, I was back on my bike in 3 weeks, but never thought of riding with my hands in my jersey pockets. -tom                   Date: 25 Oct 2007 20:36:11 From: Tim McNamara Subject: Re: Gyroscopic forces revisited In article <ffqron$lu0$1@news.Stanford.EDU >, "Tom Nakashima" <tom@slac.stanford.edu > wrote: > "Tim McNamara" <timmcn@bitstream.net> wrote in message > news:timmcn-4006AA.13593725102007@news.iphouse.com... > > In article <ffq7bn$3mm$1@news.Stanford.EDU>, "Tom Nakashima" > > <tom@slac.stanford.edu> wrote: > > > >> "Tim McNamara" <timmcn@bitstream.net> wrote in message > >> news:timmcn-9FFEAC.20423324102007@news.iphouse.com... > >> > > >> > Perhaps proper slow cornering technique ought to be added to the > >> > introductory class. You do have an introductory class for new > >> > riders, right? It would reduce the number of crashing newbies. > >> > >> As a track rider I'm surprised you don't know that most crashes on > >> the track are due to the inexperience of riding in a tight pack. > >> I've seen riders panic, stop pedaling, touch the rear wheel of the > >> bike in front of them. It isn't until one becomes experienced in > >> track riding that the newbie's become at ease. When I get back on > >> my roadbike after riding the track, everything seems like slow > >> motion. > > > > The crashes I have seen on the velodrome have come from mostly from > > grounding a pedal (probably about 5 of those, two of which took me > > down when the riders slid down the track into me), a couple from > > touching wheels and once from a blowout the cause of which I don't > > know. All the pedal strikes occurred on the roll-out before the > > start of an event. > > > > Oddly enough, the wheel touches all occurred in Cat 1/2/3 races, > > but that was just luck of the draw (and smaller fields in those > > days- there are twice as many people racing track now than 10 years > > ago). There were way more crashes in the Cat 4/5 races. > > Agree, there seems to be more crashes in the Cat 4/5. The Cat 1-3 > are more experienced, and when they see a newbie, most of the time > they try to keep you in the back of the pack. LOL. In my case that wasn't very difficult for them... I turned up at my first track race with one gear, 48 x 16. Apparently I looked highly amusing trying to keep up. > > I once did stop pedaling after a sprint and got a reminder from the > > bike to not do *that* again. Fortunately we were all lined up > > across the track and there wasn't anyone behind me to get > > drop-kicked. The official had a little chat with me after the > > race, as you might expect. > > That's usually the most common mistake of a newbie, that and weaving. > We had a few Cat 4/5 moving up to the Cat 3, their weaving for > position was dangerous. We got a lot of drills on that sort of stuff. We had to take a four week track riding class to be allowed to race at Blaine, demonstrating adequacy with a number of basic skills (like not falling off being able to look over our shoulders without weaving, riding a paceline, understanding the different lines, etc). I would imagine that most velodromes have similar programs. Bob at Blaine just didn't want to have to stop the racing to roll the injured further into the infield. > >> Try riding with no hands at 5 mph on the straights of a banked > >> track next time you're out there. -tom > > > > I haven't ridden on the velodrome since 1998 or 99 (separated my > > shoulder in one of the above-mentioned crashes. Never got back on > > the track after that, realizing that at 40 years old I wasn't going > > to heal as well as in my youth and that I couldn't afford to lose > > time from work over a hobby). I don't even have my track bike any > > more. > > Got that right, I'm still have pain from my broken clavicle and that > was two months ago. They said 6-weeks for a recovery, I was back on > my bike in 3 weeks, but never thought of riding with my hands in my > jersey pockets. -tom I've been told that broken collar bones hurt like a sonofagun whenever you try to move your arm. A friend/teammate that did one of these was sore for a long time. His doctors didn't put him in one of those figure-8 slings (or, if they did he didn't wear it) and he ended up with a pretty fine bump on his clavicle. You ever notice that bikies never wait as long as the docs say? They say "six weeks" and we're trying to get back on the bike in two. Glad you're healing up!             Date: 25 Oct 2007 05:23:04 From: Subject: Re: Gyroscopic forces revisited Tim McNamara writes: >>>> I think you ought to go to a velodrome. These tracks have sloped >>>> straights and more steeply banked circular curves at each end. >>>> Riders have no problem riding no-hands on these tracks, either on >>>> the straight sections or in the banked curve (if they are going >>>> fast enough to not strike a pedal). >>>> If you don't have a track, try a crowned paved street and notice >>>> how the angle with the pavement has essentially no effect on >>>> steering, the centerline of the contact patch moving less than >>>> 1/8" off center for a 10 degree side slope. >>> I think I may be able to answer this one. I was talking with >>> co-worker Timothy Montagne who just got back from the track >>> Nationals in Trexlertown, Pennsylvania. I forget the exact slope >>> on the straights in degrees, but he told me it was so steep that >>> you have to keep a minimum speed of 20 mph, or else you would >>> fall. So I can imagine your speed has to be greater than 20 mph >>> to be able to ride with no hands, of course which is no problem >>> with a track rider. >> I ride at Burnaby Velodrome, 200m wooden track, 47 degrees at the >> steepest part of the curve. >> It is difficult to stay on in the curve at anything under 30kph. >> We routinely see new riders slide off from going to slow. As >> recently as yesterday morning. > And yet sprinters can ride the entire lap, at any lateral position, > at literally a walking pace. So that disproves the claim that you > have to go at least 30 kph. This is magic. Just try leaning a track bicycle to 47 degrees with the pedal down at its lowest position and you'll note that it strikes the ground at far less than that angle. Grounding a pedal in the curved bankings on most tracks is a known hazard. If you believe one can ride at walking speed (bicycle vertical) I must assume you haven't tried it at speeds just under the natural pedal clearance speed. > I've ridden many times at the local wooden velodrome (Blaine MN, > 250m, 43 degree banks in corners) as slow as 10 mph. It is > necessary to tilt the bike deliberately to the inside to avoid > striking a pedal on the up-track side- failure to do so is probably > what is causing your new riders to fall off. I have seen riders > come to a complete stop in mid-corner without falling off. So you are not riding the bicycle seated as on does on a track except when accelerating. Beyond that, we are talking about riding no-hands. It requires some gymnastics to ride at moderate speeds hands-on, on steeply banked tracks and that is what was claimed. I have watched six-day racers on the Frankfurt (D) indoor track fall off the banking when traffic got too slow. It is damn hard to do your "lean the bicycle" stunt at moderate speeds with hands-on and even harder to do it no-hands. > Perhaps proper slow cornering technique ought to be added to the > introductory class. You do have an introductory class for new > riders, right? It would reduce the number of crashing newbies. Not at all. There is little purpose to that, slowness not being part of track racing. Besides, it doesn't work reliably at moderate speeds where cadence makes tilting the bicycle a hit or miss event of synchronizing the bottom of the right crank with enough lean. Jobst Brandt               Date: 25 Oct 2007 09:21:33 From: Tim McNamara Subject: Re: Gyroscopic forces revisited In article <47202838$0$14103$742ec2ed@news.sonic.net >, jobst.brandt@stanfordalumni.org wrote: > Tim McNamara writes: > > >>>> I think you ought to go to a velodrome. These tracks have > >>>> sloped straights and more steeply banked circular curves at each > >>>> end. Riders have no problem riding no-hands on these tracks, > >>>> either on the straight sections or in the banked curve (if they > >>>> are going fast enough to not strike a pedal). > > >>>> If you don't have a track, try a crowned paved street and notice > >>>> how the angle with the pavement has essentially no effect on > >>>> steering, the centerline of the contact patch moving less than > >>>> 1/8" off center for a 10 degree side slope. > > >>> I think I may be able to answer this one. I was talking with > >>> co-worker Timothy Montagne who just got back from the track > >>> Nationals in Trexlertown, Pennsylvania. I forget the exact slope > >>> on the straights in degrees, but he told me it was so steep that > >>> you have to keep a minimum speed of 20 mph, or else you would > >>> fall. So I can imagine your speed has to be greater than 20 mph > >>> to be able to ride with no hands, of course which is no problem > >>> with a track rider. > > >> I ride at Burnaby Velodrome, 200m wooden track, 47 degrees at the > >> steepest part of the curve. > > >> It is difficult to stay on in the curve at anything under 30kph. > > >> We routinely see new riders slide off from going to slow. As > >> recently as yesterday morning. > > > And yet sprinters can ride the entire lap, at any lateral position, > > at literally a walking pace. So that disproves the claim that you > > have to go at least 30 kph. > > This is magic. Just try leaning a track bicycle to 47 degrees with > the pedal down at its lowest position and you'll note that it strikes > the ground at far less than that angle. Grounding a pedal in the > curved bankings on most tracks is a known hazard. If you believe one > can ride at walking speed (bicycle vertical) I must assume you > haven't tried it at speeds just under the natural pedal clearance > speed. As noted below, I have tried it albeit on a track with a maximum banking of 43 degrees (which would be definitely different than a track with banking at 47 degrees). Leaning the bike away from the track while riding slowly is not difficult and I don't know why you would think that it is. We were taught this technique on the first day of track orientation. The more difficult thing is holding a line at slow speeds against gravity. I've also been taken off the track by two different riders who struck a pedal above me and came down like bowling pins. New or inexperienced riders tend to have to go faster to avoid these things. Skilled track riders can go slower without getting into trouble. I was only able to ride as slow as 10 mph, but I also never developed more than moderate skills at best at track racing. The Cat 1/2 riders could ride much slower when the situation demanded it. > > I've ridden many times at the local wooden velodrome (Blaine MN, > > 250m, 43 degree banks in corners) as slow as 10 mph. It is > > necessary to tilt the bike deliberately to the inside to avoid > > striking a pedal on the up-track side- failure to do so is probably > > what is causing your new riders to fall off. I have seen riders > > come to a complete stop in mid-corner without falling off. > > So you are not riding the bicycle seated as on does on a track except > when accelerating. Beyond that, we are talking about riding > no-hands. It requires some gymnastics to ride at moderate speeds > hands-on, on steeply banked tracks and that is what was claimed. I > have watched six-day racers on the Frankfurt (D) indoor track fall > off the banking when traffic got too slow. It is damn hard to do > your "lean the bicycle" stunt at moderate speeds with hands-on and > even harder to do it no-hands. Hands-off, I can't do it. When I've seen riders riding no handed around the velodrome they've been going at moderate speeds- say 18-20 mph. Hands-on is a different story. I watched national caliber sprinters ride up and down the banks at a walking pace during the strategic portion of sprint events. It can be done. Of course, the steeper the banking the harder it is and the higher the minimum speed will be. Very short tracks- under 200 m- tend to have very steep banks- the portable indoor track that circulated around the US a few years ago had 53 degree banks. 42 to 43 degrees is pretty typical on a 250 meter track. Larger tracks like Northbrook (33 m and 10 degrees IIRC) or Trexlertown (333 m and 28 degrees) tend to have shallower banking. > > Perhaps proper slow cornering technique ought to be added to the > > introductory class. You do have an introductory class for new > > riders, right? It would reduce the number of crashing newbies. > > Not at all. There is little purpose to that, slowness not being part > of track racing. Besides, it doesn't work reliably at moderate > speeds where cadence makes tilting the bicycle a hit or miss event of > synchronizing the bottom of the right crank with enough lean. You've not watched match sprints? I find that hard to believe. Slow riding is very much a part of that event. Sprint bikes also have very high bottom brackets- more so than track bikes designed for other events- which reduces the pedal strike problem somewhat. http://www.youtube.com/watch?v=BkkTSVVrPYk Great trackstand demonstration as well as good slow speed riding.                 Date: 25 Oct 2007 17:57:58 From: Subject: Re: Gyroscopic forces revisited Tim McNamara writes: >>>>>> I think you ought to go to a velodrome. These tracks have >>>>>> sloped straights and more steeply banked circular curves at >>>>>> each end. Riders have no problem riding no-hands on these >>>>>> tracks, either on the straight sections or in the banked curve >>>>>> (if they are going fast enough to not strike a pedal). >>>>>> If you don't have a track, try a crowned paved street and >>>>>> notice how the angle with the pavement has essentially no >>>>>> effect on steering, the centerline of the contact patch moving >>>>>> less than 1/8" off center for a 10 degree side slope. >>>>> I think I may be able to answer this one. I was talking with >>>>> co-worker Timothy Montagne who just got back from the track >>>>> Nationals in Trexlertown, Pennsylvania. I forget the exact >>>>> slope on the straights in degrees, but he told me it was so >>>>> steep that you have to keep a minimum speed of 20 mph, or else >>>>> you would fall. So I can imagine your speed has to be greater >>>>> than 20 mph to be able to ride with no hands, of course which is >>>>> no problem with a track rider. >>>> I ride at Burnaby Velodrome, 200m wooden track, 47 degrees at the >>>> steepest part of the curve. >>>> It is difficult to stay on in the curve at anything under 30kph. >>>> We routinely see new riders slide off from going to slow. As >>>> recently as yesterday morning. >>> And yet sprinters can ride the entire lap, at any lateral >>> position, at literally a walking pace. So that disproves the >>> claim that you have to go at least 30 kph. >> This is magic. Just try leaning a track bicycle to 47 degrees with >> the pedal down at its lowest position and you'll note that it >> strikes the ground at far less than that angle. Grounding a pedal >> in the curved bankings on most tracks is a known hazard. If you >> believe one can ride at walking speed (bicycle vertical) I must >> assume you haven't tried it at speeds just under the natural pedal >> clearance speed. > As noted below, I have tried it albeit on a track with a maximum > banking of 43 degrees (which would be definitely different than a > track with banking at 47 degrees). Leaning the bike away from the > track while riding slowly is not difficult and I don't know why you > would think that it is. We were taught this technique on the first > day of track orientation. The more difficult thing is holding a > line at slow speeds against gravity. I've also been taken off the > track by two different riders who struck a pedal above me and came > down like bowling pins. > New or inexperienced riders tend to have to go faster to avoid these > things. Skilled track riders can go slower without getting into > trouble. I was only able to ride as slow as 10 mph, but I also > never developed more than moderate skills at best at track racing. > The Cat 1/2 riders could ride much slower when the situation > demanded it. >>> I've ridden many times at the local wooden velodrome (Blaine MN, >>> 250m, 43 degree banks in corners) as slow as 10 mph. It is >>> necessary to tilt the bike deliberately to the inside to avoid >>> striking a pedal on the up-track side- failure to do so is >>> probably what is causing your new riders to fall off. I have seen >>> riders come to a complete stop in mid-corner without falling off. >> So you are not riding the bicycle seated as on does on a track >> except when accelerating. Beyond that, we are talking about riding >> no-hands. It requires some gymnastics to ride at moderate speeds >> hands-on, on steeply banked tracks and that is what was claimed. I >> have watched six-day racers on the Frankfurt (D) indoor track fall >> off the banking when traffic got too slow. It is damn hard to do >> your "lean the bicycle" stunt at moderate speeds with hands-on and >> even harder to do it no-hands. > Hands-off, I can't do it. When I've seen riders riding no handed > around the velodrome they've been going at moderate speeds- say > 18-20 mph. Hands-on is a different story. I watched national > caliber sprinters ride up and down the banks at a walking pace > during the strategic portion of sprint events. It can be done. Of > course, the steeper the banking the harder it is and the higher the > minimum speed will be. Very short tracks- under 200 m- tend to have > very steep banks- the portable indoor track that circulated around > the US a few years ago had 53 degree banks. 42 to 43 degrees is > pretty typical on a 250 meter track. Larger tracks like Northbrook > (33 m and 10 degrees IIRC) or Trexlertown (333 m and 28 degrees) > tend to have shallower banking. As I said, at moderate speeds, leaning the bicycle away from the banking doesn't work for lack of fast synchronization, and even if you can ride that way, it requires diving into the infield to start a sprint because you can't spring slowly while leaning the bicycle away from the banking. >>> Perhaps proper slow cornering technique ought to be added to the >>> introductory class. You do have an introductory class for new >>> riders, right? It would reduce the number of crashing newbies. >> Not at all. There is little purpose to that, slowness not being part >> of track racing. Besides, it doesn't work reliably at moderate >> speeds where cadence makes tilting the bicycle a hit or miss event of >> synchronizing the bottom of the right crank with enough lean. > You've not watched match sprints? I find that hard to believe. Slow > riding is very much a part of that event. Sprint bikes also have very > high bottom brackets- more so than track bikes designed for other > events- which reduces the pedal strike problem somewhat. http://www.youtube.com/watch?v=BkkTSVVrPYk > Great trackstand demonstration as well as good slow speed riding. These bicycles had pedal clearance even when standing and here I thought you were going to show how one rides on a 47 degree banking leaning the bicycle away from the banking. At about 45 degrees, traction usually fails even without grounding a pedal. Jobst Brandt                   Date: 25 Oct 2007 13:43:01 From: Tim McNamara Subject: Re: Gyroscopic forces revisited In article <4720d926$0$14112$742ec2ed@news.sonic.net >, jobst.brandt@stanfordalumni.org wrote: > Tim McNamara writes: > > >>>>>> I think you ought to go to a velodrome. These tracks have > >>>>>> sloped straights and more steeply banked circular curves at > >>>>>> each end. Riders have no problem riding no-hands on these > >>>>>> tracks, either on the straight sections or in the banked curve > >>>>>> (if they are going fast enough to not strike a pedal). > > >>>>>> If you don't have a track, try a crowned paved street and > >>>>>> notice how the angle with the pavement has essentially no > >>>>>> effect on steering, the centerline of the contact patch moving > >>>>>> less than 1/8" off center for a 10 degree side slope. > > >>>>> I think I may be able to answer this one. I was talking with > >>>>> co-worker Timothy Montagne who just got back from the track > >>>>> Nationals in Trexlertown, Pennsylvania. I forget the exact > >>>>> slope on the straights in degrees, but he told me it was so > >>>>> steep that you have to keep a minimum speed of 20 mph, or else > >>>>> you would fall. So I can imagine your speed has to be greater > >>>>> than 20 mph to be able to ride with no hands, of course which > >>>>> is no problem with a track rider. > > >>>> I ride at Burnaby Velodrome, 200m wooden track, 47 degrees at > >>>> the steepest part of the curve. > > >>>> It is difficult to stay on in the curve at anything under 30kph. > > >>>> We routinely see new riders slide off from going to slow. As > >>>> recently as yesterday morning. > > >>> And yet sprinters can ride the entire lap, at any lateral > >>> position, at literally a walking pace. So that disproves the > >>> claim that you have to go at least 30 kph. > > >> This is magic. Just try leaning a track bicycle to 47 degrees > >> with the pedal down at its lowest position and you'll note that it > >> strikes the ground at far less than that angle. Grounding a pedal > >> in the curved bankings on most tracks is a known hazard. If you > >> believe one can ride at walking speed (bicycle vertical) I must > >> assume you haven't tried it at speeds just under the natural pedal > >> clearance speed. > > > As noted below, I have tried it albeit on a track with a maximum > > banking of 43 degrees (which would be definitely different than a > > track with banking at 47 degrees). Leaning the bike away from the > > track while riding slowly is not difficult and I don't know why you > > would think that it is. We were taught this technique on the first > > day of track orientation. The more difficult thing is holding a > > line at slow speeds against gravity. I've also been taken off the > > track by two different riders who struck a pedal above me and came > > down like bowling pins. > > > New or inexperienced riders tend to have to go faster to avoid > > these things. Skilled track riders can go slower without getting > > into trouble. I was only able to ride as slow as 10 mph, but I > > also never developed more than moderate skills at best at track > > racing. The Cat 1/2 riders could ride much slower when the > > situation demanded it. > > >>> I've ridden many times at the local wooden velodrome (Blaine MN, > >>> 250m, 43 degree banks in corners) as slow as 10 mph. It is > >>> necessary to tilt the bike deliberately to the inside to avoid > >>> striking a pedal on the up-track side- failure to do so is > >>> probably what is causing your new riders to fall off. I have > >>> seen riders come to a complete stop in mid-corner without falling > >>> off. > > >> So you are not riding the bicycle seated as on does on a track > >> except when accelerating. Beyond that, we are talking about > >> riding no-hands. It requires some gymnastics to ride at moderate > >> speeds hands-on, on steeply banked tracks and that is what was > >> claimed. I have watched six-day racers on the Frankfurt (D) > >> indoor track fall off the banking when traffic got too slow. It > >> is damn hard to do your "lean the bicycle" stunt at moderate > >> speeds with hands-on and even harder to do it no-hands. > > > Hands-off, I can't do it. When I've seen riders riding no handed > > around the velodrome they've been going at moderate speeds- say > > 18-20 mph. Hands-on is a different story. I watched national > > caliber sprinters ride up and down the banks at a walking pace > > during the strategic portion of sprint events. It can be done. Of > > course, the steeper the banking the harder it is and the higher the > > minimum speed will be. Very short tracks- under 200 m- tend to > > have very steep banks- the portable indoor track that circulated > > around the US a few years ago had 53 degree banks. 42 to 43 > > degrees is pretty typical on a 250 meter track. Larger tracks like > > Northbrook (33 m and 10 degrees IIRC) or Trexlertown (333 m and 28 > > degrees) tend to have shallower banking. > > As I said, at moderate speeds, leaning the bicycle away from the > banking doesn't work for lack of fast synchronization, and even if > you can ride that way, it requires diving into the infield to start a > sprint because you can't spring slowly while leaning the bicycle away > from the banking. I don't know how you're defining moderate speeds here. It's not necessary to rock the bike back and forth to synchronize the lean with the pedal stroke. We were taught to lean the bike towards the infield and keep it that way at slow speeds. You're right that this position does hamper getting going in a sprint. I only used that technique when rolling out for the start, otherwise I made sure I was going fast enough that normal riding kept the pedals out of the track. I was always nervous about that. > >>> Perhaps proper slow cornering technique ought to be added to the > >>> introductory class. You do have an introductory class for new > >>> riders, right? It would reduce the number of crashing newbies. > > >> Not at all. There is little purpose to that, slowness not being > >> part of track racing. Besides, it doesn't work reliably at > >> moderate speeds where cadence makes tilting the bicycle a hit or > >> miss event of synchronizing the bottom of the right crank with > >> enough lean. > > > You've not watched match sprints? I find that hard to believe. > > Slow riding is very much a part of that event. Sprint bikes also > > have very high bottom brackets- more so than track bikes designed > > for other events- which reduces the pedal strike problem somewhat. > > http://www.youtube.com/watch?v=BkkTSVVrPYk > > > Great trackstand demonstration as well as good slow speed riding. > > These bicycles had pedal clearance even when standing and here I > thought you were going to show how one rides on a 47 degree banking > leaning the bicycle away from the banking. I didn't find anything demonstrating that technique in a quick search at breakfast. That clip was the first one that came up and was about the slow riding thing. The announcer mentioned that the maximum banking was 42 degrees. I'll look around a bit more this evening and post anything if I see it. I'm not particularly hopeful, but who knows. And as far as 47 degree bankings, I don't know as I've never ridden anything that steep. Our velodrome is 43 degrees, as I mentioned, and I suspect that 4 more degrees would make a big difference. > At about 45 degrees, traction usually fails even without grounding a pedal. We've talked about that in the past and you've pointed out that the slip angle is usually just past 45, maybe as much as 48 degrees, at least on asphalt. I would expect a wood surface to result in less traction than asphalt but maybe that's not actually so (although wood gets *very* slick when wet, much more so than asphalt). There are quite a few indoor velodromes that are steeper than 45 degrees.                     Date: 25 Oct 2007 19:01:58 From: Subject: Re: Gyroscopic forces revisited Tim McNamara writes: >>>>>>>> I think you ought to go to a velodrome. These tracks have >>>>>>>> sloped straights and more steeply banked circular curves at >>>>>>>> each end. Riders have no problem riding no-hands on these >>>>>>>> tracks, either on the straight sections or in the banked curve >>>>>>>> (if they are going fast enough to not strike a pedal). >>>>>>>> If you don't have a track, try a crowned paved street and >>>>>>>> notice how the angle with the pavement has essentially no >>>>>>>> effect on steering, the centerline of the contact patch moving >>>>>>>> less than 1/8" off center for a 10 degree side slope. >>>>>>> I think I may be able to answer this one. I was talking with >>>>>>> co-worker Timothy Montagne who just got back from the track >>>>>>> Nationals in Trexlertown, Pennsylvania. I forget the exact >>>>>>> slope on the straights in degrees, but he told me it was so >>>>>>> steep that you have to keep a minimum speed of 20 mph, or else >>>>>>> you would fall. So I can imagine your speed has to be greater >>>>>>> than 20 mph to be able to ride with no hands, of course which >>>>>>> is no problem with a track rider. >>>>>> I ride at Burnaby Velodrome, 200m wooden track, 47 degrees at >>>>>> the steepest part of the curve. >>>>>> It is difficult to stay on in the curve at anything under 30kph. >>>>>> We routinely see new riders slide off from going to slow. As >>>>>> recently as yesterday morning. >>>>> And yet sprinters can ride the entire lap, at any lateral >>>>> position, at literally a walking pace. So that disproves the >>>>> claim that you have to go at least 30 kph. >>>> This is magic. Just try leaning a track bicycle to 47 degrees >>>> with the pedal down at its lowest position and you'll note that it >>>> strikes the ground at far less than that angle. Grounding a pedal >>>> in the curved bankings on most tracks is a known hazard. If you >>>> believe one can ride at walking speed (bicycle vertical) I must >>>> assume you haven't tried it at speeds just under the natural pedal >>>> clearance speed. >>> As noted below, I have tried it albeit on a track with a maximum >>> banking of 43 degrees (which would be definitely different than a >>> track with banking at 47 degrees). Leaning the bike away from the >>> track while riding slowly is not difficult and I don't know why you >>> would think that it is. We were taught this technique on the first >>> day of track orientation. The more difficult thing is holding a >>> line at slow speeds against gravity. I've also been taken off the >>> track by two different riders who struck a pedal above me and came >>> down like bowling pins. >>> New or inexperienced riders tend to have to go faster to avoid >>> these things. Skilled track riders can go slower without getting >>> into trouble. I was only able to ride as slow as 10 mph, but I >>> also never developed more than moderate skills at best at track >>> racing. The Cat 1/2 riders could ride much slower when the >>> situation demanded it. >>>>> I've ridden many times at the local wooden velodrome (Blaine MN, >>>>> 250m, 43 degree banks in corners) as slow as 10 mph. It is >>>>> necessary to tilt the bike deliberately to the inside to avoid >>>>> striking a pedal on the up-track side- failure to do so is >>>>> probably what is causing your new riders to fall off. I have >>>>> seen riders come to a complete stop in mid-corner without falling >>>>> off. >>>> So you are not riding the bicycle seated as on does on a track >>>> except when accelerating. Beyond that, we are talking about >>>> riding no-hands. It requires some gymnastics to ride at moderate >>>> speeds hands-on, on steeply banked tracks and that is what was >>>> claimed. I have watched six-day racers on the Frankfurt (D) >>>> indoor track fall off the banking when traffic got too slow. It >>>> is damn hard to do your "lean the bicycle" stunt at moderate >>>> speeds with hands-on and even harder to do it no-hands. >>> Hands-off, I can't do it. When I've seen riders riding no handed >>> around the velodrome they've been going at moderate speeds- say >>> 18-20 mph. Hands-on is a different story. I watched national >>> caliber sprinters ride up and down the banks at a walking pace >>> during the strategic portion of sprint events. It can be done. Of >>> course, the steeper the banking the harder it is and the higher the >>> minimum speed will be. Very short tracks- under 200 m- tend to >>> have very steep banks- the portable indoor track that circulated >>> around the US a few years ago had 53 degree banks. 42 to 43 >>> degrees is pretty typical on a 250 meter track. Larger tracks like >>> Northbrook (33 m and 10 degrees IIRC) or Trexlertown (333 m and 28 >>> degrees) tend to have shallower banking. >> As I said, at moderate speeds, leaning the bicycle away from the >> banking doesn't work for lack of fast synchronization, and even if >> you can ride that way, it requires diving into the infield to start a >> sprint because you can't spring slowly while leaning the bicycle away >> from the banking. > I don't know how you're defining moderate speeds here. It's not > necessary to rock the bike back and forth to synchronize the lean with > the pedal stroke. We were taught to lean the bike toward the infield > and keep it that way at slow speeds. You're right that this position > does hamper getting going in a sprint. I only used that technique when > rolling out for the start, otherwise I made sure I was going fast enough > that normal riding kept the pedals out of the track. I was always > nervous about that. You don't need to use "that technique" on the start (straight) that is designed to not ground pedals. It is the banked curves that present a problem both for grounding and losing traction that fails at about 45 degrees. >>>>> Perhaps proper slow cornering technique ought to be added to the >>>>> introductory class. You do have an introductory class for new >>>>> riders, right? It would reduce the number of crashing newbies. >>>> Not at all. There is little purpose to that, slowness not being >>>> part of track racing. Besides, it doesn't work reliably at >>>> moderate speeds where cadence makes tilting the bicycle a hit or >>>> miss event of synchronizing the bottom of the right crank with >>>> enough lean. >>> You've not watched match sprints? I find that hard to believe. >>> Slow riding is very much a part of that event. Sprint bikes also >>> have very high bottom brackets- more so than track bikes designed >>> for other events- which reduces the pedal strike problem somewhat. http://www.youtube.com/watch?v=BkkTSVVrPYk >>> Great trackstand demonstration as well as good slow speed riding. >> These bicycles had pedal clearance even when standing and here I >> thought you were going to show how one rides on a 47 degree banking >> leaning the bicycle away from the banking. > I didn't find anything demonstrating that technique in a quick > search at breakfast. That clip was the first one that came up and > was about the slow riding thing. The announcer mentioned that the > maximum banking was 42 degrees. I'll look around a bit more this > evening and post anything if I see it. I'm not particularly > hopeful, but who knows. And as far as 47 degree bankings, I don't > know as I've never ridden anything that steep. Our velodrome is 43 > degrees, as I mentioned, and I suspect that 4 more degrees would > make a big difference. I'm not too sure the 42 degree value was correct judging from the pedal clearance at zero speed. I know that the Frankfurt (D) indoor track called the cigar for its long straights and steep narrow ends caused riders to fall in stalled traffic in six-day races. >> At about 45 degrees, traction usually fails even without grounding >> a pedal. > We've talked about that in the past and you've pointed out that the > slip angle is usually just past 45, maybe as much as 48 degrees, at > least on asphalt. I would expect a wood surface to result in less > traction than asphalt but maybe that's not actually so (although > wood gets *very* slick when wet, much more so than asphalt). There > are quite a few indoor velodromes that are steeper than 45 degrees. And on these one does not ride at zero speed. Jobst Brandt                       Date: 25 Oct 2007 20:24:51 From: Tim McNamara Subject: Re: Gyroscopic forces revisited In article <4720e826$0$14112$742ec2ed@news.sonic.net >, jobst.brandt@stanfordalumni.org wrote: > Tim McNamara writes: > > >>>>>>>> I think you ought to go to a velodrome. These tracks have > >>>>>>>> sloped straights and more steeply banked circular curves at > >>>>>>>> each end. Riders have no problem riding no-hands on these > >>>>>>>> tracks, either on the straight sections or in the banked curve > >>>>>>>> (if they are going fast enough to not strike a pedal). > > >>>>>>>> If you don't have a track, try a crowned paved street and > >>>>>>>> notice how the angle with the pavement has essentially no > >>>>>>>> effect on steering, the centerline of the contact patch moving > >>>>>>>> less than 1/8" off center for a 10 degree side slope. > > >>>>>>> I think I may be able to answer this one. I was talking with > >>>>>>> co-worker Timothy Montagne who just got back from the track > >>>>>>> Nationals in Trexlertown, Pennsylvania. I forget the exact > >>>>>>> slope on the straights in degrees, but he told me it was so > >>>>>>> steep that you have to keep a minimum speed of 20 mph, or else > >>>>>>> you would fall. So I can imagine your speed has to be greater > >>>>>>> than 20 mph to be able to ride with no hands, of course which > >>>>>>> is no problem with a track rider. > > >>>>>> I ride at Burnaby Velodrome, 200m wooden track, 47 degrees at > >>>>>> the steepest part of the curve. > > >>>>>> It is difficult to stay on in the curve at anything under 30kph. > > >>>>>> We routinely see new riders slide off from going to slow. As > >>>>>> recently as yesterday morning. > > >>>>> And yet sprinters can ride the entire lap, at any lateral > >>>>> position, at literally a walking pace. So that disproves the > >>>>> claim that you have to go at least 30 kph. > > >>>> This is magic. Just try leaning a track bicycle to 47 degrees > >>>> with the pedal down at its lowest position and you'll note that it > >>>> strikes the ground at far less than that angle. Grounding a pedal > >>>> in the curved bankings on most tracks is a known hazard. If you > >>>> believe one can ride at walking speed (bicycle vertical) I must > >>>> assume you haven't tried it at speeds just under the natural pedal > >>>> clearance speed. > > >>> As noted below, I have tried it albeit on a track with a maximum > >>> banking of 43 degrees (which would be definitely different than a > >>> track with banking at 47 degrees). Leaning the bike away from the > >>> track while riding slowly is not difficult and I don't know why you > >>> would think that it is. We were taught this technique on the first > >>> day of track orientation. The more difficult thing is holding a > >>> line at slow speeds against gravity. I've also been taken off the > >>> track by two different riders who struck a pedal above me and came > >>> down like bowling pins. > > >>> New or inexperienced riders tend to have to go faster to avoid > >>> these things. Skilled track riders can go slower without getting > >>> into trouble. I was only able to ride as slow as 10 mph, but I > >>> also never developed more than moderate skills at best at track > >>> racing. The Cat 1/2 riders could ride much slower when the > >>> situation demanded it. > > >>>>> I've ridden many times at the local wooden velodrome (Blaine MN, > >>>>> 250m, 43 degree banks in corners) as slow as 10 mph. It is > >>>>> necessary to tilt the bike deliberately to the inside to avoid > >>>>> striking a pedal on the up-track side- failure to do so is > >>>>> probably what is causing your new riders to fall off. I have > >>>>> seen riders come to a complete stop in mid-corner without falling > >>>>> off. > > >>>> So you are not riding the bicycle seated as on does on a track > >>>> except when accelerating. Beyond that, we are talking about > >>>> riding no-hands. It requires some gymnastics to ride at moderate > >>>> speeds hands-on, on steeply banked tracks and that is what was > >>>> claimed. I have watched six-day racers on the Frankfurt (D) > >>>> indoor track fall off the banking when traffic got too slow. It > >>>> is damn hard to do your "lean the bicycle" stunt at moderate > >>>> speeds with hands-on and even harder to do it no-hands. > > >>> Hands-off, I can't do it. When I've seen riders riding no handed > >>> around the velodrome they've been going at moderate speeds- say > >>> 18-20 mph. Hands-on is a different story. I watched national > >>> caliber sprinters ride up and down the banks at a walking pace > >>> during the strategic portion of sprint events. It can be done. Of > >>> course, the steeper the banking the harder it is and the higher the > >>> minimum speed will be. Very short tracks- under 200 m- tend to > >>> have very steep banks- the portable indoor track that circulated > >>> around the US a few years ago had 53 degree banks. 42 to 43 > >>> degrees is pretty typical on a 250 meter track. Larger tracks like > >>> Northbrook (33 m and 10 degrees IIRC) or Trexlertown (333 m and 28 > >>> degrees) tend to have shallower banking. > > >> As I said, at moderate speeds, leaning the bicycle away from the > >> banking doesn't work for lack of fast synchronization, and even if > >> you can ride that way, it requires diving into the infield to start a > >> sprint because you can't spring slowly while leaning the bicycle away > >> from the banking. > > > I don't know how you're defining moderate speeds here. It's not > > necessary to rock the bike back and forth to synchronize the lean with > > the pedal stroke. We were taught to lean the bike toward the infield > > and keep it that way at slow speeds. You're right that this position > > does hamper getting going in a sprint. I only used that technique when > > rolling out for the start, otherwise I made sure I was going fast enough > > that normal riding kept the pedals out of the track. I was always > > nervous about that. > > You don't need to use "that technique" on the start (straight) that is > designed to not ground pedals. It is the banked curves that present a > problem both for grounding and losing traction that fails at about 45 > degrees. Yup. That's when you lean the bike towards the infield- in the corners. As you say, it's not necessary on the straights. > >>>>> Perhaps proper slow cornering technique ought to be added to the > >>>>> introductory class. You do have an introductory class for new > >>>>> riders, right? It would reduce the number of crashing newbies. > > >>>> Not at all. There is little purpose to that, slowness not being > >>>> part of track racing. Besides, it doesn't work reliably at > >>>> moderate speeds where cadence makes tilting the bicycle a hit or > >>>> miss event of synchronizing the bottom of the right crank with > >>>> enough lean. > > >>> You've not watched match sprints? I find that hard to believe. > >>> Slow riding is very much a part of that event. Sprint bikes also > >>> have very high bottom brackets- more so than track bikes designed > >>> for other events- which reduces the pedal strike problem somewhat. > > http://www.youtube.com/watch?v=BkkTSVVrPYk > > >>> Great trackstand demonstration as well as good slow speed riding. > > >> These bicycles had pedal clearance even when standing and here I > >> thought you were going to show how one rides on a 47 degree banking > >> leaning the bicycle away from the banking. > > > I didn't find anything demonstrating that technique in a quick > > search at breakfast. That clip was the first one that came up and > > was about the slow riding thing. The announcer mentioned that the > > maximum banking was 42 degrees. I'll look around a bit more this > > evening and post anything if I see it. I'm not particularly > > hopeful, but who knows. And as far as 47 degree bankings, I don't > > know as I've never ridden anything that steep. Our velodrome is 43 > > degrees, as I mentioned, and I suspect that 4 more degrees would > > make a big difference. > > I'm not too sure the 42 degree value was correct judging from the > pedal clearance at zero speed. I know that the Frankfurt (D) indoor > track called the cigar for its long straights and steep narrow ends > caused riders to fall in stalled traffic in six-day races. I recall seeing pictures of this track, it's rather striking in it's dimensions. I couldn't find any data about the banking. It's apparently 286 meters, which seems rather long for an indoor track. > >> At about 45 degrees, traction usually fails even without grounding > >> a pedal. > > > We've talked about that in the past and you've pointed out that the > > slip angle is usually just past 45, maybe as much as 48 degrees, at > > least on asphalt. I would expect a wood surface to result in less > > traction than asphalt but maybe that's not actually so (although > > wood gets *very* slick when wet, much more so than asphalt). There > > are quite a few indoor velodromes that are steeper than 45 degrees. > > And on these one does not ride at zero speed. Presumably not.         Date: 24 Oct 2007 16:12:50 From: Subject: Re: Gyroscopic forces revisited Tom Nakashima writes: >> I think you ought to go to a velodrome. These tracks have sloped >> straights and more steeply banked circular curves at each end. >> Riders have no problem riding no-hands on these tracks, either on >> the straight sections or in the banked curve (if they are going >> fast enough to not strike a pedal). >> If you don't have a track, try a crowned paved street and notice >> how the angle with the pavement has essentially no effect on >> steering, the centerline of the contact patch moving less than 1/8" >> off center for a 10? side slope. > I think I may be able to answer this one. I was talking with > co-worker Timothy Montagne who just got back from the track > Nationals in Trexlertown, Pennsylvania. I forget the exact slope on > the straights in degrees, but he told me it was so steep that you > have to keep a minimum speed of 20 mph. or else you would fall > over. So I can imagine your speed has to be greater than 20 mph to > be able to ride with no hands, of course which is no problem with a > track rider. I think there was a misunderstanding there. The speed is required to keep from grounding a pedal on steep bankings (as I mentioned) but has nothing to do with riding no-hands on the straights, which is like riding no-hands on a flat surface. The difference being that turns upslope respond slightly differently from turns downslope. > When I ride the San Jose velodrome, I have no problem riding with no > hands, but the slope is mild. I'm sure a minimum speed is needed, > just never crossed my mind. That's because it isn't any different than on a level course. > I'm riding with Tim today after work, so I can pick his brains more > on the banked track. Jobst Brandt         Date: 24 Oct 2007 09:17:31 From: Tim McNamara Subject: Re: Gyroscopic forces revisited In article <ffnhg2$n2$1@news.Stanford.EDU >, "Tom Nakashima" <tom@slac.stanford.edu > wrote: > I was talking with co-worker Timothy Montagne who just got back from > the track Nationals in Trexlertown, Pennsylvania. I forget the exact > slope on the straights in degrees, but he told me it was so steep > that you have to keep a minimum speed of 20 mph. or else you would > fall over. So I can imagine your speed has to be greater than 20 mph > to be able to ride with no hands, of course which is no problem with > a track rider. Timothy sold you a line of malarkey. If he was right, then the sprinters couldn't do track stands or pace each other at 2-3 mph up and down the track. But they can and Timothy is wrong.           Date: 24 Oct 2007 07:36:02 From: Tom Nakashima Subject: Re: Gyroscopic forces revisited "Tim McNamara" <timmcn@bitstream.net > wrote in message news:timmcn-C6BB21.09172624102007@news.iphouse.com... > In article <ffnhg2$n2$1@news.Stanford.EDU>, > "Tom Nakashima" <tom@slac.stanford.edu> wrote: > >> I was talking with co-worker Timothy Montagne who just got back from >> the track Nationals in Trexlertown, Pennsylvania. I forget the exact >> slope on the straights in degrees, but he told me it was so steep >> that you have to keep a minimum speed of 20 mph. or else you would >> fall over. So I can imagine your speed has to be greater than 20 mph >> to be able to ride with no hands, of course which is no problem with >> a track rider. > > Timothy sold you a line of malarkey. If he was right, then the > sprinters couldn't do track stands or pace each other at 2-3 mph up and > down the track. But they can and Timothy is wrong. Timothy Montagne is pretty straight forward, never caught him in a lie. However, I can't say that about reading some of your post. You can read about Montagne here: http://today.slac.stanford.edu/a/2007/10-01.htm -tom             Date: 24 Oct 2007 20:57:14 From: Tim McNamara Subject: Re: Gyroscopic forces revisited In article <ffnl8j$3uh$1@news.Stanford.EDU >, "Tom Nakashima" <tom@slac.stanford.edu > wrote: > "Tim McNamara" <timmcn@bitstream.net> wrote in message > news:timmcn-C6BB21.09172624102007@news.iphouse.com... > > In article <ffnhg2$n2$1@news.Stanford.EDU>, "Tom Nakashima" > > <tom@slac.stanford.edu> wrote: > > > >> I was talking with co-worker Timothy Montagne who just got back > >> from the track Nationals in Trexlertown, Pennsylvania. I forget > >> the exact slope on the straights in degrees, but he told me it was > >> so steep that you have to keep a minimum speed of 20 mph. or else > >> you would fall over. So I can imagine your speed has to be greater > >> than 20 mph to be able to ride with no hands, of course which is > >> no problem with a track rider. > > > > Timothy sold you a line of malarkey. If he was right, then the > > sprinters couldn't do track stands or pace each other at 2-3 mph up > > and down the track. But they can and Timothy is wrong. > > Timothy Montagne is pretty straight forward, never caught him in a > lie. However, I can't say that about reading some of your post. What do you think I have lied about, Tom? Is it a lie that sprinters can do track stands on a velodrome? Is it a lie that they pace each other at very low speeds while jockeying for advantage, without falling off the track (usually)? If you're going to accuse me of lying, have the courtesy to back it up. Unfortunately the Lehigh Valley Velodrome Web site doesn't contain the specs for the track. The local track here, the national Sports Center velodrome in Blaine MN, is a wooden 250 m track with 43 degree banking in the corners. It's quite possible on that track to ride below 10 mph if you know how to keep your pedals from striking the banking on the uptrack side. I've seen riders come to a halt in the corners. I've seen sprinters do track stands. According to: http://www.geocities.com/lvvelodrome/Velodrome.html The Lehigh Valley Velodrome has a maximum banking of 28 degrees. There is no reason whatsoever for a minimum speed of 20 mph on a track that flat, unless the surface is particularly slippery. Since that velodrome is concrete, that seems unlikely. > You can read about Montagne here: > http://today.slac.stanford.edu/a/2007/10-01.htm Thanks for the link.               Date: 25 Oct 2007 23:00:13 From: sl Subject: Re: Gyroscopic forces revisited In article <timmcn-205F17.20571424102007@news.iphouse.com >, Tim McNamara <timmcn@bitstream.net > wrote: >In article <ffnl8j$3uh$1@news.Stanford.EDU>, > "Tom Nakashima" <tom@slac.stanford.edu> wrote: > >> "Tim McNamara" <timmcn@bitstream.net> wrote in message >> news:timmcn-C6BB21.09172624102007@news.iphouse.com... >> > In article <ffnhg2$n2$1@news.Stanford.EDU>, "Tom Nakashima" >> > <tom@slac.stanford.edu> wrote: >> > >> >> I was talking with co-worker Timothy Montagne who just got back >> >> from the track Nationals in Trexlertown, Pennsylvania. I forget >> >> the exact slope on the straights in degrees, but he told me it was >> >> so steep that you have to keep a minimum speed of 20 mph. or else >> >> you would fall over. So I can imagine your speed has to be greater >> >> than 20 mph to be able to ride with no hands, of course which is >> >> no problem with a track rider. >> > >> > Timothy sold you a line of malarkey. If he was right, then the >> > sprinters couldn't do track stands or pace each other at 2-3 mph up >> > and down the track. But they can and Timothy is wrong. >> >> Timothy Montagne is pretty straight forward, never caught him in a >> lie. However, I can't say that about reading some of your post. > >What do you think I have lied about, Tom? Is it a lie that sprinters >can do track stands on a velodrome? Is it a lie that they pace each >other at very low speeds while jockeying for advantage, without falling >off the track (usually)? If you're going to accuse me of lying, have >the courtesy to back it up. Track stands and riding slow are possible, but if you watch a track sprint where they are doing it you'll notice that they angle the wheels to compensate. Sliding off happens when you don't compensate for the lack of centrifugal force by doing something else.               Date: 25 Oct 2007 07:16:09 From: Tom Nakashima Subject: Re: Gyroscopic forces revisited "Tim McNamara" <timmcn@bitstream.net > wrote in message news:timmcn-205F17.20571424102007@news.iphouse.com... > In article <ffnl8j$3uh$1@news.Stanford.EDU>, > "Tom Nakashima" <tom@slac.stanford.edu> wrote: > >> "Tim McNamara" <timmcn@bitstream.net> wrote in message >> news:timmcn-C6BB21.09172624102007@news.iphouse.com... >> > In article <ffnhg2$n2$1@news.Stanford.EDU>, "Tom Nakashima" >> > <tom@slac.stanford.edu> wrote: >> > >> >> I was talking with co-worker Timothy Montagne who just got back >> >> from the track Nationals in Trexlertown, Pennsylvania. I forget >> >> the exact slope on the straights in degrees, but he told me it was >> >> so steep that you have to keep a minimum speed of 20 mph. or else >> >> you would fall over. So I can imagine your speed has to be greater >> >> than 20 mph to be able to ride with no hands, of course which is >> >> no problem with a track rider. >> > >> > Timothy sold you a line of malarkey. If he was right, then the >> > sprinters couldn't do track stands or pace each other at 2-3 mph up >> > and down the track. But they can and Timothy is wrong. >> >> Timothy Montagne is pretty straight forward, never caught him in a >> lie. However, I can't say that about reading some of your post. > > What do you think I have lied about, Tom? Is it a lie that sprinters > can do track stands on a velodrome? Is it a lie that they pace each > other at very low speeds while jockeying for advantage, without falling > off the track (usually)? If you're going to accuse me of lying, have > the courtesy to back it up. We're talking about riding with "no-hands" on the straight section of a steep banked track. I would love to see you ride with "no-hands" at 2-3mph on the straight section of a steep banked track. Make a video and post it! -tom                 Date: 25 Oct 2007 14:07:01 From: Tim McNamara Subject: Re: Gyroscopic forces revisited In article <ffq8f9$4t1$1@news.Stanford.EDU >, "Tom Nakashima" <tom@slac.stanford.edu > wrote: > "Tim McNamara" <timmcn@bitstream.net> wrote in message > news:timmcn-205F17.20571424102007@news.iphouse.com... > > In article <ffnl8j$3uh$1@news.Stanford.EDU>, "Tom Nakashima" > > <tom@slac.stanford.edu> wrote: > > > >> "Tim McNamara" <timmcn@bitstream.net> wrote in message > >> news:timmcn-C6BB21.09172624102007@news.iphouse.com... > >> > In article <ffnhg2$n2$1@news.Stanford.EDU>, "Tom Nakashima" > >> > <tom@slac.stanford.edu> wrote: > >> > > >> >> I was talking with co-worker Timothy Montagne who just got > >> >> back from the track Nationals in Trexlertown, Pennsylvania. I > >> >> forget the exact slope on the straights in degrees, but he told > >> >> me it was so steep that you have to keep a minimum speed of 20 > >> >> mph. or else you would fall over. So I can imagine your speed > >> >> has to be greater than 20 mph to be able to ride with no hands, > >> >> of course which is no problem with a track rider. > >> > > >> > Timothy sold you a line of malarkey. If he was right, then the > >> > sprinters couldn't do track stands or pace each other at 2-3 mph > >> > up and down the track. But they can and Timothy is wrong. > >> > >> Timothy Montagne is pretty straight forward, never caught him in a > >> lie. However, I can't say that about reading some of your post. > > > > What do you think I have lied about, Tom? Is it a lie that > > sprinters can do track stands on a velodrome? Is it a lie that > > they pace each other at very low speeds while jockeying for > > advantage, without falling off the track (usually)? If you're > > going to accuse me of lying, have the courtesy to back it up. > > We're talking about riding with "no-hands" on the straight section of > a steep banked track. I missed that as being the main concern of the discussion and interpreted your post as saying "you can't ride slower than 20 mph at Trexlertown or you'll fall off." Sorry for any confusion. > I would love to see you ride with "no-hands" at 2-3mph on the > straight section of a steep banked track. > Make a video and post it! It would be entertaining what with the blood and the screaming and the splinters. Did I mention the blood? I haven't ridden on the velodrome in 8-9 years and don't even have my track bike any more. After spending weeks in a sling with a separated shoulder from crashing into the infield, I realized that at 40 and being basically self-employed I was too old for that stuff. I haven't gotten any younger in the interim.                   Date: 25 Oct 2007 12:48:06 From: Tom Nakashima Subject: Re: Gyroscopic forces revisited "Tim McNamara" <timmcn@bitstream.net > wrote in message news:timmcn-BF479E.14070125102007@news.iphouse.com... > In article <ffq8f9$4t1$1@news.Stanford.EDU>, > "Tom Nakashima" <tom@slac.stanford.edu> wrote: > >> "Tim McNamara" <timmcn@bitstream.net> wrote in message >> news:timmcn-205F17.20571424102007@news.iphouse.com... >> > In article <ffnl8j$3uh$1@news.Stanford.EDU>, "Tom Nakashima" >> > <tom@slac.stanford.edu> wrote: >> > >> >> "Tim McNamara" <timmcn@bitstream.net> wrote in message >> >> news:timmcn-C6BB21.09172624102007@news.iphouse.com... >> >> > In article <ffnhg2$n2$1@news.Stanford.EDU>, "Tom Nakashima" >> >> > <tom@slac.stanford.edu> wrote: >> >> > >> >> >> I was talking with co-worker Timothy Montagne who just got >> >> >> back from the track Nationals in Trexlertown, Pennsylvania. I >> >> >> forget the exact slope on the straights in degrees, but he told >> >> >> me it was so steep that you have to keep a minimum speed of 20 >> >> >> mph. or else you would fall over. So I can imagine your speed >> >> >> has to be greater than 20 mph to be able to ride with no hands, >> >> >> of course which is no problem with a track rider. >> >> > >> >> > Timothy sold you a line of malarkey. If he was right, then the >> >> > sprinters couldn't do track stands or pace each other at 2-3 mph >> >> > up and down the track. But they can and Timothy is wrong. >> >> >> >> Timothy Montagne is pretty straight forward, never caught him in a >> >> lie. However, I can't say that about reading some of your post. >> > >> > What do you think I have lied about, Tom? Is it a lie that >> > sprinters can do track stands on a velodrome? Is it a lie that >> > they pace each other at very low speeds while jockeying for >> > advantage, without falling off the track (usually)? If you're >> > going to accuse me of lying, have the courtesy to back it up. >> >> We're talking about riding with "no-hands" on the straight section of >> a steep banked track. > > I missed that as being the main concern of the discussion and > interpreted your post as saying "you can't ride slower than 20 mph at > Trexlertown or you'll fall off." Sorry for any confusion. No problem, just a misunderstanding. > >> I would love to see you ride with "no-hands" at 2-3mph on the >> straight section of a steep banked track. >> Make a video and post it! > > It would be entertaining what with the blood and the screaming and the > splinters. Did I mention the blood? I haven't ridden on the velodrome > in 8-9 years and don't even have my track bike any more. After spending > weeks in a sling with a separated shoulder from crashing into the > infield, I realized that at 40 and being basically self-employed I was > too old for that stuff. I haven't gotten any younger in the interim. Glad to see you recovered to keep riding. -tom     Date: 23 Oct 2007 15:08:28 From: A Muzi Subject: Re: Gyroscopic forces revisited Jon_C wrote: > How about a wheel on it's own? If you roll a wheel and it leans over > it turns in that direction. The reason is that the contact patch of a > leaning tyre is parallel to the ground and so at an angle to the > wheel's axle so the wheel acts as a rolling cone and turns into the > corner (which maintains it's balance). I'm pretty sure that's the main > steering force involved in no-handed riding. > (I seem to remember reading about someone testing this on a motorbike > and finding that on a prolonged corners the handlebars are actually > turned slightly outward instead of inward as you;d expect. I think > that was due to rake/trail but I don;t have time to think about that > right now :) David Jones' paper titled 'unrideable bicycle' discusses all that. I could not find a link (mine's paper) but a search shows 15+ sites with identical phrasing, entertaining for plagiarism buffs. -- Andrew Muzi www.yellowjersey.org Open every day since 1 April, 1971   Date: 23 Oct 2007 18:54:35 From: Subject: Re: Gyroscopic forces revisited Joseph Santaniello writes: > A while ago there was a thread discussing no-handed riding and the > role gyroscopic forces play in bike turning. > I hypothesized that gyroscopic forces were not that important, and > several things were pointed out to me that almost convinced me. A > good dose of insomnia allowed me to think about this for several > hours and I am now convinced that gyroscopic forces are virtually > irrelevant for riding bikes, hands or no-hands. > It was postulated that a clown-bike with tiny wheel would be > difficult to ride no hands due to the small gyroscopic forces. > Since I don't have a clown-bike to play with, this remains a > mystery. But I have observed little kids riding small wheel (10") > bikes at below walking pace. I have also spun some 12" wheels in my > hand a various speeds to feel what sort of gyroscopic forces are > there. Not much. A little kid with a sense of balance not as > developed as an adult can ride one of these bikes. I do not believe > a kid could keep one of these bikes upright by manual correction > alone. These small bikes are stable by themselves, and since the > gyroscopic forces are so low, there must be something else at work > here. This isn't proof or anything, this is just what got me > thinking. You should have seen the demo at InterBike where an engineer built a front wheel with a forward rotating flywheel (brass disk) between the spokes driven by a small motor at about the speed you expect from a 27" wheel at 10-15mph. He rode this bicycle no-hands at below 2mph, steady as a rock, up and down the isles. Without the flywheel turning independently, the bicycle was as difficult to ride no-hands as any other bicycle with wheels that size. > I finally pulled out a big plywood board (much to the chagrin of my > wife who imagines there are myriad things I could be better spending > my time on) and propped it up at an angle and put a bike on it to > simulate what happens when a bike is in a turn. It was suggested > that in a turn a pendulum hung from the top-tube would hang parallel > to the seat tube. When riding no hands in a constant radius turn, > this cannot be the case, and I suspect that it is not the case with > hands on the bars either. Experiments with my plywood board show > that when force is applied straight down through the bike the > steering remains straight no matter what the angle of the board > (simulated angle of lean). In a turn, the steering cannot be > straight, otherwise it wouldn't be called a turn, it would be called > a crash. So in a turn (a no-handed one in particular) something has > to be holding the steering at a non-straight angle. The only thing > it can be is that the center of mass is moved to the side of the > plane that is the centerline of the bike. This makes the steering > flop into the turn. The combined force of gravity and the > acceleration of the turn act from the center of mass through the > contact patches of the tires. Since the center of mass is not in > the plane of the bike's centerline, this means that the combined > force is not parallel to the seat tube and thus a pendulum hanging > from the top- tube could not be parallel with the seat tube. Riding > with hands on the bars I suspect is the same. But a rider could > force the bike to be in the same plane, but then they would need to > hold the steering at the proper angle manually. This would no doubt > require quite a bit of skill, and I believe in practice to be > virtually impossible. But perhaps it is just this skill which > separates the good from the great. Please discover why one should use paragraphs. You must have come across this in school. > So what does all that mean? It means that "flop" from an off-center > center of mass is what makes a bike turn, and thus while gyroscopic > forces make help the initial turn of the steering due to induced > lean, it is an unnecessary component that is ultimately irrelevant > to turning a bike. > The whole COM argument was brought about by thinking about how a > radio controlled motorcycle I used to have worked. I think your research came up with the wrong result. An easily repeatable exercise of coasting down a smooth road at more than 20mph riding no-hands, is to shake one knee from side to side while resting the other one against the top tube for stability. I think there is where you will see the effect the best. In addition, shimmy on a bicycle cannot occur without gyroscopic forces of the front wheel. Jobst Brandt     Date: 24 Oct 2007 05:56:52 From: Ron Hardin Subject: Re: Gyroscopic forces revisited The gyroscopic action stabilizes the front wheel, so you need it ; however it actually works against the countersteer needed for balancing. Since there's more mass of the front wheel ahead of the steering tube line than behind it, if you move the top of the bike sharply _left_ (walking or riding) the front wheel first steers _right_ and then moves back left. Which produces the necessary countersteer. It becomes harder and harder to go anywhere but straight as the speed increases, as the gyroscopic action overrides countersteer. It becomes harder and harder to balance as the speed decreases, as the steering continues to turn in the direction its started, there being diminishing stabilizing gyroscopic action. There's a range of speeds in the middle where no-handed riding is easy. -- rhhardin@mindspring.com On the internet, nobody knows you're a jerk.     Date: 23 Oct 2007 20:42:28 From: jim beam Subject: Re: Gyroscopic forces revisited jobst.brandt@stanfordalumni.org wrote: <snip for clarity > > In addition, > shimmy on a bicycle cannot occur without gyroscopic forces of the > front wheel. that's totally untrue. gyro-less, wheel-less ski bikes can shimmy. shimmy is a dynamic interaction between turn reaction and frame flex - no gyro forces required.       Date: 24 Oct 2007 01:18:04 From: Jambo Subject: Re: Gyroscopic forces revisited "jim beam" <spamvortex@bad.example.net > wrote in message news:feydnbYx3IW4IoPanZ2dnUVZ_ofinZ2d@speakeasy.net... > jobst.brandt@stanfordalumni.org wrote: > <snip for clarity> > >> In addition, >> shimmy on a bicycle cannot occur without gyroscopic forces of the >> front wheel. > > that's totally untrue. gyro-less, wheel-less ski bikes can shimmy. shimmy > is a dynamic interaction between turn reaction and frame flex - no gyro > forces required. That's totally bullshit. "Turn reaction"? Please. Shimmy cannot occur on skis - no matter where they are mounted, bikes or snow mobiles. Interaction with the ground cannot induce resonant vibration on anything other than rotating components, unless travelling at superhuman speed. But you wouldn't know this from your "metarials skool", would you? Idiot.         Date: 29 Oct 2007 00:52:01 From: Jasper Janssen Subject: Re: Gyroscopic forces revisited On Wed, 24 Oct 2007 01:18:04 -0400, "Jambo" <-@-.- > wrote: >Shimmy cannot occur on skis - no matter where they are mounted, bikes or >snow mobiles. Interaction with the ground cannot induce resonant vibration >on anything other than rotating components, unless travelling at superhuman >speed. But you wouldn't know this from your "metarials skool", would you? >Idiot. Why would a ski not be able to resonate if some surface irregularity manages to hit a resonance at a certain speed? Jasper           Date: 29 Oct 2007 00:45:20 From: Michael Press Subject: Re: Gyroscopic forces revisited In article <0kbai3dkv8033cd9b9ugekkmu79a1af2t3@4ax.com >, Jasper Janssen <jasper@jjanssen.org > wrote: > On Wed, 24 Oct 2007 01:18:04 -0400, "Jambo" <-@-.-> wrote: > > >Shimmy cannot occur on skis - no matter where they are mounted, bikes or > >snow mobiles. Interaction with the ground cannot induce resonant vibration > >on anything other than rotating components, unless travelling at superhuman > >speed. But you wouldn't know this from your "metarials skool", would you? > >Idiot. > > Why would a ski not be able to resonate if some surface irregularity > manages to hit a resonance at a certain speed? Because skis are designed to be over damped at all frequencies. -- Michael Press             Date: 30 Oct 2007 17:47:55 From: Jasper Janssen Subject: Re: Gyroscopic forces revisited On Mon, 29 Oct 2007 00:45:20 -0700, Michael Press <rubrum@pacbell.net > wrote: >In article ><0kbai3dkv8033cd9b9ugekkmu79a1af2t3@4ax.com>, > Jasper Janssen <jasper@jjanssen.org> wrote: > >> On Wed, 24 Oct 2007 01:18:04 -0400, "Jambo" <-@-.-> wrote: >> >> >Shimmy cannot occur on skis - no matter where they are mounted, bikes or >> >snow mobiles. Interaction with the ground cannot induce resonant vibration >> >on anything other than rotating components, unless travelling at superhuman >> >speed. But you wouldn't know this from your "metarials skool", would you? >> >Idiot. >> >> Why would a ski not be able to resonate if some surface irregularity >> manages to hit a resonance at a certain speed? > >Because skis are designed to be over damped at all frequencies. We were talking about a bike with skies instead of wheels, though. Bikes aren't damped at all frequencies that could be developed by either a rotating wheel *or* a ski at typical bicycle speeds. Jasper