| |
Main
Date: 28 Sep 2007 16:17:31
From:
Subject: some tire drop data
|
Some recent threads and an email today led me to measure how much a fairly typical tire rose as I inflated it. The results suggest that some people are fooling themselves. First, some data from dial indicator measurements of a loaded tire's height rise in thousandths of an inch from 40 to 130 psi, measured from top of nominal 700x25c tire. Tire measured 1.020" to 1.030" wide at top between 80 to 130 psi. Total load on tire was ~85 pounds, resting on a scale, but the tire stood on a concrete floor for the test. inch mm inch psi rise rise change --- ----- ---- ------ 40 0.000 0.00 n/a 50 0.021 0.53 0.021 60 0.038 0.97 0.017 70 0.052 1.32 0.014 80 0.063 1.60 0.011 90 0.076 1.93 0.013 100 0.088 2.24 0.012 110 0.100 2.54 0.012 120 0.110 2.79 0.010 130 0.120 3.05 0.010 My quick and dirty table shows only that the top of a roughly 1.020" wide tire rose ~0.120 inches from 40 to 130 psi under an 88 pound load. For those interested in arbitrary figures, 15% of a tire width of 1.020" is 0.153", or 3.9 mm. Presumably, someone could measure the diameter of a tire hanging in mid-air and inflated to some arbitrary pressure, load it with some weights, measure the change, and call it tire drop. But this would require such an elaborate test rig that it would be extremely difficult for any normal poster to even _try_ to make a practical measurement of the difference between 90 psi and 100 psi. After all, the difference between 90 and 100 psi with an elaborate test rig using a dial indicator was only 0.012", about a hundredth of an inch. Incidentally, it is highly unlikely that pressure and a 700x25 tire's rise have the straight linear relationship that some people suggest, at least not over a 40 to 130 psi range with an 88 pound load. I expect that beautifully smooth graphs of pressure and rise are just as theoretical, oversimplified, and mistaken as similar predictions for contact patch area, which does not follow the often-mentioned tire-pressure/load = area prediction. (Actual measurements of contact patches show that they fail to shrink as much as predicted with higher pressures and fail to expand as much as predicted with lower pressures. They stubbornly stick to a preferred size.) Cheers, Carl Fogel
|
|
| |
Date: 29 Sep 2007 23:43:10
From:
Subject: Re: some tire drop data
|
On Fri, 28 Sep 2007 16:17:31 -0600, carlfogel@comcast.net wrote: >Some recent threads and an email today led me to measure how much a >fairly typical tire rose as I inflated it. > >The results suggest that some people are fooling themselves. > >First, some data from dial indicator measurements of a loaded tire's >height rise in thousandths of an inch from 40 to 130 psi, measured >from top of nominal 700x25c tire. > >Tire measured 1.020" to 1.030" wide at top between 80 to 130 psi. > >Total load on tire was ~85 pounds, resting on a scale, but the tire >stood on a concrete floor for the test. > > inch mm inch > psi rise rise change > --- ----- ---- ------ > 40 0.000 0.00 n/a > 50 0.021 0.53 0.021 > 60 0.038 0.97 0.017 > 70 0.052 1.32 0.014 > 80 0.063 1.60 0.011 > 90 0.076 1.93 0.013 > 100 0.088 2.24 0.012 > 110 0.100 2.54 0.012 > 120 0.110 2.79 0.010 > 130 0.120 3.05 0.010 > >My quick and dirty table shows only that the top of a roughly 1.020" >wide tire rose ~0.120 inches from 40 to 130 psi under an 88 pound >load. > >For those interested in arbitrary figures, 15% of a tire width of >1.020" is 0.153", or 3.9 mm. > >Presumably, someone could measure the diameter of a tire hanging in >mid-air and inflated to some arbitrary pressure, load it with some >weights, measure the change, and call it tire drop. > >But this would require such an elaborate test rig that it would be >extremely difficult for any normal poster to even _try_ to make a >practical measurement of the difference between 90 psi and 100 psi. > >After all, the difference between 90 and 100 psi with an elaborate >test rig using a dial indicator was only 0.012", about a hundredth of >an inch. > >Incidentally, it is highly unlikely that pressure and a 700x25 tire's >rise have the straight linear relationship that some people suggest, >at least not over a 40 to 130 psi range with an 88 pound load. > >I expect that beautifully smooth graphs of pressure and rise are just >as theoretical, oversimplified, and mistaken as similar predictions >for contact patch area, which does not follow the often-mentioned >tire-pressure/load = area prediction. > >(Actual measurements of contact patches show that they fail to shrink >as much as predicted with higher pressures and fail to expand as much >as predicted with lower pressures. They stubbornly stick to a >preferred size.) > >Cheers, > >Carl Fogel An odd example of measurement problems. I took a few minutes and set up my dial calipers to hang and measure the same tire's width as it sat in a truing stand. As I expected, the tire widened as it was pumped up psi width 30 0.975 40 0.985 50 0.992 60 0.996 70 0.998 80 1.007 90 1.014 100 1.018 110 1.025 120 1.029 130 1.034 140 1.040 But trying to get a good reading by rolling a dial caliper's jaws shut on squishy rubber _felt_ a little squishy and inaccurate, even though the results looked good. Rather than adjusting the dial caliper to slip at less and less resistance as its jaws closed on the rubber, I decided to try a dial indicator, which supplies its own very gentle spring action. Naturally, I got sidetracked and first set the dial indicator up to measure the expansion of the top of the tire in the truing stand, not its sideways bulge. But things turned out pretty well: top of psi tire ? 1.000 as low as I could get it and still feel pressure 10 0.990 probably seated a bit, moving the wrong way 20 0.994 30 0.998 40 1.002 50 1.007 60 1.011 70 1.016 80 1.023 90 1.028 100 1.033 110 1.038 120 1.042 130 1.050 140 pump head blew off Then I set things up again in the truing stand, but turned the dial indicator rig on the side of the tire to measure how far one side of the tire bulged sideways as I pumped it up. I zeroed the dial indicator, checked that everything was at right angles, and pumped . . . Nothing. No change at 10 psi. I pumped to 20 psi, but still nothing happened. At 30 psi, the needle still sat on 0.000. I went up to 80 psi without any movement. At 120 psi, there was a little movement, about 0.003 inches. Mystified, I grabbed my dial calipers, closed their jaws on the stupid tire, and let the air out of the dumb thing. A gratifying gap promptly opened between the side of the tire and the jaws, so I knew that I wasn't crazy--tires really do expand when inflated. But another attempt with the dial _indicator_ produced the same failure. At first, I thought that the problem was just that the tire was expanding outward, as well as sideways, and the dial indicator was slipping down the side of the expanding tire, just enough for the loss to cancel the gain. Some of the failure was due to this movement in two directions, but I suspect that real problem was the tiny bit of stiction in the dial indicator rod as the rubber moved outward as well as sideways may have made things worse, too, jamming the delicate mechanism. Whatever the cause, the dial indicator utterly failed to measure the sideways expansion clearly indicated by dial calipers. Cheers, Carl Fogel
|
| |
Date: 30 Sep 2007 16:42:47
From: anth
Subject: Re: some tire drop data
|
On Sat, 2007-09-29 at 12:23 -0500, Tim McNamara wrote: > Heine measure loads on three bicycles with a rider seated on the bike > (randonneur bike, racing bike and "city" bike) and found that the weight > distribution (% front/rear) was 45/55, 30/60 and 35/65 respectively. The middle figures don't add up. I'm guessing they should have been 40/60?
|
| |
Date: 29 Sep 2007 15:57:50
From:
Subject: Re: some tire drop data
|
On Fri, 28 Sep 2007 16:17:31 -0600, carlfogel@comcast.net wrote: >Some recent threads and an email today led me to measure how much a >fairly typical tire rose as I inflated it. > >The results suggest that some people are fooling themselves. > >First, some data from dial indicator measurements of a loaded tire's >height rise in thousandths of an inch from 40 to 130 psi, measured >from top of nominal 700x25c tire. > >Tire measured 1.020" to 1.030" wide at top between 80 to 130 psi. > >Total load on tire was ~85 pounds, resting on a scale, but the tire >stood on a concrete floor for the test. > > inch mm inch > psi rise rise change > --- ----- ---- ------ > 40 0.000 0.00 n/a > 50 0.021 0.53 0.021 > 60 0.038 0.97 0.017 > 70 0.052 1.32 0.014 > 80 0.063 1.60 0.011 > 90 0.076 1.93 0.013 > 100 0.088 2.24 0.012 > 110 0.100 2.54 0.012 > 120 0.110 2.79 0.010 > 130 0.120 3.05 0.010 > >My quick and dirty table shows only that the top of a roughly 1.020" >wide tire rose ~0.120 inches from 40 to 130 psi under an 88 pound >load. > >For those interested in arbitrary figures, 15% of a tire width of >1.020" is 0.153", or 3.9 mm. > >Presumably, someone could measure the diameter of a tire hanging in >mid-air and inflated to some arbitrary pressure, load it with some >weights, measure the change, and call it tire drop. > >But this would require such an elaborate test rig that it would be >extremely difficult for any normal poster to even _try_ to make a >practical measurement of the difference between 90 psi and 100 psi. > >After all, the difference between 90 and 100 psi with an elaborate >test rig using a dial indicator was only 0.012", about a hundredth of >an inch. > >Incidentally, it is highly unlikely that pressure and a 700x25 tire's >rise have the straight linear relationship that some people suggest, >at least not over a 40 to 130 psi range with an 88 pound load. > >I expect that beautifully smooth graphs of pressure and rise are just >as theoretical, oversimplified, and mistaken as similar predictions >for contact patch area, which does not follow the often-mentioned >tire-pressure/load = area prediction. > >(Actual measurements of contact patches show that they fail to shrink >as much as predicted with higher pressures and fail to expand as much >as predicted with lower pressures. They stubbornly stick to a >preferred size.) > >Cheers, > >Carl Fogel An email asked for details of the testing. I wonder if other posters get as many emails as I do? Often it seems as if people are accidentally hitting the email instead of the post button on their newsreaders. I don't mind, but it would more fun if they were all beautiful but lonely women. Luckily, the test rig had been left untouched in hopes of gaining Historic Place status, so here's a picture: http://i24.tinypic.com/2iw4zn5.jpg The white bathroom scales are not legal for trade, but showed ~85 pounds when shoved under the tire. They were removed for testing, lest the tire inflation increased the contact patch area enough to cause the scale surface to rise--remember, the measurements involved changes as small as ten thousandths of an inch. The yellow floor pump reads in alleged 2-lb increments. Zip ties secure the pump head, which tended to come off during lengthy testing. Fogel Labs has budgeted for a new O-ring. The seat post sticks up between the jaws of the bench vise, which are slightly open, but stop everything from toppling over. If the post had been clamped, it might have held the front end of the bicycle down slightly as the tire was pumped up--again, the measurements were so tiny that such precautions were necessary. The board was clamped to the red test frame for long-ago experiments. Later, the weight bar was added for even stranger rituals. The dial indicator from Sears is directly over the front axle and contact patch. It rests on the weight bar, not the top of the tire, partly because it was just about the right height, but mostly because the slight expansion of the top of the tire during inflation would have exaggerated the rise of the bottom of the tire. A carefully machined indentation in the round weight bar assures that the dial indicator does not slip off to the side and give false readings. (It looks remarkably like a dent whacked with a blunt punch and a hammer.) The dial indicator is held by vise grips, which in turn are held in place under a 10-pound weight. Vise grips were required because half the convenient mounting lug on the back of the dial indicator was hacksawed off in 1972 in order to fit into an awkward spot while timing a motorcycle. I may spend ten bucks at Harbor Freight and get a replacement, or maybe even a few bucks more for one with an adjustable arm. Cheers, Carl Fogel
|
| |
Date: 29 Sep 2007 14:23:23
From: Gary Young
Subject: Re: some tire drop data
|
On Fri, 28 Sep 2007 21:24:36 -0600, carlfogel wrote: > On Fri, 28 Sep 2007 21:27:22 -0500, Tim McNamara > <timmcn@bitstream.net> wrote: > >>In article <oltqf3td4ftpiiishe2kcimc8vhn5jfq6n@4ax.com>, >> carlfogel@comcast.net wrote: >> >>> Some recent threads and an email today led me to measure how much a >>> fairly typical tire rose as I inflated it. >>> >>> The results suggest that some people are fooling themselves. >>> >>> First, some data from dial indicator measurements of a loaded tire's >>> height rise in thousandths of an inch from 40 to 130 psi, measured >>> from top of nominal 700x25c tire. >>> >>> Tire measured 1.020" to 1.030" wide at top between 80 to 130 psi. >>> >>> Total load on tire was ~85 pounds, resting on a scale, but the tire >>> stood on a concrete floor for the test. >>> >>> inch mm inch >>> psi rise rise change >>> --- ----- ---- ------ >>> 40 0.000 0.00 n/a >>> 50 0.021 0.53 0.021 >>> 60 0.038 0.97 0.017 >>> 70 0.052 1.32 0.014 >>> 80 0.063 1.60 0.011 >>> 90 0.076 1.93 0.013 >>> 100 0.088 2.24 0.012 >>> 110 0.100 2.54 0.012 >>> 120 0.110 2.79 0.010 >>> 130 0.120 3.05 0.010 >>> >>> My quick and dirty table shows only that the top of a roughly 1.020" >>> wide tire rose ~0.120 inches from 40 to 130 psi under an 88 pound >>> load. >>> >>> For those interested in arbitrary figures, 15% of a tire width of >>> 1.020" is 0.153", or 3.9 mm. >>> >>> Presumably, someone could measure the diameter of a tire hanging in >>> mid-air and inflated to some arbitrary pressure, load it with some >>> weights, measure the change, and call it tire drop. >>> >>> But this would require such an elaborate test rig that it would be >>> extremely difficult for any normal poster to even _try_ to make a >>> practical measurement of the difference between 90 psi and 100 psi. >>> >>> After all, the difference between 90 and 100 psi with an elaborate >>> test rig using a dial indicator was only 0.012", about a hundredth of >>> an inch. >>> >>> Incidentally, it is highly unlikely that pressure and a 700x25 tire's >>> rise have the straight linear relationship that some people suggest, >>> at least not over a 40 to 130 psi range with an 88 pound load. >>> >>> I expect that beautifully smooth graphs of pressure and rise are just >>> as theoretical, oversimplified, and mistaken as similar predictions >>> for contact patch area, which does not follow the often-mentioned >>> tire-pressure/load = area prediction. >>> >>> (Actual measurements of contact patches show that they fail to shrink >>> as much as predicted with higher pressures and fail to expand as much >>> as predicted with lower pressures. They stubbornly stick to a >>> preferred size.) >> >>Interesting data from Fogel Labs again; thanks, Carl. I don't have the >>original article about the 15% drop figure right here to read again. >>IIRC the 15% drop figure came from research by Jim Papodopolous, but how >>it was arrived at I don't know. It's a number that looks suspiciously >>arbitrary (how many things in the world work out to such a nice even >>number?). > > Dear Tim, > > I don't know where the popular 15% figure came from or from what point > it was measured--just the weight of the front end of a bike resting on > the tire is about 10% of the typical load. > > Maybe there's a clever measuring technique that was somehow confirmed > by rolling resistance tests, but it's going to have to be awfully > clever to handle differences of about a hundredth of an inch between > 90 and 100 psi. > > Here's a link that came in my email to a chart that graphs some kind > of 15% drop correlated with pressure for various tire sizes: > > > http://bp1.blogger.com/_d-Yj0VDKhWQ/RnwLUoA9sVI/AAAAAAAAAJ4/225Ym5qYKTo/s1600-h/BQ_berto_inflationgraph.jpg > > The graph seems to show some simple linear calculations, not actual > data points from measurements. > > I suspect that actual measurements would show the straight line theory > is simply wrong. > > If a tire were a piston in a car engine, then adding a 100 pound > weight (including piston) would drive a piston with an area of 1 > square inch down until the pressure rose to 100 psi. > > But a bicycle tire is significantly different from a metal piston in a > metal chamber. > > When we load a tire, the bottom of the "piston chamber" is the > reasonably unyielding metal rim. > > But on either side of the tire, the C-shaped tire walls deform outward > against the air pressure trying to keep them in their current shape > about an inch wide on a 25 mm tire, the minor axis of the toroid (your > thumb and finger going around the tire). > > Meanwhile, the tire also deforms, much less against the gigantic "O" > of the tire viewed the long way around, the major axis of the toroid. > > In effect, the tire is more like a long, narrow trampoline than the > top of a metal piston, and it distorts in a much more complicated > fashion. > > This is why the simple area = load / pressure load equation fails to > predict what happens at high and low pressures for 700c x 25 tires. > > The same explanation probably applies to how high a tire rises or > falls according to pressure. The sides of the tire act as scissors > jacks, applying increasing pressure at the edge of the contact patch > as pressure drops and the sides deform more and more. > > As I've mentioned before, what made me suspicious of simple > straight-line predictions was knowing that a 4-inch wide trials tire > can carry a 400-pound load in a wheelstand down a paved road with only > 4 psi and no way to spread out more than 4 inches. The oversimplified > equation predicts: > > 400 lbs / 4 (lbs/inch^2) = 100 ^ 2 = 4 x 25 inches > > But a 4 x 18 trials tire is only about 26 inches in diameter, so a 25 > inch long contact patch is impossible. The sidewalls had to be > exerting considerable force as springs, which seems reasonable, since > it takes considerable force to pull (or push) the wall of an inflated > tire out of shape. > > Cheers, > > Carl Fogel Jan Heine, publisher of the magazine where the graph appeared, posted a response on the ibob mailing list: "The graph with the linear relationships quoted in the link below [i.e., Carl's link, above] came from Bicycle Quarterly. It was based on actual measurements by Frank Berto, and his measurements did line up roughly on a straight line (I have seen the original measurements). He built a test rig, which he described many years ago in Bicycling Magazine. The graphs are extrapolated toward the ends... The discrepancy between Carl Fogel's measurements and Frank Berto's might be easy to explain. They measured different things. Carl measured the rise of the tire with increasing pressure. I am not surprised that this is not linear - the tire cannot expand indefinitely. Frank Berto measured how much pressure he needed for different weights to achieve the same tire deflection of 15%. The 15% came from tire manufacturers' recommendations. In Bicycle Quarterly's real road tests, we found that performance decreased significantly once the "tire drop" was more than about 13-15%. With less tire drop, higher pressures did not yield significant performance benefits. Please feel free to forward this clarification to rec.bike.tech. Jan Heine Editor Bicycle Quarterly 140 Lakeside Ave #C Seattle WA 98122 www.bikequarterly.com" http://search.bikelist.org/getmsg.asp?Filename=internet-bob.10709.1869.eml
|
| | |
Date: 29 Sep 2007 14:36:01
From: * * Chas
Subject: Re: some tire drop data
|
"Gary Young" <garyyoung3@gmail.com > wrote in message news:U4ydnbBef9e2O2PbnZ2dnUVZ_ovinZ2d@giganews.com... > On Fri, 28 Sep 2007 21:24:36 -0600, carlfogel wrote: <snip > >.....In Bicycle Quarterly's real road tests, we found that > performance decreased significantly once the "tire drop" > was more than about 13-15%. With less tire drop, higher > pressures did not yield significant performance benefits. Performance? Define performance. Was it an increase in rolling resistance? More pedal force required? Or was it anecdotal subjective opinions of the riders. My personal opinion is that some riders like very hard tires and have fooled themselves into thinking that tires run at high pressures will MAKE them go faster. If I'm going to ride on a rough surface road I ride Panaracer Paselas at 85-90PSI. If it's a smooth road surface I like Continental Grand Prixs at 95-110 PSI. I ride sewups at 90 PSI. That's MY preference based on ride comfort and puncture potential. Chas.
|
| | |
Date: 29 Sep 2007 14:49:01
From:
Subject: Re: some tire drop data
|
On Sat, 29 Sep 2007 14:23:23 -0500, Gary Young <garyyoung3@gmail.com > wrote: >On Fri, 28 Sep 2007 21:24:36 -0600, carlfogel wrote: > >> On Fri, 28 Sep 2007 21:27:22 -0500, Tim McNamara >> <timmcn@bitstream.net> wrote: >> >>>In article <oltqf3td4ftpiiishe2kcimc8vhn5jfq6n@4ax.com>, >>> carlfogel@comcast.net wrote: >>> >>>> Some recent threads and an email today led me to measure how much a >>>> fairly typical tire rose as I inflated it. >>>> >>>> The results suggest that some people are fooling themselves. >>>> >>>> First, some data from dial indicator measurements of a loaded tire's >>>> height rise in thousandths of an inch from 40 to 130 psi, measured >>>> from top of nominal 700x25c tire. >>>> >>>> Tire measured 1.020" to 1.030" wide at top between 80 to 130 psi. >>>> >>>> Total load on tire was ~85 pounds, resting on a scale, but the tire >>>> stood on a concrete floor for the test. >>>> >>>> inch mm inch >>>> psi rise rise change >>>> --- ----- ---- ------ >>>> 40 0.000 0.00 n/a >>>> 50 0.021 0.53 0.021 >>>> 60 0.038 0.97 0.017 >>>> 70 0.052 1.32 0.014 >>>> 80 0.063 1.60 0.011 >>>> 90 0.076 1.93 0.013 >>>> 100 0.088 2.24 0.012 >>>> 110 0.100 2.54 0.012 >>>> 120 0.110 2.79 0.010 >>>> 130 0.120 3.05 0.010 >>>> >>>> My quick and dirty table shows only that the top of a roughly 1.020" >>>> wide tire rose ~0.120 inches from 40 to 130 psi under an 88 pound >>>> load. >>>> >>>> For those interested in arbitrary figures, 15% of a tire width of >>>> 1.020" is 0.153", or 3.9 mm. >>>> >>>> Presumably, someone could measure the diameter of a tire hanging in >>>> mid-air and inflated to some arbitrary pressure, load it with some >>>> weights, measure the change, and call it tire drop. >>>> >>>> But this would require such an elaborate test rig that it would be >>>> extremely difficult for any normal poster to even _try_ to make a >>>> practical measurement of the difference between 90 psi and 100 psi. >>>> >>>> After all, the difference between 90 and 100 psi with an elaborate >>>> test rig using a dial indicator was only 0.012", about a hundredth of >>>> an inch. >>>> >>>> Incidentally, it is highly unlikely that pressure and a 700x25 tire's >>>> rise have the straight linear relationship that some people suggest, >>>> at least not over a 40 to 130 psi range with an 88 pound load. >>>> >>>> I expect that beautifully smooth graphs of pressure and rise are just >>>> as theoretical, oversimplified, and mistaken as similar predictions >>>> for contact patch area, which does not follow the often-mentioned >>>> tire-pressure/load = area prediction. >>>> >>>> (Actual measurements of contact patches show that they fail to shrink >>>> as much as predicted with higher pressures and fail to expand as much >>>> as predicted with lower pressures. They stubbornly stick to a >>>> preferred size.) >>> >>>Interesting data from Fogel Labs again; thanks, Carl. I don't have the >>>original article about the 15% drop figure right here to read again. >>>IIRC the 15% drop figure came from research by Jim Papodopolous, but how >>>it was arrived at I don't know. It's a number that looks suspiciously >>>arbitrary (how many things in the world work out to such a nice even >>>number?). >> >> Dear Tim, >> >> I don't know where the popular 15% figure came from or from what point >> it was measured--just the weight of the front end of a bike resting on >> the tire is about 10% of the typical load. >> >> Maybe there's a clever measuring technique that was somehow confirmed >> by rolling resistance tests, but it's going to have to be awfully >> clever to handle differences of about a hundredth of an inch between >> 90 and 100 psi. >> >> Here's a link that came in my email to a chart that graphs some kind >> of 15% drop correlated with pressure for various tire sizes: >> >> >> http://bp1.blogger.com/_d-Yj0VDKhWQ/RnwLUoA9sVI/AAAAAAAAAJ4/225Ym5qYKTo/s1600-h/BQ_berto_inflationgraph.jpg >> >> The graph seems to show some simple linear calculations, not actual >> data points from measurements. >> >> I suspect that actual measurements would show the straight line theory >> is simply wrong. >> >> If a tire were a piston in a car engine, then adding a 100 pound >> weight (including piston) would drive a piston with an area of 1 >> square inch down until the pressure rose to 100 psi. >> >> But a bicycle tire is significantly different from a metal piston in a >> metal chamber. >> >> When we load a tire, the bottom of the "piston chamber" is the >> reasonably unyielding metal rim. >> >> But on either side of the tire, the C-shaped tire walls deform outward >> against the air pressure trying to keep them in their current shape >> about an inch wide on a 25 mm tire, the minor axis of the toroid (your >> thumb and finger going around the tire). >> >> Meanwhile, the tire also deforms, much less against the gigantic "O" >> of the tire viewed the long way around, the major axis of the toroid. >> >> In effect, the tire is more like a long, narrow trampoline than the >> top of a metal piston, and it distorts in a much more complicated >> fashion. >> >> This is why the simple area = load / pressure load equation fails to >> predict what happens at high and low pressures for 700c x 25 tires. >> >> The same explanation probably applies to how high a tire rises or >> falls according to pressure. The sides of the tire act as scissors >> jacks, applying increasing pressure at the edge of the contact patch >> as pressure drops and the sides deform more and more. >> >> As I've mentioned before, what made me suspicious of simple >> straight-line predictions was knowing that a 4-inch wide trials tire >> can carry a 400-pound load in a wheelstand down a paved road with only >> 4 psi and no way to spread out more than 4 inches. The oversimplified >> equation predicts: >> >> 400 lbs / 4 (lbs/inch^2) = 100 ^ 2 = 4 x 25 inches >> >> But a 4 x 18 trials tire is only about 26 inches in diameter, so a 25 >> inch long contact patch is impossible. The sidewalls had to be >> exerting considerable force as springs, which seems reasonable, since >> it takes considerable force to pull (or push) the wall of an inflated >> tire out of shape. >> >> Cheers, >> >> Carl Fogel > >Jan Heine, publisher of the magazine where the graph appeared, posted a >response on the ibob mailing list: > >"The graph with the linear relationships quoted in the link below >[i.e., Carl's link, above] came from Bicycle Quarterly. It was based on >actual measurements by Frank Berto, and his measurements did line up >roughly on a straight line (I have seen the original measurements). He >built a test rig, which he described many years ago in Bicycling Magazine. >The graphs are extrapolated toward the ends... > >The discrepancy between Carl Fogel's measurements and Frank Berto's >might be easy to explain. They measured different things. Carl >measured the rise of the tire with increasing pressure. I am not >surprised that this is not linear - the tire cannot expand >indefinitely. > >Frank Berto measured how much pressure he needed for different >weights to achieve the same tire deflection of 15%. The 15% came from >tire manufacturers' recommendations. In Bicycle Quarterly's real road >tests, we found that performance decreased significantly once the >"tire drop" was more than about 13-15%. With less tire drop, higher >pressures did not yield significant performance benefits. > >Please feel free to forward this clarification to rec.bike.tech. > >Jan Heine >Editor >Bicycle Quarterly >140 Lakeside Ave #C >Seattle WA 98122 >www.bikequarterly.com" > >http://search.bikelist.org/getmsg.asp?Filename=internet-bob.10709.1869.eml Dear Gary, Anyone can load a tire, put a dial indicator on the top of the tire, and measure the rise as inflation is increased in 10 psi increments. It makes no difference whether the measurements are taken in increasing or decreasing order. The rise will _not_ be linear, with the effect particularly noticeable at low pressures. If a straight line was extrapolated, that could account for the discrepancy. It would be fascinating to see the details of how the measurements were achieved, since the tire deflects _most_ with the initial load. In other words, the drop for 8 pounds of load is greatest for the first 8 pounds (roughly the load of just a 20-lb bike on the front tire). Measuring differences in thousandths of an inch gets awfully tricky in practical terms out toward zero. A 13%-15% absolute tire drop range is mentioned. For a 1.000" tire, that's 0.130" to 0.150", a difference of 0.020". On a ~1-inch tire, I found that a relative drop of 0.020" at ~90 psi with an 85-lb load corresponded to a ~20 psi change in inflation. Any bicyclist with a floor pump, a dial indicator, and some weights can check my measurements. You can pick up a dial indicator from Harbor Freight for about ten bucks, less than the cost of a floor pump. Cheers, Carl Fogel
|
| | | |
Date: 30 Sep 2007 05:10:12
From: Ryan Cousineau
Subject: Re: some tire drop data
|
In article <kddtf3drng10jfmnl02pcifrlmif8fk2u3@4ax.com >, carlfogel@comcast.net wrote: > On Sat, 29 Sep 2007 14:23:23 -0500, Gary Young <garyyoung3@gmail.com> > wrote: > > >On Fri, 28 Sep 2007 21:24:36 -0600, carlfogel wrote: > > > >> On Fri, 28 Sep 2007 21:27:22 -0500, Tim McNamara > >> <timmcn@bitstream.net> wrote: > >> > >>>In article <oltqf3td4ftpiiishe2kcimc8vhn5jfq6n@4ax.com>, > >>> carlfogel@comcast.net wrote: [long correspondence trimmed; interested viewers are invited to check the archives] > Any bicyclist with a floor pump, a dial indicator, and some weights > can check my measurements. > > You can pick up a dial indicator from Harbor Freight for about ten > bucks, less than the cost of a floor pump. Aha! I am on to your craven attempts to encourage the readers of rbt to engage in the oft-forsaken cornerstone of the Scientific Method: reproducibility! Well, we scholastics are on to you. And it's a shame. I always assumed that a former literature explicator like yourself would have had a naturally scholastic bent. Silliness aside, I should take up your challenge, but my excuse is that I have too many bikes a-building right now for others. -- Ryan Cousineau rcousine@sfu.ca http://www.wiredcola.com/ "I don't want kids who are thinking about going into mathematics to think that they have to take drugs to succeed." -Paul Erdos
|
| | | | |
Date: 29 Sep 2007 23:55:00
From:
Subject: Re: some tire drop data
|
On Sun, 30 Sep 2007 05:10:12 GMT, Ryan Cousineau <rcousine@sfu.ca > wrote: [snip] >Silliness aside, I should take up your challenge, but my excuse is that >I have too many bikes a-building right now for others. Dear Ryan, Just remember that it could be a very bad thing if a bike with 80 pounds of weights at handlebar height topples over onto you. Safety rope the rig to a work bench or arrange for a hefty vise's jaws to cuddle a stout part of the bike frame. But do _something_ to make sure that your carefully balanced test bike rig can't fall over and hurt you. I haven't had any accidents so far, and I don't want to read about anyone else having different results. Er, I mean anyone else having accidents. Different results are fine. Cheers, Carl Fogel
|
| |
Date: 28 Sep 2007 20:45:21
From:
Subject: Re: some tire drop data
|
On Sep 28, 9:27 pm, Tim McNamara <tim...@bitstream.net > wrote: > In article <oltqf3td4ftpiiishe2kcimc8vhn5jf...@4ax.com>, > > > > > > carlfo...@comcast.net wrote: > > Some recent threads and an email today led me to measure how much a > > fairly typical tire rose as I inflated it. > > > The results suggest that some people are fooling themselves. > > > First, some data from dial indicator measurements of a loaded tire's > > height rise in thousandths of an inch from 40 to 130 psi, measured > > from top of nominal 700x25c tire. > > > Tire measured 1.020" to 1.030" wide at top between 80 to 130 psi. > > > Total load on tire was ~85 pounds, resting on a scale, but the tire > > stood on a concrete floor for the test. > > > inch mm inch > > psi rise rise change > > --- ----- ---- ------ > > 40 0.000 0.00 n/a > > 50 0.021 0.53 0.021 > > 60 0.038 0.97 0.017 > > 70 0.052 1.32 0.014 > > 80 0.063 1.60 0.011 > > 90 0.076 1.93 0.013 > > 100 0.088 2.24 0.012 > > 110 0.100 2.54 0.012 > > 120 0.110 2.79 0.010 > > 130 0.120 3.05 0.010 > > > My quick and dirty table shows only that the top of a roughly 1.020" > > wide tire rose ~0.120 inches from 40 to 130 psi under an 88 pound > > load. > > > For those interested in arbitrary figures, 15% of a tire width of > > 1.020" is 0.153", or 3.9 mm. > > > Presumably, someone could measure the diameter of a tire hanging in > > mid-air and inflated to some arbitrary pressure, load it with some > > weights, measure the change, and call it tire drop. > > > But this would require such an elaborate test rig that it would be > > extremely difficult for any normal poster to even _try_ to make a > > practical measurement of the difference between 90 psi and 100 psi. > > > After all, the difference between 90 and 100 psi with an elaborate > > test rig using a dial indicator was only 0.012", about a hundredth of > > an inch. > > > Incidentally, it is highly unlikely that pressure and a 700x25 tire's > > rise have the straight linear relationship that some people suggest, > > at least not over a 40 to 130 psi range with an 88 pound load. > > > I expect that beautifully smooth graphs of pressure and rise are just > > as theoretical, oversimplified, and mistaken as similar predictions > > for contact patch area, which does not follow the often-mentioned > > tire-pressure/load = area prediction. > > > (Actual measurements of contact patches show that they fail to shrink > > as much as predicted with higher pressures and fail to expand as much > > as predicted with lower pressures. They stubbornly stick to a > > preferred size.) > > Interesting data from Fogel Labs again; thanks, Carl. I don't have the > original article about the 15% drop figure right here to read again. > IIRC the 15% drop figure came from research by Jim Papodopolous, but how > it was arrived at I don't know. It's a number that looks suspiciously > arbitrary (how many things in the world work out to such a nice even > number?).- Hide quoted text - > > - Show quoted text - Thanks for the information, Carl. I think the original article came from Bicycling magazine. Currently you can get a version of it on the roadbikerider.com website in the member's section. Smokey
|
| |
Date: 28 Sep 2007 21:27:22
From: Tim McNamara
Subject: Re: some tire drop data
|
In article <oltqf3td4ftpiiishe2kcimc8vhn5jfq6n@4ax.com >, carlfogel@comcast.net wrote: > Some recent threads and an email today led me to measure how much a > fairly typical tire rose as I inflated it. > > The results suggest that some people are fooling themselves. > > First, some data from dial indicator measurements of a loaded tire's > height rise in thousandths of an inch from 40 to 130 psi, measured > from top of nominal 700x25c tire. > > Tire measured 1.020" to 1.030" wide at top between 80 to 130 psi. > > Total load on tire was ~85 pounds, resting on a scale, but the tire > stood on a concrete floor for the test. > > inch mm inch > psi rise rise change > --- ----- ---- ------ > 40 0.000 0.00 n/a > 50 0.021 0.53 0.021 > 60 0.038 0.97 0.017 > 70 0.052 1.32 0.014 > 80 0.063 1.60 0.011 > 90 0.076 1.93 0.013 > 100 0.088 2.24 0.012 > 110 0.100 2.54 0.012 > 120 0.110 2.79 0.010 > 130 0.120 3.05 0.010 > > My quick and dirty table shows only that the top of a roughly 1.020" > wide tire rose ~0.120 inches from 40 to 130 psi under an 88 pound > load. > > For those interested in arbitrary figures, 15% of a tire width of > 1.020" is 0.153", or 3.9 mm. > > Presumably, someone could measure the diameter of a tire hanging in > mid-air and inflated to some arbitrary pressure, load it with some > weights, measure the change, and call it tire drop. > > But this would require such an elaborate test rig that it would be > extremely difficult for any normal poster to even _try_ to make a > practical measurement of the difference between 90 psi and 100 psi. > > After all, the difference between 90 and 100 psi with an elaborate > test rig using a dial indicator was only 0.012", about a hundredth of > an inch. > > Incidentally, it is highly unlikely that pressure and a 700x25 tire's > rise have the straight linear relationship that some people suggest, > at least not over a 40 to 130 psi range with an 88 pound load. > > I expect that beautifully smooth graphs of pressure and rise are just > as theoretical, oversimplified, and mistaken as similar predictions > for contact patch area, which does not follow the often-mentioned > tire-pressure/load = area prediction. > > (Actual measurements of contact patches show that they fail to shrink > as much as predicted with higher pressures and fail to expand as much > as predicted with lower pressures. They stubbornly stick to a > preferred size.) Interesting data from Fogel Labs again; thanks, Carl. I don't have the original article about the 15% drop figure right here to read again. IIRC the 15% drop figure came from research by Jim Papodopolous, but how it was arrived at I don't know. It's a number that looks suspiciously arbitrary (how many things in the world work out to such a nice even number?).
|
| | |
Date: 29 Sep 2007 10:43:42
From: Jean
Subject: Re: some tire drop data
|
"Tim McNamara" <timmcn@bitstream.net > wrote in message news:timmcn-C026A2.21271728092007@news.iphouse.com...
|
| | |
Date: 28 Sep 2007 21:24:36
From:
Subject: Re: some tire drop data
|
On Fri, 28 Sep 2007 21:27:22 -0500, Tim McNamara <timmcn@bitstream.net > wrote: >In article <oltqf3td4ftpiiishe2kcimc8vhn5jfq6n@4ax.com>, > carlfogel@comcast.net wrote: > >> Some recent threads and an email today led me to measure how much a >> fairly typical tire rose as I inflated it. >> >> The results suggest that some people are fooling themselves. >> >> First, some data from dial indicator measurements of a loaded tire's >> height rise in thousandths of an inch from 40 to 130 psi, measured >> from top of nominal 700x25c tire. >> >> Tire measured 1.020" to 1.030" wide at top between 80 to 130 psi. >> >> Total load on tire was ~85 pounds, resting on a scale, but the tire >> stood on a concrete floor for the test. >> >> inch mm inch >> psi rise rise change >> --- ----- ---- ------ >> 40 0.000 0.00 n/a >> 50 0.021 0.53 0.021 >> 60 0.038 0.97 0.017 >> 70 0.052 1.32 0.014 >> 80 0.063 1.60 0.011 >> 90 0.076 1.93 0.013 >> 100 0.088 2.24 0.012 >> 110 0.100 2.54 0.012 >> 120 0.110 2.79 0.010 >> 130 0.120 3.05 0.010 >> >> My quick and dirty table shows only that the top of a roughly 1.020" >> wide tire rose ~0.120 inches from 40 to 130 psi under an 88 pound >> load. >> >> For those interested in arbitrary figures, 15% of a tire width of >> 1.020" is 0.153", or 3.9 mm. >> >> Presumably, someone could measure the diameter of a tire hanging in >> mid-air and inflated to some arbitrary pressure, load it with some >> weights, measure the change, and call it tire drop. >> >> But this would require such an elaborate test rig that it would be >> extremely difficult for any normal poster to even _try_ to make a >> practical measurement of the difference between 90 psi and 100 psi. >> >> After all, the difference between 90 and 100 psi with an elaborate >> test rig using a dial indicator was only 0.012", about a hundredth of >> an inch. >> >> Incidentally, it is highly unlikely that pressure and a 700x25 tire's >> rise have the straight linear relationship that some people suggest, >> at least not over a 40 to 130 psi range with an 88 pound load. >> >> I expect that beautifully smooth graphs of pressure and rise are just >> as theoretical, oversimplified, and mistaken as similar predictions >> for contact patch area, which does not follow the often-mentioned >> tire-pressure/load = area prediction. >> >> (Actual measurements of contact patches show that they fail to shrink >> as much as predicted with higher pressures and fail to expand as much >> as predicted with lower pressures. They stubbornly stick to a >> preferred size.) > >Interesting data from Fogel Labs again; thanks, Carl. I don't have the >original article about the 15% drop figure right here to read again. >IIRC the 15% drop figure came from research by Jim Papodopolous, but how >it was arrived at I don't know. It's a number that looks suspiciously >arbitrary (how many things in the world work out to such a nice even >number?). Dear Tim, I don't know where the popular 15% figure came from or from what point it was measured--just the weight of the front end of a bike resting on the tire is about 10% of the typical load. Maybe there's a clever measuring technique that was somehow confirmed by rolling resistance tests, but it's going to have to be awfully clever to handle differences of about a hundredth of an inch between 90 and 100 psi. Here's a link that came in my email to a chart that graphs some kind of 15% drop correlated with pressure for various tire sizes: http://bp1.blogger.com/_d-Yj0VDKhWQ/RnwLUoA9sVI/AAAAAAAAAJ4/225Ym5qYKTo/s1600-h/BQ_berto_inflationgraph.jpg The graph seems to show some simple linear calculations, not actual data points from measurements. I suspect that actual measurements would show the straight line theory is simply wrong. If a tire were a piston in a car engine, then adding a 100 pound weight (including piston) would drive a piston with an area of 1 square inch down until the pressure rose to 100 psi. But a bicycle tire is significantly different from a metal piston in a metal chamber. When we load a tire, the bottom of the "piston chamber" is the reasonably unyielding metal rim. But on either side of the tire, the C-shaped tire walls deform outward against the air pressure trying to keep them in their current shape about an inch wide on a 25 mm tire, the minor axis of the toroid (your thumb and finger going around the tire). Meanwhile, the tire also deforms, much less against the gigantic "O" of the tire viewed the long way around, the major axis of the toroid. In effect, the tire is more like a long, narrow trampoline than the top of a metal piston, and it distorts in a much more complicated fashion. This is why the simple area = load / pressure load equation fails to predict what happens at high and low pressures for 700c x 25 tires. The same explanation probably applies to how high a tire rises or falls according to pressure. The sides of the tire act as scissors jacks, applying increasing pressure at the edge of the contact patch as pressure drops and the sides deform more and more. As I've mentioned before, what made me suspicious of simple straight-line predictions was knowing that a 4-inch wide trials tire can carry a 400-pound load in a wheelstand down a paved road with only 4 psi and no way to spread out more than 4 inches. The oversimplified equation predicts: 400 lbs / 4 (lbs/inch^2) = 100 ^ 2 = 4 x 25 inches But a 4 x 18 trials tire is only about 26 inches in diameter, so a 25 inch long contact patch is impossible. The sidewalls had to be exerting considerable force as springs, which seems reasonable, since it takes considerable force to pull (or push) the wall of an inflated tire out of shape. Cheers, Carl Fogel
|
| | | |
Date: 29 Sep 2007 12:23:39
From: Tim McNamara
Subject: Re: some tire drop data
|
In article <upfrf3pifcnmdij94geh0vv2o6hi171p4r@4ax.com >, carlfogel@comcast.net wrote: > On Fri, 28 Sep 2007 21:27:22 -0500, Tim McNamara > <timmcn@bitstream.net> wrote: > > >In article <oltqf3td4ftpiiishe2kcimc8vhn5jfq6n@4ax.com>, > > carlfogel@comcast.net wrote: > > > >> Some recent threads and an email today led me to measure how much > >> a fairly typical tire rose as I inflated it. > >> > >> The results suggest that some people are fooling themselves. > >> > >> First, some data from dial indicator measurements of a loaded > >> tire's height rise in thousandths of an inch from 40 to 130 psi, > >> measured from top of nominal 700x25c tire. > >> > >> Tire measured 1.020" to 1.030" wide at top between 80 to 130 psi. > >> > >> Total load on tire was ~85 pounds, resting on a scale, but the > >> tire stood on a concrete floor for the test. > >> > >> inch mm inch > >> psi rise rise change > >> --- ----- ---- ------ > >> 40 0.000 0.00 n/a > >> 50 0.021 0.53 0.021 > >> 60 0.038 0.97 0.017 > >> 70 0.052 1.32 0.014 > >> 80 0.063 1.60 0.011 > >> 90 0.076 1.93 0.013 > >> 100 0.088 2.24 0.012 > >> 110 0.100 2.54 0.012 > >> 120 0.110 2.79 0.010 > >> 130 0.120 3.05 0.010 > >> > >> My quick and dirty table shows only that the top of a roughly > >> 1.020" wide tire rose ~0.120 inches from 40 to 130 psi under an 88 > >> pound load. > >> > >> For those interested in arbitrary figures, 15% of a tire width of > >> 1.020" is 0.153", or 3.9 mm. > >> > >> Presumably, someone could measure the diameter of a tire hanging > >> in mid-air and inflated to some arbitrary pressure, load it with > >> some weights, measure the change, and call it tire drop. > >> > >> But this would require such an elaborate test rig that it would be > >> extremely difficult for any normal poster to even _try_ to make a > >> practical measurement of the difference between 90 psi and 100 > >> psi. > >> > >> After all, the difference between 90 and 100 psi with an elaborate > >> test rig using a dial indicator was only 0.012", about a hundredth > >> of an inch. > >> > >> Incidentally, it is highly unlikely that pressure and a 700x25 > >> tire's rise have the straight linear relationship that some people > >> suggest, at least not over a 40 to 130 psi range with an 88 pound > >> load. > >> > >> I expect that beautifully smooth graphs of pressure and rise are > >> just as theoretical, oversimplified, and mistaken as similar > >> predictions for contact patch area, which does not follow the > >> often-mentioned tire-pressure/load = area prediction. > >> > >> (Actual measurements of contact patches show that they fail to > >> shrink as much as predicted with higher pressures and fail to > >> expand as much as predicted with lower pressures. They stubbornly > >> stick to a preferred size.) > > > >Interesting data from Fogel Labs again; thanks, Carl. I don't have > >the original article about the 15% drop figure right here to read > >again. IIRC the 15% drop figure came from research by Jim > >Papodopolous, but how it was arrived at I don't know. It's a number > >that looks suspiciously arbitrary (how many things in the world work > >out to such a nice even number?). > > Dear Tim, > > I don't know where the popular 15% figure came from or from what > point it was measured--just the weight of the front end of a bike > resting on the tire is about 10% of the typical load. Having reviewed the article in Bicycle Quarterly ("Optimizing your tire pressure for your weight," BQ Vol 5, Issue 4), the 15% figure was referenced from an article by Frank Berto (2004. Under Pressure. Australian Cyclist March/April 2004, p. 48). I was not able to locate the article on the Australian Cyclist Web site. Heine measure loads on three bicycles with a rider seated on the bike (randonneur bike, racing bike and "city" bike) and found that the weight distribution (% front/rear) was 45/55, 30/60 and 35/65 respectively. Heine's article references that a 15% "tire drop" is recommended by "several tire manufacturers." That statement is cited to Berto's original article and no more specifics are given. The chart with the nice linear relationships is also attributed to Berto. > Maybe there's a clever measuring technique that was somehow confirmed > by rolling resistance tests, but it's going to have to be awfully > clever to handle differences of about a hundredth of an inch between > 90 and 100 psi. My guess would be to load the wheel through a frame or fork and measure the drop from the top of the tire to the ground, with an arrangement similar to how a doctor measures your height. It would indeed need to be carefully calibrated, perhaps some dial arrangement could be used. > Here's a link that came in my email to a chart that graphs some kind > of 15% drop correlated with pressure for various tire sizes: > > > http://bp1.blogger.com/_d-Yj0VDKhWQ/RnwLUoA9sVI/AAAAAAAAAJ4/225Ym5qYKT > o/s1600- h/BQ_berto_inflationgraph.jpg > > The graph seems to show some simple linear calculations, not actual > data points from measurements. > > I suspect that actual measurements would show the straight line > theory is simply wrong. That is the graph included on p 29 of BQ 5,4. It is attributed to Berto's article in Australian Cyclist. Without being able to read the Berto article, there isn't any way to understand how Berto came to create that graph. I agree that such a linear relationship between inflation and tire drop is probably unlikely, although it's conceivable that the relationship might be close to linear within the normal range in inflation pressures. If you look at the Avocet rolling resistance data, the curve of the relationship between inflation pressure and rolling resistance flattened quite a bit at higher pressures. Maybe you can Google up the source article by Berto, or perhaps there is some Australian or New Zealand or Tasmanian participant in the newsgroup who can provide that information.
|
| | | | |
Date: 29 Sep 2007 14:24:21
From:
Subject: Re: some tire drop data
|
On Sat, 29 Sep 2007 12:23:39 -0500, Tim McNamara <timmcn@bitstream.net > wrote: [snip] >> Maybe there's a clever measuring technique that was somehow confirmed >> by rolling resistance tests, but it's going to have to be awfully >> clever to handle differences of about a hundredth of an inch between >> 90 and 100 psi. > >My guess would be to load the wheel through a frame or fork and measure >the drop from the top of the tire to the ground, with an arrangement >similar to how a doctor measures your height. It would indeed need to >be carefully calibrated, perhaps some dial arrangement could be used. [snip] Dear Tim, In this case, the more careful the measurement, the sillier the results will be. The resolution is already too fine for the subject. There's no medical point, for example, to weighing normal adults to within even a pound, since our weight easily varies that much according to our last meal and visit to the bathroom. (The doctor does not bother to have you remove your wallet, much less strip naked, when he weighs you on his balance scales, accurate to 4 ounces.) The "15%" tire drop figure implies resolution to at least 5% increments, meaning that we think that 15% is what we want, not 10% or 20%. But when compared to 10% or 20% drops, a "15% drop" means only that somewhere between 70 and 110 psi is a good idea for tire inflation. We hardly need elaborate charts and dial indicators to tell us that. Here's how it works. If a 1.000" wide tire has an absolute 15% drop of 0.150" at 90 psi under an 85-lb load (don't ask me how to actually measure that with repeatable accuracy), then we can apply my relative rise measurements for a ~1" tire under the same conditions: 10 psi theoretical measured absolute relative psi drop drops --- -------- -------- x 20% 0.200" 20% 130 0.194" +0.010" 120 0.184" +0.010" 110 0.174" +0.012" \ 100 0.162" +0.012" \ 90 15% 0.150" 0 > 15% range 80 0.137" -0.013" / 70 0.126" -0.011" / 60 0.112" -0.014" x 10% 0.100" 10% 50 0.091" -0.021" Here's my test data again: inch inch psi rise change --- ----- ------ 40 0.000 n/a 50 0.021 0.021 60 0.038 0.017 70 0.052 0.014 80 0.063 0.011 90 0.076 0.013 100 0.088 0.012 110 0.100 0.012 120 0.110 0.010 130 0.120 0.010 In any case, inflated tires are not round to within a hundredth of an inch. They're just slapped on aluminum rims and inflated, with the beads creeping out roughly regularly, and then the sides bend as the tire is loaded. What's bending in clinchers are bias-laid plies of threads, with 66 to 170 threads per inch: 66 tpi 0.015" 127 tpi 0.008" 170 tpi 0.006" A single thread is roughly as thick as a measured 10 psi difference in tire drop. Cheers, Carl Fogel
|
| | | | |
Date: 29 Sep 2007 11:58:36
From: * * Chas
Subject: Re: some tire drop data
|
"Tim McNamara" <timmcn@bitstream.net > wrote in message news:timmcn-2DA059.12233429092007@news.iphouse.com... > In article <upfrf3pifcnmdij94geh0vv2o6hi171p4r@4ax.com>, > carlfogel@comcast.net wrote: > > > On Fri, 28 Sep 2007 21:27:22 -0500, Tim McNamara > > <timmcn@bitstream.net> wrote: > > > > >In article <oltqf3td4ftpiiishe2kcimc8vhn5jfq6n@4ax.com>, > > > carlfogel@comcast.net wrote: > > > > > >> Some recent threads and an email today led me to measure how much > > >> a fairly typical tire rose as I inflated it. > > >> > > >> The results suggest that some people are fooling themselves. > > >> > > >> First, some data from dial indicator measurements of a loaded > > >> tire's height rise in thousandths of an inch from 40 to 130 psi, > > >> measured from top of nominal 700x25c tire. > > >> > > >> Tire measured 1.020" to 1.030" wide at top between 80 to 130 psi. > > >> > > >> Total load on tire was ~85 pounds, resting on a scale, but the > > >> tire stood on a concrete floor for the test. > > >> > > >> inch mm inch > > >> psi rise rise change > > >> --- ----- ---- ------ > > >> 40 0.000 0.00 n/a > > >> 50 0.021 0.53 0.021 > > >> 60 0.038 0.97 0.017 > > >> 70 0.052 1.32 0.014 > > >> 80 0.063 1.60 0.011 > > >> 90 0.076 1.93 0.013 > > >> 100 0.088 2.24 0.012 > > >> 110 0.100 2.54 0.012 > > >> 120 0.110 2.79 0.010 > > >> 130 0.120 3.05 0.010 > > >> > > >> My quick and dirty table shows only that the top of a roughly > > >> 1.020" wide tire rose ~0.120 inches from 40 to 130 psi under an 88 > > >> pound load. > > >> > > >> For those interested in arbitrary figures, 15% of a tire width of > > >> 1.020" is 0.153", or 3.9 mm. > > >> > > >> Presumably, someone could measure the diameter of a tire hanging > > >> in mid-air and inflated to some arbitrary pressure, load it with > > >> some weights, measure the change, and call it tire drop. > > >> > > >> But this would require such an elaborate test rig that it would be > > >> extremely difficult for any normal poster to even _try_ to make a > > >> practical measurement of the difference between 90 psi and 100 > > >> psi. > > >> > > >> After all, the difference between 90 and 100 psi with an elaborate > > >> test rig using a dial indicator was only 0.012", about a hundredth > > >> of an inch. > > >> > > >> Incidentally, it is highly unlikely that pressure and a 700x25 > > >> tire's rise have the straight linear relationship that some people > > >> suggest, at least not over a 40 to 130 psi range with an 88 pound > > >> load. > > >> > > >> I expect that beautifully smooth graphs of pressure and rise are > > >> just as theoretical, oversimplified, and mistaken as similar > > >> predictions for contact patch area, which does not follow the > > >> often-mentioned tire-pressure/load = area prediction. > > >> > > >> (Actual measurements of contact patches show that they fail to > > >> shrink as much as predicted with higher pressures and fail to > > >> expand as much as predicted with lower pressures. They stubbornly > > >> stick to a preferred size.) > > > > > >Interesting data from Fogel Labs again; thanks, Carl. I don't have > > >the original article about the 15% drop figure right here to read > > >again. IIRC the 15% drop figure came from research by Jim > > >Papodopolous, but how it was arrived at I don't know. It's a number > > >that looks suspiciously arbitrary (how many things in the world work > > >out to such a nice even number?). > > > > Dear Tim, > > > > I don't know where the popular 15% figure came from or from what > > point it was measured--just the weight of the front end of a bike > > resting on the tire is about 10% of the typical load. > > Having reviewed the article in Bicycle Quarterly ("Optimizing your tire > pressure for your weight," BQ Vol 5, Issue 4), the 15% figure was > referenced from an article by Frank Berto (2004. Under Pressure. > Australian Cyclist March/April 2004, p. 48). I was not able to locate > the article on the Australian Cyclist Web site. > > Heine measure loads on three bicycles with a rider seated on the bike > (randonneur bike, racing bike and "city" bike) and found that the weight > distribution (% front/rear) was 45/55, 30/60 and 35/65 respectively. > > Heine's article references that a 15% "tire drop" is recommended by > "several tire manufacturers." That statement is cited to Berto's > original article and no more specifics are given. The chart with the > nice linear relationships is also attributed to Berto. > > > Maybe there's a clever measuring technique that was somehow confirmed > > by rolling resistance tests, but it's going to have to be awfully > > clever to handle differences of about a hundredth of an inch between > > 90 and 100 psi. > > My guess would be to load the wheel through a frame or fork and measure > the drop from the top of the tire to the ground, with an arrangement > similar to how a doctor measures your height. It would indeed need to > be carefully calibrated, perhaps some dial arrangement could be used. > > > Here's a link that came in my email to a chart that graphs some kind > > of 15% drop correlated with pressure for various tire sizes: > > > > > > http://bp1.blogger.com/_d-Yj0VDKhWQ/RnwLUoA9sVI/AAAAAAAAAJ4/225Ym5qYKT > > o/s1600- h/BQ_berto_inflationgraph.jpg > > > > The graph seems to show some simple linear calculations, not actual > > data points from measurements. > > > > I suspect that actual measurements would show the straight line > > theory is simply wrong. > > That is the graph included on p 29 of BQ 5,4. It is attributed to > Berto's article in Australian Cyclist. Without being able to read the > Berto article, there isn't any way to understand how Berto came to > create that graph. I agree that such a linear relationship between > inflation and tire drop is probably unlikely, although it's conceivable > that the relationship might be close to linear within the normal range > in inflation pressures. If you look at the Avocet rolling resistance > data, the curve of the relationship between inflation pressure and > rolling resistance flattened quite a bit at higher pressures. > > Maybe you can Google up the source article by Berto, or perhaps there is > some Australian or New Zealand or Tasmanian participant in the newsgroup > who can provide that information. Years ago I remember seeing a study that showed only a 5%-7% reduction in rolling resistance between 85 PSI and 110 PSI in a variety of tires. Since most of us rode sewups in goathead country we opted to run at 85 PSI to 90 PSI and got a lot less flats ( refer to Carl Fogel's test of tire puncture vs. pressure ). Chas.
|
|