Tyre pressure?

Discussion in 'Cycling Equipment' started by Vo2, Aug 12, 2001.

  1. ScienceIsCool

    ScienceIsCool New Member

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    Nope, not nonsense. :) I've been trying to oversimplify things a bit. I'm not actually trying to suggest that all 0.4 Joules gets distributed as "lost" energy. You get some of it back (most? I don't know.) by rolling back down the bump. What I'm trying to say and should have said from the beginning is that you lose a portion of that energy. This is in addition to rolling resistance. I believe there is a certain pressure at which the total resistance (Crr + losses due to travelling over uneven surfaces) is optimized.

    I'm willing to discuss this. I'm willing to learn and to admit my mistakes. But your tone seems a bit harsh. I'm not willing to continue if you're feeling defensive and angry. Besides, we seem to actually agree on most of the issues.

    One last point is that both of us may be wrong about the model we choose to describe the behaviour, but I believe the conclusions are correct. Higher pressures for smoother surfaces and an optimum pressure exists to minimize total rolling resistance.

    John Swanson
    www.bikephysics.com
     


  2. Dietmar

    Dietmar New Member

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    Allright allright, not nonsense. Just wrong.:p
    Still not happy? How about: Inaccurate? ;)

    Yeah, that is perfectly fine with me. However, how much of that energy you might lose is anyone's guess. What is clear is that it is a relatively small portion (I am intentionally vague here, but it must be less, maybe much less, than 10%), at which point you need to talk about where that portion goes. As long as you do not have a mechanism, and a plausible estimate for the amount of energy that this mechanism dissipates, there is nothing to talk about.

    While that may well be true, another question is whether this happens at a realistic pressure. So, if you find out that there is an optimal pressure of 376.5 psi for tire X, riding on a surface with Gaussian-distributed roughness elements of a specified shape, with an RMS roughness height of 0.5 mm, then that would be of limited interest... :D

    So, you admit that your example was "a mistake"? :D

    I do apologize if I came across that way. I assure you that I am not at all angry. Probably should have put a generous sprinkling of smilies in my post above.

    Yes, I think so, too.
     
  3. ScienceIsCool

    ScienceIsCool New Member

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    Thanks, Dietmar. I wanted to be sure everything is cool and I'm glad it is.

    You know, I bet you and I could collaborate on some really awesome experiments. Something along the lines of rolling resistance vs tire pressure for a fixed surface and/or optimum tire pressure vs Ra (Ra = average surface roughness) for a fixed load. What do you think?

    Oh, and I still think that additional tire deflection (plus vibration of the entire system) could account for the losses I described. Where would be the fun if we agreed on everything?!

    John Swanson
    www.bikephysics.com
     
  4. Dietmar

    Dietmar New Member

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    Well, yes, in principle, but:

    - I really don't have the time to spare for anything but some token support right now
    - we are talking about some highly non-trivial work if it's going to be of any use at all. I think we pretty much agree that the rolling resistance caused by the effects we were debating above depends not only on the tire itself, but also on the roughness properties of the road, the wheel, fork/frame, and even the rider (and his/her position on the bike...). That is one hell of a set of experiments we are talking about there. You could cut that short by testing a wheel on different surfaces only, of course, but in that case I really doubt the results would be of much practical use.
    - On practical use, I also suspect that the results will be of little interest anyway. Let's say we even come up with something like: "For a rider of such and such weight, the optimal pressure is x (with x a small number) psi lower than the maximal tire pressure. Compared to C_rr for maximum tire pressure, this will reduce the rolling resistance of the wheel by 0.1%." Who is going to care? Who is going to care once we tell people "BUT, other factors that we have not looked at will probably have a larger effect on your C_rr"? Closing the zipper on your jersey will probably have a several times larger effect than this modification of the tire pressure...
     
  5. ScienceIsCool

    ScienceIsCool New Member

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    Dietmar, you are very correct. Excellent points. Still, I might be interested in performing a few tests just out of curiosity. If I do, will you mind if I get your opinion on the setup, etc? If nothing else, it'll give us some numbers to debate. :)


     
  6. Dietmar

    Dietmar New Member

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    No, not at all. It will be my pleasure.
     
  7. 87Trek

    87Trek New Member

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    I dont have a floor pump (i know I should get one). But I pump my tires up to 85psi and check the tires with a thumb press before every ride. I think I have a slow leak in one of my tubes (need to pump up often) even though it is a brand new tube.
     
  8. Dietmar

    Dietmar New Member

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    How slow? What kind of tube? Latex tubes typically can slowly loose some of their pressure over a few days.
     
  9. 87Trek

    87Trek New Member

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    I am not exactly sure what type of tube it is because my buddy gave it to me and installed it. I believe it is a Bontrager with a long stem if that helps. If I fill the tube up fully, it is stiff and I can barely push the tube with my thumb. The next day, it will be really easy to press the tube in.
     
  10. Dietmar

    Dietmar New Member

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    That's too much loss of air. You should replace the tube.
     
  11. ajo

    ajo New Member

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    No Dietmar, ScienceIsCool is correct:
    - You will not recover exactly the same energy. (Just try to first ride on the asphalt, and the on the graved shoulder of the road.) Jumping up and down will always slow you down, much more than deflection of tires does.
    - Deflection over bumps is the reason why rubber tyres filled with air has been soo popular amongst car drivers and cyclists the last 100 years. You loose this advantage if you pump of the tires too hard. (Only trains uses solid wheels - running on iron track.)

    But it feels faster with more air. (Just as it feels faster to drive an old car with harsh suspension in 80km/h than a new Mercedes in 90km/h)
     
  12. mikesbytes

    mikesbytes New Member

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    I seem to be much less affected by tyre pressure than I was even a year ago. Same tyres and bike but a fitter and slightly lighter rider.
     
  13. 11ring

    11ring New Member

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    Do you guys know that Schwalbe have tested this theory for off road tyres and found lower rolling resistance on wider (not sure the relevance of wider, but my theory is the wider the tyre and contact patch the smoother the road will be if you take the ground height as the average of the contact patch), and lower psi tyres. Using a power meter to measure work and riding the same distances at slow speeds, over and over again. And also on controlled surfaces.

    Another interesting point to look at is the angle at which bumps hit the tyre. The bigger the wheel, the more the bump will transmit force upwards, as opposed to backwards. Or create lift as opposed to drag. This is obviously more significant as the bumps get bigger. When the size of the bump equals the radius of the wheel there is no lift-just a very big drag force which brings you to a sudden stop. Lift is slightly recoverable- drag i dont think is at all. Hence the theory of the 29'er being faster than the 26 inch.

    This angle of force also helps determine the recovery of energy. If you ride up a bump (very small ramp) then you are converting forward kinetic energy into potential energy stored in the tyre and potential energy in lifting the rider. To recover that potential energy you need to ride back down something like a ramp as well. But falling back onto a flat surface does not transfer the upward force from the ground on impact into thrust.

    The force will be roughly perpendicular to the surface at the contact point, so only negative gradients (downramps) create thrust.

    To test this ride off a kerb. That reduction in potential energy from falling does not increase your forward velocity in the same way (at all) as riding down a ramp/hill.

    Now why do we always hit ramps and fall down onto flats, not ride up ramps and ride down ramps? Well actually we don't.

    But to get lift you have to have a positive gradient of the surface you are momentarily riding. I.e all bumps have a positive gradient when you are hitting them.

    But there is nothing special about points to land. The tyre may come back down on the flat, on another bump (positive gradient), or, if your lucky, onto a downward ramp. The average of all these gradients will be close to zero as all three posibilities are almost equally probable at high speeds. (The smaller the bump and the lower the velocity and the shallower the gradient the greater chance you will roll back down the other side of the ramp/bump, but at high speeds and with sharp bumps you will always clear the back edge of the bump and therefore land on a random gradient. (actually the up ramp of the next bump seems like the most likely point to hit)

    Note- more compliant tryes will create less lift and will deform more and therefore increase the chance of contacting with the other side of the bump (the down ramp) which is the only object which can be used to recover the potential energy transferred by lifing the bike and the rider.

    Also energy recovery would also seem to be higher at lower pressures, as an elastic body like the tyre is going to have less of a damping effect than a body like a loose rider which probably has close to 100% damping.

    On another note vertical compliance in your frame should also help reduce rolling resistance, as bump impacts will be at lower force levels and transmit less energy into lifting the big mass, the rider.
     
  14. Dietmar

    Dietmar New Member

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    Uhmm, you are a bit late to this particular party; ScienceIsCool and I have pretty much sorted out our differences. There is no doubt about the accuracy of the statements I made. Specifically, for there to be any effect at all, there must be damping present in the system of tire, wheel, frame, and rider. If tire damping itself is reduced by increasing air pressure, total rolling resistance will decrease unless some other component has increased damping.

    Of course, in your gravel example, we now have the road added as an active component of the system, which now also has non-trivial mechanical properties, and adds damping if gravel is moved against frictional forces as you roll over it.

    More generally speaking, it is clear that such effects become more important as you ride over very rough terrain, such as the one typically encountered by mountain bikers in an off-road setting. In that case it is fairly well established that lowering the tire pressure can indeed reduce total rolling friction. Overall, if a mountain biker rides over extremely rough terrain, typically out of the saddle, and actively smoothing out really large bumps, then the picture becomes significantly more complex. In that case, one might even start a heated discussion about the effects of the popular spring-damped suspensions that many mountain bikes now feature. I'd rather not go there...
     
  15. artmichalek

    artmichalek New Member

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    The statements are only accurate if the system is massless.
     
  16. Dietmar

    Dietmar New Member

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    That's a rather curious statement, given that in that case there is no energy to conserve.
    Would you like to elaborate and point out any inaccuracies you perceive?
     
  17. artmichalek

    artmichalek New Member

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    Energy has to go into accelerating the mass of the system upward when the tire hits a bump. As stated previously, this is not necessarily recovered by gravitational acceleration and turned back into forward kinetic energy on the way back down, particularly if the tire loses contact with the road. Your comments thus far have also neglected the energy required to accelerate the mass of the system laterally when the tire contacts a bump obliquely. My apologies if I'm wrong, but your denial of these effects led me to assume that you're working with a massles system.
     
  18. Dietmar

    Dietmar New Member

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    Yeah, you're wrong, and apology accepted. :D

    See, as I am sure you know and appreciate, the beauty of energy conservation laws is that I don't need to worry about the kinds of details you mention. Just as you can present the most wondrously complicated contraption to me, claiming it's a perpetuum mobile, and I can tell you straight to your face that you are wrong, without wasting my time having to examine the details of what all those cute little levers, wheels and pulleys in your machine may do.

    So, in the case of our bicycle, if we have the following scenario: At time t0 a bicycle with rider, total mass m, speed of their center of gravity v, the effective inertial moment of the rotating wheels I, altitude of the center of gravity h0. This system will have the total energy E0=1/2 m v^2 + 1/2 I \omega^2 + m g h0, with g the gravitational acceleration, and \omega the rotational speed of the wheels (\omega=v/r, r the radius of the wheel, if both wheels are identical).

    Now we imagine the bicycle has traveled some distance on a level road, so that at time t1, we still have the same altitude h1=h0. The conclusion is that, unless the system of bicycle plus rider has some internal damping mechanism that dissipates energy, at time t1 the system will still have the energy E1=E0. Note that this statement is exact, and does not depend on any details of the motion of the bicycle between t0 and t1. In the above, all I have left out are additional degrees of freedom of the bike motion, such as vibration. Thus, the implicit assumption is that those contributions do not change between t0 and t1, which is again exact under the assumption of a steady-state system (so, the bicycle does not somehow vibrate more at t1 than it did at t0; this seems to correspond to reality, at least for my bike ;)).
     
  19. artmichalek

    artmichalek New Member

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    1) I thought that time was irrelevant?
    2) In order for this to correspond to reality for your bike, it's either magical or just not moving very quickly. In reality, energy is added to the vibration of the system every time your tires hit a bump. Fortunately, the bike has a big viscoelastic blob on the saddle to dissipate this.
     
  20. Dietmar

    Dietmar New Member

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    Yes, of course it is. I use times t0 and t1 just as labels.

    That is not correct; it depends on the details of the motion of the bike and structure of the road whether or not energy is added to or subtracted from the vibrational modes. I am not prepared to present a detailed analysis of the statistics of this process right now. Feel free to do so, if you are so inclined, and have the time to spare...
    However, depending on the structure of the eigenvalue spectrum of your system, random excitation (with a finite-energy excitation spectrum) will not lead to unbounded growth in vibrational energy in general. You will find that for just about all real mechanical systems (rather than theoretical idealisations) this will be true.

    So, how is this different from what I have been saying? Of course, if you do have visco-elastic (or other forms of) damping, then energy is not conserved.
     
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