Tyre pressure?

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

  1. Vo2

    Vo2 Member

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    What is the correct pressure for road racing tyres in general?
     
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  2. Guest

    Guest Guest

    G'day VO2 good to see another post.

    Firstly do you have a pressure guage? Next question is do you have a good pump- because without a good pump you will not be able to acheive a high pressure.
    Also check the rating on the tread wall- on good quality tyres the pressure is a lot higher than cheaper tyres.
    In my experience, if training, the tyre will be ok if hard to the touch ie- can't push tyre in with the fingers. When racing it would be a good idea to check tyre pressure.
    But don't become paranoid about checking each time you ride/train, a quick pinch will be tell if tyres are hard.
    As tyres age they become softer and will feel softer even at the right pressure. Only very occasionally will you overinflate a tyre, and if your still pumping when it explodes then wow! are you in for a shock. :D

    Another thing to remember is that hard tyres will almost stop pinch flats.
    Any probs then just post, OK? Will check back and see if this helps.
    regards, sillystorm
     
  3. Vo2

    Vo2 Member

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    Hiya, sillystorm. Yeah, I bought myself a floorpump with gauage yesterday and was surprised to find both my tyres @ about 50 psi (350 kpa). And this after I've been pumping myself breathless with a handpump. I've been told that about 70 psi (490 kpa) is the general pressure that roadies go for. My Bianchi is fitted with Vredenstein tyres and tubes. The tyres have a max pressure rating of 125 psi.
    I've always used the thumb-press test to check tyre pressure after a good handpump session, but now I know it's way below what it should be.
     
  4. Guest

    Guest Guest

    A good track pump is a must have for any cyclist ;)
    Around 120psi is a good pressure for training and racing IMO.

    cheers!
     
  5. Vo2

    Vo2 Member

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    WHOAW! 120? That's rock hard! Doesn't the ride become a bit too bumpy then?
     
  6. Guest

    Guest Guest

    110-130 psi is what mine are normally at, 110psi for 23mm tyers 130+psi for 20mm tyers.

    later!
    johno
     
  7. oldbobcat

    oldbobcat Well-Known Member

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    Deepends on tire, the road surface, and the rider's weight, guys. For most 700c road wheels, I'd recommend somewhere between 90 and 125 psi. Big guys, smooth roads, and light, narrow tires at the high end of the range, and small women, rough roads, and stiff tires at the low end. And the front is always about 5 psi less than the rear.
     
  8. Evo

    Evo New Member

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    Why is that?
     
  9. rek

    rek New Member

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    Most of the weight is borne by the rear wheel.
     
  10. pj_s

    pj_s New Member

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    my tires - 7.5 bar rear and 7.2 bar front
    no idea what that would be in psi.
    sometimes for racing I put in a little more, although I have heard conflicting stories on whether this actually reduces rolling resistance.

    I've read some tests that show that some tire types have a sort of optimal pressure (around 7-8 bar). Pump them up harder and the rolling resistance increases - I would expect it to decrease - Anyone have ideas on this?
     
  11. JSCORNO

    JSCORNO New Member

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  12. dhk

    dhk New Member

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    One Bar equals 14.5 psi, so you're at 104/108 psi F/R. My current tire, Michelin Pro Race, says 6 bar min, 8 max (87/116 psi). I used to weigh 210, and ran all my tires to max or beyond. Since I dropped 40 lbs this year, discovered the ride is better at 100-105 psi. Doesn't seem to hurt rolling resistance at all.

    Concerning resistance increasing with increasing pressure, possible explanation would be that on rough (real world) road surfaces, hard tire would be "buzzing" on the coarse surface rather than smoothly rolling over the texture. If the hard tire is inducing vibration energy into your fork/frame, maybe that would cause extra drag.....OK, I'm reaching here.....

    Dan
     
  13. RussB

    RussB New Member

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  14. ScienceIsCool

    ScienceIsCool New Member

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    Rolling resistance is caused by the flexing of the tire casing. The more it has to flex, the more energy you lose. Therefore, in a perfect environment (i.e., something smooth like the surface of a track) higher pressure is better. This is why track cyclists run at ridiculously high pressures.

    However, the roads we all ride on are not smooth like a track. They have cracks, bumps, grit, pebbles, etc. For a second, let's assume your tires are pumped way up to minimize rolling resistance. As you roll over the small bumps, the tire does not deflect. It's rock hard. You did not lose any energy to tire flex. You and the bike were pushed upwards maybe a half of a millimeter. Not much, that's true. But the energy required to push you upwards is equal to your mass times the acceleration due to gravity times the distance.

    If you and your bike weigh 80 kg in total, and the acceleration due to gravity is ~10 m/s^2 and the distance is 0.0005 meters, then you just lost 0.4 Joules of energy. That's quite a lot. In fact, it's a lot more energy lost than if you let some air out of your tires and let the tires flex a bit to soak up that 0.5 mm bump.

    So depending on the roads you ride and the rolling resistance of your tires, there's an optimum tire pressure to use.

    I hope that helps. If it's not clear, it's because I didn't explain it very well so I'll clarify things if anyone wishes.

    John Swanson
    www.bikephysics.com
     
  15. Dietmar

    Dietmar New Member

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    That's pretty much nonsense, given the fact that after you have cleared the crest of the bump in the above example, you will exactly recover the energy you just calculated above (unless the bump is a step, in which case no amount of tire pressure will help you recover the energy "lost", which has been converted to potential energy). On the more general point (of increasing rolling resistance beyond an "optimal" pressure), I have never heard of that, and I strongly doubt that that is correct. All the data I have ever seen suggest a monotonous decrease of C_rr with tire pressure.
     
  16. ScienceIsCool

    ScienceIsCool New Member

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    Nope, not nonsense. What I failed to mention in the first post is that it's your forward momentum (kinetic energy) that is used to clear the bump. On the way back down, that potential energy does not get converted back into forward momentum. It is lost in deflections and vibrations of you and the bike. The result is a net loss of speed. Since the variations and deflections are smal, it simply feels like added rolling resistance.

    And yes, you are right that Crr will reach a minimum monotonically (for those not math inclined, it means that it will approach a certain value without ever quite getting there) with increasing air pressure. I suspect the reason is that the rubber itself becomes the limiting factor and dominates the term associated with a flexing tire case.

    John Swanson
    www.bikephysics.com

     
  17. artmichalek

    artmichalek New Member

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    You're defining asymptotic, not monotonic. In this case, the behavior is neither. On an ideally smooth surface, rolling resistance will decrease asymptotically to zero. In the real world resistance decreases until the tire starts riding up and down on surface irregularities, at which point it begins to increase again. You and Deitmar are arguing in circles around this one because you're both missing the time component. Just throwing out forward momentum doesn't mean anything if you can't say where it goes. While the net potential energy that goes into riding up onto a bump is fully recoverable, it needs to be recovered in the same period of time it was expended in order to result in zero net power loss. When an overinflated tire hits a surface irregularity, it accelerates upward much more quickly than gravity can pull it back down as it clears the top of the bump.
     
  18. DMF

    DMF New Member

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  19. ScienceIsCool

    ScienceIsCool New Member

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    Absolutely right in that the function is both asymptotic and monotonic (i.e., increasing or decreasing in the same direction). I screwed that one up. Can I blame that on lack of coffee? No? Okay, then. I'll just feel really dumb for the rest of the day. :)

    However, we can still talk about this in terms of energy and momentum conservation. We don't necessarily have to talk about it in terms of power. The net result is still that you are changing KE to PE, back to KE which is not lossless in that each has an efficiency associated with it. Energy is still conserved, but it is transduced to vibration, heat and noise. You lose some forward momentum. We donot have to express this as power transfer since that is misleading.

    Hope I didn't stick my foot in my mouth again...

    John Swanson
    www.bikephysics.com

     
  20. Dietmar

    Dietmar New Member

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    Yup, complete nonsense. You know, what you wrote is wrong on a number of levels. Why don't we just take your example at face value: You claim that, with an overinflated tire, the rider in your example looses about half a Joule on each half-millimeter bump. I really wonder if you ever thought about what this would mean if it were true: First off, a half millimeter is really very little, so you are talking about a perfectly smooth road; even on that "perfect" road, we would expect certainly a couple of hundred of such "bumps" for every meter of the road covered. So, let's say it's just 100 bpm (bumps per meter). That means that you loose 40 J/m, or the equivalent of a 5 cm climb per meter, or roughly a 5% incline. Now let's extend that argument to more realistic bumps, which are of the order of a few millimeters, and your rider would feel like his tires are glued to the road...

    Of course, that is obvious. So what?

    That is just speculation. First of all, yes, it is clear that for energy to get lost in this scenario, you need some damping in the system (so, those mysterious "deflections" and "vibrations" have to be damped out entirely), be that through inelastic deformations (of the tire usually, but we have ruled that out, or frame or wheels, none of which have significant damping at the deformations that we are talking about), or inherent damping in gelatinous materials (the rider...). The latter one cannot be significant either: Let's take your example again, using 100 bpm of the road, and a rider going 10 m/s (roughly 22 mph). You would then be arguing an energy dissipation of a measly 400 W in the rider. No further comment necessary, I hope...

    So, we agree?

    If you are trying to say that for the overinflated tire, more energy is dissipated in the rubber, that could indeed be a possibility leading to non-monotonous behavior.

    That is irrelevant for the argument. All we need is energy conservation.
     
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