Originally posted by oldbobcat
And the front is always about 5 psi less than the rear.
Originally posted by pj_s
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?
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.pj_s said:my tires - 7.5 bar rear and 7.2 bar front
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?
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.ScienceIsCool said: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.
Dietmar said: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.
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.ScienceIsCool said: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.
That link is no good. Here's the new one:JSCORNO said:Check out this link from FITWERX. They explain it quite well.
http://www.fitwerx.com/NewFiles/BicycleRollingResistance.html
artmichalek said: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.
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...ScienceIsCool said:Nope, not nonsense.
Of course, that is obvious. So what?ScienceIsCool said: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.
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...ScienceIsCool said: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.
So, we agree?ScienceIsCool said: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.
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.ScienceIsCool said:I suspect the reason is that the rubber itself becomes the limiting factor and dominates the term associated with a flexing tire case.
That is irrelevant for the argument. All we need is energy conservation.artmichalek said:You and Deitmar are arguing in circles around this one because you're both missing the time component.