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Everyone knows (citation needed) that (at fixed rim diameter) tyres with smaller section require less effort to move around (at least on a paved road).

TL; DR. Why?

A bit of context.

I always thought this phenomenon is due to the size of the contact patch between each tyre and the road. Notably, the area of this patch decreases when you decrease the tyre cross section. [This assumes that a tyre with smaller cross section will have a higher minimum pressure, which is generally true]. Friction is proportional to the contact area, QED.

However, I recently considered that friction (as almost everyone knows) would generate a torque that causes the wheel to spin faster [note 1]. So I had to come up with an alternative explanation. My best attempt is this. If tyre A has a smaller cross section than tyre B, tyre A will have a higher manufacturer pressure than tyre B, and therefore will be less subject to deforming. In terms of conservation of energy, deforming a tyre continuously requires a fair amount of energy, and it's here that our precious kinetic energy goes when we stop pedalling and our expensive toys (or in my actual case, inexpensive toy) come to a sad halt. So tyre A sucks less energy than tyre B, and therefore requires less effort. How can we put this in terms of forces? There must be an asymmetry in the forces near the contact point, causing a torque that slows down the spin of the wheel. Can you describe this asymmetry?

[Note 1. Friction is a force applied at the contact patch, with direction opposite to the direction of motion. Therefore, friction generates a torque that spins the wheel faster. For instance, in absence of any friction, a wheel would slide seamlessly without rolling.]

[EDIT. I apolgise for the sloppy formulation of the question. I edited to add clarifications where necessary. I opted for adding text rather than removing because some comments would otherwise look out of place]

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    What do you mean by "friction [...] would generate a torque that causes the wheel to spin faster"? Could you elaborate a bit more on how you mean that? – Benedikt Bauer Dec 7 '15 at 7:54
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    @Carel Equal pressure has equal contact. P/SI X SI = P – paparazzo Dec 7 '15 at 9:34
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    Skinny tires generally have thinner, less knobby treads, and knobby treads are a real drag on power. – Daniel R Hicks Dec 7 '15 at 13:05
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    @ frisbee: schwalbe.com/en/rollwiderstand.html – Carel Dec 7 '15 at 17:09
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    It explains the correlations between rolling resistance, tyre width and pressure. – Carel Dec 7 '15 at 18:37
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Your opening claim:

Everyone knows (citation needed) that (at fixed rim diameter) tyres with smaller section require less effort to move around (at least on a paved road).

Is actually not true. Your next claim isn't true either. The contact patch area for a tire will be nearly the same regardless of what width tire is used, for a given pressure. If I have a tire that is pumped up to 100psi, and the bike and me weigh 200lbs, the contact patch between both wheels will be nearly 2 square inches. Pressure is force divided by area.

Your third claim is also not true, as static friction is roughly proportional only to the normal force, and it doesn't even matter, as when the contact patch is touching the ground, it's not even moving!

By varying the width of the tire, while the contact area stays constant, the shape is different, notably the patch gets shorter as the tire gets wider. This translates to less vertical displacement of the rubber in the tire as the wheel rolls. Ross Millikans answer describes, the deflection in rubber is the primary source of tire-related rolling resistance.

It turns out, that for fixed pressure, wider tires have less rolling resistance than narrow tires.

The key element isn't the width of the tire, it's the higher pressure. It is much easier to make a small high-pressure tire than a wide one, for reasonable weights and costs. Furthermore, because making a high-pressure wide tire is more challenging, it may require measures such as thicker treads that increase the energy cost of tire deflections such that it is a net negative over more deflection of software rubber, and so on.

I'm not even going to address your torque theory because you are obviously confused as to the forces at play.

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    I'm not even going to address your torque theory because you are obviously confused as to the forces at play. Would you expand more on that if I said please? – astabada Dec 8 '15 at 2:55
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    Come on from the beginning the question has had "If tyre A has a smaller cross section than tyre B, tyre A will have a higher manufacturer pressure than tyre B, and therefore will be less subject to deforming." OP is clearly states a high pressure tire deforms less "therefore will be less subject to deforming". The question is about an asymmetry of forces and this answer does nothing to address that. – paparazzo Dec 8 '15 at 15:03
  • As for high pressure big tire - it the bead and casing that hold the pressure - not more rubber. Force is pressure times volume. Neither the rim not bead could handle large high pressure tire. And there is no purpose to a large high pressure tire. – paparazzo Dec 8 '15 at 15:05
  • @Frisbee: a larger pressure vessel requires stronger or thicker material, i.e. you would need more rubber. See wiki page. You would have to address everything to make a high pressure fat tire. – whatsisname Dec 8 '15 at 18:37
  • Material - not rubber. Rubber is a terrible material for containing force as it stretches (e.g. balloon). My comment was very clear on that "it is the bead and casing that hold the pressure - not more rubber". A tire worn to the cord still holds pressure just fine. And again, there is no purpose to large high pressure tire. – paparazzo Dec 8 '15 at 18:54
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The loss is friction in deformation of the rubber of the tire. When you flex the tire from round (when not on the ground) to flat (when touching the ground) there is heat generated. If the tire is narrower and the pressure is higher, there is less rubber involved in the flexing. It is true that higher pressure means less contact patch, but that is not important here. This is an energy, not a torque argument.

If you want to talk about torques, I don't understand why you think friction should accelerate the wheel. I would say the part of the tire just coming into contact produces a retarding torque and the part leaving contact produces an accelerating torque. Both torques are reduced because the (almost vertical) line of action is almost along the wheel radius. The retarding torque is greater due to the losses mentioned above. This is really the same as the energy argument above, as it must be. The losses are the same, whichever way you look at it.

  • There are probably also arguments to be made for higher pressures lending itself to less deformation. – Deleted User Dec 7 '15 at 15:30
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One answer I've never seen for this question is the moment of inertia of the fatter tire is greater than that of a skinny tire.

Even at low pressure, a fatter tire will hold more air than a skinny tire. This means there is a larger mass of air that needs to be spun - this is why fatter tires require more effort. At equal pressure there will be even more air in the fat tire compared to the skinny one.

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    Moment of inertia is only a factor when (de)accelerating and even then it's not even that much of a factor. Fatter tires require more effort because of the pressure. – whatsisname Dec 7 '15 at 23:13
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    @whatsisname more correctly "Fatter tires require more effort because of the [lack of] pressure" – Aron Dec 8 '15 at 1:24
  • It depends on the pressure. If you look at recommend pressure you would find there is an equal amount of air in various size tires. – paparazzo Dec 8 '15 at 13:47
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    This comment train has extended long beyond what it should have. If you want to discuss it further goto Bicycles Chat. – whatsisname Dec 10 '15 at 17:52
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    Comments are not for extended discussion; this conversation has been moved to chat. – jimchristie Dec 11 '15 at 12:53
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This question is confusing with some misstatements but I am going to try your "asymmetry of forces". If this is not what you mean on "asymmetry of forces" then I suggest you clarify as many people have tired to answer the question in a number of ways.

Work is done on the rubber to deform it. Work is the integral force in the direction of the work. Work and heat have the same unit of measure.

Yes conservation of energy applies.

On the front edge work (integral of f x ds) is performed on the tire to displace it. The tire is not like a spring where the work is given back on tailing edge. Virtually all the work is translated to heat (friction in the rubber). Heat is more intense than you would think - it takes a lot of work to produce a heat. You get a very small spring back from the rubber back on the tail end. The heat does not just accumulate as temperature - the heat is then transferred to the atmosphere.

The deformation on the front end also does work on the air in the tire. On the tail end the air in the tire does work tire to push it back out. Most of this work is given back. Some of this work on the air in the tire is translated to random kinetic energy (temperature) and that equates to work not given back on the tailing edge. The other problem there is on the tail end due to the delay most of this work pushing back is not even against the road to propel the tire. It is is work done on the rubber and atmosphere.

A tire with more displacement has more resistance. That resistance is more more heat. There is conservation of energy.

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    Energy and work have the same units (joule for SI), and work is the integral of force times displacement along a path. Work and energy are very closely related, as work is a form of energy, as is heat. – whatsisname Dec 8 '15 at 21:05
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At constant velocity, instantaneous equilibrium, the moment of inertia indeed won't have any effect on pedal effort. However, if there are any accelerations then the moment of inertia of a fat heavy tire will play a significant role. Since velocity changes are common even and especially with the Eddy Merckx types, in general the moment of inertia of a bicycle tire+wheel matters with respect to pedal effort.

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