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Everyone knows that racing bikes have narrower wheels and that, by definition, friction reduces speed. What I don't understand is that since at no time is the wheel of the bike in horizontal motion relative to the road, surely any friction between the wheel and the road will have no negative effect and in fact aid the forward propulsion of the bike as a whole?

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    $\begingroup$ I don't understand your question. $\endgroup$ – PiKindOfGuy Mar 10 '19 at 19:22
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    $\begingroup$ You're conflating friction and rolling resistance. $\endgroup$ – Gert Mar 10 '19 at 19:24
  • $\begingroup$ It's worth pointing out (see my comment to an answer) that the behaviour of tyres in contact with the road is pretty complicated, and that generally parts of the 'contact patch' -- the bit of the tyre in contact with the road -- are moving with respect to the road in general. $\endgroup$ – tfb Mar 11 '19 at 0:00
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when a rubber tire's tread surface comes into contact with the ground as it rotates, the rubber deforms or "squishes" against the pavement surface and then "unsquishes" when that portion of the tire rotates out of contact with the pavement. This means that as any tire rotates, the rubber in it is getting dynamically flexed continuously, and since rubber is not perfectly elastic, part of that deformation work gets transformed into heat.

This in turn means that it takes work to rotate a rubber tire, to make up for the hysteresis losses in the rubber itself.

How to minimize these losses? First, if you pump more air into the tire, you reduce the rolling deformation and hence the hysteresis losses. Second, if you reduce the amount of rubber in contact with the road by narrowing down the tire, you can reduce those losses. This is why tires on racing bikes are very skinny and have 120PSI air inside them.

Third, if you design the tire so as to minimize the squish, you also minimize the losses. This is why radial-belted tires furnish better gas mileage than bias-ply tires. This is also why high-speed tires have large diameter and narrow width, compared to (for example) tractor or wagon tires which have small diameter and broad width (this also increases their load-carrying capacity- in an application where high-speed flexure is not a problem).

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  • $\begingroup$ That's very helpful. Just to clarify, does that mean that the reason for the reduced wheel width is in fact nothing to do with reducing friction? $\endgroup$ – Avi Brynin Mar 10 '19 at 21:21
  • $\begingroup$ +1 for the nice answer! $\endgroup$ – the_candyman Mar 10 '19 at 22:10
  • $\begingroup$ This is only one of the (at least) two mechanisms involved: another is that, as the material of the tyre deforms, parts pf the contact patch get scrubbed against the road surface. This causes losses which are directly frictional. The fix is the same: reduce the size of the contact patch by reducing the deformation of the tyre. These scrubbing losses are directly frictional though (I'd argue that hysteresis losses really are too in fact). $\endgroup$ – tfb Mar 10 '19 at 23:56
  • $\begingroup$ @AviBrynin, no, the narrow wheel is to reduce rolling friction too. $\endgroup$ – niels nielsen Mar 11 '19 at 0:27
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Suppose that $\vec{F}_1$ is the force produced by the biker and $\vec{F}_2$ is the friction due to the floor.

Let $m$ be the mass of the wheel, $I$ its moment of inertia, $R$ its radius, $a$ its acceleration along $x$ and $\alpha$ its angular acceleration.

We know that:

$$\begin{cases} ma =F_{1,x} - F_{2,x}\\ I\alpha = -RF_{2,x}\\ a + R\alpha = 0 \end{cases}.$$

The equation $a + R\alpha = 0$ means that the wheel does not slip on the floor.

We can find the three unknown of such system, namely $a$, $\alpha$ and $F_{2,x}$:

$$\begin{cases} a =\displaystyle\frac{F_{1,x}R^2}{mR^2 +I}\\ \alpha = -\displaystyle\frac{F_{1,x}R}{mR^2 +I}\\ F_{2,x} = \displaystyle\frac{F_{1,x}I}{mR^2 +I}\\ \end{cases}.$$

As expected, a "narrower" wheel, i.e., smaller $m$ and hence a smaller $I$, implies an higher acceleration $a$. Indeed, the denominator of $a$ depends on both $m$ and $I$.

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    $\begingroup$ The last statement seems a bit misleading as it suggests that $m$ and $I$ independently affect acceleration. I think you should remove "smaller I" from it because a narrower wheel does not imply a smaller $I$, or maybe specify that the smaller $I$ is because of smaller $m$. As an example, a hollow cylinder and a ring with same radius and same mass will have the same $I$. $\endgroup$ – Eagle Mar 10 '19 at 20:26
  • $\begingroup$ @Natasha I mean "a smaller $m$ and hence a smaller $I$". Thanks for the remark, I'm going to fix it. $\endgroup$ – the_candyman Mar 10 '19 at 20:35

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