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While I was riding my bike funnily enough I wondered if there was an optimal size for a bike wheel. So I suppose that is my question, What is the optimal size for a bike wheel? If the bike accelarates from rest over a bumpy road?

Here's my thinking: It is going to take a certain amount of energy to accelerate the wheels from rest, given by the equation: $$E_r=\frac{1}{2}Iw^2$$ So if we were to increase the size of the radius by two in order to determine the effect of increasing or decreasing the radius in general. This would do two things:

It would decrease the angular velocity ($w$) by two for the same linear velocity because...

$$v=wr$$

$$v=w(2)$$

$$w=\frac{1}{2}v$$

Although the angular velcocity decreases, this is more than made up for from the increase in the moment of inertia. It would increase the moment of intertia by about 8 because $I$ is given by $$I=mr^2$$ And the raidus would increase by two, increasing the circumference by two, so doubling the mass. $$I=(2)(2)^2$$ $$I=8$$ If we put this into our original equation: $$E_r=\frac{1}{2}Iw^2$$ $$E_r=\frac{1}{2}(8)(\frac{1}{2})^2$$ $$E_r=1$$ In comparison to the initial $$E_r=\frac{1}{2}(1)(1)^2$$ $$E_r=\frac{1}{2}$$

So the energy required to accelerate the bike has a direct relationship with the radius.

So bike wheels should be made smaller smaller for the most efficient increase in speed, but the smaller the wheels get the more energy is lost whn gong over bumps (or at least I assume so). But how do you calculate this energy lost and how do you compare it with the energy to accelerate in oder to optimize it? Thanks!!

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    $\begingroup$ There is no energy lost due to bumps if both the bike and the cyclist are rigid. Going over a bump reduces the kinetic energy of the bike and the cyclist on the way up and increases it by the same amount on the way down. The only loss mechanism is the compression of the cyclist's tissue, which, for a well trained athlete is probably not a major problem. One could minimize that with a suitable elastic carriage with high quality springs, but I have a feeling that no serious cyclist would want to ride that kind of bike. $\endgroup$ – CuriousOne May 1 '15 at 8:11
  • $\begingroup$ I suspect wheels are sized to accommodate the largest size of bumps and potholes (i.e. surface irregularities) they are expected to encounter successfully. Otherwise they'd be as tiny as they can economically be made. Rider comfort is a factor that may be harder to model mathematically but I guess you could plug in some empirical values for tolerable vertical accellerations. If your model has perfectly smooth surfaces it won't determine optimal sizing in the real world. $\endgroup$ – RedGrittyBrick May 1 '15 at 9:36
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    $\begingroup$ The usual symbol for angular velocity is a Greek omega ($\omega$), which is different from the Latin double-you ($w$). $\endgroup$ – rob May 1 '15 at 10:36
  • $\begingroup$ There's also a gear-ratio effect between the rear cog and the wheel diameter. A smaller wheel would require a larger chainring to get the overall gear ratio from the cranks to the wheels back into the same range as conventional bikes --- which seem to be pretty well optimized. $\endgroup$ – The Photon May 2 '15 at 5:53
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Without a guideline on what you're wanting to optimize for specifically, this question is hard to answer. Different types of tires/wheels are better for different things, and differ in several different ways. Some general considerations to take into account, then:

When riding on a hard surface, a narrower tire is preferred due to its lighter weight, more aerodynamic profile and lesser deformation (which saps energy from the forward motion). A narrow tire tends to cut into soft surfaces, though, and requires a higher internal pressure to support the weight of the rider, so for trail/mountain biking a wider tire is preferred, as the larger contact patch handles a soft surface better and the lower filling pressure allows the tube to take shocks and obstacles without bursting.

A wheel with a larger diameter transitions over bumps and dips more easily and covers more ground per revolution, and so is better for most applications where speed and stamina are major contributing factors to success. A wheel with a larger diameter also has more inertia and generates stronger gyroscopic forces, so particularly in stunt bikes smaller diameter wheels are used.

A thicker tire (rim-to-face measurement) allows for more deformation at the same pressure compared to other tires, which contributes to a larger contact patch and thus better traction and control, particularly on slick or bumpy surfaces. For this reason you'll see them often on bikes designed for leisurely cruises. The large deformation reduces the efficiency of converting the rider's power into forward motion though, so you'll only see them on such leisure bikes; performance-oriented designs will use a thinner tire.

And finally, some tires are "flat-free," meaning they are permanently filled with a rubber foam to support the shape rather than pressurized air, and thus cannot go flat. They are much heavier than pneumatic tires, so they take more effort to accelerate and turn. That's why performance bikes still use the slightly more problematic but lighter pneumatic tires.

A final note: the optimal tire size for all purposes is infinite. With infinitely large tires, all the other factors are irrelevant, since as soon as you climb on the bike, your bike is at your destination and you can climb off again. Such tires are prohibitively costly though.

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  • $\begingroup$ I disagree with this. If you have an infinitely large tire then it will take an infinite amount of energy to accelerate... I'm looking for an equation that takes both the bumpiness factor and the acceleration factor into account. You could use some constant to describe the relationship between the bumpy ground and the tire. $\endgroup$ – Me2 May 2 '15 at 20:46

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