# How does one change distance travelled through wheel size?

To increase the velocity of/distance travelled by an object with wheels, let's say for example a toy car *, which is more preferable: increasing or decreasing wheel size? The aim is to make the object travel at a greater speed (and therefore a greater distance) in a given time frame**.

Imagine the scenario on a frictionless surface (and no other factors that may affect the object other than gravity).

I know that increasing the wheel size will allow the object to cover a greater distance with each revolution of its wheels and therefore the object will travel further in the timeframe. I also know that smaller wheel size means a smaller circumference and more revolutions per minute (rpm), which will increase the acceleration of the object, allowing it to travel faster, which could also in turn increase the distance travelled in the given timeframe (?).

Which of these methods will increase the distance travelled in the timeframe, and if both will, which one will be more effective?

Notes:

*In other words, an object with wheels only; no motor, as the force applied will be from a human source
**I only have the resources to test one of these options so I am calculating the better option rather than testing both. I will conduct that experiment by allowing the object to travel until it stops naturally, then I will scale the distance to m/s.

• @CountTo10 Sorry I don't understand what you mean - could you please elaborate? – BPA-Free Plastic Water Bottle Sep 6 '16 at 8:02
• @CountTo10 Sorry for probably not elaborating properly, but the scenario is just a flat surface with the vehicle/object rolling/moving across it. – BPA-Free Plastic Water Bottle Sep 6 '16 at 8:32
• @CountTo10 Ah yes, that is what I meant. Sorry for the ambiguity :| – BPA-Free Plastic Water Bottle Sep 6 '16 at 8:51
• One effect of having larger/heavier wheels is that they will store more rotational momentum (much like a flywheel) when being pushed at the same velocity. This should help the car travel further, at least on a rough surface. – Jeff Sep 6 '16 at 9:37
• "Imagine the scenario on a frictionless surface"... "no other factors that may affect the object other than gravity"... In this case, you don't need wheels at all! If the surface is level with no friction or air resistance then the car will never stop regardless of wheel size. – James Sep 6 '16 at 12:37

You should decrease the wheel size and try to make them lighter. Why?
When you input power into the vehicle, then you have a trade-off between whether to have more kinetic energy in the rotation of the wheels or in the translation of the vehicle. You obviously want the translation energy to be higher so you should choose wheels of less moment of inertia, i.e., less mass and less radius.

$K.E. = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

This is precisely the reason why when a hoop($I=MR^2$), a solid sphere($I=\frac{2}{5}MR^2$) and a hollow sphere($I=\frac{2}{3}MR^2$) all of equal mass and radius are made to run down an incline the solid sphere reaches the bottom first and the hoop reaches last.

You don't state the time frame, but you could apply the kinematic equations of motions for both wheel sizes, assuming you know the ratio of acceleration between the two wheels.

I have assumed a level surface and a fixed gear ratio.

I only have the resources to test one of these options so I am calculating the better option rather than testing both. I will conduct that experiment by allowing the object to travel until it stops naturally, then I will scale the distance to m/s.

It really all depends on how accurately you want to measure the effect of wheel diameter. But if you try various values of acceleration, it may be clear that here is an obvious advantage in one wheel size from your calculations.

So if you obtain the acceleration figures, you can do it all on the equations, also F = ma will feature as well.

The above laws apply, and the angular version of them, is listed below.

${\displaystyle \omega _{\mathrm {f} }=\omega _{\mathrm {i} }+\alpha t\!}$

${\displaystyle \theta _{\mathrm {f} }-\theta _{\mathrm {i} }=\omega _{\mathrm {i} }t+{\tfrac {1}{2}}\alpha t^{2}}$

${\displaystyle \theta _{\mathrm {f} }-\theta _{\mathrm {i} }={\tfrac {1}{2}}(\omega _{\mathrm {f} }+\omega _{\mathrm {i} })t}$

${\displaystyle \omega _{\mathrm {f} }^{2}=\omega _{\mathrm {i} }^{2}+2\alpha (\theta _{\mathrm {f} }-\theta _{\mathrm {i} }).}$

I am sure there is a easy equation that links wheel diameter to acceleration, but it might be more complicated than that if weight is an issue, as moments of inertia may then be involved. It depend on the scale/mass of the car

Best of luck with it.

This can be described by following equations:

$a$ of car = tangential $a$ at wheels = angular $a \cdot r$

Replacing alfa (angular $a$) with $\frac{\mathrm{Torque}}{I}$ gives us, $a= \frac{T}{I} \cdot r$ , notice changing $r$ would change both $r$ and $I$, since $I$ is directly proportional to $r^2$. So the final answer would be increasing $r$ will increase $a$ , but value of $I$ should be taken into account for optimal acceleration.

• Welcome on Physics SE :) Thank you for your input. You might want to see here physics.stackexchange.com/help/notation for help with typesetting formulas. – Sanya Oct 16 '16 at 20:53
• I've tried to fix up the mathjax because you haven't done so after 2 years. Please take the effort to write good mathjax, it really improves readability. – user191954 Sep 13 '18 at 9:50
• "notice changing $r$ would change both $r$ and $I$, since $I$ is directly proportional to $r^2$" does not make any sense at all; that's the strangest math I have ever seen. – user191954 Sep 13 '18 at 9:50

## protected by AccidentalFourierTransformJun 14 '18 at 1:50

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