10
$\begingroup$

I was having a discussion with a friend about rolling various shapes, in particular what shape is the worst at rolling. I thought that a triangular wheel might be particularly bad at rolling while he suggested a thin rectangular wheel, approaching a plate, might be even worse as a wheel.

We discussed this rather informally for a while, but we wanted to back it up with some maths, however we couldn't really think of a measure that captures the idea of "How easily an wheel roll". So I thought I would ask some people that are better at maths. What measure might be used to express the idea of being easy or hard to roll? Is there any literature on rolling/wheel construction?

To get it started a bit I'll cover all the ideas discussed between my friend and myself. If these are unhelpful or off topic tell me and I'd be happy to edit them out.

The only way we could think to measure how easy it is too roll was to put a shaped wheel on a flat surface with a very high coefficient of friction and tilt it until the wheel begins to roll. We would then take the angle as the measurement of roll-ability. This seems pretty nice because it can be verified to some degree of accuracy empirically (however it seems quite difficult to get the conditions pristine enough that the shapes will roll rather than slide). In addition circular wheels will begin to roll pretty much at angle 0, and everything will roll by the time the surface is vertical, this is pretty nice because it gives a finite range allowed. The problem we saw with this is that it only measures the difficulty required to get the shape to begin to roll.

For example in the plate idea suggested by my colleague takes quite a deal of effort to start up but once it begins to roll it will roll a half rotation quite easily because the thin edge is very unstable, at the same time the triangular shape will probably take significantly less effort to start rolling, but once it moves it will only roll a 3rd of a rotation before returning to the start.

An additional discussion we had was on what should be kept constant when comparing two shapes. Our initial thought was to keep the cross sectional are of the wheels constant when comparing them, however as we thought about it more we thought it might be a good idea to compare shapes with equal perimeter, because shapes with larger perimeter tend to roll farther. My friend pointed out however that concave shapes have perimeter that does not contribute to how far it rolls, we perhaps distance rolled per rotation would be a better measure.

$\endgroup$
10
  • $\begingroup$ It may help to consider the moment of inertia of a particular shape, and see what shapes have the greatest M.o.I. $\endgroup$ Commented Jul 3, 2017 at 15:36
  • 1
    $\begingroup$ Perhaps another measure of difficulty-to-roll is how much the center of mass of the object moves up and down when rolled. A pin-like shape may be very hard to roll under this metric. $\endgroup$ Commented Jul 3, 2017 at 15:39
  • $\begingroup$ Also, I would suggest setting certain parameters, as an object with arbitrarily large mass and size would not roll at all. $\endgroup$ Commented Jul 3, 2017 at 15:40
  • 2
    $\begingroup$ Reminds me of a cartoon of cavemen about to invent the wheel. The old version uses squares as wheels, but then the great inventor switches to triangular wheels: One bump less, much better. $\endgroup$ Commented Jul 3, 2017 at 15:40
  • 2
    $\begingroup$ The cartoon was "B.C.", drawn by Johnny Hart. That's long been one of my favorites ever to appear in that strip. $\endgroup$ Commented Jul 3, 2017 at 15:54

1 Answer 1

Reset to default
6
$\begingroup$

Summary: Your friend is correct in that a rectangular wheel with their sides having ratio 2.95241:1 or larger, requires even more energy to push per unit length than a triangle, if after every push the wheel turns only a quarter turn (rectangular wheel) or third turn (equilateral triangle wheel), and comes to a full stop. A flat bar (one side essentially zero) requires the most energy to push per unit length.

For simplicity, let's assume the case where the shape only turns to the next vertex, and does not continuously roll. This means we can ignore the forward momentum, and only need to look at the potential energy changes.

(In practice, some of the potential energy is converted to forward kinetic energy, meaning only the initial push needs to overcome the potential energy barrier, and the following pushes only need to "top up" the potential/kinetic energy to keep the average speed constant. This is a bit too complex to go into detail here; consider asking this at physics.stackexchange.com to go "beast mode" into the actual details.)

Let $r$ be the minimum and $R$ the maximum radius of the wheel. (For regular polygons, these are the radii of the inscribed and circumscribed circles, with all three centers the same.)

At rest, the height of the axle from ground is $r$. To push the wheel to the next vertex, we need to push the center of weight over the vertex, and that means we need to push so that the axle lifts to $R$ from ground. After the tipping point, the wheel turns, but slow enough to stop at the next minimum energy state, turning all that potential energy into waste heat.

The change in potential energy in a near Earth gravity field is $\Delta E = m g h$, where $m$ is the mass, $g$ is the free fall acceleration, and $h$ is the change in altitude (distance from the center of Earth).

If we define our wheel so that it has perimeter 1, to push the wheel forwards 1 unit we need to do work $$E = m g n (R - r)$$ where $n$ is the number of vertices in the wheel.

(For a circular wheel, $R = r$, so there is no potential energy change. Remember, we are ignoring the forward momentum, friction, and recovery of forward kinetic energy from potential energy; in this sense, pushing a circular wheel is "effortless", $E = 0$.)

For an equilateral triangle with side length $1/3$ (thus perimeter $1$), $R = 1/\sqrt{27}$ and $r = 1/\sqrt{108}$. Thus, $$E_3 = m g \; 3 \left ( \frac{1}{\sqrt{27}} - \frac{1}{\sqrt{108}} \right ) = m g \frac{1}{\sqrt{12}} \approx 0.288675 m g$$

For a square wheel with side length $1/4$ (thus perimeter $1$), $R = 1/\sqrt{32}$ and $r = 1/8$, $$E_4 = m g \; 4 \left ( \frac{1}{\sqrt{32}} - \frac{1}{8} \right ) = m g \left ( \frac{1}{\sqrt{2}} - \frac{1}{2} \right ) \approx 0.207107 m g$$

A pentagonal wheel has side length $1/5$ (thus perimeter $1$), $R = 2/\sqrt{250 - \sqrt{12500}} \approx 0.170130$, $r = 1/\sqrt{500 - \sqrt{200000}} \approx 0.137638$, and $$E_5 = m g \; 5 \left ( \frac{2}{\sqrt{250 - \sqrt{12500}}} - \frac{1}{\sqrt{500 - \sqrt{200000}}} \right ) \approx 0.162460 m g$$

For a rectangular wheel with one side $w$, $0 \le w \le 1/2$, and the other side $1/2 - h$ (so perimeter is $2 h + 2 w = 1$), $R = \sqrt{w^2 + h^2}/4$, $r_1 = w/2$, $r_2 = h/2$, $E(0) = E_2$, i.e. $$E_4(w) = \left ( \sqrt{ 8 w^2 - 4 w + 1 } - \frac{1}{2} \right ) m g$$ which reaches maximum at $w = 1/2$, $h = 0$ (or $w = 0$, $h = 1/2$), i.e. a flat bar as the wheel, with $$E_2 = m g \frac{1}{2} = 0.5 m g$$and minimum at $w = h = 1/4$, a square wheel, $E_4 = (1/sqrt(2) - 1/2 ) m g \approx 0.207105 m g$.

Interestingly enough, a rectangular wheel with sides $1/4 \pm \sqrt{\sqrt{3} - 1} / \sqrt{48}$ ($\approx 0.373495, 0.126505$) requires the same effort as an equilateral triangle wheel does. This means that a rectangular wheel with side ratio larger than $(1/4 + \sqrt{\sqrt{3} - 1}/\sqrt{48})$:$(1/4 - \sqrt{\sqrt{3} - 1}/\sqrt{48})$ requires more energy to push per unit length, than a equilateral triangle wheel does.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.