I am trying to understand quite naively wheels in a specific framework.


We normally think of squares and circles as different concepts of shapes, but I am reframing it that both are polygons with equal sides. The difference being that a circle is a polygon with infinite equal sides and a square is a polygon with four equal sides.

So imagine you have a horse and cart. The cart and its weight is 100kg. The cart has two wheels. In the first case, the cart has square wheels, in the second case round wheels.

Let's say the horse pulls the cart over a distance which would make the square wheel do ten complete turns. Call that distance D.


1) The force that that horse has to pull with in order to make the cart move at a consistent speed across D with

A) the square wheel = polygon, equal sides, number of sides =4

B) the round wheel = = polygon, equal sides, number of sides = infinite

What would that formula be?

I'd like to know what the universal formula is, referencing the number of sides of the polygon.

Then in the case of the round wheel with its infinite sides, simplification of the formula would give I assume a more well-known formula for wheel motion (I don't know what this is)

(Intuitively, it's clear that there's a far greater load on the square wheel and that also if we were to plot the graph of the force that the horse needs to pull, it would be a maximum right when there is a transition to the side of the square being flat on the ground and it would be a minimum when the line between the center of the square and the corner of the square is 90 degrees to the ground surface. So I imagine the formula would give a wave, and the more/ less sides of the polygon, the more "intense" the wave.. as the number of sides tends to infinity- ie. the wheel is more circular in shape- the wave would dampen to be an absolute number).
Guidance would be greatly appreciated!).

2) For the above, what the relation to the required maximal axle load? Conceptually and /or formulaically.

(I'd be happy to know what the formula is but I'd really like to understand the concept better).

***Intuitively, it's clear that at any point the surface area of the wheel touching the ground- whether the wheel is a circle or square- is exactly the same.

NB I am not a physicist- I do understand mathematical concepts and high school physics. Having said that, I'd really be grateful for any explanation that is built ground up rather than quoting things like torque.


  • $\begingroup$ "What would that formula be? I'd assume it's the same formula and that with with the round wheel it's simplified"....the limiting case of a polygon is a circle, as you say, and the limiting case of employing square wheels, which should be incorporated into the square wheel equation, must include a term to allow for the fact that the faster you move on square wheels, the closer you approach the motion of circular wheels, that is, the ride gets smoother. $\endgroup$
    – user226006
    Jun 10, 2019 at 7:27
  • 1
    $\begingroup$ Thanks! Yes.. expanding that.. is that suggesting that even with an equilateral triangle wheel... as the speed approaches the speed of light (?! ) very fast, plotting the force that the horse needs will converge to the same absolute number as for the case of a circular wheel. And are you suggesting there is a formula linking the relative speed and equivalent number of sides of the polygon. i.e. N(1) sides at speed S(1) produces the same line representing force the hrose pulls at for N (2) sides at some speed S(2)? $\endgroup$ Jun 10, 2019 at 7:38
  • $\begingroup$ The Step / Heaviside function, and the Dirac Delta function, (which you might look into on Wikipedia,) are designed to mathematically model impulses rather than continuous forces. I am possibly using a sledgehammer to crack a walnut here, in that there is possibly a simpler equation, but it might be interesting to check these out. Also your equation must have another limiting case, in that the length of the sides approach a circle as they reduce. Later today, (with more time) I will attempt an answer, but I would be interested to see if other answers incorporate the above. $\endgroup$
    – user226006
    Jun 10, 2019 at 7:51
  • $\begingroup$ Finally, google.com/amp/s/mathenchant.wordpress.com/2015/07/15/… you may have already read this link, but it brings up another limiting case, the centre of mass distribution and associated moments of inertia. Regards $\endgroup$
    – user226006
    Jun 10, 2019 at 8:00

4 Answers 4


You don't need any force to keep the wheel moving, no matter if it is circle or square - at least in the idealized case when energy is conserved. In real situation, the square has much bigger energy losses then circle, but this depends on many factors, not only on the shape of the wheel.

In the case of circular wheel, if there is no friction there is nothing to slow down the wheel and the cart will move on without horses needs.

In the case of a square wheel the speed will vary with time, because in order for the square to go through the diagonal it must first use some of the kinetic energy to raise the cart up (and therefore needs some minimal initial impulse to be moving), after that the cart will start falling down and returning the kinetic energy back. In the case of no friction, this will go on and on periodically without horses help.

  • $\begingroup$ Thanks! So thinking about it this way (please let me know if this makes sense and I if understood too), the number over time which is the force that the horse has to pull with can be split into : (1) The initial starting force. (2) The force which is required to overcome friction. Also, as you say in the case of the square wheel/ polygon the force will vary over time. I guess another way of looking it might be- what force over time would be needed so that the speed is constant and equal to the speed when the wheel is circular. $\endgroup$ Jun 10, 2019 at 20:49
  • $\begingroup$ @Josie Peanut Yael "what force over time would be needed so that the speed is constant" in case of polygon, the horse would need to pull the cart to accelerate it when the cart is going up and push the cart to slow it down when the cart is falling down. Seems to me like unnecesary abuse of animals, but if the horse enjoys useless work, why not:) In the frictionless case, this is actually easy to compute, i can do it in an edit if you wish. $\endgroup$
    – Umaxo
    Jun 11, 2019 at 1:02

1) The force that that horse has to pull with in order to make the cart move at a consistent speed across D with

There is no such thing, the force is changing continually and the integrated force is zero (ignoring losses in the axle and such).

I think it's more useful to calculate the maximum force. In the case of a square wheel that occurs when one of the sides of the wheel is flat on the ground. You can imagine it I'm sure. Now I will add a slight twist to get the case you're actually looking for - the friction between the surface of the wheel and the ground is infinite - we want our wheel to turn, not skid.

Consider the cart when it is balanced on one corner of the wheel. In this case the force needed to move the cart is zero - even a little touch will cause it to start rotating forward on it's own. That continues, a negative force, until the side of the wheel is flat on the ground again. At that point you have the maximum force again.

So what is that maximum? It's simply the amount of force needed to lift the cart along the incline from the axle to the "leading" corner. Think of a square, draw a line from the center to a corner. What is that angle? 45 degrees. So the force is F = W(cart) sin(45) = 70 N.

So then the general formula is simply the sine of the maximum angle between the corner and the axle. That is 90 - (internal angle of corner/2). So for a eq triangle it's 90 - (60/2) = 60, and for a hexagon it's 90 - (120/2) = 30. For a circle its 90 - (180/2) = 0. So a triangle is harder, a hexagon less so, and a circle zero.

The total force is zero because the amount you need to lift the cart is the same as what you get back when it comes back down the other side.

And as others have noted, this only refers to the force needed to turn the wheel, its ignoring sliding effects, friction in the axle, and acceleration of the mass. For what you're asking, zero is reasonable for all of these.


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Let's take first case, when wheel is square. With in this case there will be two sub-cases. (i) Slipping between surface and wheel (ii) No sleep condition.

(i)If there is slipping between wheel and surface, it doesn't matter if wheel is square or round or polygon. It might rotate or not as well. Also there won't be any different formula apart from simple force balance on wheel.

(ii) If there is no slipping between square wheel and surface, torque will come into the picture. There are two force acting on the wheel, one is static friction and other is force on axis of the wheel. If you consider rotation at the corner point of the wheel, torque due to friction will become zero. Only torque due to axis will be there which will rotate wheel as per the equation $\tau = I \alpha$. Maximum value of force can be calculated in this case by applying condition of topple. To topple the square wheel you will need torque in opposite side of torque generated by normal reaction. Maximum value of torque would be $\tau = Nr$.

N = Normal reaction r = length of one side of square/2

Now as you increase number of sides in polygon, length of side of polygon will keep on reducing. Because maximum force required to topple the wheel is $Nr$, this will also keep on reducing.

In case of wheel, as number of sides tends to infinity or in other words length of side of a polygon with infinite sides tends to zero. (Wheel is a polygon with each side of 0 length). So in case of wheel, zero force is required to topple the wheel or in other words a slight push will also be able to rotate the wheel, no matter how much weight is there on the wheel. (Please note that we're assuming that wheel is a perfect rigid body here, i.e. there are no deformation in wheel).

In real life, there is no perfect body and because of this normal reaction and friction will change and so the value of maximum force required will change in case of circular wheel.

So in fact, there is no need to create a generic formula for this, we can just apply force balance equation and can calculate maximum force required.


Roughly speaking, when trying to pull a square wheel over its corner, the force comes in because you are lifting up the whole car to the diagonal length $r$ and then letting it naturally “crash down to the ground” to inradius $r\sqrt2/2$, where $r$ is the circumradius of the square.

The total forward distance that you get each time you do this is one side length $r\sqrt2$. And now, an energy (work) divided by a distance traveled is an average force. But because the energy is $mg~\Delta h$ where $\Delta h$ is the change in height above the ground that the car experiences, you have a force proportional to a normal force and so you essentially have a coefficient of dynamic friction, which in this case is $(1-\sqrt{2}/2)/\sqrt{2}$, around 0.2, which a table online suggests is something like wood on wood, it's a substantial drag for something you were nominally trying to not drag. (There are also considerations about whether you are driving this with an engine or trying to drag it, do the wheels even turn, what is the maximum force you need to move this thing forward, as opposed to this calculation of an average force...)

Generalizing to $N$ sides, the inradius is $r\cos(\pi/N)$ and the side length is actually $2 r \sin(\pi/N)$, and your general formula is therefore $$\mu=\frac{1-\cos\left(\frac\pi N\right)}{ 2 \sin\left(\frac\pi N\right)},$$which in the limit as $N$ gets large looks like $\mu=\pi/(4N),$ going to zero as desired.


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