My question has to do with the geometry of the square-wheeled tricycle ride Pedal on the Petals at the National Musuem of Mathematics in New York (MoMath).

The tricycles ride on a circular track (see video or pictures below), whose surface forms what I might call a fanned catenary. Note that the inner wheels of the tricycles are smaller squares than the outer wheels, in order that they fit the smaller track length there. Note also that the two rear wheels each have their own axle, and the smaller inner wheel axle is correspondingly lower, although they are geared so that they turn together at the same angular rate. Thus, the two wheels turn in unison, with their corners sinking into the crevices between the catenary pieces at the same time. Another detail is that the turn angle of the tricycle is fixed — the rider cannot steer.

The point of the exhibit, of course, is that despite the square wheels and bumpy road, the tricycles offer a smooth ride around the course. Each axle stays the same distance above the floor as it turns, because the uneven track exactly makes up for the change in the radius of the square as it turns.

My question has to do with the observation that each tricycle operates properly only at a certain fixed radius from the center of the track, namely, the radius at which the circumference of the wheels matches that of the track surface. If you look in the photos, you can see that this radius is marked on the track with a solid red center line for the front wheel and dashed lines for the rear wheels. The wheels have a rubber surface that does not slip on the surface of the track.

Question. Is the geometry of this arrangement stable under small perturbations?

In other words, it seems inevitable that the tricycles will get bumped in some way or pushed a little off their line, and then the wheels will no longer exactly fit the track. Will the precise way that they don't fit tend to force them back onto the track properly? And what if the fixed angle of the steering is a little off? Will the operation of the tricycle simply force it back into place?

Looking at the operation of the exhibit in practice, it seems to be fairly stable, and I have never seen a rider simply get stuck. I asked the guy running the exhibit at MoMath, but he made only a non-committal answer, although he did suggest that it might be at least a little stable.

I am hoping that someone with a strong geometrical sense will be able to explain to me why the tricycles seem to operate smoothly even when subject to random forces of kids jostling them.

  • $\begingroup$ believe it is unstable, simply because the arc length of each piece of curve (cycloid rather than catenary, I would expect, unless thay clearly said different) needs to agree with the edge of the square. Move over a little and both sides show slippage. $\endgroup$ – Will Jagy Nov 4 '14 at 19:32
  • $\begingroup$ @WillJagy You are saying that if it gets a little off track, then it should tend to get even further off track? That is what I would mean by unstable. $\endgroup$ – JDH Nov 4 '14 at 19:35
  • $\begingroup$ Good point, I am not sure to that extent. In practice, an off course child would get a slightly bumpy ride, friction (or lack of it) playing a major part. If you made a track out of ice things might be different. Yes, from the picture it is clear that friction is important, most of the time the square touches only one petal. $\endgroup$ – Will Jagy Nov 4 '14 at 19:37
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    $\begingroup$ en.wikipedia.org/wiki/Square_wheel $\endgroup$ – Will Jagy Nov 4 '14 at 20:00
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    $\begingroup$ To a first approximation you may want to consider the case where radius of the track is taken to the $\infty$ limit, so you have a bicycle on catenary cross $\mathbb{R}$. The question now is whether if you start the bicycle in a direction slightly transversal to the catenary the bicycle will self-right. $\endgroup$ – Willie Wong Nov 5 '14 at 9:37

Interesting discussion. As a former MoMath employee, I think I can shed some light here. The short answer is that the trikes' direction and radius are self-maintaining and self-correcting to an extent. The key is that there is some slight slippage of the wheels against the track. The rear wheels will tend to fall into their groove at the same time. If everything is running perfectly, they simply roll into the groove. If not, one or the other will slide a bit to stay in sync. Now imagine the case of being too far from the center of the track. When the rear wheels fall into alignment with a groove, the fixed angle of the steering will cause the front wheel to be pointed not on a tangent to a circular path, but slightly toward the center, setting course for a smaller radius. In the case of the trike riding on too small a radius, often the corners of the rear wheels will not a get a chance to fall into their groove because they meet the next catenary (they are catenaries, by the way.) prematurely. If it's not too far off, the wheel will spin forward and fall into the groove. Usually, if it gets past this point, the self-correcting effect described above will take over (In this case, this means that the fixed-angle steering will the trike away from the center). However, sometimes the corner will hit too high on the curve to slip back and the trike will stop.

  • $\begingroup$ The observation that the rear wheels typically set together into the bottom of their crevice is what solves the issue mentioned by Will Jagy, of a rotating cone finding a new center point of rotation. That aligning feature, together with the fixed turn angle, tends to correct the motion, and thereby makes the path stable. $\endgroup$ – JDH Nov 12 '14 at 17:00

Sorry for misleading about the catenary; it is easy enough to confirm that a line, rotated without slipping around the catenary $y = \cosh x,$ carries the origin along the $x$-axis. The catenary is one of those rare curves where arc-length can be calculated in closed form.

enter image description here

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    $\begingroup$ Thanks! I guess in the circular version, we similarly have piecewise catenary section along each circular path around the course. So the course would be a fanned catenary, where the amplitude of the catenary is proportional to the distance from the center. $\endgroup$ – JDH Nov 4 '14 at 20:33
  • $\begingroup$ @JDH, that seems right. Note that stability is the sort of thing that a physics graduate student could do, Lagrangian mechanics or related. I think there is also a physics stackexchange site. $\endgroup$ – Will Jagy Nov 4 '14 at 20:42
  • $\begingroup$ Yes, you may be right that it is more of an engineering question, where even an experimental answer would be a correct approach. But in my mind, the question has to do with the slopes of the tangents planes to those surfaces when the tricycle is displaced, and whether the slippage tends to nudge the thing in the right direction. $\endgroup$ – JDH Nov 4 '14 at 20:45
  • $\begingroup$ @JDH, sure. Many ingredients to the analysis. $\endgroup$ – Will Jagy Nov 4 '14 at 20:50

Since I made no further progress, I copy here my comments above:

To a first approximation you may want to consider the case where radius of the track is taken to the $\infty$ limit, so you have a bicycle on catenary cross $\mathbb{R}$. The question now is whether if you start the bicycle in a direction slightly transversal to the catenary the bicycle will self-right. And I claim that there is a restorative force in this model from gravity. This can be seen easily by noting that when the bicycles are sitting with corner straight down, the "aligned" state where the corners fit exactly into the groove minimizes potential energy. So with sufficiently small friction, gravity will cause a stationary bike (ignore for now the problem of it falling over on its side; this can be avoided by using a trike and and restricting the perturbations to be sufficiently small) to self-right, with the perturbations dissipating from friction.

Similarly, if you put the trike as in the picture sitting in a groove, it is not too hard to see that any small perturbation increases the potential energy of the bike.

The problem with this argument is that if the rubber wheels are sufficiently sticky, static friction may be sufficient to keep the trike from righting itself. This can be seen already by Will Jagy's argument: in the case that the wheels are round and the ground is flat: moving the trike a little bit will result in it orbiting around a different center point. Something similar would happen if you only allow strict "rolling and no slipping" conditions on the wheels.

I expect in the actual physical model some sliding would be involved. As you can see the answer on stability depends on what model you are using and what configurations you assume.


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