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The mathematician Stan Wagon built a tricycle with square wheels that can ride smoothly on a carefully crafted curved surface:

enter image description here

Now I have some puzzles for you:

  1. Is there a surface that allows one to ride on triangular wheels?
  2. Which wheel can ride smoothly on the following surface? The peaks of this surface are at right angles. If the wheel is sitting on top of the first peak then after one complete rotation it will cross the next 3 peaks and end up sitting on the 5th peak.

enter image description here

  1. Which wheel can ride smoothly on the following surface? This surface is an inverted cycloid. If the wheel is sitting on top of the first peak then after one complete rotation it will end up sitting on the 2nd peak.

enter image description here

Hint for 2 and 3:

What shape should the bottom half of the wheel have to allow it to sit on top of a peak?

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    $\begingroup$ The answers are not unique unless you also specify how many "humps" the wheel covers in a single rotation. In extremis you could use a very large wheel with a rubber tire with a profile that matches any of these surfaces, so that the size of the wheel makes the effect of curvature negligible. $\endgroup$ Mar 22 at 9:10
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    $\begingroup$ ah yes good point. I'll try to fix it. $\endgroup$ Mar 22 at 9:16
  • $\begingroup$ The bicycle with square wheels of course does four humps per rotation. $\endgroup$ Mar 22 at 9:24
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    $\begingroup$ I don't know. It is an interesting mathematical question but to be honest, unless there are particularly pleasing right answers I'm not sure it qualifies as a puzzle. $\endgroup$ Mar 22 at 10:22
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    $\begingroup$ If the "hints" are required to ensure a unique solution, they're not hints, but instead integral parts of the puzzle, and as such shouldn't be spoilered. $\endgroup$
    – bobble
    Mar 22 at 14:07
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Interesting. I hadn't looked this up before. But it seems from here that:

There's a general way to convert from ground to wheel and vice versa. If you have a wheel in polar co-ordinates $(\rho,\theta)$ and ground in cartesian co-ordinates $(x,y)$, you can convert from one to the other:
Wheel $\rightarrow$ Ground: $y = \rho$ and $x=\int\rho d\theta$.
Ground $\rightarrow$ Wheel: $\rho = y$ and $\theta=\int\frac{1}{y}dx$

This gives the general idea, and it's little more than a math exercise after this.

  1. Driving a triangle:

Wheel is $\rho=1/\cos(\theta)$ for $\pi/3\leq\theta\leq\pi/3$ (for a suitably scaled triangle). So you can parametricize the curve with $y = 1/\cos(\theta)$ and $x = \ln(\sec(\theta)+\tan(\theta))$. I don't know if that's a special curve. The square is a repeating inverted catenary. This one looks similar. Obviously the angle at the bottom should meet at $\pi/3$ instead of $\pi/4$, but it's qualitatively similar. It's here.
Triangle ground

  1. Driving on triangles:

Your ground is $y=x$ (+ reflections and periodicity). So you get $\rho=e^\theta$. Do that for $0\leq\theta\leq\pi/4$ and then reflect and repeat. You'll end up with a four-pointed star, something like this:
Driving on triangles

  1. Inverted cycloids:

I'll have to look up the formula for a cycloid, but it will be something like a cardioid. Edit. Oh, yeah, so the math works out neatly. Cycloid is $x=(\theta-\sin(\theta))$ and $y=(1-\cos(\theta))$ (actually parametricized by $t$, but I'm cheating a bit because I know how it will work out). So the wheel is $\rho=(1-cos(\theta))$ and $\theta=\int\frac{1}{1-\cos(\theta)}dx = \theta$. Justifying my cheat. So it ends up as $\rho=(1-cos(\theta))$:
cardioid

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  • $\begingroup$ This is spot on so well done! Thank you for the links too. I didn't realise that 1 is possible... $\endgroup$ Mar 24 at 22:06
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    $\begingroup$ I got the idea for this puzzle from the following link. According to that page, triangular wheels don't work. mathtourist.blogspot.com/2011/05/… $\endgroup$ Mar 24 at 22:47
  • $\begingroup$ Oh, that's interesting. I could see how those curves could collide in practice. I will have to investigate further... $\endgroup$
    – Dr Xorile
    Mar 25 at 0:34
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    $\begingroup$ I think why 1 is not possible is partially explained by Retudin's answer, about the angles needing to be more than 90 degrees. $\endgroup$
    – justhalf
    Mar 25 at 11:50
  • $\begingroup$ note that for similar reasons, 3 is also not possible. $\endgroup$
    – Retudin
    Mar 26 at 7:14
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At least I have an interesting answer for 3)

A solution for 3) would be a circular wheel that has its axis on the circumference.

It starts standing on a peak with the axis touching the ground.
The wheel first turns 1/4 turn without the axis moving forward.
Then the wheel rolls across the valley with the axis moving horizontally from one peak to the other. It needs half a turn. At that point the axis reaches the next peak.
The axis then stops for 1/2 turn while the wheel "rolls" over the peak.
And so on.

It requires the valleys to be perfect half-circles. The wheel would have half the diameter of these.

Note: By "smooth ride" you have to understand a perfectly horizontal movement of the vehicle. It would do a stop-and-go motion in a most unacceptable way for public transport.

PS: Actually that stop-and-go motion can be put to good use. If you build a tramway track with this shape, with a distance between peaks matching the distance between stations, and the wheels sized accordingly of course, then you would have a tramway that stops at every station while the wheels turn at constant speed.

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My answers:

enter image description here

intermezzo:
If you have a flat surface with a peak (see picture), one can use a circle part.
1/At the top we see a half circle. It can balance on the peak without the circle center moving until it touches ground, then it can roll to the next peak; the circle center staying at the same height the entire time.
2/If the peaks are further apart, you need a bigger slice of the circle.
3/However if the circle part is more than 3/4th, the peak will block forward motion.
4/This even works with multiple heights, but the angle rotation will be faster when moving forward on higher ground.

enter image description here

This works even for an infinite number of depths, i.e. any curve.

With a sharp peak: - The wheel must temporarily stand still
- Therefore the axis must be at the outside
- Then the peak to peak rotation can be calculated
- If more than 270 degrees, the wheel will block
Dr Xorile already did the math, the rotation is 360 decrees for the cycloid, and there is indeed a sharp inward in their picture.
Thus: Answer for 3: There is no wheel that can run smoothly on the cycloid, the form that would be needed to keep constant height will be blocked by the peak.

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