Riding on non-circular wheels

Question

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

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.

3. 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.

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?

Answer 1

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.

2. 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:

3. 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 $\rho=(1-cos(\theta))$:

Answer 2

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.

Answer 3

My answers:

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.

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.