By messing around, I found the formula for a square of inner radius
(the radius of the largest circle fitting inside, lying tangent with all the edges) to be of the form:
and the cube of inner radius
(the radius of the largest sphere fitting inside, lying tangent with all the faces) to be of the form:
If one takes
to be less than infinity, the corners of the object are rounded.
The reason the exponent is
is because one of these shapes centered at the origin is only a closed curve/surface if our equation is an even function of each variable.
Finding these forms made me wonder what the forms for the regular N-gons (there are infinity of these) and N-hedra (there are five of these called the Platonic Solids) are.
I noticed that my form for the square fit a more generalized form. This is how I first found the form for the even N-gons, the 2N-gons.
are the unit normals to the edges of the polygon. Also note that
The equation is implicit, and difficult to deal with. It can easily be put into polar explicit form.
Using the trig identity:
, one can write the equation for the curve describing a 2N-gon as
This form only holds when N is even. If I input odd N, the form is exactly equal to if I input twice that N, returning a polygon with 2N number of sides instead of N. This can be fixed with a few steps.
First, I will replace the N in the previous equation with 2N, so that, no matter what N you input, you will get the formula for the corresponding 2N-gon, whether N is odd or even.
Now, we have the form for the 2N-gons when we input N. We can easily cut the figure in half and connect the edges by dividing
by 2. The resulting figure will have, correctly, N number of sides, but will be warped. This can be corrected by increasing the exponentiation on each term in the sum by a factor of 4, or by decreasing the exponentiation of the final sum by a factor of 4.
In the limit that h approaches infinity, this is equivalent to the expression
So, this is the form for any regular N-gon in polar. I do not know how to get a cartesian form for the odd (2N+1)-gons.
Here are some example plots (of course, taking h to be less than infinity):
Now, I’ll follow the same logic, minus the last few steps, to arrive at the form for four of the five platonic solids.
Beginning at the form for the surface of a cube (written above, near top), we see that it also fits a more generalized, 3D version of the previous relationship
are the unit normals to the faces of the polyhedron. Also note that
Putting it into spherical, we have
is not as easy in 3D. Find an axis of symmetry that points through the center of your solid and either a vertex or the center of a face. Make that axis the z axis, and work in spherical coordinates.
The form above only works for solids which have a particular property (mentioned below). However, 4 of the 5 platonic solids have this property, so it is close to the solution that I want. Here is an example plot of a dodecahedron:
I’m not the best at MATLAB, so I’m sure there are better ways to plot surfaces. Anyway, here is an edge-traced version:
Here are two properties that are interesting about the platonic solids:
1. They can all be divided into two identical interlocking chunks that contain half the total number of faces each.
2. In all but the tetrahedron, a
in one half-chunk is opposite one corresponding
in the other half-chunk (the
belonging to the face directly across from it’s face)
This second property makes the octahedron, cube, dodecahedron, and icosahedron similar to the even 2N-gons from before when we were in 2D. However, it makes the tetrahedron like the (2N+1)-gons, and I don’t understand how to go through a similar process to arrive at the form for a tetrahedron.
So, my question is: What is the formula for the surface of a tetrahedron?