The old Mathstronomy thread was getting a little cumbersome, since I figured out a lot of stuff as I went along in the answers. So I'm restarting the thread with some new results I found. To restate the problem, given any three non-colinear stars in space, from what direction would the stars appear to form an equilateral triangle? I now have a generalization for all isosceles triangles, not just the "skinny" ones.

To start off with, choose a coordinate system such that the stars lie in the x-y plane with the triangle's centroid at the origin and the stars at the following polar coordinates:

(a, 225°) ,  (1, 45° -  θ ) ,  (1, 45° + θ )

By the way, the centroid is a triangle's center of mass, which is always two thirds the distance from any vertex to the midpoint of the opposite side.

When 0 < a < 1 [a "fat" isosceles triangle), the equilateral viewing vector is

x  =  - y  =  ± t √[2 (1 - a²) ÷ 3a²]  ,   z  =  t

When 1 ≤ a (a "skinny" isosceles triangle), the equilateral viewing vector is

x  =  y  =  ± t √[2 (a² - 1) ÷ (4 - a²)]  ,   z  =  t

The viewing vectors are given in standard parametic notation. In either case, you don't need the angle θ to compute the viewing vector, just a (the ratio of the vertices' distances from the centroid).

At this point, I'm starting to wonder if a solution even exists for scalene triangles. I tried drawing one on paper and turning it until it looks equilateral, but I can't seem to make it work. Maybe somebody with a good 3-D rendering program can move around a scalene triangle on the computer and see if it works. Unfortunately, proving there isn't a solution would probably be even harder than finding one if there is one. Other enthusiasts are still welcome to jump in on the scalene solution (or non-solution).

This message has been edited. Last edited by: silenteuphony,

After playing around with scalene triangles, I'm once again convinced that there is a solution for them, it's just going to be a bear to find it. Unlike the isosceles cases, you might need to use your angles as well your distances to calculate it. To facilitate generalization of the isosceles solution, I'm including the angles, generalizing the coordinate system, and putting the solution in cylindrical coordinates.

I'm also doing away with the symmetry about y = x that I used before, and putting one of my points at (1, 0°) for simplicity. I only used the y = x symmetry before this because I started this whole problem in the old thread with an example that happened to have that symmetry.

To start off with, choose a coordinate system such that the stars lie in the x-y plane with the triangle's centroid at the origin and the stars at the following polar coordinates:

(1, 0°) ,  (a, φ_a° ) ,  (b, φ_b° )   ,  where 1 ≤ a ≤ b

When a = b (a "fat" isosceles triangle), the equilateral viewing vector is

r  =  ± 2t √[(b² - 1) ÷ 3]  ,   θ  =  90°  ,   z  =  t

When 1 = a (a "skinny" isosceles triangle), the equilateral viewing vector is

r  =  ± 2t √[(b² - 1) ÷ (4 - b²)]  ,   θ  =  φ_b°  ,   z  =  t

In this form, if you set z = t, then r is a direct indicator of how irregular your triangle is. In the trivial case (an equilateral triangle), the viewing vector is just the z-axis, and r = 0. The more irregular the triangle becomes (the greater a and b are, in numerical terms), the greater r becomes, as your viewing vector becomes more skewed from the z-axis.

Another way of looking at this problem is to think of it as compressing the triangle instead of changing the viewing axis. Any change in your viewing angle from the z-axis is equivalent to compressing the original triangle in the x-y plane. You can think of this as drawing a line through the middle of the triangle, and then moving every point in the triangle proportionally closer to that line. Figure out how to find this "axis of compression" (along with the scale of compression) which transforms any given triangle into an equilateral triangle, and you've just solved the problem from another angle.

This message has been edited. Last edited by: silenteuphony,

I don't know if anyone's still reading this, but after much sweat and tears, I have the general solution. I'm using the letter v to stand for viewing vector, so I'll use the distance v and the angle φ_v in my solution.

Unfortunately, I wasn't able to isolate these variables, but I was able to find a simple system of equations which would enable a computer (and a decent math program) to find their values for any given triangle.

To start off with (once again), choose a coordinate system such that the stars lie in the x-y plane with the triangle's centroid at the origin and the stars at the following polar coordinates:

(1, 0°) ,  (a, φ_a° ) ,  (b, φ_b° )   ,  where 1 < a < b

The equilateral viewing vector is (in parametric notation with cylindrical coordinates)

r  =  ± v t  ,   θ  =  φ_v°  ,   z  =  t

where v and φ_v are given by the following system of equations:

1 + v² sin²φ_v   =   a² [1 + v² sin²(φ_v- φ_a)]   =   b² [1 + v² sin²(φ_v- φ_b)]

Note that a, b, φ_a , and φ_b are constants for a given triangle, so you're only solving for the two variables, v and φ_v , which would be difficult by hand, but any good math program should be able to do it.

This message has been edited. Last edited by: silenteuphony,

quote:

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Originally posted by silenteuphony:
I don't know if anyone's still reading this, but after much sweat and tears, I have the general solution.

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Silenteuphony,
Although this level of math is a bit beyond my understanding, I certainly appreciate the effort and calibre of your contributions. Keep up the excellent work!