5 ms·

>(0,0,0), A, and B are the three points of a right-triangle. Your next line suggests you already know this, but just so it's clear it's not, they are three poi

>(0,0,0), A, and B are the three points of a right-triangle. Your next line suggests you already know this, but just so it's clear it's not, they are three points in an arbitrary triangle. Each vector has a right triangle between the origin and (two of?) it's axis, but I'm not sure that's helpful. Yeah, I agree in that it feels like you should be able to do this, but I just know that I've been stumped in the past, and can't figure it out now either.

dragontamer 6y ago
Ah right. Law of Sines sin(a) / A = sin(b) / B = sin(c) / C Apparently, MY trigonometry is awful and needs more practice! Law of sines is basic 'non-right triangles' stuff. There are three legs: A-0, B-0, and B-A. Hmmm... B-A seems like a vector operation (but its a really easy one, so... probably can be figured out without formally studying linear algebra).. That gives us the "A", "B" and "C" for lengths. Aaaannnnd I'm stuck again. I have the lengths but I don't have the angles. I only need to find one angle and the other two are solved with law of sines. Maybe a system of equations at this point (lol, another subject where linear algebra helps). The law of sines gives 3-unknowns, but only 2-equations. So I still need one more constraint before I can math-out a solution. ------ EDIT: Law of sines is fully: sin(a) / A = sin(b) / B = sin(c) / C = d Where "d" is the diameter of the circumcircle of the triangle. So the circumcircle's diameter gives us the 3rd equation needed to solve the above system of equations (using non-matrix math). So yeah, I think its possible. Its just waaayyyy harder to do. I also am handwaving a difficult part. I guess... take the average of 3 points, which would give the center of the circumcircle. Then sqrt(x, y, z) from the center to find the radius, then 2*radius = d. Wow, that's a lot of work compared to an easy projection matrix...
- Arelius 6y ago
  Yeah, sure is messy.
- Pyrious 6y ago
  The average of the 3 points would give you the triangle's centroid, not necessarily its circumcenter. You can still calculate the circumradius from a, b, and c: https://mathworld.wolfram.com/Circumradius.html https://mathworld.wolfram.com/Circumradius.html. You could also use the law of cosines to get a triangle's angles pretty easily when you know its edge lengths: https://mathworld.wolfram.com/LawofCosines.html https://mathworld.wolfram.com/LawofCosines.html.

dragontamer 6y ago

Ah right. Law of Sines sin(a) / A = sin(b) / B = sin(c) / C Apparently, MY trigonometry is awful and needs more practice! Law of sines is basic 'non-right triangles' stuff. There are three legs: A-0, B-0, and B-A. Hmmm... B-A seems like a vector operation (but its a really easy one, so... probably can be figured out without formally studying linear algebra).. That gives us the "A", "B" and "C" for lengths. Aaaannnnd I'm stuck again. I have the lengths but I don't have the angles. I only need to find one angle and the other two are solved with law of sines. Maybe a system of equations at this point (lol, another subject where linear algebra helps). The law of sines gives 3-unknowns, but only 2-equations. So I still need one more constraint before I can math-out a solution. ------ EDIT: Law of sines is fully: sin(a) / A = sin(b) / B = sin(c) / C = d Where "d" is the diameter of the circumcircle of the triangle. So the circumcircle's diameter gives us the 3rd equation needed to solve the above system of equations (using non-matrix math). So yeah, I think its possible. Its just waaayyyy harder to do. I also am handwaving a difficult part. I guess... take the average of 3 points, which would give the center of the circumcircle. Then sqrt(x, y, z) from the center to find the radius, then 2*radius = d. Wow, that's a lot of work compared to an easy projection matrix...

Arelius 6y ago
Yeah, sure is messy.
Pyrious 6y ago
The average of the 3 points would give you the triangle's centroid, not necessarily its circumcenter. You can still calculate the circumradius from a, b, and c: https://mathworld.wolfram.com/Circumradius.html https://mathworld.wolfram.com/Circumradius.html. You could also use the law of cosines to get a triangle's angles pretty easily when you know its edge lengths: https://mathworld.wolfram.com/LawofCosines.html https://mathworld.wolfram.com/LawofCosines.html.