Finding the centroid of a triangle in hyperspherical polar coordinates

Question

I have three points:

$A = (r{_0},\psi{_0},\theta{_0},\phi{_0})$

$B = (r{_1},\psi{_1},\theta{_1},\phi{_1})$

$C = (r{_2},\psi{_2},\theta{_2},\phi{_2})$

Actually, they are orthogonal, so that:

$A = {(r_0,\psi_0,0,0)}$

$B = {(r_1,0,\theta_1,0)}$

$C = {(r_2,0,0,\phi_2)}$

These form a triangle. I'd like to find the centroid (centre of mass):

$cog(A,B,C) = {(r_{cog},\psi_{cog},\theta_{cog},\phi_{cog})}$

where the radius represents the mass, so r=42 means that the weight of that point m=42. The triangle is assumed massless, apart from the masses at the vertices.

Is there a simple formula for this?

I'd like, if possible, to avoid converting it all to cartesian coordinates.

Answer 1

For calculations, see the attached diagram. The solution is, in cartesian coordinates:

Cog=(w,x,y,z)

w = ( R sin(ψ) ) / 3

x = ( G cos(θ) + R cos(ψ) ) / 3

y = ( G sin(θ) + B sin(ϕ) ) / 3

z = ( B cos(ϕ) ) / 3

In polar (hyperspherical coordinates):

Cog = (r,ψ,θ,ϕ)

$r ={\sqrt{w^2+x^2+y^2+z^2}}$

ψ = atan2(w,x)

θ = atan2(y,x)

ϕ = atan2(y,z)

Corrections welcome.