- #1
AwooOOoo
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Hi,
I have been referencing this (https://www.physicsforums.com/threads/radius-of-ellipsoid.251321/) previous post to calculate the radius of a Triaxial Ellipsoid (a>b>c), but I'm running into some issues.
Let
0 ≤ ϕ ≤ π
0 ≤ θ ≤ 2π
and
x=r * cos(θ) * sin(ϕ) (1)
y=r * sin(θ) * sin(ϕ) (2)
z=r * cos(ϕ) (3)
sub into
Ellipse: (x/a)2+(y/b)2+(z/c)2=1 (4)
and solve for r
##r=\sqrt{\frac{a^2}{{cos^2\theta}{sin^2\phi}}+\frac{b^2}{{sin^2\theta}{sin^2\phi}}+\frac{c^2}{cos^2\phi}}## (5)
The issue is that the products squares of the cosines and sines drive a division by zero.
For instance
if a = 100, θ = 0, ϕ = 0
the first part of eqn (5) is
##\frac{a^2}{{cos^2\theta}{sin^2\phi}} = \frac{10000}{cos(0) * cos(0) * sin(0) * sin(0)} = \frac{100^2}{1 * 1 * 0 * 0}=\frac{10000}{0} = DIV / 0##
I'm missing something obvious I'm sure, what is it?
Thanks, Paul.
I have been referencing this (https://www.physicsforums.com/threads/radius-of-ellipsoid.251321/) previous post to calculate the radius of a Triaxial Ellipsoid (a>b>c), but I'm running into some issues.
Let
0 ≤ ϕ ≤ π
0 ≤ θ ≤ 2π
and
x=r * cos(θ) * sin(ϕ) (1)
y=r * sin(θ) * sin(ϕ) (2)
z=r * cos(ϕ) (3)
sub into
Ellipse: (x/a)2+(y/b)2+(z/c)2=1 (4)
and solve for r
##r=\sqrt{\frac{a^2}{{cos^2\theta}{sin^2\phi}}+\frac{b^2}{{sin^2\theta}{sin^2\phi}}+\frac{c^2}{cos^2\phi}}## (5)
The issue is that the products squares of the cosines and sines drive a division by zero.
For instance
if a = 100, θ = 0, ϕ = 0
the first part of eqn (5) is
##\frac{a^2}{{cos^2\theta}{sin^2\phi}} = \frac{10000}{cos(0) * cos(0) * sin(0) * sin(0)} = \frac{100^2}{1 * 1 * 0 * 0}=\frac{10000}{0} = DIV / 0##
I'm missing something obvious I'm sure, what is it?
Thanks, Paul.