Radius of Triaxial / Scalene Ellipsoid

  • Thread starter AwooOOoo
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Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
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Fixed it.... equation 5 was wrong

##r = \sqrt{\frac{a^2b^2c^2}{b^2c^2{sin^2\theta}{cos^2\phi}+a^2c^2{sin^2\theta}{sin^2\phi}+a^2b^2{cos^2\phi}}} ##
 

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