Radius of Triaxial / Scalene Ellipsoid

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The discussion centers on calculating the radius of a triaxial ellipsoid using spherical coordinates, where the original equation led to a division by zero issue. The user initially encountered problems when substituting values into the equation, particularly when angles resulted in sine or cosine values of zero. After identifying the error, the user corrected the equation for the radius, ensuring it no longer produced undefined results. The revised formula accounts for the dimensions of the ellipsoid and avoids division by zero. The focus remains on accurately determining the radius without encountering mathematical inconsistencies.
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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.
 
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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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