I am looking for a formula to calculate the radii of an ellipsoid from coordinates of noncoplanar locii on its surface.
If we know the x, y and z coordinates of n noncoplanar points on the surface of an ellipsoid, I would like a formula to calculate three radii of the ellipsoid (a, b and c), therefore determining a, b and c in the following equation for an ellipsoid:
$\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}=1$
In this, I define n to be the least number of noncoplanar points on the surface of the ellipsoid needed to determine radii a b and c. (See attached image for an example of these surface points connected by a line.)
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Three surface points is likely to provide infinite circumellipsoids of any radii. Three points can determine the radius of a unique 2D inscribed circle (radius = abc / 4A) but not an ellipsoid. 4 points (tetrahedral pyramid) can be used to determine the circumradius of a sphere (https://en.m.wikipedia.org/wiki/Tetrahedron#Circumradius). For an oblate or prolate ellipsoid, we likely require 5 or more points on the surface to be able to determine unique radii of a, b and c in the ellipsoid formula

It's centered on the origin though. I'll think about it a bit more and then come back.

For comparison, if you know that the circle is of the form x^2 + y^2 = r^2, you only need one point to determine which circle is it, because you have only one variable, r.

Also for comparison, here's the solution with 2 points for a 2D ellipse: https://www.wolframalpha.com/input?i=solve+x%5E2%2Fa%2By%5E2%2Fb+%3D+w%5E2%2Fa%2Bz%5E2%2Fb+%3D+1%2C+for+a%2Cb

Note that Wolfram cannot handle the 3D case. I stand by my solution, I think it is correct. In the future, I strongly recommend expressing your doubts in a comment rather than contesting the answer, as mistakes (that can indeed happen) often occur from misunderstandings, and the users on this website are usually efficient at fixing them.

The question is a whole other thing if you want to find the radius of a generic ellipsoid, not knowing the rotation angles and the center. In that case, I think you need 10 points. That is a different question though.

Thanks for the explanation! Yes, 10 points might be right! We don't know the rotation angle or the center, but, we know the x, y and z coordinates of a cluster of spots on the surface. It's like a satellite knowing exactly where 10 cities are in USA, and calculating the radii unique to earth as an oblate ellipsoid. Apologies for contesting the incorrect answer before commenting. It's my first time here, and I don't know how this works yet.

No worries, feel free to open another question with the correct statement (either removing the equation or specifying "after rotation and translation"). I will consider taking it. Also note that 25$ might be low for that question, it's not difficult but it's computationally expensive and timeconsuming. Most users here have doctoral degrees, so when it comes to pricing take into account what a reasonable hourly rate for someone with similar qualifications would be.

That's a good idea! Thanks! I'll wait for the dispute resolution process to complete before reposting. Out of curiosity, how many hours of time would this be, and what is the hourly rate of a postdoc where you are? Where I am from, $25 pays for 40+ meals for an average citizen, so I worry our perspectives might vary a bit on the hourly rate!

Oof, with 25$ here you can get maybe one meal at some average restaurant. I normally value my time at 30$/hour (including the website's cut, PayPal fees and conversion fees), which is on the low end. I think the range varies, but anything below 20$/hour is considered borderline offensive, and some people ask for 50$/hour or more. It also depends on the difficulty of the question, this is bachelor level, more advanced stuff tends to get more expensive.

Since you already blew up some money on this question, if you open the new one at the same price I'll give you a sketch of the solution, plus instructions on how to do the computations on the computer. This should take a reasonable amount of time and let you get to the solution eventually. Sounds good?

Perfect! I'll do that in a few hours! Thanks again!
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