Le 21/09/2020 à 00:24, Bald Eagle a écrit :
> I just stumbled across this while trying to regain my sanity,
> and if even Ramanujan couldn't come up with anything but an approximation, than
> I think the rest of us are Fuuuuuuuuuuuuuuuuuuuuuumbling around in the dark.
> 
> :)
> 
> https://www.youtube.com/watch?v=5nW3nJhBHL0
> 
> But don't anyone think that that will stop us from making another attempt, using
> new inspiration and mathematical tools!
> 
> I think the above implies that due to the interesting properties of ellipses,
> that a workable solution will use numerical methods.
> 

I tried to play a bit with the hope to transform the quartic equation of
circular torus into elliptical torus.

From (sqrt(x²+y²)-R)²+z²=r², replacing x²+y²=R² with (x/a)²+(y/b)²=1

And it is deception.

x^4/a^4
+ (2 x^2 y^2)/(a^2 b²)
- (2 r^2 x^2)/a²- (2 x^2)/a²
+ (2 x^2 z^2)/a²
+ y^4/b⁴
- (2 r^2 y^2)/b²- (2 y^2)/b²
+ (2 y^2 z^2)/b²
+ r⁴- 2 r²+ 1
- 2 r^2 z²+ 2 z²
+ z⁴


Because the scaling of x & y also influence the radius of the minor circle.

But it must be possible to have it with a 4th degree poly.


If F( Point ) is such polynomial, we have the following requirement

F( 0, +/- B , +/- Minor  ) = 0  (point B)
F( +/- A, 0, +/- Minor ) = 0    (point A)
F( 0, +/- B +/- Minor, 0 ) = 0  (points M & N)
F( +/- A +/- Minor, 0, 0 ) = 0  (points K & L)

I guess, but cannot yet assert, that to avoid compressing the minor
circle, the z⁴, x⁴ and y⁴ must be 1.
And there might be some non-traditional power of x,y and z (aka odd) in
the non-null coefficient, which would disappear when A=B to have a
continuity with the circular torus.

Also, any point W On the central ellipse with z=+/- Minor is also F(W)=0

And if any more point is needed, there is the two minor circle at A & B
to explore more deeply than K, L, M & N.

Is that enough to try some spreadsheet solver, and how to do it ?

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