1
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Clear["'.*"]
{ecc, p, r1, t1} = {0.35, 1., 1.4, 0.8}
EQ = {p/r2 == 1 - ecc Cos[t2], p/r3 == 1 - ecc Cos[t3], 
  r1^2 + r2^2 - 2 r1 r2 Cos[t1 - t2] == 
   r3^2 + r2^2 - 2 r3 r2 Cos[t3 - t2], 
  r3^2 + r1^2 - 2 r3 r1 Cos[t3 - t1] == 
   r3^2 + r2^2 - 2 r3 r2 Cos[t3 - t2]}
NSolve[EQ, {r2, r3, t2, t3}]

First radius vector is given/defined in the first line. Also given are ellipse properties ( $ecc, p =$ eccentricity and semi-latus rectum).

I am trying to inscribe an equilateral triangle in order to find two more vector arms $r2,r3$ along with their polar angles $ {t2,t3}. $

What assumptions or essential changes did I miss that makes this to hang? Is there a version problem?

The question has relevance to this MSE link.Thanks in advance.

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4
  • $\begingroup$ I suspect I know what ecc and your other variables represent, but you might want to indicate what they're supposed to be for reference. See this as well. $\endgroup$
    – J. M.'s torpor
    Mar 30 '18 at 15:21
  • $\begingroup$ Yes sir, classical symbols of Newton planetary motion focal conic used. $\endgroup$
    – Narasimham
    Mar 30 '18 at 19:51
  • $\begingroup$ Perhaps you did not see the image I just deleted or gone through the program. There are three radial arms. First one has length $1.4.$ Other two $(r2,r3)$ are unknown at start of computation. It would be also evident from the program. Feel free to say if my problem expression is in some way lacking. $\endgroup$
    – Narasimham
    Mar 31 '18 at 17:46
  • $\begingroup$ Because I chose origin / ellipse focus as reference of vectors starting , not its center of gravity. The link has an image of sorts. $\endgroup$
    – Narasimham
    Mar 31 '18 at 18:39
1
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You can use NMinimize

NMinimize[{1, EQ}, {r2, r3, t2, t3}]
(*{1., {r2 -> 1.343, r3 -> 1.31612, t2 -> 0.75289, t3 -> 0.814458}}*)
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4
  • $\begingroup$ you might try FindInstance as well.. $\endgroup$
    – george2079
    Mar 30 '18 at 21:27
  • $\begingroup$ Thanks,shall try it $\endgroup$
    – Narasimham
    Mar 31 '18 at 6:15
  • $\begingroup$ @Ulrich Neumann I wianted to plot a solution triangle. When plotted three points they do not form vertices of an equilateral triangle. Why so? Are inverse functions giving wrong choices? $\endgroup$
    – Narasimham
    Mar 31 '18 at 14:53
  • $\begingroup$ @Narasimham: I don't think so. Please show your code, how you plot it... $\endgroup$ Apr 1 '18 at 12:22

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