0
$\begingroup$

I know the radius $R$ of the circle and the area $A$ of the segment.

How can I solve for central angle $\alpha^{\circ}$ in this (or some other) equation:

$$A=\frac{R^{2}}{2} \left( \frac{\alpha \pi}{180}-\sin \alpha^{\circ} \right)$$

?

Here Newton's algorithm is recommended, but with an initial guess of

$$x(0) = (6k)^{1/3}$$

Why is this the initial guess?

$\endgroup$
  • $\begingroup$ you after numerical methods I suppose ... $\endgroup$ – Math-fun Mar 30 '16 at 10:13
  • $\begingroup$ @Math-fun Why?? $\endgroup$ – mavavilj Mar 30 '16 at 10:15
  • $\begingroup$ your equation is a mixture of a polynomial and trigonometric function this does not in general admit a closed form solution and one has to rely on numerical methods. $\endgroup$ – Math-fun Mar 30 '16 at 10:17
  • $\begingroup$ I'm looking for if there's some other equation I could use, given this info that I have. Or a pointer into what kind of numerical method to use for the solution? $\endgroup$ – mavavilj Mar 30 '16 at 10:18
  • $\begingroup$ You would get $\alpha -\sin \alpha$= something, use the Taylor approx. for $\sin \alpha$. $\endgroup$ – Nikunj Mar 30 '16 at 10:19
2
$\begingroup$

Your equation is transcendental,closed form solution is not possible. Newton -Raphson numerical iteration method is often used. If an approximate solution is acceptable, a graphical solution is also one method.

EDIT1:

By series expansion upto 2 terms we get a good approximation

$$ 2 A /R^2 = k \approx \alpha - \sin \alpha = \alpha ^3 /6$$

so we can choose a reasonably accurate value for starting iteration as:

$$ \alpha_{initial}= (6 k)^ { \frac13} .$$

$\endgroup$
  • $\begingroup$ I really wonder why you received a downvote for a good answer ! $\endgroup$ – Claude Leibovici Mar 31 '16 at 5:58
  • $\begingroup$ @ClaudeLeibovici I initially downvoted the answer before the edit, as I felt the first paragraph, while true, is somewhat redundant to what the OP said or didnt answer what the OP was asking for... The OP didn't want a graphical solution and already knew that Newtons method is used. However, the edit is fantastic, and i have thus removed my down vote and in fact up voted it :) $\endgroup$ – Brevan Ellefsen Mar 31 '16 at 21:15
  • $\begingroup$ Thanks, gentlemen, there was backlash from my side also.. it took time to realize what OP was looking for. $\endgroup$ – Narasimham Mar 31 '16 at 21:22
  • $\begingroup$ @BrevanEllefsen. It is very elegant to explain ! Thanks for acting this way. $\endgroup$ – Claude Leibovici Apr 1 '16 at 4:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.