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I want to solve this equation, determining y (all others letters are constants) :

$$2 \arccos(3+(1.6y-80)/R) - \sin(2\arccos(3+(1.6y-80)/R)) = 2π(1-P)$$

I've try to use some automatic solvers but they failed. If this equation doesen't have any determinable solution, I would like to have some approximative form of it...

(For the context of the problem, $P$ is a percentage, that must correspond to the percentage filled of a $R$ radius circle. This circle is filled by moving a rectangle shape on the $y$ abscissa, that is going from 80 for 0% to -80 for 100%.)

Thank !!

Edit : with your help, I have now this : $$with : X = 3 + (1.6y - 80)/R$$

$$ \arccos(X) - X\sqrt(1-X²) = π(1-P)$$

And I'm afraid I do not know how to solve it either.

FitzFish
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    Perhaps this interests you. – Git Gud May 19 '14 at 20:09
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    Also apply the double-angle formula $\sin 2\theta = 2 \sin \theta \cos \theta$ so that $\sin(2 \arccos(\cdots))$ becomes $2 \sin(\arccos(\cdots)) \cos(\arccos(\cdots))$. Evaluating that cosine is trivial, and your link should help with the sine. – David K May 19 '14 at 20:18
  • Thank you for your usefull comments ! But this is not won yet... I've edit my question. – FitzFish May 20 '14 at 14:36

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