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ris8_allo_zen0 · #9 January 14th, 2008, 05:13 AM

Ok, I think I'm on the right way now.

Let's call $\alpha_1$ the angle between x and y, and $\alpha_2$ the one between y and z.
x and z join the "external" corners with the observer, and y joins the one which is farthest from the obs. Thus, y is between the two and is shared between $\alpha_1$ and $\alpha_2$ .
$\gamma_1$ and $\gamma_2$ are the angles opposite to x and z, respectively. Since we're working on a half-square, their sum is $\frac{\pi}{2}$ .

These are the calculations to be done, exploting theorems of sines and cosines:
$\gamma_1=\arcsin(x \sin(\alpha_1))$
$y=\sqrt{x^2+1-2x \cos(\pi-\alpha_1-\gamma_1)}$
$\gamma_2=\frac{\pi}{2}-\gamma_1$
$z=\sqrt{y^2+1-2y \cos(\gamma_2)}$
$\alpha_2=\arcsin \left( \frac{\sin \left( \gamma_2 \right) }{z} \right)$

(am I correct until here?)

So the the computer's mission is to find an x value such as the calculated $\alpha_2$ is near enough to the given $\alpha_2$ . This can be done in a way I'll investigate soon

Thanks mr fantastic to pointing me to the right direction!!!