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Tuufless · #3 November 7th, 2009, 09:38 PM

Ah yes, I forgot about that condition. Thanks for pointing it out!

On a similar note, do you think it would it be necessary to account for all values of $\phi$ in similar fashion?

If $\phi$ were to fall in say, the second quadrant, then $\sin\phi > 0$ and $-\cos\phi > 0$ .
Still, it's not immediately obvious to me given that $\frac{v_T}{v_P} < 1$ by default (well, barring one heckuva fast tank), that the entire expression $\frac{v_T(\triangle x\sin\phi - \triangle y \cos\phi)}{v_P\sqrt{(\triangle x)^2 + (\triangle y)^2}}$ will evaluate to a value whose magnitude is less than 1, thus allowing us to evaluate $\sin^{-1}\frac{v_T(\triangle x\sin\phi - \triangle y \cos\phi)}{v_P\sqrt{(\triangle x)^2 + (\triangle y)^2}}$ when $\phi$ lies in the second quadrant.

I'll try coding what you've done once the weekend is over. With any luck, it should work. ^^

Edit: I forgot to divide the final expression by $\sqrt{(\triangle x)^2 + (\triangle y)^2}$ both here and in the original post. Now, at least it's a lot more plausible that $\frac{v_T(\triangle x\sin\phi - \triangle y \cos\phi)}{v_P\sqrt{(\triangle x)^2 + (\triangle y)^2}} < 1$ in all quadrants.