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Thread: Please help derive angle theta_s in terms of angle theta

  1. #1
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    Derive angle under spiral radius in terms of unit circle angle

    Please see the attached thumbnails. I have here this lovely archimedean spiral inscribed within a unit circle and I want to derive angle $\displaystyle \theta_s$ in terms of angle $\displaystyle \theta$. $\displaystyle \theta$ is the angle under the pink line, $\displaystyle \theta_s$ is the angle under the spiral radius, green line.

    Known: $\displaystyle \theta$ and the equation for the spiral radius $\displaystyle r_s$ which is just the polar equation for said spiral curve, and b, the y intercept of this curve:

    $\displaystyle r_s=1+\frac{2\theta_s(b-1)}{\pi}$

    $\displaystyle 1\ge{b}>0$

    Again, the spiral is inscribed within a unit circle. I want to express $\displaystyle \theta_s$ in terms of $\displaystyle \theta$.

    (If $\displaystyle r_s$ can be solved for in terms of $\displaystyle \theta$ that would be fine too.)

    So far Ive taken the following approach:

    1) I convert the $\displaystyle r_s$ curve from polar to Cartesian form:

    $\displaystyle \sqrt{x^2+y^2}=1+\frac{2\arctan(\frac{y}{x})(b-1)}{\pi}$


    2) I find the intercept between the spiral and the line $\displaystyle y=x\tan\theta_s$ by substituting in $\displaystyle x\tan\theta_s$ for y:

    $\displaystyle \sqrt{x^2+x^2\tan^2\theta_s}=1+\frac{2\arctan(\fra c{x\tan\theta_s}{x})(b-1)}{\pi}$

    Which simplifies to:


    $\displaystyle x=\sqrt{\frac{\left(1+\frac{2\theta_s(b-1)}{\pi}\right)^2}{\tan^2\theta_s+1}}$

    3) This, then, is equal to $\displaystyle \cos\theta$, so we can express $\displaystyle \theta$ in terms of $\displaystyle \theta_s$:

    $\displaystyle \theta=\arccos\sqrt{\frac{\left(1+\frac{2\theta_s( b-1)}{\pi}\right)^2}{\tan^2\theta_s+1}}$

    But, alas, the reverse is not so simple. I cannot find a way to solve this equation for $\displaystyle \theta_s$ in terms of $\displaystyle \theta$.

    Perhaps a whole different approach is necessary. Your help is much appreciated.
    Attached Thumbnails Attached Thumbnails Please help derive angle theta_s in terms of angle theta-solving-angle-theta_s.jpg   Please help derive angle theta_s in terms of angle theta-2.jpg  
    Last edited by rainer; Sep 10th 2009 at 11:02 AM. Reason: snazzier title
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  2. #2
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    Since that equation involve $\displaystyle \theta_s$ inside a transcendental function (tangent) as well as just squared, there probably is no simple formula for it.
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