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Math Help - 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 \theta_s in terms of angle \theta. \theta is the angle under the pink line, \theta_s is the angle under the spiral radius, green line.

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

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

    1\ge{b}>0

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

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

    So far Ive taken the following approach:

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

    \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 y=x\tan\theta_s by substituting in x\tan\theta_s for y:

    \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:


    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 \cos\theta, so we can express \theta in terms of \theta_s:

    \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 \theta_s in terms of \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; September 10th 2009 at 11:02 AM. Reason: snazzier title
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  2. #2
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    Since that equation involve \theta_s inside a transcendental function (tangent) as well as just squared, there probably is no simple formula for it.
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