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Parametric equation for a cycloid

  1. Jan 28, 2008 #1
    Hi, I am having trouble reversing the formula [tex] x=R(\theta - \sin(\theta))[/tex] to get [tex]\theta[/tex] in terms of x. Am I missing something obvious or is it just impossible?

    To put it into context this is part of the parametric equation for a cycloid. The other part of the parametric equation is [tex] y = R (1- \cos(\theta))[/tex]. Setting R to 1 (the radius of the rolling wheel) does not seem to help. The non parametric equation for the cycloid is [tex] \pm \cos^{-1}((R-y)/R) \pm \sqrt{2 R y -y^2}[/tex]. I would also like to reverse this full equation to get y in terms of x but I am having trouble with that too. The reason I am trying to reverse the equations is that I am trying to get the intersection of two loci (the cycloid locus and the perimeter of a ellipse).

    Any help appreciated. Thanks :)
     
  2. jcsd
  3. Jan 28, 2008 #2

    HallsofIvy

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    In general, there is no "formula" for solving an equation in which the unknown number occurs both inside a transcendental function (such as cosine) and outside it. typically, the best that can be done is a numerical solution.

    I ahve no idea what you mean by "The non parametric equation for the cycloid is [tex] \pm \cos^{-1}((R-y)/R) \pm \sqrt{2 R y -y^2}[/tex]" because that is not an equation. Did you leave something out?
     
  4. Jan 28, 2008 #3
    Yes I did! That should read "The non parametric equation for the cycloid is [tex] x = \pm \cos^{-1}((R-y)/R) \pm \sqrt{2 R y -y^2}[/tex]" which is obtained by substituting [tex] \pm \cos^{-1}((R-y)/R) [/tex] for [tex]\theta[/tex] in [tex] x=R(\theta - \sin(\theta))[/tex].


    It seems ridiculous that there is no easy solution to the question "If the point on the perimeter of a wheel has advanced linearly by x then what angle has the wheel rotated through?"
     
    Last edited: Jan 28, 2008
  5. Jan 29, 2008 #4
    Can the Lambert W function help here?
     
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