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Hypocycloid: A Mysterious Curve.

  1. Feb 28, 2013 #1
    1. The problem statement, all variables and given/known data
    The hypocycloid is the plane curve generated by a point ##P## on the circumference of a circle ##C##, as this circle rolls with out sliding on the interior of the fixed circle ##C_0##. If ##C## has a fixed radius of ##r## and ##C_0## is at the origin with radius ##r_0## and the initial location of the point ##P## is at ##(r_{0}, 0)##, what is the representation of a hypocycloid


    2. Relevant equations



    3. The attempt at a solution

    I didn't know how to start but to draw a picture. then I thought about the Epicycloid... these curves are related? I didn't know what to do so I played around, and I don't know if they are right or not. s.th. ##\delta_{0} = ((r+r_{0})/(r_{0}))\theta## ##(r+r_{0})cos(\delta_{0})-r cos(\delta_{0}) + (r+r_{0})sin(\delta_{0}) + r sin(\delta_{0})## I know this isn't right but just what I was thinking they could be...
     
  2. jcsd
  3. Mar 1, 2013 #2

    haruspex

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  4. Mar 1, 2013 #3
    That isn't much help, =/ I thought someone would be able to work out the problem and tell me how they got the answer because it doesn't make sense, I am not looking for the answer I am looking for a solution... I know it sounds redundant, but look... I know the answer already but I do not know how to reach the solution, so if anyone can help me out that would be awesome
     
  5. Mar 1, 2013 #4
    Well, first you can try to express it parametrically. Let t be a time parameter (letting the circle roll with constant angular velocity,) and try to find x and y in terms of t. Then it's possible to just convert back to rectangular by solving for a relationship between x and y.
     
  6. Mar 1, 2013 #5

    haruspex

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    Ok, that was not clear. In the equations at http://en.wikipedia.org/wiki/Hypocycloid#Properties, the centre of the small circle, O', is at location (R-r, θ) in polar coordinates. Suppose P is at (r, ψ) in polar coordinates relative to O'. rψ = (R-r)θ (do you see why?). Converting to Cartesian and adding up the x and y coordinates separately gives the equations.
     
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