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Osculating circle for 3D curve

  1. Feb 11, 2010 #1

    jks

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    1. The problem statement, all variables and given/known data
    Given the space curve r(t)=<t^2,(2t^3)/3,t>, give the parametrized equation for the osculating circle along the trajectory at r(p).

    I should note this is so I can give a plot of the circle in Winplot, which is a free grapher/plotter type program.

    2. Relevant equations
    From r(t), the Frenet frame is:
    T(p)=<2p,2p^2,1>/(2p^2+1)
    N(p)=<1-2p^2,2p,-2p>/(2p^2+1)
    B(p)=<-2p,1,2p^2>/(2p^2+1)

    the radius of curvature is c(p)=((2p^2+1)^2)/2
    so the expression for centers of curvature is:

    <p^2,(2p^3)/3,p> + <1-2p^2,2p,-2p>(2p+1)/2 = r(p) + N(p)c(p)


    3. The attempt at a solution

    Ok my first crack at this was to try and solve it as the intersection of the osculating sphere and osculating plane, which would give a circle. So the osculating plane's equation would be

    (x-p^2)(-2p)/(2p^2+1) + (y-(2p^3)/3)/(2p^2+1) + (z-p)(2p^2)/(2p^2+1) = 0

    then I chose to represent the sphere as a parametric surface:
    defining x0 = (1-p^2)(2p^2+1)/2
    y0 = (2p^3+6p)(2p^2+1)/6
    z0 = (-p)(2p^2+1)/2
    are the x, y, and z components of the center of curvature respectively.

    x = c(p)cos(θ)sin(φ) + x0
    y = c(p)sin(θ)sin(φ) + y0
    z = c(p)cos(φ) + z0

    so then I rearranged the plane's equation for x, y, and z separately, and set them equal to the parametric equation. I don't think I have to say that the equations were quite big, but I noticed that they were a linear system, so I solved it by using matrices. I'm going to stop right here because this approach didn't give me anything useful, problem #1 of which was that the parametric representation of the intersection would necessarily have been an expression of 2 parameters, giving a surface, while only 1 parameter is required for a curve, which is what I need.

    I have a nagging feeling that I'm overlooking something very simple that could make this problem a lot easier. Might I be able to construct the circle, then rotate it so that it fits into place? I'm not too well versed on how to do 3d rotations.
     
  2. jcsd
  3. Dec 16, 2015 #2
    I have an idea. You already have the following quantities:
    - The point r(p)
    - The unit normal n(p)
    - The radius of curvature c(p)
    Then the center of the osculating circle is r(p)+c(p)N(p).
    So now you have a center, radius and normal B(p) to the circle,
    it shouldn't be hard if you find the right formulas...
     
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