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Circle of Curvature

  1. Mar 17, 2009 #1
    1. The problem statement, all variables and given/known data
    I am given a space curve r(t)= t i + sin(t) j and point (pi/2,1). They ask me to find an equation for the circle of curvature.

    2. Relevant equations

    Kappa, T, N, not sure

    3. The attempt at a solution

    So I have found the radius of curvature which is row= 1/kappa= 1 at the given point. I have also found out T which is the unit tangent vector. Do I have to find the unit normal vector also ?

    How do I write the equation ? will it just be (x-pi/2)^2 + (y-1)^2 = 1 ??
  2. jcsd
  3. Mar 17, 2009 #2


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    Hi nns91

    what do you need to define a circle in the plane?
    - a radius
    - a circle centre
    The equation of a cirlce of centre (a,b) and radius r is
    [tex] (x-a)^2 + (y-b)^2 = r^2 [/tex]

    You have your radius from
    [tex] r = \frac{1}{\kappa} [/tex]
    Also are you sure \kappa is 1? how did you get this?

    how do you find the centre of the circle? It is not (pi/2,1). This is a point on your curve. Think about the unit normal direction.
  4. Mar 17, 2009 #3


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    you should also consider whether you curve is unit speed or whether you need to normalise...
  5. Mar 17, 2009 #4
    I calculate Kappa and got like sqrt(1+cos^2(t)) / sin(t). Then I substitute pi/2 for t and get 1. Am I right ??

    I still don't get the relationship between center of circle and the unit normal direction.

    My curve is not a unite speed curve I think since it is r(t) instead of v(t)
  6. Mar 17, 2009 #5


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    try drawing your curve and imagine where the circle sits, the tangent of the curve will match the tangent of the circle

    the normal direction will point toward (or away) from the centre of the circle

    you curve is unit speed instanaeously at that point (calulate |dr/dt|, but not in general so you may need to be careful with your calcs and check what this affects... i'm not totally sure without looking back at the equations which I don't have handy

    Also i'm not too sure how you got it, but I think your curvature value is correct, if you are speaking about absolute curvature. If we are talking about signed curvature need to be careful as it could be + or - depending on how things are defined.
    Last edited: Mar 17, 2009
  7. Mar 17, 2009 #6
    WIll the normal direction point toward the center of the circle ?
  8. Mar 17, 2009 #7


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    either directly toward or directly away depending on your parameterisation & curvature definition
  9. Mar 17, 2009 #8
    Then how do I find the center of the circle base on the normal vector ?
  10. Mar 18, 2009 #9


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    you have a point on the circle (ie the point on your curve) a direction to the centre (normal vector) and a radius (from you curvature).... should have everything you need
  11. Mar 18, 2009 #10
    so the radius is the distance from the center to the point, so do I use the distance formula then ?
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