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Math Help - How to find normal vector of quadatic bezier curve, curvature of bezier curve.

  1. #1
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    How to find normal vector of quadatic bezier curve, curvature of bezier curve.

    Hello. How to find normal vector of quadatic bezier curve, curvature of quadratic bezier curve.
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  2. #2
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    Re: How to find normal vector of quadatic bezier curve, curvature of bezier curve.

    According to Wikipedia, a quadratic Bezier curves are given by begin/end points \vec{P_0}, \vec{P_2}, and a control point, \vec{P_1}:

    \vec{B}(t) = (1-t)^2\vec{P_0} + 2(1-t)t\vec{P_1} + t^2\vec{P_2}, \ t \in [0,1].

    From there, I let \vec{u} = \vec{P_0} - \vec{P_1}, and \vec{v} = \vec{P_2} - \vec{P_1}, giving

    \vec{B}(t) = \vec{P_1} + (1-t)^2\vec{u} + t^2\vec{v}, \ t \in [0,1], and

    \frac{d\vec{B}(t)}{dt} = -2(1-t)\vec{u} + 2t\vec{v}, \ t \in [0,1].

    Thus one normal would be \vec{N}(t) = (-(1-t)\vec{u}_y + t\vec{v}_y)\hat{i} - (-(1-t)\vec{u}_x + t\vec{v}_x) )\hat{j}, \ t \in [0,1] , and

    the signed curvature is k = \frac{<\vec{u} \times \vec{v}, \hat{k}>}{2 \left[ \ \lVert\vec{u}+\vec{v}\rVert^2 \ t^2 - 2 (\Vert\vec{v}\rVert^2 + <\vec{u}, \vec{v}>) \ t + \Vert\vec{v}\rVert^2 \ \right]^{3/2}}.

    Note: <\vec{u} \times \vec{v}, \hat{k}> is just a vector way of writing \vec{u}_x \vec{v}_y - \vec{u}_y \vec{v}_x.

    I also pulled the formula for the signed curvature off wikipedia, though that should be easy to find everywhere.

    I certainly could have made a mistake somewhere, so I'd suggest you work it out yourself. If you get the same thing, then it's a good bet neither of us made a mistake.
    Last edited by johnsomeone; September 28th 2012 at 08:35 PM.
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