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Differential of Bezier curve

  1. Apr 16, 2007 #1
    Hi guys,
    I am plotting a Bezier curve in a computer program I've written. The curve is created using four control points to generate the coefficients to the following equations:

    x = ax*t^3 + bx*t^2 + cx * t + dx
    y = ay*t^3 + by*t^2 + cy * t + dy

    Where t a value between 0 - 1 to evaluate along the Bezier curve.

    However, I need to know what the differential of this curve is, not with respect to t, but with respect to x. i.e. dy/dx at a value of x that you choose.

    The reason for this is that I'm plotting a Bezier through some experiemental data and I need a mathematical representation of that data so I can get a smooth differential and double differential. But I've now come across this problem.
    The only way I've thought of to do this so far is to, at your given x value, involves solving the cubic for t, which is quite expensive and diffucult.

    Any suggestions?
     
  2. jcsd
  3. Apr 16, 2007 #2

    HallsofIvy

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    If x= f(t) and y= g(t), then dx/dt= f'(t), dy/dt= g'(t) and, by the chain rule,
    dy/dx= (dy/dt)/(dx/dt)= g'(t)/f'(t).
     
  4. Apr 18, 2007 #3
    Thanks Halls, thats the stuff I was looking for.
    Furthermore, in order to calculate the differentials, I need to find t. I used the method illustrated over at www.nr.com (specifically http://www.nrbook.com/a/bookcpdf/c5-6.pdf) which seems to work well. It may be quicker to solve using bisection or something, but I didnt want to enter the realm of tolerance limits and the like. Explicit is better and performance is quite reasonable.
     
  5. Apr 18, 2007 #4

    hotvette

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    I'm also trying to fit a Bezier (cubic) to data points, but have so far not been able to figure it out. I'm trying a least squares approach but can't figure out how to formulate the problem. It is easy if the x-values of the 4 control points are already known, but in general they aren't.

    Can you tell me how you are doing it?

    By the way, in terms of finding the value of t for a specific x, Newton-Raphson works well (the higher derivative methods even better).
     
    Last edited: Apr 18, 2007
  6. Apr 18, 2007 #5
    I'm taking a slightly perculiar approach to this one; the cubic Bezier fitting is all done interactively with the user. The user clicks the mouse to add/delete Bezier knots. Each knot can be moved and each knot also has handles to change the Bezier shape. Its not mathematically rigorous, but I think its the best way to do it, at least for my particular case.
     
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