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How to form a bezier curve?

  1. Aug 30, 2013 #1
    Sorry if i post this in the wrong spot. I am trying to form the curve of the half quadrant of a circle. And i wonder that how do we know which or where is our control point? For cubic bezier, the 2nd control point should be on the tangent line of the starting point and the 3rd control point should be on the tangent of the end point. But how do i know how far should i take?

    And how to combine them together? Do the controls points position change?
    Last edited: Aug 30, 2013
  2. jcsd
  3. Aug 30, 2013 #2


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  4. Aug 31, 2013 #3
    I have read some of the tutorials and now i still don't understand how to merge or join 2 curves become 1 without its shape deviate so much? I have tried a simple test with 2 degree-3 bezier curve which if they connected properly, they would form a half circle.

    lets say small curve a(t) is the curve from the second quadrant of the circle, b(t) is the first quadrant of a circle, And my c(t) is the combination of the a(t) and b(t), all of them are in degree-3. And how to satisfy all the conditions below

    a(0) = c(0),
    a(1) = b(0) = c(1/2),
    b(1) = c(1),

    a'(t = 0..1) = c'(t = 0..1/2),
    b'(t = 0..1) = c'(t = 1/2..1),

    a''(t = 0..1) = c''(t = 0..1/2),
    b''(t = 0..1) = c''(t = 1/2..1)

    All i can figure out right now is making 2 new unknown control points for c(t) which is (P1, P2) and the P0 and P3 is the starting and ending point which is same as the a(0) and b(1). Then form 2 equation with the following condition,

    a(1) = b(0) = c(1/2)...i usee a(1) = c(1/2), b(t) isn't involve, and i think this shouldn't affect the curve


    a'(1) = c(1/2)

    this 2 condition to find the P1 and P2, and the result is

    Cx(t) = (-10)(1-t)3+(-5)(3t)(1-t)2+(5)(3t^2)(1-t)+(10)(t3)
    Cy(t) = (-10)(1-t)3+(10/3)(3t)(1-t)2+(10/3)(3t^2)(1-t)+(-10)(t3)

    the result isn't very desirable, because the curve is only correct at the center like how it shown in this page, http://www.fooplot.com/#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-, the blue curve is the c(t), and the red curve is the real circle

    for a(t) control point,
    A0 = (-10,-10)
    A1 = (-10,-5)
    A2 = (-7.5,0)
    A3 = (0,0)

    for b(t) control point,
    A0 = (10, 0)
    A1 = (7.5, 0)
    A2 = (10, -5)
    A3 = (10, -10)

    they re just approximation, not very "like" a circle yet.

    and c(t), control point
    P1 = (-10,-10)
    P2 = (-5, 10/3)
    P3 = ( 5, 10/3)
    P4 = (10, -10)

    I believe that if my c(t) is expressed in higher degree, with more control point, i would probably get a more accurate curve, but is it possible to get an accurate resultant curve in degree-3
    Last edited: Aug 31, 2013
  5. Sep 10, 2013 #4
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