寻找贝塞尔曲线的顶点
我正在使用 Flex，尽管我认为这是一个独立于语言的问题。我正在尝试使用 3 个点绘制一条曲线，使用 curveTo （二次贝塞尔函数，我不相信 Flex 有任何其他函数，如果有，请纠正我！）点 1 和 3 是“节点”，点2 是拖动手柄。
我想要的不是线向点 2 弯曲，而是实际上穿过它。我设法通过侥幸让它工作 - 通过将（点 1 和 3 之间的线的中点之间的距离）和点 2 加倍。
但这并没有把它放在线的顶点上，只是在附近的某个地方到它。
有人有什么想法吗？
安德鲁
I'm working in flex, although I reckon this is a language independent problem. I'm trying to draw a curve using 3 points, using curveTo (a quadratic bezier function, I don't believe Flex has any other, if it does, please correct me!) Points 1 and 3 are "nodes", with point 2 being a drag handle.
What I want is not for the line to curve towards point 2 but in fact pass through it. I've managed to get this working by fluking it - by doubling the (distance between the midpoint of a line between Points 1 and 3) and Point 2.
This doesn't put it on the Apex of the line though, just somewhere close to it.
Anyone any ideas?
Andrew
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二次贝塞尔曲线使用以下公式计算，
其中 P0、P1 和 P2 是您指定的 3 个点。曲线从 P0 开始，到 P2 结束
t 的范围为 0 到 1
应在 t = 0.5 时达到顶点
因此尝试将 P0、P2 和 t = 0.5 插入公式中，使其等于以下点：
你想要的顶点是并从公式中提取 P1
the quadric bezier curve is calculate using the formula
where P0,P1 and P2 are the 3 points you specify. The curve starts in P0 and ends in P2
t ranges from 0 to 1
the apex should be reached at t = 0.5
so try to insert P0, P2 and t = 0.5 into the formula set it equal to the point where
you want the apex to be and extract P1 from the formula
使用以下公式：
B'(t) = 3 (1 - t) 2 (P1 - P0) + 6 (1 - t) t (P2 - P1) + 3 t2 (P3 - P2)您可以使用导数来查找最大值和最小值。
Us this formula:
B'(t) = 3 (1 - t) 2 (P1 - P0) + 6 (1 - t) t (P2 - P1) + 3 t2 (P3 - P2)You can use the derivative to find maximums and minimums.
Bezier 样条线不会通过其控制点，但 Catmull Rom 样条线会通过。
B(t) = ((2*P1)+(-P0+P2)*t + (2*P0-5*P1+4*P2-P3)*t*t + (-P0+3* P1-3*P2+P3)*t*t*t )) / 2
虽然这是三次样条而不是二次样条。你可以尝试制作
P1=P2A Bezier spline will not pass through its control points, but a Catmull Rom spline will.
B(t) = ((2*P1)+(-P0+P2)*t + (2*P0-5*P1+4*P2-P3)*t*t + (-P0+3*P1-3*P2+P3)*t*t*t )) / 2Although this is a cubic rather than quadratic spline. You could try making
P1=P2