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Is there an explicit form for cubic Bézier curves?

I want to calculate Y for any given X of a bezier to help me chart a graph.

X represents time and Y represents distance from an object.

I got this formula for cubic beziers:

$$X(t) = (1-t)^3 X_0 + 3(1-t)^2 t X_1 + 3(1-t) t^2 X_2 + t^3 X_3$$

$$Y(t) = (1-t)^3 Y_0 + 3(1-t)^2 t Y_1 + 3(1-t) t^2 Y_2 + t^3 Y_3$$

I need to know what $t$ is for $X$ to I can work out what $Y$ would be for $t$... right?

So I need to rearrange this formula with $t$ on the left, I think:

$$X(t) = (1-t)^3 X_0 + 3(1-t)^2 t X_1 + 3(1-t) t^2 X_2 + t^3 X_3$$

Ummmm... is this right?


marked as duplicate by J. M. is a poor mathematician, William, rschwieb, sdcvvc, Emily Sep 27 '12 at 18:42

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  • $\begingroup$ I think that normally you'd start with t, find x and find y. I am not sure why you are starting with x? $\endgroup$ – NoChance Sep 6 '12 at 1:40

One way to graph $Y$ in terms of $X$ would be to choose { X0, X1, X2, X3 } so that $X( t ) = t$, so you can choose the control points { 0, 1/3, 2/3, 1 }.

  • $\begingroup$ I'm sorry that doesn't make sense to me. Can you possibly link me to a page with more detailed explanation? $\endgroup$ – Beakie Sep 6 '12 at 3:26
  • $\begingroup$ I found this... degrafa.googlecode.com/svn/branches/Origin/Degrafa/com/degrafa/… what technique do they use? (see xAtY) $\endgroup$ – Beakie Sep 6 '12 at 3:44
  • $\begingroup$ Sorry, I misunderstood your question then. The function xAtY first converts a quadratic bezier to power basis form. The resulting quadratic equation is solved to find the t corresponding to the given Y. The bezier curve is then evaluated at the two roots to get the X. $\endgroup$ – SKP Sep 6 '12 at 4:07

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