How do I convert the 2 control points of a cubic curve to the single control point of a quadratic curve?
bezier, geometry
Solution
As mentioned, going from 4 control points to 3 is normally going to be an approximation. There's only one case where it will be exact - when the cubic bezier curve is actually a degree-elevated quadratic bezier curve.
You can use the degree elevation equations to come up with an approximation. It's simple, and the results are usually pretty good.
Let's call the control points of the cubic Q0..Q3 and the control points of the quadratic P0..P2. Then for degree elevation, the equations are:
Q0 = P0
Q1 = 1/3 P0 + 2/3 P1
Q2 = 2/3 P1 + 1/3 P2
Q3 = P2
In your case you have Q0..Q3 and you're solving for P0..P2. There are two ways to compute P1 from the equations above:
P1 = 3/2 Q1 - 1/2 Q0
P1 = 3/2 Q2 - 1/2 Q3
If this is a degree-elevated cubic, then both equations will give the same answer for P1. Since it's likely not, your best bet is to average them. So,
P1 = -1/4 Q0 + 3/4 Q1 + 3/4 Q2 - 1/4 Q3
To translate to your terms:
controlX = -0.25*startX + .75*control1X + .75*control2X -0.25*endX
Y is computed similarly - the dimensions are independent, so this works for 3d (or n-d).
This will be an approximation. If you need a better approximation, one way to get it is by subdividing the initial cubic using the deCastlejau algorithm, and then degree-reduce each segment. If you need better continuity, there are other approximation methods that are less quick and dirty.
Problem
Having searched the web, I see various people in various forums alluding to approximating a cubic curve with a quadratic one. But I can't find the formula. What I want is this: input: startX, startY, control1X, control1Y, control2X, control2Y, endX, endY output: startX, startY, controlX, controlY, endX, endY Actually, since the starting and ending points will be the same, all I really need is... input: startX, startY, control1X, control1Y, control2X, control2Y, endX, endY output: controlX, controlY