However Mortenson points out that we are better if we minimize the error rather than allow all the error to be on one side of the circle.
The if the naive geometry solution is the same we get:
And finally we have Mortensen's solution to it, by making the metric for closest be the average error across the entire graph.
Well, solving this for 1 control point means doing the same thing he did. Which isn't super-trivial because it means calculating min and max error and adjusting various things. So I wrote a program to do it.
Your C points in Quad Bezier curves are:
Having the program I refigured it for Cubic:
Having a deviation in both directions of: ±0.00019607646987687817401874512914923 ....
Mortensen gave this as,
Since we might well be using doubles I'd give it as:
C0.5519150244935106,1 1,0.5519150244935106 1,0
C1,-0.5519150244935106 0.5519150244935106,-1 0,-1
C-0.5519150244935106,-1 -1,-0.5519150244935106 -1,0
C-1,0.5519150244935106 -0.5519150244935106,1 0,1
And the c value for the quad bezier curve as:
Q0.9251982088362565 0.9251982088362565 1,0
Q0.9251982088362565 -0.9251982088362565 0,-1
Q-0.9251982088362565 -0.9251982088362565 -1,0
Q-0.9251982088362565 0.9251982088362565 0,1
This is much better than the more naive value: 0.91421356237 which is effectively unusable. While Mortensen's use for the cubic is great, it changes the quad naive to almost usable, generally not, but *almost* usable. The Mortensen-optimized value for quads is off by max 0.007767318 whereas the naive value is off by 0.010781424258 which is 28% better. Hm. That's the same value Mortensen got for the cubic.
Naive Quad: For comparison.