Hi Jim. > With respect to finding a cubic root, currently you are doing that in > 2 dimensions, but what if we converted to 1 dimension? > Consider that the control polygon is "fairly linear". What if we > rotated our perspective so that it was horizontal and then squashed it > flat? Consider instead a 1 dimensional bezier with control values > of: > (where |mn| is the length of the m->n control polygon of the original > curve - sum of all segments from point m to point n) > 0.0, |01|, |02|, |03|
I had thought of something like this but I was afraid that the loss of Curve.java:141-152 would hurt accuracy. I implemented this though, and testing shows that that's not a problem. This should also double the performance of the computation since we only run one cubic root finder, and that was the major bottleneck. I updated the webrev. Should I remove some no longer needed methods, like getTCloseTo? > Solve that 1 dimensional bezier for v=(leaflen*polylen)/linelen... Don't you mean (targetLength - lenAtLastT) * polylen / leaflen? Regards, Denis.
