3 ms·
I see, so basically assume the two end points are at zero and there is some rotation accounting for the endpoint offset in real space. It still doesn't seem fu
by saynsedit 10y ago
I see, so basically assume the two end points are at zero and there is some rotation accounting for the endpoint offset in real space.
It still doesn't seem fully accurate as I can imagine a non-rotated cubic curve with endpoints at an offset, but I assume your simplification works well enough.