Comment on Curvature Beziers: Improving on a timeless recipeparentComments−unconed13dIf you like that, then try this.Take a row of pascal's triangle, add a zero in front (shift it to the right by one place), and divide it by the row below it. 0 1 4 6 4 1 / 1 5 10 10 5 1 = ... From this you can prove that a linear bezier line (of any degree) must have equally spaced control points.−dahart13dBy line you mean a Bézier curve with no acceleration? As opposed to (say) a cubic with control points [(1,1),(2,1),(8,1),(9,1)] which makes a straight line but has a non-uniform parameterization?−unconed13dYes, a linearly parametrized line.
Comments
If you like that, then try this.
Take a row of pascal's triangle, add a zero in front (shift it to the right by one place), and divide it by the row below it.
From this you can prove that a linear bezier line (of any degree) must have equally spaced control points.By line you mean a Bézier curve with no acceleration? As opposed to (say) a cubic with control points [(1,1),(2,1),(8,1),(9,1)] which makes a straight line but has a non-uniform parameterization?
Yes, a linearly parametrized line.