I have a task to join two bezier curves, so that the resulting curve is two-times continuously differentiable.
I have the cubic bezier Curve C with control points:
c0 = (1,1)
c1 = (3,4)
c2 = (7,5)
c3 = (8,2)
I shall continue this curve C with Curve D from control point c3 to a control point d3 = (12,1) so that this curve is two-times continuously differentiable.
First Task: Determine control points d0, d1, d1 for the new curve. Second Task: Specify a pecewise defined formula for the new curve G(v) with v out of [0,1] that passes through c0,c3 and d3. Thus connect curves C and D in v = 1/2. Third Task: Prove by calculation that the transition between C and D is two-times continuously differentiable.

Regarding first task: I don't know how to determine the points. Can someone help to do this? The rest of the tasks is then maybe something easier to do for me.


For 2 bezier curves to connect and be continuous at all, the last control point of the first curve (C in your case) must be the same as the first control point of the next curve (D). So c3 has to equal d0. That gives you C0 continuity or positional continuity.

The next step is to make sure that the tangents where the 2 curves meet are the same. That will ensure c1 continuity. (That is continuity of the first derivatives of the 2 curves.) To do this, the slope must be equal at both points. In order to do that, you need to ensure that the line from c2 to c3/d0 is continuous with the line from c3/d0 to d1.

I don't know of a good way to describe C2 continuity, but this Stack Overflow answer describes it better than I could. That answer should also explain the calculation you need for your proof.


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