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:
$c_0 = (1,1)$
$c_1 = (3,4)$
$c_2 = (7,5)$
$c_3 = (8,2)$
I shall continue this curve $C$ with Curve $D$ from control point $c_3$ to a control point $d_3 = (12,1)$ so that this curve is two-times continuously differentiable.
First Task: Determine control points $d_0, d_1, d_1$ for the new curve. Second Task: Specify a piecewise defined formula for the new curve $G(v)$ with $v$ out of $[0,1]$ that passes through $c_0,c_3$ and $d_3$. 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. Tank You!
The second derivative at the end of the red curve is $6T = 6(S-R)$. The second derivative at the start of the green curve is $6U = 6(V-W)$. So we just have to arrange things so that $T=U$.
The details: compute as follows:
R = C2 - C1 S = C3 - C2 T = S - R
Obviously we need to set $D0 = C3$. Also, to get continuity of first derivative, it's easiest if we have $D1 - D0 = C3 - C2$. So we do this:
D0 = C3 W = S D1 = D0 + W U = T V = U + W D2 = D1 + V