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!
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$\begingroup$ Use mathjax. Click on this tutorial $\endgroup$ – Arbuja May 7 '17 at 13:23
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Consider the following diagram:
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
