C2 continuous Bezier contour.

Question

I am searching a way to make c2 continuous CLOSED bezier spline via given points. I found a perfect description for OPENED spline here. Briefly it is: If were describe path needed as $i$ segments, where each segment is cubic Bezier curve, with control points from $P_{0,i}$ to $P_{3,i}$, then C2 continuity conditions are:

$P_{0,i}=P_{3,i-1}$ ; ($C0$ continuity)

$2P_{0,i}=P_{1,i}+P_{2,i-1}$; ($C1$ continuity)

and

$-2P_{1,i}+P_{2,i}=P_{1,i-1}-2P_{2,i-1}$ ; ($C2$ continuity)

Until this moment everything is clear for me (even with my poor math background).
Later they use fact that, the spline becomes linear at the end points. But I need to create a contour:

$P_{0,0}=P_{3,n}$, and this complication burn my brain. I have no ideas how to deal with such equations.

Could you help me to find control point for c2 continuous closed bezier spline?

Answer 1

Here is a simple way for building a $C^2$ closed spline obtained by assembling cubic Bezier curves. See Fig. 1.

1) Begin by considering a "scaffolding" polygon $P_0P_1P_2\cdots P_{n-1}P_n$ that will give the general shape of the curve to come (we take the cyclic convention: $P_n=P_0$ that will as well be used for families of points $A_k$, $B_k$, $C_k$ that are to be defined).

2) On each side $[P_kP_{k+1}]$, consider the $2$ points situated resp. at the third and two-thirds of this side:

$$B_k = \tfrac13(2 P_k + P_{k+1}) \ \ \ \text{and} \ \ \ \ C_k = \tfrac13(P_k + 2 P_{k+1}).$$

3) Then define midpoints $A_{k+1} = \tfrac12(B_k + C_k).$

4) At last, draw the $n$ Bezier curves defined by points $(A_k, B_k, C_k, A_{k+1})$.

At junction points $A_k$, it is almost immediate to verify that the speed vectors of the two joined arcs are the same; it not difficult to prove either that the acceleration vectors of the two arcs are the same too.

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There is more to do. The process just described is convenient for curve approximations: for example, a designer who wants to build an egg-shaped curve has a satisfying answer.

But for "technical drawing", where the curve is constrained to pass through pre-determined points $A_k$, we have to build (if possible) points $P_k$ from points $A_k$. This is curve interpolation.

It is always possible. How ? We have just to understand Fig. 2 and the resulting system of equations (3).

Let us start from $6A_k$: this point can be expressed like this:

$$6A_k = 3C_{k-1}+ 3B_k = (P_{k-1} + 2P_k)+(2P_k + P_{k+1}) $$

(using definitions of $B_k$ and $C_k$), i.e.,

$$\tag{2}6A_k = P_{k-1} + 4P_k+ P_{k+1}$$

Gathering equations (2) for all values of $k$, and taking into account the cyclicity property, we get a system that, in the case of a 5 points, is:

$$\tag{3}\begin{cases}&4P_0 &+& P_1&+ &&&&& P_4 &=& 6A_0\\ &P_0 &+& 4P_1&+& P_2 &&&&&=& 6A_1\\ &&&P_1 &+& 4P_2&+& P_3 &&&=& 6A_2\\ &&&&&P_2 &+& 4P_3&+& P_4 &=& 6A_3\\ &P_0 &+&&&&& P_3&+& 4P_4 &=& 6A_4\\ \end{cases}$$

System (3) is invertible (diagonally dominant matrix) ; therefore, being given points $A_k$, we can obtain all points $P_k$.

Fig. 3 shows the versatility of this method : the generated curve can have as well concave parts or even loops.

Here is a Matlab implementation of the method just detailed:

clear all;close all;axis tight;hold on;
z=5;axis(z*[-1,1,-1,1]);
[X,Y]=getpts(); % mouse selection of points Ak ...
n=length(X);A=[X,Y]; % ... coords placed into a n x 2 array
u=ones(n-1,1);mat=4*eye(n)+diag(u,1)+diag(u,-1);
mat(1,n)=1;mat(n,1)=1;mat=mat/6;P=inv(mat)*A;
P=[P;P(1,:)];A=[A;A(1,:)];
B=zeros(n,2);C=zeros(n,2);
for k=1:n;
   B(k,:)=(2*P(k,:)+P(k+1,:))/3;
   C(k,:)=(P(k,:)+2*P(k+1,:))/3;
end;
step=0.01;t=(0:step:1)';s=1-t;
s3=s.^3;s2t=3*s.^2.*t;t2s=3*t.^2.*s;t3=t.^3;
for k=1:n
   a=A(k,:);b=B(k,:);c=C(k,:);d=A(k+1,:);
   bez=s3*a+s2t*b+t2s*c+t3*d;
   plot(bez(:,1),bez(:,2),'color',.4+.4*rand(1,3),'linewidth',3)
end

Fig. 1.

Fig. 2.

Fig. 3.