DERIVE source

The length of a Be\'zier curve is an elliptic integral. This is a Be\'zier curve.

x=a3 t3+a2 t2+a1 t+x0

y=b3 t3+b2 t2+b1 t+y0

The control points of a Be\'zier curve are:

(x0,y0)    (x1,y1)    (x2,y2)    (x3,y3)

This is the curve for those control points.

a1=3 (x1-x0)    a2=3 (x0-2 x1+x2)    a3=-x0+3 x1-3 x2+x3

b1=3 (y1-y0)    b2=3 (y0-2 y1+y2)    b3=-y0+3 y1-3 y2+y3

The length of the Be\'zier curve from 0 to u is:

ó õ u 0  Ö   (a1+2 a2 t+3 a3 t2)2+(b1+2 b2 t+3 b3 t2)2    dt

The control points of an example Be\'zier curve are:

(x0=6,y0=8)    (x1=1,y1=10)    (x2=7,y2=3)    (x3=4,y3=4)

Solve for a₁,a₂,a₃,b₁,b₂,b₃.

a1=-15    a2=33    a3=-20    b1=6    b2=-27    b3=17

The Be\'zier curve for this example is:

x=-20 t3+33 t2-15 t+6    y=17 t3-27 t2+6 t+8

For this example the length is:

3  ó õ 1 0  Ö   689 t4-1492 t3+1076 t2-292 t+29    dt

Numerical integration gives 7.237223368328592885826619210956022413967067986017504163362886516

DERIVE factored the polynomial:

689 t4-1492 t3+1076 t2-292 t+29=

689  æ ç ç è t+   æ Ö Ö 6005 1378 +  23377 474721   -  373 689 +i  æ ç ç è   æ Ö Ö 6005 1378 -  23377 474721   +  7 689 ö ÷ ÷ ø ö ÷ ÷ ø

æ ç ç è t+   æ Ö Ö 6005 1378 +  23377 474721   -  373 689 -i  æ ç ç è   æ Ö Ö 6005 1378 -  23377 474721   +  7 689 ö ÷ ÷ ø ö ÷ ÷ ø

æ ç ç è t-   æ Ö Ö 6005 1378 +  23377 474721   -  373 689 +i  æ ç ç è   æ Ö Ö 6005 1378 -  23377 474721   -  7 689 ö ÷ ÷ ø ö ÷ ÷ ø

æ ç ç è t-   æ Ö Ö 6005 1378 +  23377 474721   -  373 689 +i  æ ç ç è  7 689 -   æ Ö Ö 6005 1378 -  23377 474721   ö ÷ ÷ ø ö ÷ ÷ ø

Now define values k, m, a, b:

k=3  Ö   689      m=[1,1,1,1]    b=[-1,-1,-1,-1]    e=[e1,e2,e3,e4]

a=[0.216589-0.0937743 i,0.216589+0.0937743 i, 0.866139-0.0734550 i,0.866139+0.0734550 i]

Do a partial fraction expansion.

I(m)=D219(1,m,4,a,b,e)

I(m)=I(4 e1)+(1.29909+0.375097 i) I(3 e1)+(0.383342+0.365466 i) I(2 e1)

-(0.0228475-0.0784923 i) I(e1)

Reduce I(4 e₁).

R35(0,1,4,a,b,e)=0

A(2 e1+e2+e3+e4)+3 I(4 e1)+(3.24774+0.937743 i) I(3 e1)+(0.766684+0.730933 i) I(2 e1)

-(0.0342712-0.117738 i) I(e1)=0

A(2 e1+e2+e3+e4)=-0.138815+0.00794130 i

Substitute I(4 e₁).

I(m)=(0.216516+0.0625162 i) I(3 e1)+(0.127780+0.121822 i) I(2 e1)

-(0.0114237-0.0392461 i) I(e1)+0.0462716-0.00264710 i

Reduce I(3 e₁).

R35(-1,1,4,a,b,e)=0

A(e1+e2+e3+e4)+2 I(3 e1)+(1.94864+0.562645 i) I(2 e1)

+(0.383342+0.365466 i) I(e1)-(0.0114237-0.0392461 i) I(0)=0

A(e1+e2+e3+e4)=-0.0846852

Substitute I(3 e₁).

I(m)=-0.0655894 I(2 e1)-(0.0415000+0.0123012 i) I(e1)

+(0.00246348-0.00389163 i) I(0)+0.0554395

Reduce I(2 e₁).

R35(-2,1,4,a,b,e)=0

A(e2+e3+e4)+I(2 e1)+(0.0114237-0.0392461 i) I(-e1)

+(0.649549+0.187548 i) I(e1)=0

A(e2+e3+e4)=-0.949300-0.327220 i

Substitute I(2 e₁).

I(m)=(0.000749279-0.00257413 i) I(-e1)+0.00110358 I(e1)

+(0.00246348-0.00389163 i) I(0)-0.00682448-0.0214622 i

Calculate the basic integrals numerically.

I(-e1)=-17.2709+33.5142 i    I(0)=12.0414   I(e1)=-3.86310-1.12917 i

Substitute the values of the integrals.

I(m)=0.0919054       k I(m)=7.23722

That is a good match.

Jim FitzSimons Mailto:cherry@neta.com