T_{E} X

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

x = a_{3} t^{3} + a_{2} t^{2} + a_{1} t + x_{0}

y = b_{3} t^{3} + b_{2} t^{2} + b_{1} t + y_{0}

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

(x_{0}, y_{0}) (x_{1}, y_{1}) (x_{2}, y_{2}) (x_{3}, y_{3})

This is the curve for those control points.

a_{1} = 3 (x_{1} - x_{0}) a_{2} = 3 (x_{0} - 2 x_{1} + x_{2}) a_{3} = - x_{0} + 3 x_{1} - 3 x_{2} + x_{3}

b_{1} = 3 (y_{1} - y_{0}) b_{2} = 3 (y_{0} - 2 y_{1} + y_{2}) b_{3} = - y_{0} + 3 y_{1} - 3 y_{2} + y_{3}

The length of the Be\'zier curve from

0

u

is:

\int_{0}^{u} \sqrt{(a_{1} + 2 a_{2} t + 3 a_{3} t^{2})^{2} + (b_{1} + 2 b_{2} t + 3 b_{3} t^{2})^{2}} dt

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

(x_{0} = 6, y_{0} = 8) (x_{1} = 1, y_{1} = 10) (x_{2} = 7, y_{2} = 3) (x_{3} = 4, y_{3} = 4)

Solve for

a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}

a_{1} = - 15 a_{2} = 33 a_{3} = - 20 b_{1} = 6 b_{2} = - 27 b_{3} = 17

The Be\'zier curve for this example is:

x = - 20 t^{3} + 33 t^{2} - 15 t + 6 y = 17 t^{3} - 27 t^{2} + 6 t + 8

For this example the length is:

3 \int_{0}^{1} \sqrt{689 t^{4} - 1492 t^{3} + 1076 t^{2} - 292 t + 29} dt

Numerical integration gives

7.237223368328592885826619210956022413967067986017504163362886516

DERIVE factored the polynomial:

689 t^{4} - 1492 t^{3} + 1076 t^{2} - 292 t + 29 =

689 (t + \sqrt{\frac{\sqrt{6005}}{1378} + \frac{23377}{474721}} - \frac{373}{689} + i (\sqrt{\frac{\sqrt{6005}}{1378} - \frac{23377}{474721}} + \frac{7}{689}))

(t + \sqrt{\frac{\sqrt{6005}}{1378} + \frac{23377}{474721}} - \frac{373}{689} - i (\sqrt{\frac{\sqrt{6005}}{1378} - \frac{23377}{474721}} + \frac{7}{689}))

(t - \sqrt{\frac{\sqrt{6005}}{1378} + \frac{23377}{474721}} - \frac{373}{689} + i (\sqrt{\frac{\sqrt{6005}}{1378} - \frac{23377}{474721}} - \frac{7}{689}))

(t - \sqrt{\frac{\sqrt{6005}}{1378} + \frac{23377}{474721}} - \frac{373}{689} + i (\frac{7}{689} - \sqrt{\frac{\sqrt{6005}}{1378} - \frac{23377}{474721}}))

Now define values

k, m, a, b

k = 3 \sqrt{689} m = [1, 1, 1, 1] b = [- 1, - 1, - 1, - 1] e = [e_{1}, e_{2}, e_{3}, e_{4}]

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) = D 219 (1, m, 4, a, b, e)

I (m) = I (4 e_{1}) + (1.29909 + 0.375097 i) I (3 e_{1}) + (0.383342 + 0.365466 i) I (2 e_{1})

- (0.0228475 - 0.0784923 i) I (e_{1})

Reduce

I (4 e_{1})

R 35 (0, 1, 4, a, b, e) = 0

A (2 e_{1} + e_{2} + e_{3} + e_{4}) + 3 I (4 e_{1}) + (3.24774 + 0.937743 i) I (3 e_{1}) + (0.766684 + 0.730933 i) I (2 e_{1})

- (0.0342712 - 0.117738 i) I (e_{1}) = 0

A (2 e_{1} + e_{2} + e_{3} + e_{4}) = - 0.138815 + 0.00794130 i

Substitute

I (4 e_{1})

I (m) = (0.216516 + 0.0625162 i) I (3 e_{1}) + (0.127780 + 0.121822 i) I (2 e_{1})

- (0.0114237 - 0.0392461 i) I (e_{1}) + 0.0462716 - 0.00264710 i

Reduce

I (3 e_{1})

R 35 (- 1, 1, 4, a, b, e) = 0

A (e_{1} + e_{2} + e_{3} + e_{4}) + 2 I (3 e_{1}) + (1.94864 + 0.562645 i) I (2 e_{1})

+ (0.383342 + 0.365466 i) I (e_{1}) - (0.0114237 - 0.0392461 i) I (0) = 0

A (e_{1} + e_{2} + e_{3} + e_{4}) = - 0.0846852

Substitute

I (3 e_{1})

I (m) = - 0.0655894 I (2 e_{1}) - (0.0415000 + 0.0123012 i) I (e_{1})

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

Reduce

I (2 e_{1})

R 35 (- 2, 1, 4, a, b, e) = 0

A (e_{2} + e_{3} + e_{4}) + I (2 e_{1}) + (0.0114237 - 0.0392461 i) I (- e_{1})

+ (0.649549 + 0.187548 i) I (e_{1}) = 0

A (e_{2} + e_{3} + e_{4}) = - 0.949300 - 0.327220 i

Substitute

I (2 e_{1})

I (m) = (0.000749279 - 0.00257413 i) I (- e_{1}) + 0.00110358 I (e_{1})

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

Calculate the basic integrals numerically.

I (- e_{1}) = - 17.2709 + 33.5142 i I (0) = 12.0414 I (e_{1}) = - 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

File translated from T_EX by T_TM, version 3.01.
On 5 Jul 2001, 00:33.