To split at a general s=t, it is easier to manipulate coefficient rather than control points. We can construct a bezier for the [0,t] portion by substituting s=ut to obtain coefficients (a₀, a₁t, a₂t², a₃t³). The [t,1] portion is given by the coefficients of u in a(t + u(1-t)).

Bezier Approximations


To Cubic
    The cubic a₃s³ + a₂s² + a₁s + a₀ is precisely equivalent to a Bezier with control points
    p₃ = (a₀+a₁+a₂+a₃)/4 ; p₂ = (3a₀+2a₁+a₂)/3 ; p₁ = (3a₀+a₁)/3 ; p₀ = a₁ .

To Circular Arc
   With acknowledgment to the work of David Seal
    Suppose we wish to construct a 2D bezier curve   which approximates the unit circular arc from ( cosa,- sina) to ( cosa, sina).
    If we impose a second order fit at the arc endpoints then our bezier must have control points
    p₀ = ( cosa,- sina) ; p₁ = p₀ + b( sina, cosa) ; p₂ = p₃ + b( sina,- cosa) ; p₃ = ( cosa, sina) ;
    whence the curve is given by

p1(s)	=	3b sinas(1-s) + cosa
p2(s)	=	(-4 sina +6b cosa)s3   + (6 sina - 9b cosa)s2 + 3b cosas - sina
	=	(2s-1)((2 sina-3b cosa)s(1-s) + sina)
	=	(2s-1)((1+2s(1-s)) sina -3b cosas(1-s))
		Writing z = s(1-s)
p1(s)	=	3b sinaz + cosa
p2(s)	=	(2s-1)((1+2z) sina -3b cosaz)

    Consider the error terms g₂(s) = p(s)² - 1 and g(s) = |p(s)| - 1.
    g₂(s)= p(s)²-1 = (|p(s)|+1)g(s) = (2+g(s))g(s) whence g(s) » g₂(s)/2.

g2(s)	=	(3b sinaz + cosa)2 +   (2s-1)2((1+2z) sina -3b cosaz)2 -1
	=	(9b2sin2az2 + cos2a) -1 + (1-4z)((1+2z)2sin2a -6b cosaz(1+2z) sina +9b2cos2az2)
	=	(9b2sin2az2 + cos2a) -1 + (1-4z)(z2(4sin2a - 12b cosa sina + 9b2cos2a) +z(4sin2a -6b cosa sina) + sin2a)
	=	z3(48b cosa sina - 16sin2a - 36b2cos2a)
		+z2( 9b2sin2a + (4sin2a - 12b cosa sina + 9b2cos2a) - 4 ( 4 sin2a -6b cosa sina ))
	=	-4z3(3b cosa-2 sina)2 + z2(9b2 + 12b cosa sina - 12sin2a)

dg2/ds	=	(1-2s)(-12z2(3b cosa-2 sina)2 +2z(9b2 + 12b cosa sina - 12sin2a))
	=	(1-2s)z(-12z(3b cosa-2 sina)2 +2(9b2 + 12b cosa sina - 12sin2a))
	=	6(1-2s)z(-2z(3b cosa-2 sina)2 + (3b2 + 4b cosa sina - 4sin2a))

    Hence dg₂/ds = 0 when s=0,1, ½, or when z = (3b² + 4b cosa sina - 4sin²a))/(-2z(3b cosa-2 sina)².     These give g₂(s) = 0, 0, (-(3b cosa-2 sina)²+(9b²+12b cosa sina-12sin²a))/16 = (9b²sin²a+24b cosa sina-16sin²a)/16, and
(3b²+4b cosa sina-4sin²a)³/(4(3b cosa-2 sina)⁴) repsectively.  

    We can ensure g₂(½)=0 by choosing b₀ such that p₁(½)=1 Þ 3b₀ sina½(1-½) + cosa = 1 Þ b₀ = 4(1- cosa)/(3 sina).
    The maximal error is then

g2max	=	(3b02+4b0 cosa sina-4sin2a)3/ (4(3b0 cosa-2 sina)4)
	=	(16(1- cosa)2/(3sin2a)+16(1- cosa) cosa/3-4sin2a)3 / (4(4(1- cosa) cosa/ sina-2 sina)4)
	=	...
	=	(1- cosa)3/(27(1+ cosa))
Þ gmax	=	(1- cosa)3/(54(1+ cosa))
	»	(1- cosa)3/108   when a small

a	gmax	Pixel radius for > ½ pixel error
p/2	0.019	27
p/3	.0015	324
p/4	2.8 x 10-4	1831
p/8	4.24 x 10-6	117770
p/16	6.6 x 10-8	7538744
d	d6/864	432 d-6

To General 1-D Curve
    Suppose we wish to construct a 1D bezier curve q(s)=s³q₃ + s²q₂ + sq₁ + q₀ with given values q(0), q(1) and gradients q'(0), q'(1) .

æ

1

1

1

1

ö

æ

q0

ö

=

æ

q(1)

ö

ç

0

1

2

3

÷

ç

q1

÷

ç

q'(1)

÷

ç

0

1

0

0

÷

ç

q2

÷

ç

q'(0)

÷

è

1

0

0

0

ø

è

q3

ø

è

q(0)

ø

is easily solved by q₀ = q(0) ; q₁ = q'(0) ; q₂ = 1/3(q'(1)-2q₁-q₀) = 1/3(q'(1)-2q'(0)-q(0)) = ; q₃ = q(0)-q₀-q₁-q₂ = -(q₁+q₂) = -1/3(q'(1)+q'(0)-q(0)) .

To General N-D Curve
    Suppose we wish to construct a 2D bezier curve q(s)=s³q₃ + s²q₂ + sq₁ + q₀ through given points q(0), q(t) and q(1).
    Suppose we also   require q'(0) = m₀ g₀ , q'(1)= m₁g₁ where g₀ and g₁ are given. If g₀ and g₁ are non parallel this may be achieved. If g₀ and g₁ are (close to) parallel then a (viable) fit is only possible if q(0), q(t) and q(1) are colinear and in accordance with the desired gradient.

æ

1

t

t2

t3

ö

æ

q0

ö

=

æ

q(t)

ö

ç

1

1

1

1

÷

ç

q1

÷

ç

q(1)

÷

ç

0

1

2

3

÷

ç

q2

÷

ç

m1g1

÷

ç

0

1

0

0

÷

è

q3

ø

ç

m0g0

÷

è

1

0

0

0

ø

è

q(0)

ø

Substituting q₀ = q(0), then q₁ = m₀g₀, and finally q₂ = q(1)-q₀-m₀g₀ - q₃ we obtain:

æ	t	t2	t3	ö	æ	q1	ö	=	æ	q(t) - q0	ö
ç	1	1	1	÷	ç	q2	÷		ç	q(1)-q0	÷
ç	1	2	3	÷	è	q3	ø		ç	m1g1	÷
è	1	0	0	ø					è	m0g0	ø

Substituting q₁ = m₀g₀ :

æ	t2	t3	ö	æ	q2	ö	=	æ	q(t) - q0 - tm0g0	ö
ç	1	1	÷	è	q3	ø		ç	q(1)-q0-m0g0	÷
è	2	3	ø					è	m1g1-m0g0	ø

Substituting q₂ = q(1)-q₀-m₀g₀ - q₃ and then eliminating q₃:

	t2(q(1)-q0-m0g0 - q3) + t3q3	=	q(t) - q0 - tm0g0
	2(q(1)-q0-m0g0 - q3) + 3q3	=	m1g1-m0g0
Þ	m0g0(1-t)/t - m1g1	=	q(t)/t2(1-t) + q(1)(2t-3)/(1-t) - q0(2t3 - 3t2 +1)/t2(1-t))

    If t is known and g₀ and g₁ are not parallel this may be solved for m₀,m₁ whereupon we can set

• q₀ = q(0)
• q₁ = m₀g₀
• q₃ = m₁g₁ + q₁ -2q(1) +2q₀
• q₂ = q(1) - q₃ - q₁ - q₀

    What is a sound choice for t ?
    t =½ is an obvious candidate, the equations for m₀ and m₁ becoming
m₀g₀ - m₁g₁ = 8q(½) - 4q(1) - 4q₀
    The point of least curvature is given by minimising q"(t)² = (6q₃t + 2q₂)² which is minimised by t = -q₂.q₃/3 so if this lies comfortably inside (0,1) it may be a good choice. Please let me know if progress is made in this direction.