Path: utzoo!attcan!utgpu!jarvis.csri.toronto.edu!mailrus!wuarchive!gem.mps.ohio-state.edu!apple!motcsd!hpda!hpcupt1!schi
From: schi@hpcupt1.HP.COM (Shan Chi)
Newsgroups: comp.graphics
Subject: Re: Bezier curves
Message-ID: <6350003@hpcupt1.HP.COM>
Date: 14 Sep 89 16:57:46 GMT
References: <273@marvin.moncam.co.uk>
Organization: Hewlett Packard, Cupertino
Lines: 40

I must have been dreaming while I was reading the question.  The previous
response by me was only for the case when the splitting point is at t=0.5.  Now
here is what I think the right answer.

-----------------------------------------------------------------------------
Let the Bezier curve be P(t) = T M P,  0 <= t <= 1
where
	T = [t^3 t^2 t 1],   M = [-1  3  -3  1]   and P = [ P0 ]
                                 [3  -6   3  0]           [ P1 ]
                                 [-3  3   0  0]           [ P2 ]
                                 [ 1  0   0  0]           [ P3 ]


Suppose we split at t=x, then the two new curves are:

Q(t) = P(xt) = T Dq M P, 0 <= t <= 1, and
R(t) = P((1-x)t+x) = T Dr M P, 0 <= t <= 1

where  Dq = [x^3 0   0  0]   and Rq = [(1-x)^3    0        0       0] 
	    [0  x^2  0  0] 	      [3x(1-x)^2  (1-x)^2  0       0] 
	    [0   0   x  0] 	      [3(1-x)x^2  2x(1-x)  (1-x)   0]
	    [0   0   0  1] 	      [x^3        x^2      x0      1] 

Now we rearrange them into Bezier form.

Q(t) = T M M^-1 Dq M P = T M (M^-1 Dq M P),  and
R(t) = T M M^-1 Dr M P = T M (M^-1 Dr M P)

where M^-1 is the inverse matrix of M.
Therefore,  

Pq = M^-1 Dq M P,   and
Pr = M^-1 Dr M P

are the two sets of new control points.

Note that (M^-1 Dq M) and (M^-1 Dr M) are functions of x and can be
pre-computed if x is relatively constant.

Hope this time I am not dreaming. :-)