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Hi,
Thanks for your response my friend. Actually i cant tell you the link. Its just for fun. However if somebody else ask me i would surely guide him. Actually i dont want to leave this thread incomplete.
Zulfi.
Hi,
I did the google and i found my answer is correct.
Thanks for your advice. Actually it was not that difficult that i goto SE.
Zulfi.
Hi,
Thanks for your response. I dont know whats "SE". I have provided you my solution. If its wrong plz let me know.
Zulfi.
Hi,
Thanks for your response again. I dont think that this question is beyond the limits of your forum. I hope some body must have the notes for the solution of this eq.
Zulfi.
Hi,
Thanks for your response. I have not studied Hermite curve. We have jmped directly on to the Bezeir curves. I would tell you what the book says about Pk
Suppose we are given n+1 control -point positions: PK=(xK, xk, zk) with k varying from 0 to n. These coordinate points can be blended to produce the position vector P(u) (in the image its C(u)) which describes the path of an approximating Bezier polynomial function between p0 and pn.
In my view Pk is a point in 3d. Plz guide me about my work. If you familiar with parametric eq of line you can guide me.
Zulfi.
Yes. BEZn,k are the Berntein Polynomials but i dont know about Pk. Some body plz check whether my work is correct or not?
Zulfi.
Hi,
I have tried to solve this eq. for n=1 (straight line):
P(u) = Σ Pk BEZk,n (u)
k=0 to n
BEZk,n(u) = C(n, k) uk (1-u)n-k, For n=1
BEZ0,1(u) = C(1, 0)u0 (1-u)1-0 = C(1,0) * (1-u) = 1!/(0!(1-0)! ) * (1-u) = 1-u)
BEZ1,1(u) = C(1,1)u1 (1-u)1-1 = C(1,1) u * (1-u)0 = 1!/(1! (1-1)!) * u = u
Now
P(u) = P0 BEZ0,1(u) + P1 BEZ1,1 (u) = (1-u)P0 + uP1
Plz guide me if this is correct or not. Also tell me what this answer tells me?? How its sum would be 1? plz show me.
Zulfi.
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