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Lecture 7: Intro to Geometry, Splines, and Bezier Curves (76)
Unicorn53547

I am assuming Bezier is just a transformation of the hermite. So why do we choose the coefficient 3 (i.e., $3(p_1 - p_0)$ as tangent, but not $2, 4$ or other numbers)? Any particular reason that $3$ will make the curve fit better?

jacklishufan

I believe this can be derived from the Linear Interpolation of quadratic bezier. Namely for lines we define L_1=(1-t)p_0+t p_1; L_2=(1-t)p_1+t p_2 ;L_3=(1-t)p_2+t p_3; Then for quadratic we define B_(0,1,2)=(1-t)L_1(t)+t L_2(t);B_(1,2,3)=(1-t)L_2(t)+t L_3(t); Then the cubic is the linear interpolation of the two quadratic bezier, which can be explicitly written as (1-t)^3p0+3(1-t)^2t p1+3(1-t)t^2p2+t^3p3. Taking the first order differential and evaluate at t=0 and t=1 we have 3p1-3p0 and 3p3-3p2

jacklishufan

Also can the stuff confirm if there are typo on this page. In the graph the points are indexed as p0,p1,p2,p3 , but in the differential equation we have 3(p4-p3), what is p4 (the 5th point???)

Staffjamesfobrien

Sorry, p3-p2.

The 3 could be any number except zero, but by using 3 the basis functions end up summing to 1 and we get convex hull property, which is nice.

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