It might come from the value of t used for this particular example?
jgforsberg
I think the professor said he was defining the tangent vectors as being 3 times the difference p(i+1) and p(i).
If you recall from discussion, the curve is defined as
B(t) = (s^3)p0 + 3(ts^2)p1 + 3(st^2)p2 + (t^3)p3 where s = 1-t.
Maybe the 3's come from the coefficients of the equation?
x-fa19
I agree with jgforsberg; tacking a little more onto their answer, this is because the derivative is the slope of the tangent.
Courtesy of Wikipedia: "A straight line is said to be a tangent of a curve y = f (x) at a point x = c on the curve if the line passes through the point (c, f (c)) on the curve and has slope f'(c) where f' is the derivative of f."
sphindle1
I'm a little bit confused about this part, since it seems the bezier curve doesn't go through p1 and p2. Correct me if I'm wrong, but I thought we wanted a curve that goes through all four points?
orkun1675
@sphindle1 When we were doing Catmull-Rom we said that the curve was not defined for p0 and p4, by definition. Those points only helped set the tangent at p1,p3. In a way, they were "meta-points". Here, the idea is the same. Its just swapped around, so now the p1,p2 are "meta-points". I think this is a better design choice because as you see on the next slides it is much more intuitive to edit in a vector drawing program such as Illustrator or CAD.
What is the reasoning behind the 3 coefficient?
It might come from the value of t used for this particular example?
I think the professor said he was defining the tangent vectors as being 3 times the difference p(i+1) and p(i). If you recall from discussion, the curve is defined as B(t) = (s^3)p0 + 3(ts^2)p1 + 3(st^2)p2 + (t^3)p3 where s = 1-t. Maybe the 3's come from the coefficients of the equation?
I agree with jgforsberg; tacking a little more onto their answer, this is because the derivative is the slope of the tangent.
Courtesy of Wikipedia: "A straight line is said to be a tangent of a curve y = f (x) at a point x = c on the curve if the line passes through the point (c, f (c)) on the curve and has slope f'(c) where f' is the derivative of f."
I'm a little bit confused about this part, since it seems the bezier curve doesn't go through p1 and p2. Correct me if I'm wrong, but I thought we wanted a curve that goes through all four points?
@sphindle1 When we were doing Catmull-Rom we said that the curve was not defined for p0 and p4, by definition. Those points only helped set the tangent at p1,p3. In a way, they were "meta-points". Here, the idea is the same. Its just swapped around, so now the p1,p2 are "meta-points". I think this is a better design choice because as you see on the next slides it is much more intuitive to edit in a vector drawing program such as Illustrator or CAD.