jnxd wrote: ↑Fri Jun 24, 2022 9:29 pm
Thing is, it seems you're mixing up the concepts of "Bezier curves" and "B-Splines". Bernstein polynomials form the basis for Bezier curves only. When weights are applied, we get what is called a
rational Bezier curve.
I think you make a mistake here. The polynomials of a B-spline that is formed out of e.g. 3 Bézier segments. We know it is not just a linear combination of the 3 polynomials defining the Béziers, but it is nevertheless a Bernstein polynomial -> the structure of the polynomial did not change.
As I wrote other basis functions would loose the properties the Bézier curves provide. So a B-spline with a basis does not consist of Bernstein polynomials is not that useful.
When you look at
rational Bézier curve you see that there the b(t) functions are the Bernstein polynomials, see their definition:
https://en.wikipedia.org/wiki/B%C3%A9zi ... erminology). The additional factor like the weight does not change this.
jnxd wrote: ↑Fri Jun 24, 2022 9:29 pm
In this case, the first curve spans a 0<=t<0.5, and the second spans 0.5<=t<=1.0. We define the functions to be zero outside of their respective ranges. Now, if we consider the pole at t = 0.5 to be the same for both curves, we have 7 poles, and this is equivalent to a cubic spline with a 3-multiplicity knot at t=0.5. However, if we want continuity at higher derivatives, we have to impose more restrictions on these poles. This effectively reduces the number of poles, since one new point dictates two other points at every derivative continuity imposed. I will not go into the details of how, but the lectures we have been discussing can shed some light on that.
But that does not prove the basis are no longer Bernstein polynomials. Let's look at a standard quadratic polynomial:
f(t) = A*x^2 + B*x + C
When you add a factor W (weight) to it you get:
f(t) = W*A*t^2 + W*B*t + W*C
this is still a quadratic polynomial:
f'(t) = A'*t^2 + B'*t + C'
Where A' = W*A
moving it to a different interval is easy:
f(t) = A'*(t-s)^2 + B'*(t-s) + C'
where s is the amount the curve is shifted to greater t values.
The result is nevertheless a quadratic polynomial.
I had again a look and I cannot see neither in the Wikipedia nor in the videos that the basis functions are no longer Bernstein polynomials. They are still in the form
b(t) = (n over i) * t^i *(1 - t)^(n - i)
Where do you see that this form is no longer there?
Edit: the B(t) function in
https://en.wikipedia.org/wiki/B%C3%A9zi ... ier_curves is of course no longer a Bernstein polynomials, but the b(t) are still ones.