"The linear interpolations (aka lerps) can be summarized into a single compact formula, e.g. for 4 control points (A,B,C,D):
γ(t)=A⋅(1−t)^3+B⋅3(1−t)^2⋅t+C⋅3(1−t)t^2+D⋅t^3
The rule is simple: descending powers of (1−t), ascending powers of t, with coefficients taken from the n'th row of Pascal's triangle.
I've never seen the connection between the equation for Bezier Curves (more specifically the linear equations of curves with N control points aka "binding points" / "points of stability" / "fixed points" / "immovable points", etc.) and Pascal's Triangle before!
By line you mean a Bézier curve with no acceleration? As opposed to (say) a cubic with control points [(1,1),(2,1),(8,1),(9,1)] which makes a straight line but has a non-uniform parameterization?
"(more specifically the linear equations of curves with N control points aka "binding points" / "points of stability" / "fixed points" / "immovable points", etc.)" -- there's a terminology mix-up here -- the control points (A, B, C, D) are
(obviously) variable / flexible in nature, but the values from Pascal's Triangle are (obviously!) bound/stable/fixed to their respective row!
So that's what I meant to say...
So my apologies for having messed-up the terminology there!
(You know what I meant! :-) )
Anyway, some very interesting responses and connections! Thanks to everyone who responded!
This comes from the binomial expansion of (X+y)^n, where X^k y^(n-k) has coefficient (n choose k). This is since you "choose" X in K of the (X+y)'s (in the Bezier case it's kind of writing (t+(1-t))^n, the abcd are to not make it equal to 1).
The choice function is also exactly the same as the pascal triangle!
Comments
γ(t)=A⋅(1−t)^3+B⋅3(1−t)^2⋅t+C⋅3(1−t)t^2+D⋅t^3
The rule is simple: descending powers of (1−t), ascending powers of t, with coefficients taken from the n'th row of Pascal's triangle.
I've never seen the connection between the equation for Bezier Curves (more specifically the linear equations of curves with N control points aka "binding points" / "points of stability" / "fixed points" / "immovable points", etc.) and Pascal's Triangle before!
Brilliant!
Great article, too!
If you like that, then try this.
Take a row of pascal's triangle, add a zero in front (shift it to the right by one place), and divide it by the row below it.
From this you can prove that a linear bezier line (of any degree) must have equally spaced control points.By line you mean a Bézier curve with no acceleration? As opposed to (say) a cubic with control points [(1,1),(2,1),(8,1),(9,1)] which makes a straight line but has a non-uniform parameterization?
Yes, a linearly parametrized line.
Quick update:
When I said:
"(more specifically the linear equations of curves with N control points aka "binding points" / "points of stability" / "fixed points" / "immovable points", etc.)" -- there's a terminology mix-up here -- the control points (A, B, C, D) are (obviously) variable / flexible in nature, but the values from Pascal's Triangle are (obviously!) bound/stable/fixed to their respective row!
So that's what I meant to say...
So my apologies for having messed-up the terminology there!
(You know what I meant! :-) )
Anyway, some very interesting responses and connections! Thanks to everyone who responded!
This comes from the binomial expansion of (X+y)^n, where X^k y^(n-k) has coefficient (n choose k). This is since you "choose" X in K of the (X+y)'s (in the Bezier case it's kind of writing (t+(1-t))^n, the abcd are to not make it equal to 1).
The choice function is also exactly the same as the pascal triangle!