Re: [Devel] Problems with bbox code and cubic bezier curves

[David Turner]

Hi Tom,

> 
> Hey, you aren't the only one who has been flexing (or trying to flex) those 
> old
> rusty mathematical muscles!  Anyway, I must admit despite a degree in math, I 
> do
> not understand the new code, other than the general gist of things (solve for
> extrema, and test the bbox at the endpoints and the extrema). Can you tell me
> what reference you used for the code?  I had asked about the particular
> parametric equations used to describe the bezier arcs, but no one took the 
> bait
> and answered.
> 
> If I had that (the pararmetric equations), I could follow the code a lot 
> easier.
>

OK, no problem:

  for quadratic (conic) beziers, were P1, P2 and P3 are the control points:

    P(t) = P1 + 2t.(P2-P1) + t^2.(P3-2P2+P1)
   
  for cubics:

    P(t) = P1 + 3t.(P2-P1) + 3t^2.(P3-2P2+1) + t^3.(P4-3P3+3P2-P1)

A quick rule of thumb to reconstruct the polygon, or even to
easily draw an arbitrary point of a conic or cubic by hand:

  for t in (0..1)

  for conics:

     let P1, P2, P3 be the control points,

     define  Q1(t) = P1 + t.(P2-P1)
             Q2(t) = P2 + t.(P3-P2)

     then   P(t) = Q1 + t.(Q2-Q1)

  for cubics:

     let P1, P2, P3 and P4 be the control points,

     define  Q1(t) = P1 + t.(P2-P1)
             Q2(t) = P2 + t.(P3-P2)
             Q3(t) = P3 + t.(P4-P3)

     define  R1(t) = Q1 + t.(Q2-Q1)
             R2(t) = Q2 + t.(Q3-Q2)
     
     then    P(t) = R1 + t.(R2-R1)

  you can apply this "rule" recursively to build
  higher-order beziers too :-)

  Also note that all Q1, Q2, Q3, R1, R2 are control points
  for the sub-arcs P1..P(t) and P(t)..P3 (or P(t)..P4) !!

  Beziers have _tons_ of weird and useful properties,
  I love them :-)

Regards,

- David