ETOOBUSY 🚀 minimal blogging for the impatient
Where I re-discover a useful source for information, and that contributions might get lost.
But… wait a minute! I already knew this place from before, because… I actually contributed to it about five years ago (I still have an oline version of the old site). Alas, in the meantime it underwent some re-writing, and it seems that my old contribution got lost 🤔
The contribution was actually a minor one, but it was enough to tickle the not-so-little nit-picker in me at the time. In section Splitting curves using matrices there is:
[…] the new end point is a mixture that looks oddly similar to a Bernstein polynomial of degree two
and my point is that the new end point is a Bernstein polynomial.
The key in this insight is that $z$ is actually the free variable in the parametric equations, which ranges in $[0,1]$. For this reason, $(z-1)$ is better expressed as $-(1-z)$, because it gives you an immediate view of what’s the real sign of the expression.
For this reason, then, the following expression:\[z^2 \cdot P_3 - 2 \cdot z \cdot (z-1) \cdot P_2 + (z - 1)^2 \cdot P_1\]
is best put as:\[(1-z)^2 \cdot P_1 + 2 \cdot (1-z) \cdot z \cdot P_2 + z^2 \cdot P_3\]
which also reveals its… Bernstein nature.
So there you have it… I know (where to find info on) Bézier curves! (And now you do too).