Subject: Re: is a point 'inside' a bezier?
On 7 May 2006 20:57:03 -0700, firstname.lastname@example.org wrote:
>Anyhow that equation looks pretty complicated for a simple implicit
>quadratic, especially since the result came out so simple. Can you
>elaborate on what you're doing?
An implicit curve is a concept from algebraic geometry, the set of
points at which a polynomial, or ratio of polynomials, evaluates to
zero. In general degree we have curves that can be represented only
implicitly, not parametrically. For degree 2, we can always choose
either form. Mathematicians prefer to work with projective geometry
and homogeneous coordinates, and so do we in general, not here; but
they also prefer to work with complex numbers, where we prefer reals.
Unlike the complex numbers, the reals are ordered. So we can treat the
implicit function as defining "sides", positive and negative. That's
what you originally asked for.
The only fly in the ointment is that you provide a parametric function
but we need its implicit form. I simply grabbed a conversion formula
from the first source that occurred to me.
Could we write down a formula without the determinants? Yes, but it
turns out to be rather long and messy, and not really that helpful.
For example, we have
L01(x,y) = 2 (-x0 y + x1 y + x y0 - x1 y0 - x y1 + x0 y1)
L12(x,y) = 2 (-x1 y + x2 y + x y1 - x2 y1 - x y2 + x1 y2)
L02(x,y) = -x0 y + x2 y + x y0 - x2 y0 - x y2 + x0 y2
Sorry, I'm not going to explain the theory of implicitization today.
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