Re: NSBezierPath geometry question...

FROM : Robert Clair
DATE : Mon Apr 11 21:02:48 2005

>
> I've read this claim before but never understood it. Since a
> NSBezierPath is a cubic polynomial, why in the world would it have any
> trouble representing an arc, which is a quadratic polynomial?: just
> set the coefficients of the x^3 term to zero, and it IS a quadratic
> polynomial. (or in bezier speak: if the two off axis control points
> are in the same place, you've got a quadratic spline, which should be
> the same as a segment of arc.



You have mixed up your variables - the Bezier is cubic in its
*parameter*, not in the spatial variables X and Y. With my physicist's
hat on I suggest you do an experiment: take a drawing program and enter
a single segment of Bezier curve with end points (1, 0), (0, 1) and
both interior control points set to (1, 1) and then *look* at it. It's
not anywhere remotely close to an approximation of a 90 degree arc.


.....Bob Clair