Re: NSBezierPath geometry question...
Re: NSBezierPath geometry question...
- Subject: Re: NSBezierPath geometry question...
- From: Robert Clair <email@hidden>
- Date: Mon, 11 Apr 2005 15:02:48 -0400
<x-tad-bigger>></x-tad-bigger><x-tad-bigger>
</x-tad-bigger><x-tad-bigger>> I've read this claim before but never understood it. Since a </x-tad-bigger><x-tad-bigger>
</x-tad-bigger><x-tad-bigger>> NSBezierPath is a cubic polynomial, why in the world would it have any </x-tad-bigger><x-tad-bigger>
</x-tad-bigger><x-tad-bigger>> trouble representing an arc, which is a quadratic polynomial?: just </x-tad-bigger><x-tad-bigger>
</x-tad-bigger><x-tad-bigger>> set the coefficients of the x^3 term to zero, and it IS a quadratic </x-tad-bigger><x-tad-bigger>
</x-tad-bigger><x-tad-bigger>> polynomial. (or in bezier speak: if the two off axis control points </x-tad-bigger><x-tad-bigger>
</x-tad-bigger><x-tad-bigger>> are in the same place, you've got a quadratic spline, which should be </x-tad-bigger><x-tad-bigger>
</x-tad-bigger><x-tad-bigger>> the same as a segment of arc.</x-tad-bigger><x-tad-bigger>
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</x-tad-bigger> _______________________________________________
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