Re: NSBezierPath geometry question...
Re: NSBezierPath geometry question...
- Subject: Re: NSBezierPath geometry question...
- From: Nicko van Someren <email@hidden>
- Date: Mon, 11 Apr 2005 18:48:42 +0100
On 11 Apr 2005, at 17:46, David Phillip Oster wrote:
At 7:38 AM -0700 4/11/05, Robert Clair <email@hidden> wrote:
There is no cubic Bezier that is exactly a circular arc, the best you
can do is an approximation. However this isn't what you want - it's a
very bad approximation of an arc.
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.
The problem is that an arc is not a quadratic. The function for Y^2 is
a quadratic in X but the function for Y is not.
Nicko
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