Re: NSBezierPath geometry question...

FROM : Nicko van Someren
DATE : Mon Apr 11 19:48:42 2005

On 11 Apr 2005, at 17:46, David Phillip Oster wrote:

> At 7:38 AM -0700 4/11/05, Robert Clair <<email_removed>> 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