approximating a 2nd order bezier curve with a 3rd order bezier curve
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From: PmI on 19 May 2010 09:22 Hi, does anyone know how to derive the values for the control points of a 3rd order bezier curve that approximates a 2nd order bezier curve? I need to work in cubic bezier curves, but have some quadratic bezier data that needs to be best approximated, but I don't really know how to do this. The curve functions are relatively simple, fx(t) = x1�(t-1)� + cx�(t-1)�t + x2�(t)� fy(t) = y1�(t-1)� + cy�(t-1)�t + y2�(t)� for quadratic, and fx(t) = x1�(t-1)� + 3�cx1�t�(t-1)� + 3�cx2�(t-1)�t� x2�t� fx(t) = y1�(t-1)� + 3�cy1�t�(t-1)� + 3�cy2�(t-1)�t� y2�t� for cubic. For the curve approximation these functions will have the same values for x1/y1 and x2/y2, but different values for the control points... how do I express cx1/cy1/cx2/cy2 in terms of cx/cy (and possibly x1/y1/x2/y2)? - PmI
From: Rob Johnson on 19 May 2010 21:23
In article <3cednTMBYJZheG7WnZ2dnUVZ8hWdnZ2d(a)giganews.com>, "PmI" <pmi(a)int13h.com> wrote: >does anyone know how to derive the values for the control points of a 3rd >order bezier curve that approximates a 2nd order bezier curve? I need to work >in cubic bezier curves, but have some quadratic bezier data that needs to be >best approximated, but I don't really know how to do this. The curve >functions are relatively simple, > >fx(t) = x1�(t-1)^2 + cx�(t-1)�t + x2�(t)^2 >fy(t) = y1�(t-1)^2 + cy�(t-1)�t + y2�(t)^2 > >for quadratic, and > >fx(t) = x1�(t-1)^3 + 3�cx1�t�(t-1)^2 + 3�cx2�(t-1)�t^2 x2�t^3 >fx(t) = y1�(t-1)^3 + 3�cy1�t�(t-1)^2 + 3�cy2�(t-1)�t^2 y2�t^3 > >for cubic. > >For the curve approximation these functions will have the same values for >x1/y1 and x2/y2, but different values for the control points... how do I >express cx1/cy1/cx2/cy2 in terms of cx/cy (and possibly x1/y1/x2/y2)? Note that (1-t)^2 x1 + 2 (1-t) t cx + t^2 c2 = ((1-t)^3 + (1-t)^2 t) x1 + 2 ((1-t)^2 t + (1-t) t^2) cx + ((1-t) t^2 + t^3) x2 = (1-t)^3 x1 + 3 (1-t)^2 t (1/3 x1 + 2/3 cx) + 3 (1-t) t^2 (2/3 cx + 1/3 x2) + t^3 x2 = (1-t)^3 x1 + 3 (1-t)^2 t cx1 + 3 (1-t) t^2 cx2 + t^3 x2 If we set cx1 = (x1 + 2cx)/3 and cx2 = (x2 + 2cx)/3. Similar for y. Rob Johnson <rob(a)trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font |