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From: Willem de Boer <wdeboer@mi...>  20051013 14:15:28

A quick test with Matlab reveals that the coordinatetransformation matrix is: 1/6 * [ 0 6 0 0 ] [1 6 1 0 ] [ 0 1 6 1] [ 0 0 6 0 ]. So yeah, Danny's calculations are correct. Cheers,  Willem H. de Boer Homepage: http://www.whdeboer.com Original Message From: gdalgorithmslistadmin@... [mailto:gdalgorithmslistadmin@...] On Behalf Of Danny Kodicek Sent: 13 October 2005 14:58 To: gdalgorithmslist@... Subject: Re: [Algorithms] Bezier curve fitting In case anyone else is following this: >The matrix for CatmullRom is > 1/2 * [ 1 3 1 1] [ 2 5 4 1] [1 0 1 0] [0 2 0 0] I found what I guess was a typo here, and the first line should be [1 3 3=20 1]. By my calculations (not yet tested) the translation matrix is 1/6 [0 6 0 0] [1 6 1 0] [0 1 6 1] [0 0 6 0] This seems right, in that Bezier P0 maps to Catmull Q1 and P3 to Q2 as=20 expected. Danny=20  This SF.Net email is sponsored by: Power Architecture Resource Center: Free content, downloads, discussions, and more. http://solutions.newsforge.com/ibmarch.tmpl _______________________________________________ GDAlgorithmslist mailing list GDAlgorithmslist@... https://lists.sourceforge.net/lists/listinfo/gdalgorithmslist Archives: http://sourceforge.net/mailarchive/forum.php?forum_id=3D6188 
From: Danny Kodicek <dragon@we...>  20051013 13:57:56

In case anyone else is following this: >The matrix for CatmullRom is > 1/2 * [ 1 3 1 1] [ 2 5 4 1] [1 0 1 0] [0 2 0 0] I found what I guess was a typo here, and the first line should be [1 3 3 1]. By my calculations (not yet tested) the translation matrix is 1/6 [0 6 0 0] [1 6 1 0] [0 1 6 1] [0 0 6 0] This seems right, in that Bezier P0 maps to Catmull Q1 and P3 to Q2 as expected. Danny 
From: Danny Kodicek <dragon@we...>  20051013 13:28:27

> Alternatively (probably equivalently), how can I convert between a 2D > CatmullRom spline and a Bezier curve? Either of these will solve my > current > problem, which is to approximate a curve made up of lots of points with a > curve made up of a few Bezier segments. I think this approach won't solve your problem if you treat the initial curve as CatmullRom one (passes through all points); and in the end want a Bezier (or some other type) of curve with much fewer points. You don't want just to create a point in the output for each point in the input... That's true, but what I can do is make a Catmullrom passing through every n'th point, and convert *that* to a Bezier. It won't be a perfect approximation, but assuming my curves are reasonably smooth it should work all right. Now anyone want to email me offlist to talk about magnetic field lines? :) Danny 
From: Aras Pranckevicius <nearaz@gm...>  20051013 12:58:19

> Alternatively (probably equivalently), how can I convert between a 2D > CatmullRom spline and a Bezier curve? Either of these will solve my curr= ent > problem, which is to approximate a curve made up of lots of points with a > curve made up of a few Bezier segments. I think this approach won't solve your problem if you treat the initial curve as CatmullRom one (passes through all points); and in the end want a Bezier (or some other type) of curve with much fewer points. You don't want just to create a point in the output for each point in the input...  Aras 'NeARAZ' Pranckevicius http://nesnausk.org/nearaz  http://nearaz.blogspot.com 
From: Danny Kodicek <dragon@we...>  20051013 12:44:46

Danny Kodicek wrote: > > Alternatively (probably equivalently), how can I convert between a 2D > CatmullRom spline and a Bezier curve? >That's easy enough. You use the matrix formulation for parametric curves. >If you have a cubic parametric curve of type A and want to convert to type >B, and Pa are the control points for A, then you need the matrices for >types A and B, Ma and Mb. You then apply >Pb = Mb^1 * Ma * Pa Doh  that makes sense. Thanks. Danny 
From: Simon Fenney <simon.fenney@po...>  20051013 12:26:19

Danny Kodicek wrote: >=20 > Alternatively (probably equivalently), how can I convert between a 2D=20 > CatmullRom spline and a Bezier curve?=20 That's easy enough. You use the matrix formulation for parametric = curves. If you have a cubic parametric curve of type A and want to = convert to type B, and Pa are the control points for A, then you need = the matrices for types A and B, Ma and Mb. You then apply Pb =3D Mb^1 * Ma * Pa =20 The matrix for CatmullRom is 1/2 * [ 1 3 1 1] [ 2 5 4 1] [1 0 1 0] [0 2 0 0] and for Bezier (which you'll have to invert yourself) is=20 [1 3 3 1] [3 6 3 0] [3 3 0 0] [1 0 0 0] =20 Hope that helps. Simon 
From: Danny Kodicek <dragon@we...>  20051013 11:41:54

Hey all  been lurking for a while but this is only my second post, so Hi there. I'm dealing with what ought to be a really common algorithm, but for some reason I can't find one anywhere: given points p1, p2, q1, q2, and vectors v1 and v2, how can I find the control points p1+s*v1 and p2+t*v2 for a 2D Bezier curve p1 q1 q2 p2? (Is that clear? :) ) Alternatively (probably equivalently), how can I convert between a 2D CatmullRom spline and a Bezier curve? Either of these will solve my current problem, which is to approximate a curve made up of lots of points with a curve made up of a few Bezier segments. Thanks Danny 