From: Dr. Olaf Hoffmann <Dr.O.Hoffmann@gmx.de>

Date: Wed, 8 Nov 2006 19:04:20 +0200

To: www-svg@w3.org

Cc: nand.kumar@catalytic.com

Message-Id: <200611081804.20933.Dr.O.Hoffmann@gmx.de>

Date: Wed, 8 Nov 2006 19:04:20 +0200

To: www-svg@w3.org

Cc: nand.kumar@catalytic.com

Message-Id: <200611081804.20933.Dr.O.Hoffmann@gmx.de>

> Now, how to convert T -> Q and S -> C. > And what about the relatives path data, can that also be converted > from t -> q and s -> c. > As I was doing T -> C which was giving wrong results. > If T -> Q is done then I think my problem would be solved as I have api's > for Q. Parametrisation and Conversion (of curve fragments/segments) common parameter: u from 0 to 1 c(u) Bezier curve a) Linear Bezier P(0), P(1) initial and final point c(u) = (1-u)P(0) + u P(1) b) Quadratic Bezier: P(0), P(1) initial and final point, P(k) control point c(u) = (1 - u)^2 P(0) + 2u(1-u) P(k) + u^2 P(1) derivative dc(u)/du = 2(u-1) P(0) + 2(1 -2u) P(k) + 2u P(1) using this to get the same value for the derivative for the next curve segment gives the following, using points or control points a, b, c, d: M a Q b c T d is the same as M a Q b c Q (2c-b) d c) Cubic Bezier P(0), P(3) initial and final point, P(1), P(2) control points c(u) = (1 - u)^3 P(0) + 3u(1-u)^2 P(1) + 3u^2 (1-u)P(2) + u^3 P(3) derivative dc(u)/du = -3(1 - u)^2 P(0) + 3(1-4u +3u^2) P(1) + 3(2u-3u^2) P(2) + 3u^2 P(3) looking at the derivate for the curve segments gives with the points or control points a, b, c, d, e, f: M a C b c d S e f is the same as M a C b c d C (2d-c) e f interesting for authors is a smooth closed curve (continuous differentiable), for example: M a C b c d ... S e (2a-b) a Z d) Conversion small to large letters for example for a qubic curve with (control) points i,j,k,l,m,n,o: M i c j k m s n o is the same as M i C (j+i) (k+i) (m+i) S (n+m+i) (o+m+i) in general add the previous point in absolute coordinates, but not a previous control point. e) L -> C or Q -> C conversion L to C conversion: i, f initial and final point of the curve: M i L f = M i Q (i+f)/2, f = M i C (2i+f)/3, (i+2f)/3 f Q to C conversion (p control point): M i Q p, f = M i C (2p+i)/3, (2p+f)/3, f f) Other conversions reduce the problem to a combination of already solved problems. For example t -> C use t -> T -> Q -> C g) Not part of SVG tiny: A -> C? Conversion from elliptical arcs to cubic Beziers is not exactly possible and may cause nonsense if the path is animated - don't do it or test it very carefully. Ideas for this: Take a parametrisation of an elliptical arc, compare with the parametrisation and derivative above to get a useful approximation. h) Test with non-trivial examples or better prove, wether the given formulars are correct or not ;o)Received on Wednesday, 8 November 2006 17:34:01 GMT

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