From: Alvin Penner <penner@va...>  20090130 22:57:15

> The two Bezier quadratics that you could extract from the Inkscape file > (assuming that the curve representing the function is a single Bezier > curve segment) I am not entirely clear on which way the data is flowing here. Are you trying to do a curvefit to an existing curve (not in Inkscape) and then subsequently represent it in svg, or are you drawing a curve in svg and then trying to represent it as f(x)? In any event, just quick comment. The typical svg path element is a cubic Bezier which can only be represented parametrically as a function of t, not as an explicit f(x) relationship. However, there is a subset that might be interesting to look at. If you take a cubic Bezier curve and force the 4 control points to be equally spaced in the x axis, then what happens is that x(t) becomes linear and y(t) remains cubic. So you now have an explicit function y(x) which is still SVGcompatible, and is a cubic polynomial. So you could do a standard curve fit using cubic polynomials, and still be able to represent the result in svg.  View this message in context: http://www.nabble.com/erasertoolopinionstp21675675p21757247.html Sent from the Inkscape  User mailing list archive at Nabble.com. 