Felipe,
That was just an example - I think the family of curves are more likely to look like the attached - it would be nice if there was some automatic equation extraction method . .
Thanks,
Phil.
Felipe Sanches wrote:
well... that curve looks like half of a gaussian curve. And people usually model things using gaussian ditributions in statistics.
On Fri, Jan 30, 2009 at 10:26 AM, Philip Rhoades <phil@...2067...> wrote:
Felipe,
I have a statistical relationship in mind and I find it is convenient to use Inkscape to create the graphs of what I think it should look like under different conditions. Having done that, it would be nice to be able to extract the formula for each curve that I have drawn. The curve of the relationship that I am thinking of should have only a single y value for each x coordinate, so it should be a function.
Thanks,
Phil.
From: Felipe Sanches <felipe.sanches@...2592...> - 2009-01-30 02:02 well.. maybe.
it is a bezier curve. It is a parametric curve whose coordinates are defined by quadractic equations:
x(t) = a*t² + b*t + c y(t) = d*t² + e*t + f
finding the values of a,b,c,d,e and f involves looking at the coordinates of the control points of that curve and doing a bit of math
Not all bezier curves can be described as 1 parameter functions, though... I think that it is indeed possible to do that with the one you provided since it appears to have a single y value for each x coordinate, thus it satisfy the requirements to be called a function.
what are you trying to do with this math stuff?
On Thu, Jan 29, 2009 at 10:14 PM, Philip Rhoades <phil@...2593...> wrote:
People,
Is there a way of extracting the formula for the curve from the
attached svg
ie in something resembling a conventional y = f(x) format?
Thanks,
Phil.
-- Philip Rhoades
GPO Box 3411 Sydney NSW 2001 Australia E-mail: phil@...2593...
-- Philip Rhoades
GPO Box 3411 Sydney NSW 2001 Australia E-mail: phil@...2067...
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