Re: [PyX-user] merging graphs and paths

[Ken S. <ke...@la...>]

Reply sent privately (in error)   on  24 August

Mico Filós wrote:
> Hi,
>
> I am sorry if this is a trivial question but I cannot figure out how 
> to do this.
>
> Imagine I have a nonlinear function f(x) and its linearization at some
> point (x0,f(x0)),
> given by y(x) = f'(x0) (x - x0) + y0, where y0 = f(x0). Imagine also that
> I don't use the same scale for the x and y axes. How can I obtain a
> straight line that
> *looks* perpendicular to the tangent y(x)?
>
> Since the x and y scales are not the same, a line with slope equal to 
> -1/f'(x0)
> does not look perpendicular to y(x). I want to plot a straight line
> that passes through
> an arbitrary point of the tangent (not necessarily (x0,y0)) and that
> looks perpendicular to
> it. My idea is actually to plot the projection of an arbitrary point
> of the nonlinear function,
> (x1,f(x1)) to the subspace spanned by the tangent at x0. For me the
> most straightforward thing to do would be:
>
>    * Compute the angle theta (on the actual canvas, i.e., as it looks
> on the screen) of the path
>       defined by the tangent function y(x)
>
>    * Plot a line path L with an angle theta + pi/2 that passes through
> some particular point
>       of the tangent Q=(x0', y0'), which I specify.
>
>     * Find the intersection R of the line L with the path defined by
> the nonlinear function.
>
>     * Ideally, I would stroke only the portion of L that connects the 
> point
>        (x0',y0') with the intersection point R.
>
> Describing this in words is painful. I hope you see what I mean.
>
> Thanks a lot in for your patience.
>
> Mico
>
> -------------------------------------------------------------------------
> This SF.Net email is sponsored by the Moblin Your Move Developer's 
> challenge
> Build the coolest Linux based applications with Moblin SDK & win great 
> prizes
> Grand prize is a trip for two to an Open Source event anywhere in the 
> world
> http://moblin-contest.org/redirect.php?banner_id=100&url=/
> _______________________________________________
> PyX-user mailing list
> PyX...@li...
> https://lists.sourceforge.net/lists/listinfo/pyx-user
>
>
>   

------------------------------------------------------------------------

from pyx import *


g = graph.graphxy(width=8, x=graph.axis.linear(min=-5, max=5))
g.plot(graph.data.function("y(x)=x*x"))
# Let us plot the tangent at (x0,y0) = (2,4) slope 4
g.plot(graph.data.function("y(x)=4*(x-1)"))
"""
Let us take two more points on the tangent
"""
slope=4
deltaX=2
x1=-3 # example, you can change this
x2 = x1 + deltaX
y1 = slope * (x1 -1)
y2 = slope * (x2 - 1)
"""
Now you need to get the two points (x1,y1), (x2,y2)
into canvas-coordinates rather than scaled coordinates.

This is the crucial stage that will provide for your peculiar 
request that the line *looks* perpendicular to the tangent line. 

I shall call these U-V coordinates, and their origin is bottom left of the 
canvas (g), their units the usual user-units (e.g cm)
"""
u1,v1= g.pos(x1,y1)
u2,v2= g.pos(x2,y2)
deltaU=u2-u1
deltaV=v2-v1
# lastly a point (u3,v3) that can define your 'normal' from (u1,v1)
u3=u1-deltaV
v3=v1+deltaU
# as these are in canvas coordinates, not graph ones, we use
# stroke .. path .. etc to plot a segment from u1,v1 to u3,v3

g.stroke(path.path(path.moveto(u1,v1),path.lineto(u3,v3)))
g.writePDFfile("graph3")