Hi,
that's true... I have been playing with maxima.
I define the function
F(x,y,z):=38244.74787*%pi*(x^2+y^2+z^2)^0.125
+1615.975261*%pi*z^2/(x^2+y^2+z^2)^0.875
+(-1292.78021)*%pi*z^2/((x^2+y^2+z^2)^0.875*(1+y^2/x^2))
+1292.78021*%pi*(x^2+y^2+z^2)^0.125/(1+y^2/x^2);
then
On Mon, Jan 21, 2013 at 8:06 PM, Pau wrote:
> Hi,
>
> I am somehow new to matplotlib and I am trying to plot this function of x
> ,y ,z
>
> F(x,y,z)=
> 38244.74787*Pi*(x^2+y^2+z^2)^.125+1615.975261*Pi*z^2/(x^2+y^2+z^2)^.875-1292.780210*Pi*z^2/((x^2+y^2+z^2)^.875*(1+y^2/x^2))+1292.78*Pi*(x^2+y^2+z
Hi,
I am somehow new to matplotlib and I am trying to plot this function of x ,y ,z
F(x,y,z)=
38244.74787*Pi*(x^2+y^2+z^2)^.125+1615.975261*Pi*z^2/(x^2+y^2+z^2)^.875-1292.780210*Pi*z^2/((x^2+y^2+z^2)^.875*(1+y^2/x^2))+1292.78*Pi*(x^2+y^2+z^2)^.125/(1+y^2/x^2)
in a similar way as
http://matplot
On Tue, Dec 13, 2011 at 1:12 PM, Martella, C. wrote:
> Hello,
>
> I have 2D array where each (x, y) cell represents the height of that point
> on the Z axis (you can think of it as the map of some mountain chain).
> I'd like to get the plot_surface() of this data. What I don't understand
> in par
Hello,
I have 2D array where each (x, y) cell represents the height of that point on
the Z axis (you can think of it as the map of some mountain chain).
I'd like to get the plot_surface() of this data. What I don't understand in
particular is the content of the X, Y ,Z array parameter in my par
I have a collection of Nx3 matrices in scipy/numpy and I'd like to
make a 3 dimensional scatter of it, where the X and Y axes are
determined by the values of first and second columns of the matrix,
the height of each bar is the third column in the matrix, and the
number of bars is determined by N.
On 17 September 2010 16:26, Simon S. Clift wrote:
> I have a data set that is in the form of an irregular 2D grid with
> associated values, one for each node. I would like to plot this as a
> raised surface, with colours that indicate the z-value. Somehow I
> didn't find just quite the example
Hi folks,
I have a data set that is in the form of an irregular 2D grid with
associated values, one for each node. I would like to plot this as a
raised surface, with colours that indicate the z-value. Somehow I
didn't find just quite the example I was looking for. After digging
around in the m
Adam Mercer wrote:
> On 09/10/2007, Jeff Whitaker <[EMAIL PROTECTED]> wrote:
>
>
>> Adam: If you can convert your coordinates into latitudes and
>> longitudes, then you can plot the data with the basemap tookit on your
>> choice of map projection (see
>> http://www.scipy.org/Cookbook/Matplotlib
On Tuesday 09 October 2007 20:08:29 Adam Mercer wrote:
> On 09/10/2007, Jeff Whitaker <[EMAIL PROTECTED]> wrote:
> > Adam: If you can convert your coordinates into latitudes and
> > longitudes, then you can plot the data with the basemap tookit on your
> > choice of map projection (see
> > http://
On 09/10/2007, Jeff Whitaker <[EMAIL PROTECTED]> wrote:
> Adam: If you can convert your coordinates into latitudes and
> longitudes, then you can plot the data with the basemap tookit on your
> choice of map projection (see
> http://www.scipy.org/Cookbook/Matplotlib/Maps for an example).
Followi
On 09/10/2007, Jeff Whitaker <[EMAIL PROTECTED]> wrote:
> Adam: If you can convert your coordinates into latitudes and
> longitudes, then you can plot the data with the basemap tookit on your
> choice of map projection (see
> http://www.scipy.org/Cookbook/Matplotlib/Maps for an example).
Thats j
Adam Mercer wrote:
> Hi
>
> I have some skymap data, i.e. theta, phi and some intensity that I
> would like to plot on the surface of a sphere, does matplotlib support
> plotting on the surface of a sphere? I've looked through the examples
> and can't seem to find anything.
>
> Any help would be g
Hi
I have some skymap data, i.e. theta, phi and some intensity that I
would like to plot on the surface of a sphere, does matplotlib support
plotting on the surface of a sphere? I've looked through the examples
and can't seem to find anything.
Any help would be greatly appreciated.
Cheers
Adam
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