Re: [Matplotlib-users] Polar plot of radar data

[Benjamin R. <ben...@ou...>]

It is because the polar projection expects theta to be in radians, not
degrees. This is a somewhat common mistake as the default labelling of a
polar plot is in degrees, and so it tends to confuse new users.

Also, don't forget that your radar data is (most likely) describing
"bearing" (so, 0 azimuth is North rather than East).

Cheers!
Ben Root



On Wed, Jun 25, 2014 at 9:14 AM, Alexander Borghgraef <
ale...@gm...> wrote:

> Hi all,
>
>  I'm trying to plot some radar data I'm working on, which is presented as
> a 2048x1440 2D array of ints. This means 2048 azimuth angle values between
> 0° and 360°, and 1440 range increments. Array values represent signal
> return in that angle/range coordinate. Due to hardware problems, some angle
> ranges are missing, there the signal value is zero in the grid.
>  The obvious way of plotting this would be a polar grid, so I used a
> pcolormesh, like this:
>
>
>
>
> * import numpy as np  import matplotlib.pyplot as plt  *
>
> *  plt.ion()*
>
>
>
>
> *  fig = plt.figure()  ax = fig.add_subplot(111, projection='polar')
> theta,rad = np.meshgrid(np.arange(0., 360., 360./scan.shape[0]),
> np.arange(scan.shape[1]))   ax.pcolormesh(theta, rad, scan.T)   plt.draw()*
>
> I transposed scan because theta and rad are 1440x2048 arrays, where scan
> is a 2048x1440 array.
> The resulting image seemed ok, but there were no gaps, which I knew were
> there, so I ran the same code without the 'polar' option, and put them side
> to side. The result is this:
>
>
> ​In the rectangular plot, there is a gap between around 120° to 160°,
> which should be an empty wedge in the polar plot, but it isn't there. Same
> for the obvious echos at long range between 0° and 50°. OTOH the plot seems
> quite right, with the gating gap around the radar station in the center and
> the stronger echos at short range.
>  So I'm a bit baffled as to why this doesn't work the way it should. Does
> anyone here have an explanation?
>
> --
> Alex Borghgraef
>
>
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