Limits of polar plots axes
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    alberto tonizzo
 on 30 Jan 2023
  
    
    
    
    
    Commented: Walter Roberson
      
      
 on 31 Jan 2023
            Hello, please see the code atatched that makes a polar plot and adds transparency to it with the included (in the body) function "updateTransparency". The mat file to load is also attached (i.e. 'PolarPlotIssue_v1.mat').
The problem is that if lines 6-8 are commented out (if excluding angles large than 40 degrees), the polar plot becomes very small, i.e. it's not scaled.
How do I make the polar plot as large as the original (when lines 6-8 are NOT commented out)?
THANK YOU!
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Accepted Answer
  Walter Roberson
      
      
 on 31 Jan 2023
        You are using polar() which is the older function that uses cartesian axes. It is difficult to customize.
We would normally recommend that you switch to polarplot() but I see that you have hold on and overlay a semi-transparent contourf() which is based on cartesian coordinates. 
the = linspace(0,2*pi);
r = the.^2 - cos(the);
h = polar(the,r);
findall(h.Parent)
You can see that the older polar() throws up a bunch of text objects for labels, and draws a bunch of line() for the spokes and radii. There are no internal controls for polar() that limit angles.
2 Comments
  Walter Roberson
      
      
 on 31 Jan 2023
				contourf cannot be used with polarplot()
You would have to extract the contour boundaries and convert them to polar coordinates and polarplot those. But angles would very likely go out of range.
It would perhaps be easier to create a second axes in the same position as the first, with axes background color 'none' and possibly using alpha (transparency).
I do not know at the moment whether it is possible to use yyaxis between numeric axes and a polar axes. I would tend to suspect not.
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