Make a polar plot from data

I am trying to make a polar plot of a set of data in file called 't8m'. I am calculating parameters as:

1. theta angle = arccos (column 4)
2. bin width = 10 (making a total of 18 bins between 0 and 180deg).
3. length of each bin should be = (number of the theta angles that fall in the bin)/total number of all angles between 0 and \pi. This can be seen as length of array within the bin/length of array in that frame.

I am having problems with the for loop and the polar plot command. Can someone help out with this loop? Here is what I have now:

izm=length(t8m)
theta =   acosd(nz) ;

for i = drange(0:10:180)
  k= 0.0
  for j = 1:izm
      if theta == i
          k = k+1
      lth1 = k/length(k)
      end
  end
end
polar(theta,lth1,'--r')

Answer 1

[Thorsten]

 theta = pi*rand(1, 100); 
 th = linspace(pi/10, pi, 10);
 hi = hist(theta, th); 
 polar(th, hi/numel(theta))

[kayak]

Thanks Thorsten, but now I don't get it. It seems you replaced the theta calculation (theta = acos (nz)) completely. I still want the angle to be calculated as arccos, but the scaling should be between 0 and 1.

[Thorsten]

Sorry for the confusion. I just generated some sample data (because I do not have nz), using

theta = pi*rand(1, 100);

Of course you have to use

theta = acos(nz);

Answer 2

[Thorsten]

 theta = acos(nz);
 rose(theta)

[kayak]

thanks Thorsten, but my major problem is how to do the loop for sorting into the angle categories. Does anyone have an idea about this?

[Thorsten]

Hi Kayak, rose is an angular histogram that does the sorting for you, if theta is a bunch of angles in radiant.

[kayak]

Thanks, I get it now. After using the command:

rose(theta, 18),

I want to scale this with the length of the whole array (ie length (theta))so that the bar length lies between 0 and 1. Do you know how I can do this? I have tried plotting:

rose ((theta/length(theta)), 18)

but this is showing only one bar. Can someone help please.