polar plot bar chart combo or something similar?
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Benjamin Cowen
le 9 Jan 2017
Réponse apportée : KAE
le 27 Avr 2017
Hello,
I have 24 spaced angles such as:
[100],
[10tan(15)],
[10tan(30)]
etc.
Note that tan is in degrees.
For each of these angles I have a value. How can I plot a line of that magnitude at each angle? Can MATLAB handle something like this?
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Walter Roberson
le 9 Jan 2017
Modifié(e) : Walter Roberson
le 9 Jan 2017
deg = 0:15:359;
theta = 10*tand(deg);
rho = the values corresponding to each angle
polar(theta, rho)
or
polarplot(theta, rho) %recommended if your MATLAB is new enoug
4 commentaires
Walter Roberson
le 9 Jan 2017
The revised version
polar([theta;theta], [zeros(size(rho));rho])
does not generate zig-zagged lines.
The zig-zag lines are due to the fact that you specified that your angles are to be 10*tand(0:15:345) which give angles in radians that are all over the place, including +/- infinity.
If you have a column "a" with angles in degrees,
theta = reshape(a*pi/180, 1, []); %want a row output
rho = reshape(b, 1, []); %want a row
polar([theta;theta], [zeros(size(rho));rho])
You could add markers:
polar([theta;theta], [zeros(size(rho));rho], '-*')
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John BG
le 10 Jan 2017
Mr Cowen
simplifying for just 5 sections
with a start point set to [10 10]
with possible angles multiple of of let's say 15º
amount_angles=5;
da=15;
a_range=[0:da:360-da];
na=randi([1 numel(a_range)],1,amount_angles);
a=a_range(na); % angles
d=randi([100 1000],1,amount_angles); % section lengths
p0=[10 10]
figure(1);grid on
plot(p0(1),p0(2),'go')
hold all;
for k=1:1:numel(a)
dx=d(k)*sind(a(k))
dy=d(k)*cosd(a(k))
plot([p0(1) p0(1)+dx],[p0(2) p0(2)+dy],'b')
plot(p0(1)+dx,p0(2)+dy,'ro')
p0(1)=p0(1)+dx
p0(2)=p0(2)+dy
end
I have repeated twice and the green point is the start point while the red points are way points, and in blue each section:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/176489/image.jpeg)
.
Mr Cowen if you find this answer useful would you please be so kind to mark my answer as Accepted Answer?
To any other reader, please if you find this answer of any help solving your question,
please click on the thumbs-up vote link,
thanks in advance
John BG
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