Plotting a radial function f(r), in a 3d isotropic way, with axes (x,y,z=f(x^2+y^2)).

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Erez 2018-8-22

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https://ww2.mathworks.cn/matlabcentral/answers/415767-plotting-a-radial-function-f-r-in-a-3d-isotropic-way-with-axes-x-y-z-f-x-2-y-2

评论： Erez 2018-8-22

I have a function f(r), where r=sqrt(x^2+y^2). In my data r, is an array which takes values from [0,8], for example: f(r)=sin(r). I don't have data for an array of x or y seperately. The function is perfectly isotropic. How can I create a 3-dimensional plot, where the function f(x^2+y^2) is presented versus x and y in the correct range?

Further explanation:

The vector of the values of f(r) is given. It was calculated from some numerical simulation (I don't have its analytic form). I don't have the vectors for x and y. The function is isotropic, so one can imagine that I just draw the same curve f(r) in every direction. But the problem I encounter for example, is how to create the vector Theta, x and y, so that they will have the correct number of sites, like f(r), and that they will represent f(r) in the 3-dimensional plot.

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Answer 1

Dimitris Kalogiros 2018-8-22

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I reshaped my code.

Lets say that we have loaded values of r and f(r) from a file. Cause I haven't these files , I must create these values.

Then for every value of r , I rotate a vector of length r , in order to take values of x and y. This is a reason for the arbitrary creation of this angle, which varies from 0 to 2pi.

For any given value of r, we have f(r) from our file and a very dense set of possible pairs (x,y) which form a vecor with magnitude r. In that way we can claim that we have "transformed" our function from f(r) to f(x,y) .... in fact to f(sqrt(x^2+y^2))

clc; clear; close all;
%%input data ( predefined)
% values of r (they are predefined and stored to a file)
r=0:0.1:8;
% values of f(r) (they are predefined stored to a file) 
f=sin(r);
%%creation of x and y and connect them with f() stored values
% Angle between 0 and 2*pi, with arbitrary small step
% (small step means better accuracy)
phi=0:0.1:2*pi-0.1;
% creation of x and y
x=[];
y=[];
f_xy=[];
for n=1:length(r)
    xtemp=r(n)*cos(phi);
    ytemp=r(n)*sin(phi);
      x=[x xtemp];
      y=[y ytemp];
      f_xy=[f_xy f(n)*ones(1, length(xtemp))];
  end
%%plot
plot3(x,y,f_xy); zoom on; grid on;
xlabel('x'); ylabel('y'); zlabel('f');
grid on; zoom on;

..if you run the script you will receive this:

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Erez 2018-8-22

Fantastic! That's what I was looking for! Thank you very much.

Erez 2018-8-22

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 Quick follow-up: How can I color this curve, so that the color scheme will indicate height, like in a topographical map for instance? Thanks!

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Answer 2

Dimitris Kalogiros 2018-8-22

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Dear Erez

Here you are my suggestion:

clc; clear; close all;
% values of r
r=0:0.1:8;
% create values of x and y
phi=0:0.1:2*pi-0.1;
x=[];
y=[];
for n=1:length(r)
    xtemp=r(n)*cos(phi);
    ytemp=r(n)*sin(phi);
      x=[x xtemp];
      y=[y ytemp];
  end
%calculate values of f()
f=sin(sqrt(x.^2 + y.^2));
% plot
plot3(x,y,f); zoom on; grid on;
xlabel('x'); ylabel('y'); zlabel('z');
grid on; zoom on;

My code is not optimum from the point of view of speed , but showes clearly how to overcome your problem

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Erez 2018-8-22

编辑：Erez 2018-8-22

Dear Dimitris, thanks for you reply. However perhaps I did not explain my question correctly. In your code you have : "%calculate values of f() f=sin(sqrt(x.^2 + y.^2));". The point is, that for me, the vector of the values of f(r) is given. It was calculated from some numerical simulation (I don't have its analytic form either). What I don't have, is the vectors for x and y, nor do I have Theta. The function is isotropic, so you can imagine that I just draw the same curve f(r) in every direction. But the problem I encounter for example, is how to create the vector Theta, x and y, so that they will have the correct number of sites, like f(r), and that they will represent f(r) in the 3-dimensional plot. I hope my explanation made it more clear...

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