transform a radius into a circle

2014 12 26
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更新时间：2014 12 28
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I have some data along a line in a Cartesian coordinate. I know I can transform the Cartesian into polar coordinates using cart2pol. I would like to know how could I create a grid to map a quarter circle with the data along the initial line (to have a top view)?

thank you

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Nicolas
Nicolas 2014-12-26
编辑：Nicolas 2014-12-26
I have rho (2991*1), data (2991*1), and I created theta = (0:0.5:90)*pi/180; to represent a quarter of a circle. I can mesh rho and theta together using meshgrid. but then I don't know how to use my data to be correlated to the mesh?

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Nicolas
Nicolas 2014-12-28

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here is the solution:
%%Example data
x = (2:0.1:8);
data = sin(x);
%%Shift the values towards the center, so you get a circle instead of an annulus.
x = x-min(x);
%%Your meshgrid generation code.
rho = x;
theta = (0:0.5:90)*pi/180;
[th, r] = meshgrid(theta, rho);
%%Plotting the values
surf(r.*cos(th), ...
r.*sin(th), ...
repmat(data(:), 1, size(th,2)), ...
'linestyle', 'none');
view(2);
axis equal tight;
colorbar;

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Image Analyst
Image Analyst 2014-12-26

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Nicolas
Nicolas 2014-12-27
thank you. I'm not sure it is answering my question though. As i wrote i can create a mesh that has the shape of a quarter of a circle, however I'm not sure how to apply the data I originally have on a line to the mesh.
Image Analyst
Image Analyst 2014-12-27
So you have the coordinates of the arc (quarter circle). OK, fine. But I don't know what it means to "apply the data" to it? Do you mean like you want an image where the pixel gray levels along the arc are those specified in your "data"? Let's say you have 500 (x,y) points covering the arc, and 500 data points with some value that represents something (which I don't know what it is). Now what? Please clarify.
Nicolas
Nicolas 2014-12-28
I added an picture of my problem. I think it is easier like that as I'm not sure how to explain properly. thank you
Image Analyst
Image Analyst 2014-12-28
What do you want out? An image (2D array)? A plot/graph?
All I can think of was a not very efficient method based on the FAQ. I hesitate to show it because it's not fast, efficient, or clever. Hopefully someone else will come up with a fast and clever method.
Nicolas
Nicolas 2014-12-28
thank you, I didn't know how to explain properly what I wanted. by digging I found out the repmat option.. and that's what I was looking for.

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