How do I make 3D sphere for convolution?

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Lieke Pullen
Lieke Pullen am 12 Jan. 2022
Kommentiert: Lieke Pullen am 13 Jan. 2022
Hi all,
I want to make a 3D sphere where the x, y and z coordinates are being solved in a 3D array, so I can convolute this with a kernel I have. But, I don't seem to both save the coordinates and get the shape right at the same time. I have the following code:
figure;
half=50;
for k=1:100
for j=1:100
for i=1:100
mesh_sphere(i,j,k)=((i-half).^2)+((j-half).^2)+((k-half).^2);
end
end
surf(mesh_sphere(:,:,k));
hold on
end
hold off
After this, I also want to use other shapes that I can test my kernel on, such as a hollow sphere and an ellipsoid. How can I make those and also same them in a 3D array? Thank you in advance!

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Rik
Rik am 12 Jan. 2022
If you want a sphere in a 3D array, you should probably do something like this:
array_width=21;%must be odd
radius=5;%must be smaller than half the array width for the sphere to fit
a=(array_width-1)/2;
a=linspace(-a,a,array_width);
[X,Y,Z]=meshgrid(a);
solid_sphere= sqrt(X.^2 + Y.^2 + Z.^2) <= radius;
For a shell you can repeat this with two radii and use the or operator |. For an elipsoid you need you can scale the coordinate values in each direction according to the partial radius you want.
  9 Kommentare
Image Analyst
Image Analyst am 13 Jan. 2022
Bearbeitet: Image Analyst am 13 Jan. 2022
@Lieke Pullen like I showed in my answer below, the strel() function creates the sphere in one line of code. The convolution is also done in one line of code.
Lieke Pullen
Lieke Pullen am 13 Jan. 2022
@Image Analyst I know, and I very much appreciate the effort. But I wanted to manually also change some of the proposed parameters, and this what this does to the shape. For me, Rik's option worked best.

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Weitere Antworten (1)

Image Analyst
Image Analyst am 12 Jan. 2022
Bearbeitet: Image Analyst am 13 Jan. 2022
SE = strel('sphere',r) % creates a 3-D spherical structuring element whose radius is r pixels.
out = convn(inputImage, SE.Neighborhood);
  3 Kommentare
Rik
Rik am 12 Jan. 2022
I don't remember this object notation. Admittedly, I haven't used this function in years, but I will look up this function in the release notes to see whether I forgot, or if they changed it.
Image Analyst
Image Analyst am 12 Jan. 2022
Thanks Matt for the correction.

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