使用MATLAB将2D图像放入半球

Question

I am trying to place a 2D image into a 3D hemisphere. I have an image of size 128x128. I generate my hemisphere:

[x,y,z] = sphere(127);
x = x(64:end,:);
y = y(64:end,:);
z = z(64:end,:);

Attempt = warp(x,y,z,Img)

My image is a circle (as shown below). What I get is a strange warping around the hemisphere. However, I essentially want how it would look if I dropped the image into the hemisphere (which is not what I get).

Thanks for any help you can provide!

Answer 1

There are two general issues to consider here: The ordering of (x,y,z) coordinates generated by sphere and the way in which the image will be mapped to them:

Ordering of points

To see how sphere is generating points, I'm going to warp a sample 128-by-128 RGB image mapImage to the spherical surface:

subplot(1, 2, 1);
imshow(mapImage);
subplot(1, 2, 2);
[x, y, z] = sphere(128);
warp(x, y, z, mapImage);
axis equal

Notice that the left-most column of the image gets mapped to a line running up the left side of the sphere (from its bottom point to its top point). Each successive column from the image gets mapped in sequential segments following clock-wise (looking down) around the sphere. This shows us how the points in our (x,y,z) matrices are ordered.

If you want a hemispheric "bowl" to map an image onto, you would want to generate points running along one half of the bowls rim, wrapping in sequential segments around the bottom and to the other half of the rim. Avoiding involved geometric explanations, I'll simply tell you that you can generate your bowl by just swapping your y and z matrices in the return call from sphere , then discarding the last half of the columns from the three matrices:

[x, z, y] = sphere(128);
x = x(:, 1:65);
y = y(:, 1:65);
z = z(:, 1:65);
warp(x, y, z, mapImage)
axis equal

Mapping your image

As you can see from the figure above, the entire image is mapped to the surface, not just the circular center region. Notice how the bright-colored corners get pinched up at the top and bottom points of the bowl rim, where the generated points of the sphere are densely clustered. Perhaps you only want to map the circular center region to the surface?

To do this, you'll have to transform your image so that the circular region is stretched along each row so that it fills the image. You could probably do this sort of thing with imwarp , but I'm going to show an example of doing this with interpolation .

[Xq, Yq] = meshgrid(1:128);
Xq = 64.5+sqrt(abs(63.5^2-(Yq-64.5).^2)).*(Xq-64.5)./63.5;
centerImage(:, :, 1) = interp2(mapImage(:, :, 1), Xq, Yq);
centerImage(:, :, 2) = interp2(mapImage(:, :, 2), Xq, Yq);
centerImage(:, :, 3) = interp2(mapImage(:, :, 3), Xq, Yq);
subplot(1, 2, 1);
imshow(centerImage);
subplot(1, 2, 2);
warp(x, y, z, centerImage);
axis equal

The line where Xq is computed looks kinda ugly. To explain, the indices in each row of Xq are rescaled from -1 to 1 (instead of 1 to 128), multiplied by half the width of the circular region in that row, then shifted back up to be used as an interpolant in the range of 1 to 128. This stretches the circular region moreso at the top and bottom so it fills the entire square image and maps better to the spherical surface.

Answer 2

You should pay attention the way that the wrap function is warping the image on the given surface, and make the surface prepared somehow that the center of the image stays on the minimum of the surface.

In this case I use the following image by MATLAB (since I do not have your image, but the result would be the same):

Img = imread('ngc6543a.jpg');

Now I make a surface as follows:

fx = @(u,v) u.* cos(v);  
fz = @(u,v) u.^2;        
fy = @(u,v) 5.*u.*sin(v);
u = linspace(-10,10, 100);
v = linspace(-pi,pi, 100);
[uu,vv] = meshgrid(u, v);
X = fx(uu,vv);
Y = fy(uu,vv);
Z = fz(uu,vv);
surf(X,  Y,  Z);

Now do the warping:

warp(X,Y,Z,Img);
axis off

Note: remove the 5.0 from equation fy makes a semi-sphere instead of a praboloid.