How do I find common volume of spheres?

I have plotted some spheres using surf and want to find where they all meet. i.e. co-ordinates that are inside all of the spheres.

the code I used to generate the spheres is:

r=15;
phi=linspace(0,pi,30);
theta=linspace(0,2*pi,40);
[phi,theta]=meshgrid(phi,theta);
x=r*sin(phi).*cos(theta);
y=r*sin(phi).*sin(theta);
z=r*cos(phi);
clf
hold on

inx=3363.5;
iny=1195.5;
inz=21.5;
h=surf(x+inx, y+iny, z+inz);   % generates first sphere

inx=3363.5;
iny=1169.5;
inz=32.5;
b=surf(x+inx, y+iny, z+inz);   %generates second sphere

inx=3346.5;
iny=1177.5;
inz=33.5;
c=surf(x+inx, y+iny, z+inz);

inx=3366.5;
iny=1182.5;
inz=12.5;
e=surf(x+inx, y+iny, z+inz);

inx=3357.5;
iny=1182.5;
inz=12.5;
f=surf(x+inx, y+iny, z+inz);

shading faceted
set(h,'FaceAlpha',0.5)     %makes them transparent

set(b,'FaceAlpha',0.5)
set(c,'FaceAlpha',0.5)

set(e,'FaceAlpha',0.5)
set(f,'FaceAlpha',0.5)

I have tried simply comparing the spheres but I am finding it difficlut to see where they meet since there is soo many.

How can I make them easier to compare?

P.S. I am using r2011a

Answer 1

[Richard Brown]

Here's how I'd do it (if I was in a hurry). First pretend all the spheres are cubes, and find their intersection.

inx = [3363.5, 3363.5, 3346.5, 3366.5, 3357.5];
iny = [1195.5, 1169.5, 1177.5, 1182.5, 1182.5];
inz = [21.5,   32.5,   33.5,   12.5,   12.5];

% Find bounding box for region
xlim = [max(inx - r), min(inx + r)];
ylim = [max(iny - r), min(iny + r)];
zlim = [max(inz - r), min(inz + r)];

Then, generate a large number ( ngrid^3 ) of points inside that box

ngrid = 100;
[X, Y, Z] = ndgrid(linspace(xlim(1), xlim(2), ngrid), ...
    linspace(ylim(1), ylim(2), ngrid), ...
    linspace(zlim(1), zlim(2), ngrid));

X = X(:); Y = Y(:); Z = Z(:);

Finally, rule them out by checking every point against every sphere. If you're lucky there should be some left.

% Now rule points out
idx = true(size(X)); % All points assumed in to start with
for i = 1:numel(inx)
    idx = idx & (X - inx(i)).^2 + (Y - iny(i)).^2 + (Z - inz(i)).^2 <= r^2;
end
X = X(idx); Y = Y(idx); Z = Z(idx);

In your case, the intersection of all spheres is empty.

Answer 2

[Cameron Henry]

thanks for the quick response!

so there is no common volume? :(

what I needed to find was the maximum number of spheres that you could be in at the same time and where the co-ordinates of that is. I thought that these ones were the maximum amount, but obvisuly not.

so how would I find the common volume that includes as many spheres as possible?

There is 9 spheres, the coordinates of the spheres are the 5 from before as well as 4 more

x=[3358.5,3364.5,3353.5,3361.5]

y=[1168.5,1201.5,1190.5,1179.5]

z=[8.5,12.5,17.5,23.5]

I excluded these before because, the first two are further away so I think they won't work, and the other 2 are in close so I think that they will always be part of the common volume

[Richard Brown]

If you have only 9 spheres, then there are only 81 possible ways of choosing between 1 and 9 different spheres. I suggest you use my approach above on each of these combinations, and see what the largest number of spheres you can find that has a nonempty intersection is.