I am working in an algorithm to order a bed of close-packed spheres. In the case where I have got four spheres, I understand that the fifth sphere position and radius is determined by the positions and radii of the four other spheres. It seems that there would be different solutions: one that produces a sphere that is similar in size to the other four, and a second solution in which the four spheres are encompassed in a bigger sphere. What interests me is the first solution. Any ideas what equation system solves this? The input data for the systems would be the positions of the original four sphere centres and their respective radii, and the output would be the position and radius of the fifth sphere. It is also important to notice that the four original spheres can but do not need to be tangent between them. Thanks a lot in advance!
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Distance from center of fifth sphere to centers of $i$-th of four spheres must be equal $r_i+r_5$. Then there is 4 algebraic equations of second order for $x_5,y_5,z_5,r_5$. – Ivan Kaznacheyeu May 04 '23 at 09:13
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General case $|r_i\pm r_5|$ – Ivan Kaznacheyeu May 04 '23 at 09:15
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@Ivan Kaznacheyeu thak you four your kind reply. I already implemented this and works well. However I am contemplating some degenerate cases. One of them would be if the four spheres are in a plane and have the same radii. In this case, what comes to mind is just using 3 and setting the radius of the new one to a fixed constant. Have you got any suggestions for this or for more degenerate cases that may arise? Thank you in advance. – cosmogato May 05 '23 at 10:28
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I believe there are placements of spheres which give no solution or multiple (finite or infinite number) solutions. Result obtained will depend on numerical procedure used for solving problem. If the problem is numerical, in such degenerate case one can slightly change parameters to make this case non-degenerate. – Ivan Kaznacheyeu May 05 '23 at 16:38