Blacknet, a few comments on your explanation of virtual images created by a dome port. Most manufactures state the size of their dome ports as the diameter of the port. In your explanation you use a 4 dome as an example. If this is a 4 diameter dome, than your math is wrong. You state that the virtual imager is 3 x the radius of the dome. Then you state that a 4 dome would have a virtual image at 12. The radius of a 4 dome is not 4, its 2, and so 3 x 2 would put the virtual image at 6 not 12. Also this would be 6 from the front of the port, not the film plane. I have found there are two different accepted approximations of the distance of the virtual image. There is an exact formula, but its a bit complicated, it takes into account the inside radius of the dome, the outside radius of the dome (i.e. thickness of dome), the index of refraction of water, the index of refraction of the dome material (glass or acrylic), and index of refraction of air. This formula will give the exact distance, from the outside radius of the dome, of the virtual image. A rule-of-thumb for locating the infinity virtual image is: "three radii minus four thickness (of the dome material) from the front of the dome". Another accepted approximation is; 2 x the diameter of the dome, measured from the film plane. This for me, is a little easier to use, since the minimum focus distance of land based lenses is not measured from the front of the lens, but from the film plane. So in my case, I use an 8 (dai.) dome port. Using the 2x approximation would give me a virtual image at 16 from the film plane. If I use your calculations it would be 3 x 4 or 12 from the front of the port. 12 from the front of the port, or 16 from the film plane.
What I should have stated is a dome port of 4" radius not a 4" dome port, there's a difference 
