The reason this looks confusing is because of the numbers chosen.
For one thing, we don't care what the illumination at the surface is. We care what the illumination at the coral is.
Let's make some assumptions to simplify the math: 1) assume the MH is a point source, 2) assume the FL is a linear source with even distribution, 3) assume our reflectors are 100% efficient, 4) assume the area we are illuminating occupies the equivalent of a 4x4" square, 5) assume the area of interest is 8" below the surface and centered under the light, 6) assume 2 VHO tubes to make it a little fairer for the VHO, 7) assume other numbers you mentioned, although why you have the MH so high is beyond me. I run mine 2 3/4" off the water.
So:
If the MH is 8" above the water and coral is 8" down, we have the light going over a sphere with a 16" radius. Using the formula A = 4 * Pi * R^2, this will have a surface area of 3217 in^2. But since we assume the reflector is 100% efficient, the area is half that, or 1609. With 16000 lumens we have 9.94 lumens/in^2, or 159 lumens on our area of interest.
If the 2 VHOs are 3" over the water and coral is 8" down, we have a cylinder with an 11" radius that is 48" long. Using the formula A = L * Pi * R^2 this will have a surface area of 18246 in^2. Assume the 100% reflector again and the area is halved to 9123 in^2. With 7000 lumens we have 0.77 lumens/in^2, and our area of interest gets 12 lumens. Less than 10% as much as the MH.
Pretty big difference, no?
[This message has been edited by Gannet (edited 02-19-2000).]