So,
20kg (fat) = 2.2 kg (positive buoyancy/lead required) converted to 9 kg (fat) = 1 kg (pos/lead required)
So,
170 lbs (77kg)= 7.7 kg (positive buoyancy) = 17 lbs to be neutral in fresh or
77kg displaces 85 kg of water = 8kg of positive bouyancy to overcome, thus requiring 8kg to be neutral?
However, if fat is .9 and water is 1 then the difference being .1 * 18 = 1.8 L = 1.8 kg
He is now displacing 1.8kg or 3.96 lbs less........
40 lbs body weight = 4lbs positive buoyancy
170 lbs/40= 4.25
4.25x4lbs= 17lbs
It's just not making sense to me.
It seems like you went out of your way to unnecessarily complicate it. Let's stick to round numbers and metric units. The figuring I did is only relevant to the fat, not the whole diver. You seem to have generalized it to the whole diver, who has fat, muscle, bones, fluids, etc.—we have no idea what his specific gravity is (until we do a weight check). Hopefully we don't have a diver with 170 pounds of fat.
Let's examine
the fat portion of a diver weighing 100 kg. Let's say he has 27 kg of fat. How much water does the fat displace? Well, the specific gravity of fat is .9, so .9 kg of fat displaces 1 kg of (fresh) water. There are 30 x .9 kgs of fat, so it displaces 30 kg of water. The fat on its own would float, because it displaces a weight of water that exceeds its own weight.
Wikipedia:
Archimedes' principle is a law of physics stating that the upward buoyant force exerted on a body immersed in a fluid is equal to the weight of the fluid the body displaces. In other words, an immersed object is buoyed up by a force equal to the weight of the fluid it actually displaces.
The net buoyant force is 3 kg. The weight of the water displaced exerts a 30 kg buoyant force, and that is offset by the gravitational force of the weight of the fat, 27 kg. To achieve neutral buoyancy, we have to add 3 kg (and displace no water)*. If he loses those 27 kg of fat, he will have to take the 3 kg of weight off his belt to remain neutral. If you're looking for a simple rule of thumb, multiply the lost fat by .11 (not .1) to find the change in buoyancy.
As has been correctly noted, this is all theoretical and subject to a number of simplifications, but it might be a good starting point for an in-water weight check. And it is a good review of some basic physics.
*for simplicity's sake, we assume lead displaces no water, but of course it does. Its specific density is ~11.34, so 1 kg of lead displaces ~88 ml (g) of water.