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The thing is, it assumes a slow, linear rate of melting, i.e., viewing the ice sheets as a solid block of ice. They certainly are not that. My illustration is the breakup of the Larsen B ice shelf.

In January 2002, it appeared solid. In March 2002 it was splinters (not all of which have melted yet, obviously). The author of the posted piece would apparently take comfort in the fact that not all of that ice had melted even after the ice shelf was gone.

It's not just about melting. It's about the alteration of glacial and ice sheet dynamics caused by increased melting. Melt streams running down through more crevasses, lubricating the ice/bedrock interface. Warmer ocean temperatures eroding the underside of floating ice sheets. If a large chunk of ice breaks off an ice sheet and starts floating in the ocean, its going to melt a lot faster than if its still part of the main bulk of the ice sheet.

For later.

One thing he did not include was the (small but measurable) increase in ocean level caused by the thermal expansion of water from baseline (today) to +5 deg C (sometime later). Of course, it is small, but that 1/2 cm is about the same effect as the melting of ice.

A second item is the effect of the increase in ocean depth: Ice melts (from the Greenland and Antarctic ice fields) and goes into the oceans. Ocean levels rise. Greater ocean height means more weight on the ocean floors, so the ocean floors lower down into the mantle, and thus the final ocean ocean depth change is less. This is a geological change, and is very slow, but needs to be addressed as well.

(However, his fundamental truth that there isn’t enough energy from the sun to “melt the ice caps!!!!” (and flood Florida) is true. All of the global warming hysteria is based on extrapolated and magnified effects.

AGW/CC back of the envelope calculation that seems to be correct, but I can’t figure where he got the number noted in my comment# 1.

You should be able to get a pretty good estimate by calculating the difference in volume between a sphere of the earth's radius and one of the earth's radius plus 20 feet. Then multiple the answer by 3/4 since only 3/4 of the surface is ocean. (This ignores the increase in surface area due to the rise, but that shouldn't matter too much). (although due to precision problems, best to do a little algebra first). After the algebra:

Delta_vol = pi (3 * delta_h * Re^2 + 3 * delta_h^2 * Re + delta_h^2)

Where:
Delta_vol = change in volumne of ocean
delta_h = increase in sea height (sea level)
Re = radius of the earth.

ignoring the last two terms since delta_h << Re

Delta_vol ~= 3 * pi * delta_h * Re^2

(You'd get the same answer if you assumed the area of earth was spread out flat, and then covered to a depth of delta_h and then multiplied by 3/4. (Area of sphere = 4*p*R^2)
3/4 * area * delta_h = 3 * pi * delta_h * Re^2

Using for Re, the volumetric radius of the earth, 3,440 nm, since the earth isn't a sphere, but rather more nearly an ellipsoid. This is the radius of a sphere with the same volume as the ellipsoid that approximates the actual earth surface.

or 3 * pi * 20 ft * (3,440 nm * 6076 ft/nm)^2 / (.3048 m/ft)^3 ~= 2.9 * 10^18 m^3 ,

several orders of magnitude less than his answer of 6 * 10^24 m^3

One of us has messed up.