33 Gigatons ÷ 5 Petatons ÷ 1.5 = 4.4 ppmv ... we measured 2.2 ppmv ...
Yep. The oceans are a big CO2 sink. And you didn't know such a basic thing. You don't seem to be qualified to be in a discussion with the grownups.
The ocean surface can only hold the CO
2 that is in equilibrium with the concentration in the atmosphere ... and not one microgram more ... and you didn't know such a basic thing ... how does the CO
2 migrate down the water column? ... keep in mind, we don't have any convection in the oceans ... so we only have the top 100 meters or so of the ocean to absorb CO
2 ...
The sad part is that if you're right ... then human CO
2 emissions are meaningless, it will get absorbed by the oceans and never change the climate ... maybe you shouldn't be saying two different things out of each side of your mouth when you speak with grown-ups ...
Oh, I was wrong about forcing being at the surface instead of the top of the atmosphere. You and Flac were more wrong, however, about the magnitude. You pooched it by not dividing the incoming solar radiation by four. So, your estimates were off by a factor of four (instead of by eight, as I had said). So the solar forcing is still insignificant, and I'm still right.
Do you understand the source your error, or would you like me to explain it in more detail?
It's a vector thing ... attended if you can:
Incoming solar energy arrives
in parallel ... so when we say 1,360 W/m^2, these square meters are the Earth in cross section ... shine a flashlight on a baseball, notice only half is illuminated? ... outgoing energy leaves
radially ... call it 238 W/m^2, these square meters are the surface area of the Earth ... also, and strictly by convention, we give climate forcing values using these radial vectors ... +1.8W/m^2 is using surface area square meters ...
It's not my fault astronomers do things this way ... they have excellent reasons to do so ... vector math is very difficult for the average
peon ... 1,360 w/m^2 is measured at the top-of-atmosphere ... albedo is ≈ 0.3, so we have 950 W/m^2 at the surface as measured in cross section ... NOW we divide by 4 to find the out going energy as measured in surface area ...
See Stefan, J. (1879) and Boltzmann, L. (1884) ...