dchomak, your calculation seems reasonable. The radiation portion of the heat transfer from our lights is not trivial, thus cannot be ignored. It is just not the primary mode. In fact, in a situation like yours where you are in need of more cooling than you have, it may be reasonable to dull any shiny surfaces that you have. Emissivity for polished Al is .039-.057. "Roughened" Al is around .2 or higher, depending on how rough you go. I'm thinking a nice bead blast surface for Al would put it in the .2 range as well.
Equation for radiation heat transfer of a gray body. q = E*Boltzman Constant *(Thot^4-Tcold^4) *Area
E is emissivity. Temperatures are in Kelvin. Area is in meters squared. Answer is in Watts.
Even being very conservative and rounding up E for polished Al to .06. Rounding E (way) down for roughened aluminum to 0.10, you can see there is a significant difference. Doing the math for a pretty hot light with those values for E gives a 40% increase for radiation heat transfer just with changing the surface finish. Putting a more realistic value of .2 in for the roughened aluminum instead of .1 shows an increase of 70%. So there is a gain to be had.
I bolded "Area is in meters squared" above because our lights have very small surface areas. Just quickly, I measured a Mag D head and calculate that it's area is in the ballpark of 17 square inches That's only .011 meters squared. And it is why radiation heat transfer is the minor player of the two. Just to give you some numbers, a shiny un-anodized mag head will emit about .14 watts while the factory anodized head will emit a whopping 1.79 watts. A bead blasted head would emit about half a watt. These numbers all assume head temperature is an even 130F, which is pretty darn hot. Cut those numbers in roughly half for head temp of 110F.
Convection is a little more murky. The equation is simple:
q = hA(Th-Tc)
q is heat transfer in watts
h is the heat transfer coefficient (the murky part). It has a wide range. For instance, if you tail stand the light it might be 20. Lying on it's side on a table it might be 11.
A is in meters squared again.
Th and Tc can be either Celsius or Kelvin.
For the same .011 meter square mag light at 130F (I did the conversion to C) we get anywhere from 3.5 to 7 watts of heat transfer in still air depending on what value of h is used.
Neither of these calculations account for any conduction of heat to the body and it's subsequent heat transfer from both mechanisms, which would be considerable. Therefore, total estimated heat transfer would be higher, likely considerably.
While I'm at it, I might as well talk a little about paint. Black epoxy has an emissivity of .89. That's pretty high (Anodized is .77). Putting that into my radiation spreadsheet (yeah, I made a spreadsheet) tells me that I can expect 2.06 watts to be radiated from the same Mag D head at 130F. That's nothing to sneeze at. This begs the question; How much does the insulating property of the paint decrease the convective heat transfer?
***Disclaimer***
My math is rusty and my spreadsheet skills are waning. There are probably a mistake or several. I'm sure you will point them out. :)