Edwin Pawlowski wrote:

wrote:

Covering the ceiling with foil would give it about US R10, ie 1.76 mK/W,
with E = 0.03 and a Tc = 20 C ceiling temp and a Tf = 14 C floor temp
and a large air gap. This would reduce the radiation from the ceiling
to the floor, es((Tc+273)^4-Tf+273)^4) W/m^2, with s = 5.6697x10^-8
W/m^2-K^4. If the upper foil surface is perfectly clean, with no dust
(you are German, right? :-), this may work even better. The linearized
radiation conductance is 4esTm^3 W/m^2-K, where Tm is the approximate
mean absolute temp.

Su, what the **** does all that gibberish mean?

That's called "elementary engineering" :-) The OP Klaus is an engineer.

Nick