Quote:
Sometimes a simple question has a complex answer. Perhaps someone with a background in physics or meteorology can give me an answer I can understand.

Apparently radiant energy travels out from a warm surface until in encounters some object. Then part of it is reflected, partly absorbed and part of it is transmitted. So if I wear the typical three layers of clothing I am wondering what part of my heat loss comes from radiation through the layers. Also, would it make any difference if the first layer (or any layer for that matter) is white or black? I suppose someone could also ask what the balance of radiation incoming and out going is.

This was probably covered in my Physics 101 class and I would know the answer, had I been paying attention.
Depends on what you mean by black and white. Something that is more of a white body to the visible spectrum can be more of a black body in the infrared spectrum. I understand ice is a good example of this. It is very effective in absorbing your heat, if you are naked.

Here is the equation for radiative heat transfer:

Q = e1 x e2 x Sigma x A x ( T1^4 - T2^4 )

where:
Q = Heat transfered from body 1 to body 2
e1, e2 = emmissiviity of body 1,2 ( black body = 1, white body = 0, 0 < grey bodies < 1 )
Sigma = Stefan Boltzman Constant = 0.174 x 10^-8 [ BTU/hr.ft^2.degR^4)
A = Surface area of body 1 exposed to body 2
T1 = Temperature of body 1
T2 = Temperature of body 2

From this equation you can see that while it can be effective to reduce the emmissivity of your outer layer, it can be just as effective to reduce the temperature of your outer layer by wearing the insulative layers you should have on anyways. I prefer dark clothing for sweaters, so they will dry better in the sun and near a fire, and light bright colours for shell materials, because they look nice. <img src="/forums/images/graemlins/smile.gif" alt="" />