For what it's worth, in the front of my Aavid,
iERC, and Thermalloy heatsink catalogs, there's
some formulas and simple explanations. The
following is highly theoretical and I bow to
all sorts of argumentation.  I'm responding
to a fellow professional who's asked for
help a second time.  And leave it open to
discussion for those in the know.

With that in mind, here goes ...

Theta = (T1 - T2) / Q

Theta = thermal resistance (degrees C/Watts)
(T1 - T2) = temperature differential (degrees C)
Q = heat (Watts)

Thermal resistance is also expressed as

Theta = (rho * t)/A

rho = thermal resistivity (degrees C/Watts*inch)
t = thickness of material (inches)
A = square area (square inches)

So I plug the second Theta expression into the
first Theta expression (minding my units) and
get

Q = A * (T1 - T2) / (rho * t)

So for completely theoretical considerations,
the watts dissipated is proportional to area
and temperature differential; inversely
proportional to resistivity and thickness.
Well, looking at it, it kinda makes sense -
 the bigger the area, the more heat to dissipate.
 the bigger the temp diff, the more heat to dissipate.
 the bigger the thermal resistivity, the less heat
 to dissipate
 the bigger the thickness, the less heat to dissipate.

In other words, the above expression gives how
many watts the thing you're analyzing should
dissipate given area, temp diff, thermal
resistivity of material, thickness. Now, you
can rearrange this expression any way you
want depending upon what you do know, drop
it into an Excel spreadsheet, do some
predictions, and log in actual results.

I have absolutely no idea how this relates to
reality.  But, with some empirical testing,
it might be useful.  I don't know.

You asked for an equation, and that's all
I could find.  It's easily verified by getting
said catalogs for free and seeing if this is
really what you're looking for.

It's been my limited experience that theoretical
thermal stuff can be highly unpredictable given
real world conditions.