I searched the CD and found this data on the ability of
air to absorb heat.

I wrote a little program to calculate the area required
at 100 MPH and 120 MPH as a function in the increase in
air temp passing through the rad. Not sure about the
specific heat constant so by all means check my calculations.

As you can see in the ideal case where the air is increased
by 60 degrees exiting the rad the area required is about 200 square
inches or about 10 by 20 roughly 63% of the frontal area of a 320
square inch rad. Sounds real high so there could be something
wrong. Sounds like it is off by a factor of two.

Tracy promised to measure the air temp and air velocity
exiting his rad.
Paul Lamar

I found the mistake. It had to do with actual airflow through
the rad. You must measure the actual air velocity at a point just
inside the duct opening. You also need to measure the static
pressure at the same point and correct the air velocity measurement
for the static pressure. In other words use a pito tube system.
If the rad is too thick this number will be below aircraft
forward speed due to external diffusion. This also assumes no leaks
around the radiator.

There are also other reasons why this velocity number may be below 
forward speed. The back of the radiator could be too close to 
an obstruction. The pressure in the cowl or the exit duct
behind the rad could be higher than ambient static air pressure.
The duct shape itself could be less than ideal so not as
much air is getting through the rad building excess static pressure
in the duct.

When you measure this number and relate it to forward speed
you will get a good handle on the overall efficiency of your 
cooling airflow system.
 
Paul Lamar
 
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