Aha! I was right-my path lead me to my physics book.
In which I was able to follow most of Mark's equations.
Very interesting; among other things!

The only thing I would add to the eq.: a=(20.04)sqrt(T), is that this is in meters per second and the temperature, T, is in Kelvin.

Through a little algebra I arrived at: a=(44.89)sqrt(T) giving 'a' in mph.

To convert from Celsius to Kelvin just add 273.15.
To convert from Farenheit to Kelvin: K=5/9(F-32)+273.15.

That being said, the speed of sound in mph:
@32 deg F = 742
@50 deg F = 755
@80 deg F = 777
@90 deg F = 784
@100 deg F = 791.5

Further research as to how altitude and air density affects 'a' on my part is not necesary for me to be satisfied.

Thanks All for your inputs.
Jeff


Thanks Will. My thirst for knowledge found it before I logged back on.