[MilCom] AWACS President Support * Low Power WT Range withAirborneUnits?

[David L. Wilson]

> SQUARE ROOT OF 30,000 FT  X 1.415   GIVES LINE OF SIGHT DISTANCE

Not sure where you got the1.415.
In statute miles which more here are familiar with, it should be
square root(30,000 ft)  x 1.225   ground path distance (result in miles)

Derivation
Let d=the line of sight distance, R= earth mean radius (3959 miles),
and h=altitude (30,000 ft)

Then draw a right triangle with a leg d, a leg R, and hypotenuse R+h.
By the Pythagorean Thm.  R^2+d^2=(R+h)^2 solving,
d^2=2Rh+h^2.  Since R>>h, we can drop the second term
Thus d=square_root(2Rh).  But we want the answer in miles, R in miles, and h
in feet, so in thouse units, this becomes:
d=square_root(2R[mi]*5280 ft/mile*h[ft])/(5280 ft/mi)
=1.225*square_root(h)

(In nautical miles, the constant is 1.06 instead of 1.225 by the conversion
0.869 nmi/mi.)
--
David L. Wilson     AC4IU