The UK isn't really 115,000 sq. nm's is it? According to my map, it's 94,270 square miles, which is about 82,000 sq nautical miles. Now that's for the UK - i.e. including Scotland and Northern Island. London FIR covers less than than that, so your calculation was based on 9 towers to cover the London and Scottish FIR's. My guess would be somewhere more in the (conservitave) region of the London FIR being 60,000 sq nm.
So working on the basis of a circular transmission footpint, the area of each circle will be 60,000sq nm / 6 = 10,000sq nm = A. 6 comes from a fudge that you'll need 4+2 transmitters to cover the area in which four would fit without overlapping, so we'll assume that 6 trasmitters fit in the London FIR without over lapping, leaving 3 to cover overlap. This just simulated for the numbers of course.
The radius of said circle will be r [nm] =SQRT(A[nm^2]/PI).
r= SQRT(3183) = 56nm = 64 miles.
d [miles] = 1.23 * SQRT(h [feet]) so
h = (d / 1.23)^2
h = (64 / 1.23)^2
h = 52 ^ 2 = 2704'
Which seems more realistic.
Now, on a seperate issue. If the signals from a aircraft are re-broadcast on the other transmitters, assume the following:
A/C [A] is at 2000ft, LOS ONLY with Tx [X].
A/C [B] is at 2000ft, LOS ONLY with Tx [Y].
A/C [C] is at 12000ft, LOS with Tx [X], Tx [Y].
Now, [A] Transmits, so [B] gets to hear what was said since [Y] is \'coupled\' to [X]. That\'s cool. Now what does [C] hear? [C] recieved the same transmission from two sources [X] and [Y]. Now, it\'s not impossible that the carrier frequency of these two stations are perfectly syncronised, therefore this could be possible. But not easy.
However - let\'s be more realistic...
A/C [C] is at 12000ft, LOS with Tx [X], Tx [Y] and A/C [A] & [B].
No way will that work! - [Y] (and I guess [X]) will be broadcasting at the same time as [A] as they attempt to re-broadcast what he\'s saying.
SCRREECCHH!