Originally Posted by
Luc Lion
I find 16,000 ft above with the approximate 1/60 rule : (+2°+2°) * (1/60°) * 40 nm * (6000 ft/nm) = +16,000 ft
Are you sure that this question is not a mix of 2 different questions/answers pairs?
I found the following very similar question on the internet:
For that question, the answer is : (+4°- 2.5°) * (1/60°) * 40 nm * (6000 ft/nm) = + 6,000 ft , or 6000 ft above
Post Scriptum: bear in mind that "disappears" practically means "fades with a 50% return" as the beam width is defined as the -3 dB perimeter of the outgoing wave (-3dB = 50% signal).
This is correct. However I would do some simple mental maths instead.
Consider that 1deg at 40nm = 4000ft . Therefore 4 deg at 40nm = 4x4000= 16000ft
works for any distance. Eg 1deg at 75nm = 7500ft. So 3 deg at 75nm = 3x75(00)= 22500ft