flying at FL210 AWR wx return at range 40 NM is identifiable when the center of the beam is tilted between?

Question

+/- 1 deg and -6 deg. what is the height of the Base and top of the rain bearing cloud

Answer 1

Trig: You have two right triangles with their common sides 240,000 long (40 NM) at FL210 (21,000 ft altitude with altimeter set to 29.92).

So the top is 21,000 ft plus the arc subtended by the 1 deg plus angle. The bottom is 21,000 ft minus the arc subtended by the six deg minus angle.

Given the one side (S...along your FL) is way longer than the two arcs (above and below the FL), we can say the hypotenuses (H) of the two triangles are about the same length as the one side (S = H). So we can say, u = H sin(1 deg) = 240,000 sin(1 deg) = arc above FL210 and d = H sin(6 deg) = 240,000 sin(6 deg) = arc below.

Thus, the top T = 21,000 + 240,000 sin(1) ~ 25,000
and, the bottom B = 21,000 - 240,000 sin(6) ~ 0

Don't think I'd want to make an approach through this guy. Also, my guess is that the angles are not very precise since the B is actually negative at six degrees. Five degrees would be about zero feet AGL. All these answers are based on FL, not on the actual altitude with the altimeter set to the local barometric pressure.

Answer 2

Well, my guess is that this is a trick question of some sort. But what I did was use a simple tangent function with 40 miles as the flight level leg, a half degree to the top of the cloud and solved for the cloud top. It was approx. 19,159 feet. Doing the same for the base using 3 degrees down from flight level, I found the base to be at 9931 feet, which is pretty darn high for a rain bearing cloud except in the desert southwest where cloud bases are that height.