what is the approximate height of the cloud above or below the aircraft ?

[Mach1me]

Hello everyone
Anyone know how to solve this? I tried and i am getting 0 but there isn't an option for that answer.
The tilt angle of an Airborne Weather Radar (AWR) is set at +2°. If the beam width is 4° and the range of the cloud is 40 NM, what is the approximate height of the cloud above or below the aircraft when the weather return from the cloud just disappears from the screen?
6000 above
4000above
4000 below
6000 below

[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:
#48. The tilt angle on the AWR at which an active cloud just disappears from the screen is 4 degrees up. If the beam width is 5 degrees and the range to the cloud is 40 nm use the 1 in 60 rule to calculate the approximate height of the cloud relative to the aircraft?
4000 ft above.
4000 ft below.
6000 ft above.
6000 ft below.

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).

[Denti]

If its shaded it is more than 4000ft below the flight path. Otherwise select the different flight levels and check how high or low it is.

Sorry, Modern radar systems without direct beam control and altitude instead of angle make stuff easier, even more so if you can see the weather in the vertical profile view.

[EXDAC]

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).

-3 dB applies to the received signal as well as to the transmitted signal does it not?

[Luc Lion]

Yes, you are correct

[InSoMnIaC]

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