keneter

Hi guys I need some help with this question. Im desperately trying to comprehend the calculation in this question.

Airspeed in a decent is 90 kts. U have a 6 kts. tailwind component. The aircraft is at FL40. QNH is 1025.
Decend to level at 1000 ft. QFE 5NM before reaching the distination airfield.
QFE at the airfield is 1015. the decend will be started is 25 NM. from the airfield.

What will the ROD be?

Please tell my how you woulde solve this question.

fisbangwollop

As an airtraffiker I will have a stab!!!

FL040 with QNH 1025 = 4360ft

Therefore assuming airfield at MSL you have 3360ft to lose?

Groundspeed = Airspeed + wind =96kts or 1.6nm per minute

Distance to lose height 20nm or 12.5 minutes

therefore 3360ft to lose in 12.5 mins = 269 ft per min??


OK thats just a guess from an air traffiker....maybe I shouldnt give up the day job....let the real pilots come along soon

Curtis E Carr

First of all, calculate your groundspeed (a).

Next, calculate the distance (b) over which you have to lose your height.

Divide your distance (a) by the groundspeed (b) and multiply the quotient by 60.

You now have time (c) taken in the descent.

You now calculate height you need to lose. You are currently flying a flight level, therefore have the Standard Pressure Setting dialled up on your altimeter (1013 mb). But, you need to descend to a height with the airfield QFE set on your altimeter (the QNH in this example is not relevant). Maintaining FL40, what would your altimeter now read if you were to set QFE (1015 mb) (assume 1 mb equates to 27 feet)? You now have your current height (d). You are descending to 1000 ft therefore, the total height to lose (e) is current height (d) minus 1000 ft.

Finally, divide your total height to lose (e) by the time (c) to arrive at your answer.

Try and work it out for yourself before highlighting the rest of this post to reveal the answer.

Groundspeed is 90 + 6 = 96 Kts

Distance is 25 - 5 = 20 nm

Therefore Time in the Descent is 20/96*60 = 12.5 minutes

1015 mb (QFE) -1013 mb (SPS) = 2 mb

2 * 27 = 54 ft.

Therefore current height is 4000 + 54 = 4054 feet

Total height to lose is 4054 - 1000 = 3054 feet.

Therefore Required Rate of Descent is 3054/12.5 = 244.32 Feet per Minute.

paco

The airfield elevation is 300 ft (QNH - QFE at 30 ft per Mb) so you only have 3060 feet to lose over 12.5 mins, starting from 4360 ft AMSL. You get the same figures if you work just on QFE, so the QNH figures are irrelevant here. The exact answer would be 244.8, but one of the choices should be 250 fpm.

Phil

PS we take 27 ft per Mb only in Met questions. All others are 30.

Curtis E Carr

You're right, Phil

Must take my IRI head off.

See you Monday?

JD

fisbangwollop

PACO...Oh well was close, get your drift with regards QNH/QFE that makes more sense, was trying to look too deeply into the question....OK then its back to the day job for me

paco

Curtiss - I'll be there - coffee will be on!

Keneter - the no 1 big tip is always to draw a diagram, with the big numbers at the bottom, because they represent higher pressure.

cheers

Phil

keneter

Hi guys thank you all for your replys i see how it works now thanks alot

Lightning Mate

Cows getting bigger

Alternatively (or should I say practically?) you're travelling at about 1.5nm/min. 20nm to go gives about 13 minutes to descent. Somewhere in the region of 3000ft to descend. 3000/13 (close enough to 12) = 250. So, how about you start with 250ft/min and see how things pan out?