Hi there,

Have a question relating to Nagation progress test 4, qusetion 5 in study notes, i just dont understand how to work this one out, any one shed any light? cheers

Hutcho

This is a matter of proportion. If you are 4 minutes behind your flight planned estimate at the second marker, your ground speed must be lower than you expected. It is taking you two extra minutes to cover each of the segments. You have lost 4 minutes by the time you arrive at the second marker. It is reasonable to assume that this will keep happening because of your reduced gound speed. There are four and a half segments to go, so, if you are losing two minutes per segment, you will lose another nine minutes by the time you get to B. That's a total of thirteen minutes later than the original flight planned time interval.

cool Bob, thanks for the good explanation, heres one more for ya.....


Given the following information:

Aerodrome ELV .......... 1250 ft
Cruise level ................. A075
Area QNH ................... 1023 HPA
Climb GS .................... 90 kt
Start of climb .............. 1000 ft AGL

Your calculation of the rate of climb required to reach cruise level 36 nm from the airport is -

1 220 fpm
2 435 fpm
3 125 fpm
4 260 fpm

Just cant quite nut it out,answer is 220fpm, not sure how to arrive at that?

From 1000 feet above the aerodrome to 7500 feet the aircraft must climb through 5250 feet [the QNH is not relevant]. 36 nm at 90 kt = .4 hours. That's 24 minutes. [36 divided by 90 multiplied by 60]. 5250 feet in 24 minutes is 218.75 ft/min.

so because they state the climb is starting at 1000 AGL & not AMSL thats why the QNH is not even considered? so very simple when you see it broken down....

Thank you very much Bob!

Matt.