Learning to fly, or thinking of learning? Post your questions, comments and experiences here

Moderator: AndyR

#1688253
Hi there people, this is my first post but I've been skulking in the background for a while now.
I'm working my way through the PPL text books by Pooley's (Air Pilot's Manual) and have arrived at the Navigation book.
Firstly I will admit I'm no mathematician so you'll have to bear with me, I'm at pressure altitude / density altitude.
Pressure altitude made complete sense, if its over 1013, times the amount by 30 and take off the elevation and vise versa.
Density altitude was a little harder to grasp as I've no experience of negative numbers and so it took a while to sink in. any who I felt I'd got it.
Got to the end of the chapter, to the questions (which I always look forward to) and question 8 and 9 I failed miserably at. They were phrased in a way that hadn't been mentioned in the book i.e. FL instead of just elevation, no mention of an OAT just a seemingly vague nod to temperature structure being the same as ISA. On top of which I can't seem to work it out even though I have the correct answer which is sending me up the fricken wall!!!!!
So please please put me out of my misery:

8. An aircraft is flying at FL60 over terrain which is 1750ft AMSL where the local QNH is 998 hPa. If the temperature structure is the same as ISA,what is the aircraft's height above the terrain?

9.An aircraft is flying at FL60 over terrain which is 1750ft AMSL where the local QNH is 1020 hPa. If the temperature structure is the same as ISA,what is the aircraft's height above the terrain?

So maybe you could tell me the answer and how you arrived at it so I can try to see where I'm going wrong?

Should I also give you the multiple choice answers?

Anyway, if you got this far, thanks for taking the time to at least read this.

Regards,

Tim.
#1688265
Qu 8.
The aircraft is using an altimeter setting of 1013hPa to read FL60 (6000ft). We know the QNH is 998hPa which gives a difference of 15hPa.
1hPa = 30ft.
So 15*30=450ft.
We know the plane is flying at FL60 (6000ft). So to get the actual altitude of the plane you do;
6000ft-450ft=5550ft.
The terrain is 1750ft AMSL. So we do;
5550ft-1750ft=3800ft.

Someone beat me to it :(