Gen Nav Help
 

Hi Guys,

I need help with the following ATPL Nav question:

At 1321 UTC an a/c 500nm from point X is required to reach point X at 1445 UTS. Present cruise conditions are FL310, Mach 0.83, OAT -46'C, Wind Component -85kts, Assuming no change in the ambient conditions, what is the latest time at which speed muct be reduced to Mach 0.69 in order to reach X at 1445 UTC?

A. 14.35
B. 14.24
C. 14.11
D. 13.59

I make it D - 13:59.

This is mainly an algebra question but here's a rough cut of my workings!

MN = TAS/LSS and LSS = 39 x SQRT(T) where T is in kelvin. Therefore, T = -227 and LSS = 587.6.

Therefore, at MN=0.83, TAS = 487 and G/S = 402.
At MN = 0.69, TAS = 405 and G/S = 320.

The total time from 13:21 to 14:45 is 84 mins i.e. 1h24.

Therefore, with distance = speed x time and there are two times we need to calculate t1 and t2.

500nm = 402t1 + 320t2 here t1 + t2 = 84mins = 1.4 hours

Therefore, t1 = 1:24 - t2 and replace in first formula.

So, 500 = 402x(1.4 - t2) + 320t2
500 = 562.8 - 82t2 + 320t2
62.8 = 82t2
Therefore, t2 = 46 minutes.

14:45 less 46 minutes = 13:59.

I hope this makes sense as it's difficult to type without sub and superscripts!!

Cheers

Whirls

Put the books down and watch TV or go to bed!!:ok:

Who? Me or him??? :8 :p

Cheers

Whirls

Oh God, how much of the ATPL exams have that sort of questions?! I get lost half way through that. It's putting me off starting.:{

It's a typical question and would possibly be a two or three- marker!!

The aviation bit is the Mach number and Local Speed of Sound; everything else is speed=distance/time and manipulation of the formulae which is GCSE maths.

Cheers

Whirls

Hey Whirlygig,

Thanks for the quick reply, makes sense now i look at it again, after a rest from this :mad:.

im sure i'll have a few more for you in the next week or so.

Have the exam again in 2 weeks, failed in feb with 73%

Cheers again

Dan