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-   -   General Navigation questions. (https://www.pprune.org/professional-pilot-training-includes-ground-studies/356271-general-navigation-questions.html)

Mohit_C 30th Dec 2008 19:47

General Navigation questions.
 
Hi all,

I'm back with 4 questions from the Volare questions that I have a few doubts with.

Q378) In a remote indicating compass system the amount of
deviation caused by aircraft magnetism and electrical
circuits may be minimised by:
A) the use of repeater cards (mark scheme answer)
B) positioning the master unit in the centre of the aircraft
C) using a vertically mounted gyroscope
D) mounting the detector unit in the wingtip
-> I would say D.

Q481) The alignment time, at mid-latitudes, for an Inertial
Reference System using laser ring gyros is
approximately:
A) 5 MIN (mark scheme answer)
B) 10 MIN
C) 20 MIN
D) 2 MIN
->My GeneralNav book says for IRS: Equator = 5 min, 70º latitude = 10 min, 70-78º latitude = 17 min. My Instrumentation book says: Equator = 2.5 min, 45º latiude = 5 min, 60 º latitude = 10 min. Now is there an actual formula to calculate this time for alignment as there seems to be differences between textbooks (at least mine)?

Q247) At what approximate latitude is the length of one minute
of arc along a meridian equal to one NM (1852 m)
correct?
A) 30°
B) 45° (mark scheme answer)
C) 0°
D) 90°
-> I don't actually understand why it's 45º. All latitudes are along a meridian and one minute of arc will always equal one NM along a meridian.

Q251) In which two months of the year is the difference
between the transit of the Apparent Sun and Mean Sun
across the Greenwich Meridian the greatest?
A) June and December
B) April and August
C) February and November (mark scheme answer)
D) March and September
->Well the "True Sun" and "Apparent (fictitious) Sun" meet at the Aphelion (Cancer) and Perihelion (Capricorn) and the "Mean Sun" and "Apparent Sun" meet at the Equinox Aries and Libra. Well I would guess that the greatest difference would be in Cancer (Summer in Northern Hemisphere) and Capricorn (Winter in Northern Hemisphere) = A.

Thanks.

BRS_flyer 30th Dec 2008 20:54


Q226) Given:
TAS = 485 kt,
OAT = ISA +10°C,
FL 410.
Calculate the Mach Number?
A) 0.90
B) 0.85
C) 0.87
D) 0.825 (mark scheme answer)
-> Using the CR-3 Circular Computer and the formulas I get approximately 0.85 (C).

Using formula I get the following:

OAT at FL410 = 206.8k (assuming LR of 1.98 per 1000') +10 = 216.8k

LSS= 38.94xSquare root of temp - 38.94x14.72 = 573.38

Mach # = TAS/LSS - 485/573.38 = 0.845

So I would have said B as well

Phenom100 30th Dec 2008 22:38

It's been a while since i've done this ****, but here's the answer for Q226

485kts
OAT ISA + 10deg
FL410
Mach #: ????

1st calculate the correct temp, ISA@ FL410 = Approx -56deg + 10 = 46deg

On wizz wheel put your MACH ARROW on -46deg

On the outer scale find 485, read off the inner scale, which gives you the answer D: 0.825

Will have a look at the other ones for you tom.

PS. E-mail me, as i may have something which will interest you
[email protected]

:ok:

Nearly There 30th Dec 2008 23:04

ISA at FL410 is -56.5
Question says ISA + 10 = -46.5......so 273k - 46.5 = 226.5k

38.94 x SqrRt226.5 = 586.....485/586 = 0.827 MNo

Phenom100 30th Dec 2008 23:08

Q65

TAS: 180KTS
TRACK: 090
W/V: 045/50KTS
ENDURANCE: 1 HR

If you use the wizz wheel correct it gives you the following groundspeeds

G/S OUT: 141KTS
G/S HOME: 214KTS

Put these into the formula 1x214x141
214+141 Gives you the answer: 84.99

:ok:

Mohit_C 31st Dec 2008 10:18

Regarding Q226, when it says OAT = ISA + 10ºC does it mean that using ISA we must calculate the OAT, in this case: 15ºC at SL - (41*2) = -67ºC at FL410 but in ISA is -56.7ºC and then add 10º to this OR do we do this: 15ºC at SL + 10º C = 25ºC at SL - (41*2) = -57ºC but in ISA -56.7ºC? Using the CR-3 circular computer, when do we subtract the temperature rise from the OAT reading?

Q65) Thanks Danny C. It appeared to be that I was calculating the GSout and GSback incorrectly, not subtracting/adding HW/TW to ETAS whilst I was using TAS.

Whirlygig 31st Dec 2008 10:36

ISA at FL410 is -56.5 degrees C. Therefore, ISA +10 = -46.5 degrees C. There shouldn't be any confusion.

How much more of the Volare Question Bank have you got to through? :}

Cheers

Whirls

BRS_flyer 1st Jan 2009 10:41


ISA at FL410 is -56.5

OAT at FL410 = 206.8k (assuming LR of 1.98 per 1000') +10 = 216.8k
:O Rookie mistake!!

Mohit_C 1st Jan 2009 12:39

Thanks. 9 questions to go.


Originally Posted by Whirlygig
How much more of the Volare Question Bank have you got to through?

Well I actually have the following questions banks to do yet:
-Instrumentation (need to post the questions I have doubts on)
-Meteorology
-Mass and Balance
-Flight Planning and Monitoring
-Operational Procedures.

Once finished all the Volare question bank, I will subscribe to ATPL Feedback question bank or Bristol GS question bank to do all the new questons.

:ok:

Keith.Williams. 1st Jan 2009 15:21


Regarding Q226, when it says OAT = ISA + 10ºC does it mean that using ISA we must calculate the OAT, in this case: 15ºC at SL - (41*2) = -67ºC at FL410 but in ISA is -56.7ºC and then add 10º to this OR do we do this: 15ºC at SL + 10º C = 25ºC at SL - (41*2) = -57ºC but in ISA -56.7ºC? Using the CR-3 circular computer, when do we subtract the temperature rise from the OAT reading?
I think that you are overlooking the fact that the temp lapse rate (1.98 deg/1000 ft or approximately 2 deg/1000ft ) applies only up to the Tropopause at 36000 ft.

So at FL410 ISA Temp = 15 -(36*1.98) = -56.28)
using 15 deg at ISA msl and -1.98/1000 ft

So ISA +10 at FL410 = -56.28 +10 = -46.28

The examiners have picked a flight level above the Tropopause to try to get you to make this error.

Mohit_C 20th Jan 2009 13:57

I have a couple of new questions where I'm stuck on:

Q1) In the following Annex 061-12593A, where is the variation indicated:

http://img254.imageshack.us/img254/1943/dibujogx9.png
http://img254.imageshack.us/img254/d...png/1/w567.png

Q2) [Q49 VOLARE]
(For this question use annex 061-1818A)
Assume a North polar stereographic chart whose grid is
aligned with the Greenwich meridian.
An aircraft flies from the geographic North pole for a
distance
of 480 NM along the 110°E meridian, then follows a grid
track
of 154° for a distance of 300 NM.
Its position is now approximately:
A) 70°15'N 080°E
B) 80°00'N 080°E (mark scheme answer)
C) 78°45'N 087°E
D) 79°15'N 074°E

http://img258.imageshack.us/img258/8007/dibujo2en4.png
http://img258.imageshack.us/img258/d...png/1/w692.png

I have 4 questions from the first post pending.

Thanks.

PS. Does anyone know if it is possible to take a magnifying glass into the exam?


bajadj 20th Jan 2009 15:13

q2, came up in my exam!! apparently it is the only question in the database that has that kind of grid in it, so just remember 80 80 and you'll get it right, it's one of two questions which we were told to just remember the answer..80 80 and 6 microtesla! think microtesla was in ops, but i know there are 6 of them!

aerosumi 24th Jan 2015 16:40

hi
 
hi mohit c,
liked your consistency in what you are studying. realy apreciable. i need your help in getting the link for downloading the volare question bank.
thanks


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