DGCA CPL NAV Questions

[AZFlyUSA]

Here are a couple of NAV questions. CAN YOU PLEASE EXPLAIN how to get the answers.

1. Given the value of coefficient, A= -1, B= -2, C= -3, what shall be the deviation on a HDG 295 (C)

2.
HDG (M) HGD (C)
LC P4
275 271
000 358
090 092

Find Coefficient A, B, C & Deviation on HDG 225

[jimmygill]

Magnetic deviation of a ship(aircraft) as a function of its heading is given by following equation

http://myreckonings.com/wordpress/Images/MagneticDeviation/Equation_2.png

where delta = deviation, and the other greek letter is the ship's magentic heading.

But for our problem we don't have values of D & E, so we will just ignore the last two terms to have simple formula

DEV = A + B*sin(HDGm) + C*cos(HDGm)

deviation = -1 + -2*sin(295) -3*cos(295) = -0.5 degrees



Problem 2

[AZFlyUSA]

Thanks for that Jimmygill, the answer that is marked correct on my test question is;

Question 1= 0.46 Westerly
Question 2= A=1, B= -3, C=1, DEV=2.4

I'm working through Compass Swinging & Associated Problems, I'll post working out to above 2 questions when I get it figured out.

[AZFlyUSA]

Jimmygill,

Your formula for question 1 is perfect, below is the working out.

-1 + -2 x Sin295 + -3 x Cos295 = -0.46 (Westerly)

Thanks again.

[arba]

no FMC, huh !?

[jimmygill]

I removed the solution to problem 2, it had two errors...
both related to sign of the deviation...

I don't know which one of the following is correct convention

DEV HDG(m) HDG(c)
-2 000 358
+2 090 092
-4 275 271

This gives solution (-1,3,-1)


DEV HDG(m) HDG(c)
+2 000 358
-2 090 092
+4 275 271

This gives solution (1,-3,1) and Dev = -2.4


So I deduce Dev = HDG(m)- HDG(c)

[jimmygill]

no FMC, huh !?

Yes, no FMC and absolutely no utility to pilot...

[AZFlyUSA]

Thanks Jimmygill, here is another DGCA CPLCG question;

Thanks for taking the time in answering the last questions I posted, could you explain the following;

An AC heading towards a station of HDG (M) 060 and ADF Indiacation 030 (H), cleared to intercept radial 250 and enter into a standard hold.

A. The pilot should enter the hold through;
Direct
Parallel
Tear Drop

B. The pilot should alter the heading to_________to intercept the radial
230
250
070
130


I know the answers are Direct & 130, would just like an explanation of how the answer was derived.
Thanks for all your help.

[jimmygill]

1. Magnetic Bearing to the station is = Mag Heading + Relative Bearing = 60 + 30 = 090, that means the aircraft is on 270 radial approaching the station while headed 060

2. 250 radial lies to the right of radial 270 (keeping in mind the current heading 060), so we know we will need to make a right turn, opposite of 250 is 070 will give a course parallel to 250 radial, so add 60 or 45 (intercept angle) which will give 115 to 130 as intercept heading.

3. Even though it is appropriate to use intercept heading of 115-130, only 130 appears in the choices.

4. As per the ATC instruction you will be intercepting the radial 250 inbound before crossing the fix, this gives us a direct entry holding pattern when we cross the fix.


I hope this solves the problem.

For such questions always make a good drawing to start and answers will be obvious.

[AZFlyUSA]

Thanks Jimmygill, you're a great help!:ok:

[jimmygill]

A direction gyro on a ground test at 50 s latitude is found to
have a drift at 6/hr, reading increases, calculate the real wander?

The answer is 5.5"/hr decreases

Assumption: The direction gyro axis is parallel to local North-South during the test.


The direction gyro has teh property of rigidity in space. Which means
that it will maintain its axis direction in relation to space, and not
just in relation to earth. We know earth undergoes one full rotation
(360 degrees) every 24 hrs. Which means the earth rotates at 15 degrees/hour.



Our directions are fixed with earth, so these directions also rotate with earth.
But not all these directions at all the location on earth rotate at 15 deg/hr.

For example on equator, North and South always remain in the
same direction in relation to space, hence its rotation rate is 0 deg/hr.
While close to the poles the earth fixed North/South rotates at 15 deg/hr.

At the intermediate latitudes the formmula is 15*sin(Latitude) deg/hr
(+ Southern Hemisphere, -ve for Northern)




So at the latitude of 50s the rate at which the Local North-South direction rotates
in space is 15*sin(50) = 11.5 deg/hr


Now if the gyro didn't have any mechanical error it would be drifiting by 11.5 degrees per hour (increasing).
But the test says it is drifting by only 6 degrees per hour, this discrepancy is due the mechanical drift error also called as Real Wander, since the observed wander is less than apparent wander, the real wander must be negative.

Real Wander = 6.0 - 11.5 = -5.5 degrees/hr

[AviatorVette]

bump, subscribed.

[navneet.baron]

this is the first time im using a forum, so kinda new ...excuse for any errors/ stupidities.....
have a copule of questions::
1--on a compass swing readings obtained are
MLC P4
98 95
182 178
270 271
find ABC ?
what if north dev was alsoo given?
i have no idea HOW TO SOLVE THIS QUESTION EXCEPT FOR FORMULA A + BSIN@ + C COS@

2--what is most economical to fly? Given : specific gravity of fuel = 0.8
a- 34.5 litres/km
b- 0.065 GNm/liter
c- 16.85Kg/GNm
d- 0.20 GNm/imp gallon.

3-- what kind of questions come in composite exam, on the formula for drift and topple
!5*sin (lat)*cos (inclination of gyro axis)
and how to solve them......
help would be very much appreciated...TY

[Sperry CL2]

@navneet.baron

sorry in advance if thre's somthing wrong .... i'm also a learner
lets start-
question 1
1-first of all you have to firgure it out the deviation.
2-apply formula { Dev N + dev S } = { Dev E + dev W }
3- now u actually can put the values in and figure it out the value of "dev N"
4-now u have all the deviations appy furmula.
For Dev A - sum of all deviation divided by 4
for dev b - East-west divided by 2
for dev c - North-south divided by 2.

[t.a]

Can any one please share with the topics and if possible questions that were mentined in the last july nav-met composit attempt.

Here are few topics that i remembered
*charts - Mercator's Transvers
*ILS/ROD Formula related
*CP/PNR/PET
*Dip(M)
*Reason for Day/night/diff timeing
*DME horizontal distance(Kind of asks every time)
*convert 7mt/sec to km/hr
*Pressure instruments/ins . gyro
*IRS/INS
*scale 1:200000 1 cm to ?nm
*vhf signal strength
*DME lost signal, What would it still continue to indicate
*ADF behaviour in X'ing station/Homing
*Tropical Cyclone
*Pressure Gradiant in india
*NEI chanding storm
* Physical process of weather is accompained by?
*Invertion/Stability/DALR/SALR/ELR
*Valitdity of TAF
*ITCZ


These are the few topics that were covered in the exam can u please update it and possible to post any questions relavent.

Cheers!!
`!`

[t.a]

Briefly describing about the questions for this attempt (Oct)

I see no major aforesaid topics.

It is mostly from Radio aids and are picked from Oxford aviation publication (Oxford ATPL Books (download torrent) - TPB (http://thepiratebay.org/torrent/3784170))

Keith williams and I C Joshi didn't really work for this attempt as there were no questions on Indian Climatology and no numerical as such to solve,

No CP, PNR questions. Hardly any on magnetism. Few questions were picked up from FAA commerical question bank (Gleim/ Prep ware).

Met was also mostly form that jaa met syllabus and was in general.

For the first time i seen the question paper was of 100 questions.

Kindly post any questions if your know remember for this attempt.

Thanks..

[rahulred5]

The paper this time was a disaster imho. Lot of questions came from ATPL and ZERO questions from nav. Thats right ZERO. Half the paper was DME\IRS\INS\FMS. and gen met.
I predict single digit pass rates.

[aditya104]

The following is a question from Ground Studies for Pilots-Navigation by R.B.Underdown.

Ch-6
Pg74
q.8 The distance flown by an aircraft in 40 seconds at a groundspeed of 480kts is represented on a chart by a line 3.6cm long. The chart scale denominator is:
a) 274200
b) 1 865 000
c) 3 125 000
d) 2 742 000

The correct answer from the key at the back is (d). How is that? :confused:

[gAMbl3]

Printing error I guess. Answer should be A) 274 000.

[aditya104]

gAMbl3 to the rescue again. Thanks. :)