> >
>

# ATPL theory questions

Professional Pilot Training (includes ground studies) A forum for those on the steep path to that coveted professional licence. Whether studying for the written exams, training for the flight tests or building experience here's where you can hang out.

# ATPL theory questions

22nd Nov 2013, 12:10

Join Date: Nov 2013
Location: London
Age: 35
Posts: 1
Likes: 0
Received 0 Likes on 0 Posts
General Navigation questions not from DB

Hey guys,

I'm doing my ATPL at the moment and I'm struggling to find the right explanation to the following questions: (if any of guys could help me I would really appreciate it)

Number 1:

A Northern Lamberts conformal conic chart is overprinted with a false grid.
The constant of the cone of the chart is 0,80.
At 60°W the grid track is 090° and the true track is 010°.
At which longitude is the false grid aligned?

a) 160° W b) 140°W c) 020° E d) 040°E

Number 2:

A Northern Lamberts conformal conic chart is overprinted with a false grid which is aligned with the 25°W longitude.
The constant of the cone of the chart is 0,80.
If the true track at 120°W is 090° will the grid track
be:
a) 014° b) 166° c) 185° d) 355°

Number 3:

A Lamberts conformal conic chart is overprinted with a false grid.
The constant of the cone is 0.60.
At 40°S 70°W the grid track is 197° and the true track
is 239°.
At which longitude is the false grid aligned?
a) 112° W b) 028°W c) 000° E/W d) 030°E

Number 4:

North polar stereographic chart is overprinted with a false grid.
At 77°N 37°W the grid track is 175° and the true track is 093°.
At which longitude is the false grid aligned?
a) 045° W b) 045°E c) 082° E d) 119°W
22nd Nov 2013, 16:03

Join Date: Feb 2002
Location: Sunny Solihull
Age: 67
Posts: 276
Likes: 0
Received 0 Likes on 0 Posts
Hi Lauren, I am not surprised you want an explanation!. I quote the JAA LOS

061 01 04 05 Gridlines, isogrives
LO Explain the purpose of a Grid north (GN) based on a suitable meridian on a polar stereographic chart. (reference or datum meridian).
Grid questions should only be asked in relation to polar charts not Lamberts. You are potentially going to waste hours of time trying to sort out questions 1 to 3.

Even Q4 is somewhat obtuse but unless I am having a bad day its 45E. To understand PS questions these are best done by drawing diagrams. However there is always a relationship between GRID, TRUE, LONGITUDE. 175-93=82-37=45.

These questions usually ask what is the grid or true depending on north or south polar with grid aligned on either prime or anti-meridian. Certainly nowhere near as tricky as these questions.

Your FTOs notes should cover all that's required for PS charts and grid - refer back to them. Also try looking the the Polar 5AT chart in the Jeppesen and drawing some tracks between places, measure true then grid. Bit hard to explain in a written reply.

However if you have no joy then PM me as I have a dummies guide to PS charts you can have.

Last edited by RichardH; 22nd Nov 2013 at 17:14.
3rd Dec 2013, 14:34

Join Date: Jul 2012
Location: UK
Posts: 26
Likes: 0
Received 0 Likes on 0 Posts
Hi All,

Not ATPl question but didnt want to start a new thread:

Somebody wrote this ages ago on the forum:
They then give you 2 small written tests. The first one is again basic mathematics but applied to flying. Questions like "you want to be at 1000' 5 miles from the beacon. You are currently at 8000' and 20 miles from the beacon and your airspeed is 150kts. What descent rate do you need?" So basically, a few distance/speed/time questions. Know the 1/60 rule also because I think there's a question on that.
I make the answer 7000/150 x 60 = 2400ft per minute decent.

Can anyone confirm?

Cheers.
3rd Dec 2013, 15:00

Join Date: Jan 2011
Location: England
Posts: 661
Received 20 Likes on 13 Posts
You are at 8000 ft when 20 miles from the beacon.

You want to be at 1000 ft when 5 miles from the beacon.

So you need to lose 7000 ft while flying a distance of 15 miles.

At 150 kts you will take 15/150 = 0.1 hours (which is 6 minutes) to fly the 15 miles.

So you need to lose 7000 ft in 6 minutes.

Required ROD = 7000 ft / 6 mins = 1167 ft/min.

7000/150 x 60 = 2400ft should read 7000 / 150 x 60 = 2800ft
3rd Dec 2013, 18:33

Join Date: Jul 2012
Location: UK
Posts: 26
Likes: 0
Received 0 Likes on 0 Posts
Thanks very much,

I now get it..
Understand I need to find how long to fly the 15 miles

gives me time 6mins

Then 7000ft divide 6 mins give the 1167ft per minute, to achieve the 1000ft

Many thanks, I keep rushing my workings!
29th Dec 2013, 08:25

Join Date: Sep 2013
Location: Singapore
Age: 37
Posts: 7
Likes: 0
Received 0 Likes on 0 Posts
Enquiry

I am currently a SPL holder with 40hrs of flying at Malaysia.Currently at Navigation phase

I understand formula for lift is-->L=1/2 s*p*v*sq*Cl where Cl is coefficient lift,v is velocity of air,s is surface area of wing and p is density of air

I have to following question to ponder while revising my unusual attitude recovery

1)Why does increasing pitch decrease airspeed and why does decreasing pitch increases airspeed?

2)I understand when we are using aileron to turn,one wing lift is more than the other,hence we roll to either right/left
30th Dec 2013, 15:04

Join Date: Jul 2012
Location: Crete
Age: 42
Posts: 1
Likes: 0
Received 0 Likes on 0 Posts
1)Why does increasing pitch decrease airspeed and why does decreasing pitch increases airspeed?

2)I understand when we are using aileron to turn,one wing lift is more than the other,hence we roll to either right/left
1) although you are correct, you also have to consider drag.
Changing pitch will affect your Cl, and your Cd (Coefficient of drag), the latter being:
D = Cd * 1/2 * ρ * V^2 * s. Notice the similariles with the lift formula.
Make sense now?
To understand more, refer to the Lift and Drag diagrams as well.

2) absolutely correct. Deflecting a aileron will change the chordline of (part of) the wing (chordline being from leading edge to trailing edge) thereby changing that parts AoA (angle of attack, alpha) and its Cl and Cd.
4th Jan 2014, 23:58

Join Date: Jan 2006
Location: Europe
Posts: 404
Likes: 0
Received 0 Likes on 0 Posts

Hi all,

I'd need some help figuring out the following two questions, been working on this for too many hours now. Many thanks in advance!

Aircraft maintaining straight and level flight at speed of 1.9 VS. If a vertical gust causes a load factor of 2.9 the load factor n caused by the same gust at a speed of 2 VS would be:
a) n = 3.00 (correct answer)
b) n = 3.61
c) n = 2.85
d) irrelevant, since airplane would already be in a stalled condition at 1.9 VS

Aircraft maintaining straight and level flight at speed of 1.5 VS. If a vertical gust causes a load factor of 2.5 the load factor n caused by the same gust at a speed of 2 VS would be:
a) n = 3.00
b) n = 2.30
c) n = 2.25
d) irrelevant, since airplane would already be in a stalled condition at 1.5 VS (correct answer)

Last edited by Transsonic2000; 5th Jan 2014 at 13:36.
5th Jan 2014, 13:59

Join Date: Jan 2011
Location: England
Posts: 661
Received 20 Likes on 13 Posts
The increase in load factor caused by flying at different speeds in the same gust can be calculated using the following equation:

N increase at new speed = N increase at old speed x (V new / V old)

For the first question comparing the conditions at 1.9 VS and at 2.0 VS gives the following:

N increase at old speed = 2.9 - 1 (straight and level) = 1.9
Old speed = 1.9 VS
New speed = 2.0 VS

N increase at new speed = 1.9 x (2 / 1.9) = 2.0

Adding this increase to the initial straight and level load factor of 1 give a new load factor of 3.0

Using the same equation in the second question gives us the following:

N increase at new speed = 1.5 x (2 / 1.5) = 2

Adding this to the initial straight and level load factor of 1 g give new load factor = 3.0.

But option d in each question poses the possibility that the aircraft may stall before it attains these increased load factors, so we need to test this possibility.

We can do this using the following equation:

VS at new N = VS at old N x (square root of new N / square root of old N)

Inserting the data for question 1 gives the following:

VS at 2.9 g = VS at 1 g x (square root of 2.9 / square root of 1)

VS at 2.9 g = 1.7 VS at 1 g

The aircraft was initially flying at 1.9 VS, so it would not stall before attaining 2.9 g. So option d is incorrect.

Inserting the data for question 2 gives the following:

VS at 2.5 g = VS at 1 g x (square root of 2.5 / square root of 1)

VS at 2.5 g = 1.58 VS at 1 g

This means that the stalling speed at 2.5 g would be 1.58VS. But the aircraft is flying at only 1.5 VS, so it would have stalled before attaining 2.5 g. This means that the higher load factor of 3.33 g could not be attained. So option d is correct.

Last edited by keith williams; 5th Jan 2014 at 20:54.
5th Jan 2014, 14:39

Join Date: Jan 2006
Location: Europe
Posts: 404
Likes: 0
Received 0 Likes on 0 Posts
Hi Keith,

once again, many thanks for your excellent explanation! It is greatly appreciated! it's of much help to me, without it I'd probably just went on hopping that this type of question won't be asked in the exams or simply just guessing, choosing answer "d" and keeping the fingers crossed
6th Jan 2014, 17:10

Join Date: Dec 2011
Location: Europe
Age: 38
Posts: 14
Likes: 0
Received 0 Likes on 0 Posts

Hi All,

Does anyone of you know, why are sometimes different DA(H) for different aircraft categories specified on the approach chart? For example:

CAT 1 ILS (missed approach gradient of MIM 2,5%)

DA(H):

A: 250' (213')
B: 260' (223')
C: 270' (233')
D: 280' (243')

Usually there is only one figure for ALL aircraft categories?
7th Jan 2014, 19:04

Join Date: Jan 2013
Location: Estonia
Posts: 2
Likes: 0
Received 0 Likes on 0 Posts
What's the difference between a semi-monocoque and a reinforced shell aircraft structures? All I see in the explanations is that the reinforced shell is a development on the semi-monocoque type and has a reinforced skin supported by other strucural members. But isn't that what differentiates monocoque from semi-monocoque - longitudinal stringers?
8th Jan 2014, 15:39

Join Date: Jun 2013
Location: Singapore
Age: 36
Posts: 39
Likes: 0
Received 0 Likes on 0 Posts
a monocoque has no supporting structures... at all... Look at the diagrams again.
10th Jan 2014, 01:11

Join Date: Jan 2006
Location: Europe
Posts: 404
Likes: 0
Received 0 Likes on 0 Posts
@ Flying hog:

see Jeppesen Route Manual, TOC (Approach Chart Legend), page 113 (landing minimums)

Last edited by Transsonic2000; 10th Jan 2014 at 01:48.
10th Jan 2014, 17:13

Join Date: Nov 2007
Location: Texas
Posts: 1,924
Likes: 0
Received 1 Like on 1 Post
The one I like is the ILS 16L at SEA. The category A minimums are 60 feet higher than the B, C and D minimums. When the TERPS people did all the sums they figured a slow aircraft with a strong crosswind could be blown too close to the control tower so they set the category A mins at the height of the control tower. A faster aircraft would not be displaced as much by the crosswind and so can go lower.

Flying Hog, without knowing which airport it would be hard to say why the minimums are different.
14th Jan 2014, 17:35

Join Date: Jul 2006
Location: Ireland
Posts: 27
Likes: 0
Received 0 Likes on 0 Posts
Hi all,

Studying mass and balance at the moment and working through the Oxford book. One thing is puzzling me about one of the exam questions. The question is based off CAP MRJT but in the worked answer they use a different MZFM. The MRJT MZFM is 51300 kg but they use 53000 kg in the answer. Is there a reason for this?

For anyone who has this book, the question is on page 3-14 (Q23) and the worked answer is on page 3-19.
15th Jan 2014, 07:58

Join Date: Feb 2002
Location: Sunny Solihull
Age: 67
Posts: 276
Likes: 0
Received 0 Likes on 0 Posts
I suggest you contact your FTO (Oxford) directly as they will know why more than anybody on this forum. This goes for any material from any FTO best to contact the source first.

You are likely to get 1 of 3 responses:
1. Yes it's known a mistake in our notes - sorry
2. It's a mistake thanks for letting us know
3. You got it wrong because of x and y
19th Jan 2014, 19:42

Join Date: Jul 2006
Location: Ireland
Posts: 27
Likes: 0
Received 0 Likes on 0 Posts
Hi RichardH, apologies for the slow reply - yes, I think best bet is to contact the people who wrote the book! Thanks
29th Jan 2014, 16:01

Join Date: May 2012
Age: 29
Posts: 8
Likes: 0
Received 0 Likes on 0 Posts
P0041
What is the maximum vertical speed of an aircraft with a mass of 76 000 kg?

Given: Thrust = 160 kN

Drag = 110 kN

TAS = 190 knts

g = 10 m/s²

I get an answer of 1250 ft/min. But in the potential answers the correct is 1267 ft/min. 1250 isn't even an option. Why?
29th Jan 2014, 17:15

Join Date: May 2012