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

Hi Lauren, I am not surprised you want an explanation!. I quote the JAA LOS

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.

Hi All,

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

Somebody wrote this ages ago on the forum:
I make the answer 7000/150 x 60 = 2400ft per minute decent.

Can anyone confirm?

Cheers.

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.

And from your post

7000/150 x 60 = 2400ft should read 7000 / 150 x 60 = 2800ft

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!

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

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.

POF - Gust Load
 

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)

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.

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 :uhoh:

Question about DA(H)
 

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? :ugh:

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?

a monocoque has no supporting structures... at all... Look at the diagrams again.

@ Flying hog:

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

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.

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.

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

Hi RichardH, apologies for the slow reply - yes, I think best bet is to contact the people who wrote the book! Thanks :)

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?

Ahhhh explains it thank you :)

Require some help
 

Hi,

I am trying to revise for flight planning and mass and balance. I seem to come across quite a number of LRJT questions on question bank. Although I have been told they are unlikely to appear in the exam. Is there any truth to that. Anyone recently sat the exam and had such questions asked.

Secondly although I am trying to read materials provided on running load and distribution load but I am still encountering problems grasping the principle and if someone can provide help with that. It will be much appreciated.

This post might be better in ATPL theory questions forum.

LRJT certainly have/are been asked.

Use 50 nm instead of 30 nm. 2700 is correct for 30 nm, but the question is confusing.

two aircraft intercepting
 

hello guys.
would really really appreciate it if someone on pprune could help me with the solution to this problem on navigation:ugh:

**) aircraft takes off from 00 N 170 W at 2100 LMT on 21/12/99 at a ground speed of 300 knots. .Another aircraft takes off from 15 N 165 E on 22/12/99 at 2100 LMT to intercept the first aircraft.Find

a) time of interception
b) ground speed of second aircraft

please please help me with the solution to this problem. would really appreciate your time and help :):)

Torque
 

Which of the following is not and active force on the aircraft? lift,weight, thrust, drag, torque?

flight planning
 

If CAS is 190 kts, Altitude 9000 ft. Temp. ISA - 10°C, True Course (TC) 350°, W/V 320/40, distance from departure to destination is 350 NM, endurance 3 hours and actual time of departure is 1105 UTC. The distance from departure to Point of Equal Time (PET)

9000ft/ISA -10⁰C (OAT= -13⁰C) TAS =212 kt D = 350 NM … O = 176 kt … H =246kt Distance to PET = D x H / (O + H)…………………. = 350 x 246 / (176 + 246) = 204 NM

Answer 203nm


I am unsure about the temprature corrrection as i get is -7 { 15 - (9 x 2) }= 3 and -10 =-7
can someone just explain this -10 or +10 isa and correction made accordingly please
thank you in advance.

ISA at 9000ft is -3, however the question states ISA -10 which means the ambient OAT is 10 COLDER than -3 so should be using -13 to get your TAS of about 214 its.

If it had said ISA +10 then your +10-3 = +7.

Understanding these corrections is an important skill for a number of the subjects.

Thank you.

How did I pass myATPL writtens....
 

I have no idea how I managed to pass my ATPL written papers.


All I can say is please do not ask me any of the questions as shown above, I have wasted 1/2 day trying to get even one right.


Perhaps 45 years ago I was just clever, but now stupid.

You have a crate of mass 174 kg with dimensions of 1ft 3in x 1ft 9in x 2ft 4 in. What are the distribution and linear loads?

79.81 kg/sq.ft and 8.28 kg/in

can someone please help me break this down as i get different answer when i work out longest side for running load answer

Purpose of finding pressure height and density height
 

Hi,
I know how to find the pressure height but I dont fully understand the reason for finding it. I know that it has to do with the performance of the aircraft and terrain clearance, and also to know the take off distance. Can someone kindly explain to me the purpose of finding these two?
I am a noob
Cheers

Say you have a mountain that is 6000 feet high and you want to land your helicopter on it. Assuming you know what ISA conditions are (as a noob :)), in ISA conditions, your machine would think it was landing at 6000 feet. But say the QNH was 995 instead of 1013. There is an 18 mb difference which translates to roughly 540 feet. Now your machine thinks it is at 6540 feet, and the engine and rotors (or wings) have to work just that bit harder (in these types of questions, draw a diagram and place the larger pressures at the bottom).

Now let's look at the temperature. The ISA temperature at that height should be around 3 degrees C - but say it's actually 15 degrees - it is warmer than it should be by 12 degrees and the air is thinner (less dense), so your machine has to work harder still because it thinks it is another 1400 or so feet higher - actually around 8000 feet.

To find out density altitude, take the ISA temperature difference, multiply it by 120 and add or subtract it to the PA.

We use it to calculate performance related stuff such as take off distance, and landing distance.

At my organization, we use the pressure alt at our aerodrome to calculate the take off and landing distance for our aircraft, and add a 15 percent safety factor.

N5748E

Unfortunately ISA figures are only theoretical, in practise you will hardly be flying at altitudes or places where ISA conditions are present, personally I have been doing a lot of flying from a coastal Aerodrome, never experienced ISA conditions.

I know. It's never an ISA day...

Just get the airport elevation, the forecasted QNH, for e.g. at my airport today, the QNH was 996hpa, and airport elevation was 55ft. Gives u a pressure alt of 563ft.

Go to your POH and use the charts in section 5, performance, remembering the temperature as well...

clkorm3 - first work out the load intensity:

174 kg/(1.25 ft x 1.75ft) = 79.5 kg/sq. ft.

(Note that this answer is given in feet so you must convert from inches to feet.)

If you take the longest length according to the first part of the answer, you must calculate the running load with a length of 1 ft 9 in rather than the 2 ft 4 in. In this case, your answer has to be in inches, rather than feet.

Running load: 174 kg/1 ft 9 in= 174 kg/21 inches = 8.23 kg/in.

I'm not sure if you had other answers to select from as you can definitely choose a better configuration of running load and intensity by working with the 2 ft 4 in length. Maybe all the other answers had incorrect units or scaling?

They just needed a datum to refer and compare to so they could write the performance charts. Without a starting reference, it would be a complete mess.

The fact that it's never an ISA day is irrelevant.

thank you that was very helpful.

You also mention the reason finding Density Altitude (you said height, but meant Altitude)

This is used to correct the readings from an airspeed indicator to give the true airspeed of the aircraft.

The steps are:

Indicated Airspeed (subject to instrument and position error)

Calibrated Airspeed (subject to Compressibility error)

Equivalent airspeed (subject to density error)

True Airspeed (once you've factored IAS for all the above errors)