General Navigation Questions

[Just another student]

I am preparing to take my last two exams this March, one of them being G-Nav. Can anyone provide me with any guidance for the following questions I have? I would appreciate any assistance.


* When answering a PET/PSR question, W/V is often given in two different formats, either headwind/tailwind or +/- . When +/- is used, should I take it at face value and use - as a headwind and + as a tailwind, or as in some graphs, + as a headwind and minus as a Tailwind? Does this vary in the flight planning exam?


* When calculating Tas from Cas, which temperature should I use? If True OAT is given, should I convert to indicated using the temp rise scale etc ? They constantly vary which temp they give you in questions, that's why I ask. Also, you only alter tas (due to compressibility) when CAS is 300+? Is that correct?


* If you are given two way points and they want to know the rhumb line distance E/W between them, how do you go about it? Do you use the departure formula? How does working out Great Circle distances differ from this? (I understand about anti-meridians etc but must you allow for convergence etc ? )


* The following waypoints are entered into an INS.

1. 60N 30W
2. 60N 20W
3. 60N 10W

The INS is connected to the autopilot on the route 1-2-3.The track change on passing 2 will be?

a) 9 increase
b) 4 increase
c) zero
d) 9 decrease

Am I correct in saying that you have to calculate the conversion angle? This comes to 9 degrees, but how do you know wether it is a 9 increase or decrease?


* An aircraft is at FL140, IAS 210 and a true OAT -5. Wind is
-35.When the plane is 150nm from a reporting point, ATC request to lose 5mins by the time you get to the beacon. How much should IAS be reduced?

a) 15 knots
b) 25 knots
c) 30 knots
d) 20 knots

No matter which way I work this question, I get the answer to be 25 kts, not the given answer of 20 kts. Does anyone agree with me on this one?

* In order to maintain an accurate vertical pendulous system, an aircraft inertial platform incorporates a device.

a) Without damping period 84.4mins
b) with damping period 84.4 mins

Conflicting feedback here, which one is correct?

* what is the highest latitude at which an observer could see the rising and setting sun every day?

a) 62 degrees
b) 64
c) 66
d) 72

Again conflicting feedback, with a choice between 64 and 66 degrees. Any ideas?


I know it seems alot, but I just want to remove 99% of any doubt from my mind, before the exam.

Thanks for your help in advance.

[Genome_Junkie]

It is a long time ago that I did Gen Nav but I will have a go at some of your questions:

1) Headwind/Tailwind:
- is a Headwind.
+ is a Tailwind.
I assume you are talking about the Flight Planning graphs in the CAP where you enter from the bottom of the graph and adjust for the wind before going further upwards to altitude. If that is the case then relax. The graphs in the Flight Planning give it the other way round because they simply got it wrong. It is a misprint.

2) Temperature.
True OAT is true OAT. Use that value. What you see on your thermometer is usually reading higher because of the heating effect which depends on airspeed. So to get OAT from indicated temp you need to find out how much a temp rise you get for the speed that you are at. I think you then need to subtract that value from the Indicated Temp to get true OAT. So, true OAT is what you use for your calculations.

Speed: Yes, compressability only becomes significant above 300kts.

3) Rhumb Lines.
The departure formula will give you Rhumb Line distance.
As an aside, this is quite useful to remember for some of the exam questions because they will ask you for Great Circle distance. The great circle distance will be less than Rhumb Line. Calculate Rhumb Line distance and look for a number that is smaller.

4) INS
I hope I get this right - So take this with a large pinch of salt!

Well, yes it helps to work out conversion angle.
Draw a diagram.
Middle (20 degree) Meridian vertical, the 30 degree meridian sloping bottom left to top right and the 10 degree meridian slopin top left to bottom right. Draw a straight line across the meridians to deliniate 60 degrees north. Look at the angles and decide for yourself.


5) relative Velocities.
I don't have my CRP with me.
However, I kind of remember this.
First, work out TAS. This will most certainly be over 300 Kts.- Include Compressablilty.
Calculate Groundspeed
Work out how long you need for the 150nm without any speed alteration.
Add/subtract the 5 minutes. I guess losing 5 minutes means subtrating 5 minutes.
Calculate a new Groundspeed and then the new TAS. That will be one of the answers for sure!
However, you need to calculate:
A new IAS and DO NOT forget to include compression again!

6) With damping.

7) Solar System.
I hated it and just learned it for the exam and thats it.
Remember they are bastards. Checkt the feedback and see if it really does conflict. There is a difference in the wording of the questions:
What is the highest lattitude that the sun can be seen every day of the year.
IS DIFFERENT TO:
What is the highest lattitude that a rising and setting sun can be seen every day of the year.

The question is asking about the arctic circle I guess.
Above it you have the midnight sun. It DOES NOT set during some days of the year.
Below the circle it does.
Check it in your notes.

Good luck with the exam. It is not that hard.

Cheers

Arnie

[Just another student]

Thanks for that help Arnie. I think your right about the solar question, it is indeed worded slightly differently.

So with great circle distance you just look at the two sets of lat/long work out the number of degrees difference, and convert into miles (X 60) ?

I also see what I am doing wrong with the velocity question, it is 25 kts difference in GS not IAS.... doh!

Cheers

[Send Clowns]

Genome is correct where he has answered

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You will not be asked to calculate the great-circle distance except in cases around the equator or along meridians. The general formula for a great-circle distance is specifically excluded from the syllabus, so if the points are not on the equator and not on the same meridian (or a meridian and its anti-meridian, for which the great-circle route is via the pole) then the question is not about calculating the great-circle distance.

Note you may be asked to measure a distance. Measure on the appendix given, don't try to calculate it as some did.

------

There is no such term as 'convergence angle', this is a source of much confusion. You must distinguish between 'convergence' and 'conversion angle'. Convergence is the angle between two meridians, giving the change of bearing along a great-circle arc (for Earth convergence) or along a straight line on a chart (chart convergence). Conversion angle (C.A.) is the approximate angle between the shorter great-circle arc and the shortest rhumb line between two points.

This question requires you to recognise that flying a rhumb line between these three points the bearing would be a constant 090 degrees. Draw a diagram it will show that the great-circle track (1) to (2) measured at (2) is less than 090 degrees by the C.A. between (1) and (2), about 4.5 degrees.

The track from (2) to (3) measured at (2) is less than 090 degrees by 4.5 degrees. Therefore at (2) the aircraft track changes from 090 + 4.5 to 090 - 4.5, a left turn of 9 degrees.

This is hard to see without the diagram, if you're having problems getting that right then send me a PM with an email address and I'll send you a diagram.

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20 kts

A couple of ways to do this one. The simplest way (but it only works for a given point of slowing, not for a given new airspeed where you have to work out the point at which to slow down) is to calculate the time taken, then add the delay required and recalculate the speed from the distance and the new time. At 265 kts, 230 kts G/S this is about 38 minutes. To delay by 5 minutes then the plan is to take 43 minutes. This requires a groundspeed of 210 kts, which will require a drop in speed of 20 knots.

----

Solar system

66°, of those choices. Actually it's somewhat higher, as the centre of the sun reaches the celestial horizon every day up to about 66°30' (maximum declination of the sun) but for sunrise only the top need be seen above the visible horizon (lower than the celestial due to atmospheric refraction), so that applies at a higher latitude.

Hope this helps

Send Clowns
Gen Nav Instructor
Bournemouth Commercial Flight Training

[Just another student]

Regarding the conversion angle, I realised my mistake shortly after posting but forgot to change it.

The velocity question is fine also, I simply forgot to convert the new GS to IAS! I have the answer as 20kts with no problems now.

With the INS question, I have drawn a diagram with lines 30W and 10W converging to the N Pole and line 20W straight and a horizontal 60N line. I then drew in a concave line (towards the pole) to represent the rhumb line, then noted wether there should be an increase or decrease by judging the angles. Is that correct?

I have also noted another question where I can not match my answer to the 'correct' one.

An aeroplane is flying at TAS 108kt on a track of 090 degrees.The w/v is 045/50kt.How far can the aeroplane fly out from its base and return in one hour?

a) 56nm
b) 88
c) 85
d) 176

Using the CRP5 I get

GS Out 148
GS Home 218/219

60 x 219 148
----------- = 35.8 ------- = 2.47 = 2.47 x 35.8 = 88.426nm
148+219 60

According to the answer I have, it should be 85nm. Any idea where I may be going wrong?

Cheers

[flyfish]

J.A.S - read your PMs

[Just another student]

Cheers flyfish.

Could someone please work me through the following question.

What is the final position after the following rhumb line tracks and distances have been followed from position 60 00'N 030 oo'W? South 3600 NM, East 3600 NM, North 3600 NM, West 3600 NM. The final position is.....?

a) 59 00'N 060 00'W
b) 60 00'N 030 00'E
c) 60 00'N 090 00'W
d) 59 00'N 090 00'W

I'm probably trying to make it seem more complex than it actually is, but I've got to understand the workings for the exam.

Cheers

[Send Clowns]

Good, JAS. The convergence/conversion term confusion does cause some students a lot of trouble, it is an unfortunate similarity!

The 60N parallel is the rhumb line! It must cross each meridian at 90°, so must be curved concave to the pole, as these converge. That is a critical point. Then draw the great-circle segments (the only point to mentioning the INS or FMS is that these fly a great circle) as straight lines, and you will see what I mean. One line is from (1) to (2), one is from (2) to (3). Both ar to the north of the parallel, except at their ends.

That help?

--------


I take it that was 180 kts?

The G/S out is 140 kts. Home is 214 kts. Check your CRP-5 work! You are using the track as heading, not adjusting for drift.

Time is 1 hour, don't use 60 minutes. Remember, if the speeds are in nm/hr (kts) then times must be in hours. That gets rid of your spurious 60s that just cancel out,

correct formula though

EOH = PSR
-----
O+H

(1 x 140 x 214) / (140 + 214) = 85 nm

[Just another student]

I simply did not read the question properly:) I read it as heading 90 instead of track! That's what happens when u've been studying all day. But there's no excuses for reading them incorrectly, its points dropped!

[Send Clowns]

Will always be track - you fly the headings you won't get back to the airfield! :p :eek:

-----

60°N 090°W

South by 3600 nm is 60° ch. lat. (60 nm per degree) so after the first leg postion is 00°N/S 030°W.

3600 nm east at the equator is 60° ch. long. (equator is also a great circle, so still 60 nm per degree), taking you to 00°N/S 030°E.

3600 nm North takes you back to 60°N.

3600 nm West (a departure, remember, as it is constantly due West) is a ch. long of 120°. This is from

Departure = ch.long (') x cos lat

rearranged to give

ch.long (') = departure / cos lat
=3600 / cos 60
=7200

7200 minutes = 120° (just divide by 60)

So 120° west of 030°E is 090°W, still at 60°N

Does that clear it up?

Actually all you need is to recognise that the north and south movement bring you back to the same parallel (60°N) to discard 59°N. Then recognise that the ch.long is greater at greater latitude for the same distance of movement (a simple lambert's graticule of 2 meridians and 2 parallels should convince you) to recognise that the end point is West of the start, leaving only one answer.

[Just another student]

I knew that Ch long would be greatest at 60N, it was just the simple manipulation of the formula which caught me out.

Thanks for the help

[Send Clowns]

Anytime :)

Post anything about gen nav in here and you bring out the competitve spirit between Oxford Blue, Alex Wittingham and me. They must have been busy today :p

[timzsta]

I sat the Gen Nav exam in Dec 2003. Much was written about that paper because more people failed it then usual apparently. They failed because their CRP 5 work was not up to speed. 26 of the 54 questions were on CRP 5 and none of the questions in the paper on CRP 5 were above what you are taught for your PPL.

Know how to use your CRP 5 speedily and accurately and you will go a long way to passing this exam. If you can grasp the stuff that Send Clowns has written on this thread on the rest of the Gen Nav stuff you will be fine. The key to the form of the earth type questions is to draw a picture. The INS one is a classic - if you draw the picture you will be able to work out that only 1 of the given answers is possibly correct.

Best of luck, I got 90%.

[Just another student]

Thanks for the advice. I think the key to passing G-Nav is just practice, practice, practice. Do many PET/PSR questions pop up, where fuel flow is added to the equation?

[Send Clowns]

Not many, as far as I have been told by my students. Most are straight-forward, all-engines-operating CP/PET and PSR/PNR.