ATPL Exam Questions | Polar Stereographic Question Bank

ATPLQ: 618079
In this video I will go through a question from ATPL Questions for General Navigation focusing on Polar Stereographic charts.
There will be two methods shown, a quicker one which uses the DIID drawing and a method which helps visualise the problem better (using Polar Stereographic charts).
If you would like to suggest any particular questions for me to go through, can be from the question bank or not. Please leave them down in the comments below and i'll take a look!
If you have other topics which you are stuck on within General Navigation, for example:
1. Direction, Latitude and Longitude, Great Circles, Rhumb Lines & Directions on the Earth
2. Earth and Aircraft Magnetism, Direct Indicating Compass
3. The Flight Computer and Calculator Tips
4. 1 in 60 Rule and Mental Dead Reckoning
5. (Basic or General) General Navigation problems
6. Topographical Charts and Map Reading, Plotting and General Chart Properties
7. Convergency, Mercator Charts and Lambert's Conformal Charts
8. Polar Stereographic Charts
9. Gridded Charts
10. Time
from either the 2020 EASA Syllabus, or FAA ATPL, ICAO ATPL. Or maybe you are on your PPL and would like some help instead, please comment and subscribe!
Sharing this video with your course buddies and the communities you are within will help me create more content like this. Thanks for watching!
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[@HypoxicWaiter]

Great video! Came here from ATPLQ!

[@tofesky]

the legend of Gnav .. from Aviation exams QB

[@ethanteberrerou6933]

I cordially love you

[@smwagh123]

you are amazing

[@Dogfight2000]

Could you explain how did you get 45’ in the first method? I said something about add 10 and add 35’ what is this?

[@ATPLApproach]

We are going from 170W to 145E, so I broke that down into two steps. The first 10 degrees was to get from 170W to the antimeridian (180E/W) and then the 35 degrees is to get from there (180E/W) to 145E (180-145=35). If you look at my Polar stereographic drawing at 5:30 mins in the video, you will see the change in longitude shown with the 10 and 35 angles drawn on.

[@rashid37009]

First method is way easy but could you please elaborate how did you apply lambert equation to stareographic chart i.e. sin 90 =1

[@ATPLApproach]

If you take a look at the top 3 images of figure 1 on page 4 of this paper cdn.ymaws.com/www.osls.org/resource/resmgr/docs/Ground_Truth_handout_OSLS_20.pdf. You will see a Polar Stereographic and two Lambert's Conformal Conic (LCC) charts. The Parallel of Origin (P.O) for the first two is where the chart intersect the globe, and for the last chart, the P.O is the average of the two intersections. For the LCC chart in the middle, the parallel of origin is about 45°. if you keep moving the P.O/intersection up you will eventually get to 90°. This forces the cone to be a flat plane which is the Polar Stereographic Chart.
So the Polar Stereographic Chart is just a LCC chart with the highest possible Intersection/P.O (90°). Therefore you can use the Lamberts equation for a Polar Stereographic Chart as it is a LCC Chart, just with a high P.O.
I hope that clears things up ☺

[@mahmoodalrahbi4437]

Good explanation! Can you please explain this q 614065 . I couldn’t understand it . Thank you sir 🙏

[@ATPLApproach]

Hey there,
This question is very similar to the one I covered in this video ua-cam.com/video/PT5M7yMFYQQ/v-deo.html . If you skip to 9:40 in the video, I draw a Lamberts Conformal Conic Chart (LCC) for the Southern Hemipshere (SH). However we need to imagine that in the diagram I draw, A is 155°E and B is 170°W. Therefore the Change in Longitude is 35°. We care about the straight line in the question, and this is the Great Circle track for a LCC so we need to work out the Conversion Angle.
To work out the Conversion Angle you use the formula I used at 13:00 mins. This will be 1/2 * 35 * Sin(53), which equals 14°. Just like the diagram I drew, we can see that the inital track at 155°E is 90° + 14° = 104°, and the final track at 170°W will be 90° - 14° = 76°. To get from initial track to final track you do 104° - 28° = 76°, we minus 28° as that is the Convergency which is just two Conversion Angles.
However we measure the track at 175°W, not 170°W, so we need to use a fraction of the Convergency. We are only going 30° instead of 35° so the fraction is 30/35. To get the track at 175°W, we start from our initial track of 104° then we minus (30/35) * 28° = 80°. This is quite a tough question, but I hope this clears things up!