{"id":3871,"date":"2022-12-08T07:32:21","date_gmt":"2022-12-08T06:32:21","guid":{"rendered":"https:\/\/rent-a-pilot.nl\/?p=3871"},"modified":"2024-11-05T11:32:13","modified_gmt":"2024-11-05T10:32:13","slug":"gnav-two-not-so-random-points-4_find-the-triangle","status":"publish","type":"post","link":"https:\/\/rent-a-pilot.nl\/nl\/gnav-two-not-so-random-points-4_find-the-triangle\/","title":{"rendered":"Geval 4 - EASA ATPL General Navigation berekeningen tussen twee \"willekeurige\" punten: Vind de driehoek"},"content":{"rendered":"\n

General navigation for ATPL – Find the triangle<\/h2>\n\n\n\n

I told you<\/a> there is a system in the madness. <\/p>\n\n\n\n

If you want to calculate track and\/or distance for an EASA General navigation exam question without using the classic great circle formulas and after all the other shortcuts have failed then you must find the triangle in the lines that connect the two coordinates<\/strong>. <\/p>\n\n\n\n

How computers do it<\/h2>\n\n\n\n

Before JAR-FCL took over the ATPL exams in Europe around the year 2000, you would remember the formulas for great circles, go to the exam, insert the coordinates of A and B into them and presto! Bob’s your uncle.<\/p>\n\n\n\n

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The EASA way, how the ATPL exam does it<\/h2>\n\n\n\n

Fast forward to today. The EASA syllabus does not want you to learn the formulas for track and distance of a random great circle<\/strong> but they still want you to calculate tracks and distances. This is an interesting dilemma. The question writers must now find a way to make it look like General Navigation.<\/p>\n\n\n\n

There are however some shortcuts and they allow you to calculate tracks and distances by simplifying them along meridians and parallels. That is, if you are lucky that the coordinates are both actually on the same meridian or parallel. Like this:<\/p>\n\n\n\n

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Previously in this series<\/h3>\n\n\n\n

Case 1: Same meridian<\/a>?
Case 2: Same anti-meridian<\/a>?
Case 3: Same parallel? <\/a><\/p>\nTracks and distances between two “random” points<\/cite><\/blockquote>\n<\/div>\n\n\n\n

If these 3 cases do not apply you can use a method that I call “Find the triangle”. Usually when you end up here we are dealing with two positions that are in the polar regions. 70N or 70S or beyond.<\/p>\n\n\n\n

Where is the triangle?<\/h2>\n\n\n\n

To have a triangle you need 3 lines. The great circle track is just one of them. The other two are usually meridians. Like this:<\/p>\n\n\n\n

\"\"<\/a>
A straight line that forms a triangle with two meridians<\/figcaption><\/figure>\n\n\n\n

What is in the triangle?<\/h2>\n\n\n\n

The meridians are obvious, they are the two legs of the triangle. The dotted line is a straight line that completes it. Could be a track between two points.
Inside a triangle all the angles count up to 180 degrees and you can do some basic calculations to find the length of each leg.<\/p>\n\n\n\n

Does it look familiar yet? You may recognise this as a POLAR STEREOGRAPHIC PROJECTION<\/strong>.<\/p>\n\n\n\n

Look at this question from the European question bank:<\/h3>\n\n\n\n
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On a Polar Stereographic chart, the initial great circle course from A 70\u00baN 060\u00baW to B 70\u00baN 060\u00baE is approximately:<\/strong><\/p>\n\n\n\n

A: 330\u00ba (T)
B: 030\u00ba (T)
C: 150\u00ba (T)
D: 210\u00ba (T)<\/p>\n<\/blockquote>\n<\/div>\n\n\n\n

This is what it looks like in a drawing.
You may notice that I draw the great circle from A to B as a straight line. This is close enough and you will find the correct answer.
In reality a great circle it is not a straight line in a polar stereographic projection but in our case this is close enough. There is in fact a slight curve concave to the pole but that’s a long story that I’m not going to cover here.<\/p>\n\n\n\n

\"\"<\/a>
The parallel of 70N with the two points A and B<\/figcaption><\/figure>\n\n\n\n