ATPL GNAV theory question finding final GCT from Initial GCT
 

Hi again, I want to check a caculation of a Great cercle track GCT on a Polar strerographic chart PSC with the following: A 80 ° S 110° W and B 85°S 100° E. First question: can I deduce the GC tracks at A and B without any futher information? (I think I can deduce one of the GCT from the other, let's say the final from the initial). But with that amout of data, can I already find the GCtracks at A and B?

Many thank in adavance!

An unusual question as it has been asked. Questions with positions on different latitudes usually state that the highest track (or vertex) occurs at a particular longitude.
At the vertex the GCT must be either 270 or 090, then you can use convergency (conversation angle not available) between the given vertex longitude and wherever they want the GCT measured.
Once you have done a couple these are relatively straightforward.

For this question I drew an accurate scaled polar/circle diagram and got GCT at A 190 GCT at B 340, but the diagram is too complicated to show & explain here.

I wouldn't waste too much time on these type of questions, also not sure your source of questions is entirely reliable, been some strange wording along with an incorrect answer before.

Many thanks again Richard. If you wouldn't find following my thoughts a little further: this isn't a question I have found in an ATPL question bank. I am trying to make a ATPL question of my own. But choosing 2 points A and B, I guess the value of the GCT at A and B has a specific value. I can't simply "STATE" that the GCT at B is that much, to then calculate the GCT at A. So I wanted to find the GCT at B to deduce the one at A. I have tried using this webite: https://onboardintelligence.com/GC.
And I get:
https://cimg9.ibsrv.net/gimg/pprune....542f629997.jpg

And:
https://cimg3.ibsrv.net/gimg/pprune....d68fc6f285.jpg

So GCT arrival at B = 340°.
From that I can do my hand calculation and find GCT at A:
________________________
Given GCT(B) = 340°, let’s find the exact GCT(A): We know g° (the difference in longitude) between 110°W and 100°E = 70°+80°=150°.

The Mean Latitude is Lm = 82,5

And we know that: GCTB - GCTA = Cg = 2 CA = g. sin (Lm) = 150°. sin(82,5) = 148,72.
So the GCTA= GCT B - Cg = 340 - 148,72 = 191,28°.
Close enougt to the answer 190° from the website?

Right? Is that acceptable?

xplanefactor - Though it is good to have an understanding of your subject I think you are making life unnecessarily difficult over this.

1. Do NOT go inventing your own questions as you may not yet fully understand the background behind them, this will just lead to more confusion & frustration. You have better things to do.
2. Use a well established & popular question bank such as ATPL Questions, ATPL GS, Aviation Exam or BGS Online (there may be others) former students of mine have used these with great success in the past.
3. The GC calculation website might be interesting but a pointless diversion in my view & not available in your exam!
4. The original question was about Polar charts so IF you knew that arrival at B was 340 just take 150 off = 190.
5. You might get away with your 'method' in a multi-choice question with well spaced answers but not if it was a 'type in' answer.
6. I come back to the fact I think this is a poor question and would usually state the where the vertex was.

Regards RichardH

1) I am not an ATPL theory student, but a TKI... that is why i am trying to figure this out.
4) ok, so 340 - 148,72 = 191, 28° do you agree? With the Sin (82,5) close to 1 surely...

Thanks again :)

For a PS chart you should be using the sine of 90 = 1. The sine of mean latitude of (82.5) will give you a close answer but not technically correct.

Just try to keep it has simple as possible, the convergency method (with a given vertex position if needed) works best.