Problem finding track direction using polar stereographic

[ILS27R]

I'm having big problems finding track direction using a polar stereographic chart. I find that the 'laws' which work on one example don't seem to work on another. It all seems very messy. Ive even drawn diagrams of the damn things. I can work out convergence no problem. But I can't decide whether to add 90, take 90, add 270 or take 270 away from the convergence angle.

Can anyone help me please. I don't understand why a given track heading east is classified as heading west at the mid point on the polar steorographic and vice versa. Are there any rules that help. Please don't anybody refer me to Oxford's notes thankyou!

[Polar_stereographic]

Ah,

My favorite subject - not - I'll have you know. What is the point of studying a pointless useless chart that unless you are going be fly for the British Antartic survey (yea right, lots of them too).

Well, to answer your question, I did the national ATPL's earlier this year, and did not find them too dificult. My tips are:

1: Draw the thing out, and look at it to give you the answer.

2: Whan you draw it, remember in the north, the arrow pointing to N goes into the centre of the dart board, and in the south, they go away from it. Getting this wrong cocks it all up.

I'm no authority, but I'll happily help out if I can if you live near me.

Good luck

PS

[Alex Whittingham]

Answering this without diagrams is going to be hard.....

There are two sorts of polar stereo/grid problems you are likely to encounter. The easiest is the type found in Operational Procedures which talks of convergence as 'east' or 'west'. You can work true to grid or grid to true using a rhyme.

'convergence east true track least,
convergence west true track best'

Where 'best' means 'bigger'

For instance:

The true track at 134ºW is 305º, the convergence is 90ºE, what is the grid track?

So we say 'convergence east true track least' which means grid is bigger so the grid track is 305º + 90º = 395º. This is more than 360 having passed through north so we subtract the 360 to find grid track is 035º

The second sort of problem demands a diagram. A useful convention is to draw northern hemisphere diagrams as a circle with the greenwich meridian at the 6 o'clock position and southern hemisphere diagrams as a circle with greenwich at the 12 o'clock position. This means that easterly longitudes are always on the right and westerly longitudes on the left, just like a normal map.

For instance:

A straight line track is drawn on a polar stereographic chart from A (80ºN 120ºW) to B (80ºN 150ºE). Is the line easterly or westerly at its mid point and what is the initial track at 120ºW?

Draw a northern hemisphere diagram as described above with greenwich at the 6 o'clock. Put in the two positions. A has a westerly longitude so its on the left at about the 10 o'clock position, B is easterly on the right at about 1 o'clock. Draw a line from A to B. Mark the track direction on the line with an arrow.

Identify the half way position roughly, it should be about 11 o'clock. At that point draw a line with an arrow on it pointing to the middle of the circle as a north reference.

Now turn the piece of paper round until the north reference you have just drawn is pointing straight up the page with A on the right and B on the left. The track runs right to left. On a normal map this would be westerly, as it is here. The answer to the first bit is that the track at the mid-point is 270º.

To answer the second bit draw in a north reference at A. Rotate the paper so that the north reference is pointing straight up the page towards the centre of the circle. Notice the track AB is going up and left, this indicates a big angle, more than 270º. To find out how much more find the convergence between A and the mid point. this will be half the total change of long between A and B, which is 90 ÷ 2 = 45º. The track A to B at is therefore 270 + 45 = 315º.

gibber......drool......

[ILS27R]

Yes its all clear now!! Thank you Polar_stereographic and especially Alex for typing such a long post. Much appreciated guys thankyou!

ILS27R

[ILS27R]

I'm not sure whether I'm pushing my luck here! :D Ive got 2 more questions to ask.

When you come to calculating mid point longitudes why is it for example that if the track is A to B it is convergency that is added to A but in other examples its added to point B instead. I can't see why??

Furthermore, along the same lines why is it that convergency is either taken away or added to say point A or point B to find the mid point longitude? What are the common laws.

Please help!
Many thanks,
ILS27R

[Alex Whittingham]

The mid point longitude is just half way between A and B. In the example above we could find it by adding half the total change of longitude to A or by adding half the total change of longitude to B. I would chose to addit to A because it is simpler. 120ºW + 45º = 165ºW.

If I had added it to B it would have been 150ºE + 45º = 195º. Now you can't have a bigger longitude than 180º so this has 'gone beyond' the greenwich antimeridian by 15º. 15º the other side of greenwich is 165ºW. Same answer more work. Although the convergency and the change of longitude are the same on a polar chart its the change of logitude we are working with here, not convergency.

If you are asking why, when calculating track directions, convergency is sometimes added and sometimes subtracted the answer is that it depends on which angle you are working on. We could restate the question above in two ways to find the same answer:

1.The initial track direction from A to B measured at A is 315º, find the final track direction at B.

If you still have your piece of paper put in a north reference at B pointing to the centre of the circle. Make sure the line AB extends beyond B, i.e. it sticks out to the left. Now rotate the diagram so that the north reference at B points straight up the page. Look at the track at B. It runs from top right to bottom left, the track at B is less than it was at A. The track direction between A and B must differ by the convergency between A and B. This is the change of longitude between A and B, 90º.

We have now established that the track direction at B is smaller than it is at A, and smaller by 90º so it must be 315º - 90º = 225º.

2. The track direction at 165ºW is 270º, find the track direction at B.

Look at the track direction at B once more, notice it is less than 270º. The convergency between 165ºW and 150ºE is 45º so the track direction at B must be 270º - 45º = 225º.

Once you have the track direction at any point on a polar chart you can find the track direction at any other point by adding or subtracting the convergency between the two points. The diagrams help you decide which one to do.

A backstop position is to draw the diagrams accurately using a protractor to find the exact longitudes and simply measure the track. You will also find that, with multi-choice answers, once you have decided the track is, for instance, less than 270º but more than 180º often only one answer fits.

Who are you studying with?

[Polar_stereographic]

ILS27R,

With advice like that, I'd get youself enrolled with BGS.

I did mine full time with PPSC and I found them really good, but having seen how Alex has behaved throught all the PPSC drama and with replies to you and others, there can be only one place to do the ground school in my books.

I'd seriously consider phoning the man up and change to BGS. Just imagine the help you'll be in for when enrolled with them.

Good luck

PS