Question Polar stereographic chart

[WhiteKnight]

Can anybody show me how to calculate the answer of this question? Thanks a lot.


On a polar stereographic chart where the earth convergence between 2 points located on the parallel 60°N is 20°, the great circle maximum cross-track difference with the straight line joining the 2 points is :
9.2 NM.
4.0 NM.
30 NM.
4.0 NM.

[swh]

I get 27.34 nm ...

Colatitude is 30 degrees, degrees from the pole is 1800 nm,
ch long is 20 degrees, so mid point ch long is 10 degrees

properties of a triangle .... 1800 cos 10 is 1772 nm so 1800 - 1772 is 27 nm

:ok:

[Alex Whittingham]

It might be a bit more complicated than that. The question actually asks for the cross track distance between the great circle track and the straight line track. You have answered for the cross track distance between the rhumb line and straight line tracks.

Assuming the question means what it says, on a polar chart great circles are very nearly straight lines, the difference arising from the difference between earth and chart convergency. Chart convergency here is 20º, earth convergency 17.3º, the difference is 2.7º. The angle between the great circle and straight line at each end of the track will be roughly half that, about 1.3º. The rhumb line track is about 600NM long, the great circle distance to half way would be less than half, guess at 290NM, the cross track distance is about 290NM x sin1.3º = 6.5NM, bodge it down a bit because we’re dealing with a 3D surface, I’d guess about 3 - 4NM.

How did I do?

[swh]

Alex,

I ruled that one out as both points are not on the same meridian...thinking the question was somehow related to the old INS function..I read it as asking the diff between the start latitude and the latitude at half way.

The great circle distance between the points is 597.71 nm assuming a sphere, or 600.29 nm assuming WGS84 spheriod. Using you approach I get closer to 6.78 nm for a sphere or 6.80 nm assuming a WGS84 spheriod.

May as well asked about a lambert chart instead of a polar if they were taking your approach....who knows ..

:hmm:

[Alex Whittingham]

I read it your way at first because that's the obvious application. How did you do those impressive calculations?

[swh]

Alex,

Just using a spreadsheet and calculator

:ok:

[enicalyth]

If the earth is spherical and colat is 30 deg and angle subtended at the pole is 20 deg then the two points at lat 60 are 597.71nm apart. I've got that. Using a calculator provided by http://www.naco.faa.gov/index.asp?xml=naco/online/compsys
and WGS84 spheroid the distance is 600.29 I've got that too. But isn't the apex of the great circle found to be at latitude 60d 22.74m? Does that not indicate the max cross track error twixt rhumb line and GC? I seem to remember the red Admiralty book Navigation had a formula for finding the apex in general (and difficult) cases. Bear in mind that I am brilliant at getting hold of the wrong end of the stick.

[oxford blue]

Neither of you is answering the question as asked. It is not the change of long which is 20º, but the earth convergence.

Convergency = change of long x sine mean lat

20 = ch long x .866
ch long = 20/.866 = 23.09º

(This is itself a slight approximation, because the mean lat will actually be higher than 60N, but not by much).

This means that there is 11.547005º ch long (or chart convergence) between the start point and the mid-point of the straight-line Polar Stero track. Using SWH's method, this gives a co-latitude distance at the mid-point of the straight-line track of 1763.569 nm or, if you prefer, a latitude of N6036.43. (There is a slight inaccuracy caused because the scale expansion of a Polar Stereographic is not linear, but a function of the secant squared of half the co-latitude, which will give a very slightly different scale expansion between the N Pole and N6000 and the N Pole and N6036, but I think that we can afford to ignore that).

We now need to work out the latitude of the Great Circle track at mid-point, and this is best solved by spherical trigonometry.

Draw a triangle with A as the North Pole, B as N 6000 W 01132.82, and C as the mid-point of the Great Circle track. Label the opposite sides to angles A,B, and C as sides a, b and c.

Angle A = 11.5470005º
Angle C = 90º
side c = 30º (co-latitude of B)

sin a/ sinA = sin c/sin C

This gives side a as 5.7441º of Great Circle arc (344.65 nm, if you're interested).

We can now use Napier's rules:

sin b = tan a x co-tangent A

So side b (which is the co-latitude of the mid-point of the Great Circle) is 29.4953º, or a co-latitude distance of 1769.72 nm.

This gives the latitude of point C as N6030.28

We now know the lat of the midpoint of the straight line - N6036.43 - and the latitude of the mid-point of the Great Circle - N6030.28. So the cross track-displacement is 6.15 nm.

So none of the given answers is correct.



Are you sure you got the question right, WhiteKnight? It is not,as far as I know, a JAA question, yet you give 4 options as answers.

I suspect that it should have been the rhumb-line v straight-line cross track distance, that it's the change of longitude, not the convergency which is 20 degrees, and that this isn't a real JAA question but an FTO worksheet example.

[swh]

Paul,

It is not the change of long which is 20º, but the earth convergence.

I thought on a polar stereographic, earth convergency is equal to the change of longitude at the pole.

Is there someting I am missing ?

:hmm:

[oxford blue]

It is, at the pole, but that is not what the question said.

It says:"where the earth convergence between 2 points located on the parallel 60°N is 20°". If the earth convergence is 20° at 60N, the earth ch long is 23.09º.

If the question had said "chart convergence", I would have agreed with you.

As for dungfunnel, this forum is not "wannabes". Experienced knowledgeable pilots pursue subjects well beyond ATPL level here. A lot of people (but obviously not you) are interested in this knowledge for its own sake, not just to pass exams.

In any case, this question cannot be answered using ordinary ATPL syllabus knowledge - which is why I suspect that WhiteKnight may have it wrong - unless, of course, it comes from another source altogether and he has not got exams in mind.

[enicalyth]

I wish I wish I hadn't lost that nav manual in one of my house moves. I had so much written in the margins and jeezo you don't half forget with age. Of course I got it wrong, I knew I had the wrong end of the stick but just couldn't make the mental jump. It's fun though innit?

[No_Speed_Restriction]

No its not! :yuk:

[WhiteKnight]

Thank you very much for your comments. I finally found by myself a satisfying solution.


The given earth convergency is 20° at N60. The formula dlong = earth conv / sin lat shows a dlong of 23.1° between A and B.

Draw a circle with its centre at the North Pole( NP) and a radius of 1800 NM corresponding to the distance NP to N60, sketch a angle of +/- 11.55°. The secant represents the straight line between A and B. At half way the distance between secant and circumference is 1800 NM x (1-cos 11.55°) = 36.45 NM.


As the given earth convergency is only 20 degrees you now sketch an angle of +/- 10°. The above formula gives the distance between N60 and the great circle: 1800 NM x (1-cos 10°) = 27.35NM

Now you have the following information:

N60 and the straight line differ by 36.45 NM
N60 and the great circle differ by 27.35 NM
The straight line and the great circle differ by only 9.1 NM

[blackmail]

hello whiteknight,

following might be an answer to your nav problem.
formulae & elaboration:
(1) delta R(earth converg)=delta g(longitude)xsinlm(mean latitude)
so: 20°=delta g x sin60°

for info: a quick means to find sin & cos values

0 1 2 3 4
0 1 1.41 1.73 2 = square root
0 0.5 0.7 0.866 1 = divided by 2
0° 30° 45° 60° 90°=sin values(corresponding numbers abv)
90° 60° 45° 30° 0° =cos values(reverse order)



so again:20°=delta g x 0.866 with delta g=23.09°=23°[1]

(2)greek delta letter(pseudo givry correction)= angle between chart straight line & ortho line)= 0.5xdelta gx(1-sinlm) for stereo polaire chart.
so: greek delta=0.5x23°x(1-0.866)=1.54°[2]

(3)loxo distAB=delta gx coslm=23°xcos60°=23°x0.5=11.5°x60=690nm. corresponding ortho dist= 688nm[3] (formula ortho dist: cosM=sinlatAxsinlatB + coslatAxcoslatBxcosdelta g

(4) f(la fleche, in french)=max offset dist between chart straight line & ortho line= greek delta x ortho distAB(nm)/230

so finally putting [2]&[3]in this formula gives: f=1.54x688nm/230=4.6nm so 4nm offset seems a reasonable answer.

note: somebody suggested that for a stereo polaire chart , angle between chart meridians & delta g(longitude) are the same & this is correct as n(constant of the cone)= 1 for stereo polaire chart: delta alfa=nxdelta g with n=1.

i stop here before getting a real headache.