G Nav question
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The great circle vearing of position F longitude 24E from position 27N 8W measured at longitude 8E is 084T
The great circle bearing of E from F, measured at F is 272t
What is the bearing of F from E measured at E and what latitude
is F??
Please help.
also, a quick one....
The rhumb line distance from A at 57N 12*30' W to B at 57N 167*30E is :
a. 3690 nm
b. 5882 nm <--- answer given
c. 10800 nm
d. 9057 nm
nothing ive tried so far gives me B .. please explain.
The great circle bearing of E from F, measured at F is 272t
What is the bearing of F from E measured at E and what latitude
is F??
Please help.
also, a quick one....
The rhumb line distance from A at 57N 12*30' W to B at 57N 167*30E is :
a. 3690 nm
b. 5882 nm <--- answer given
c. 10800 nm
d. 9057 nm
nothing ive tried so far gives me B .. please explain.
PPRuNe Handmaiden
This is pushing my ol memory.
The second one. Isn't the trick to add up the longitudinal points and see if you get 180. In this case you do. Don't you then just go "over the top" as it were?
Just read Alex's reasoning. Told you it was a while ago.
The second one. Isn't the trick to add up the longitudinal points and see if you get 180. In this case you do. Don't you then just go "over the top" as it were?
Just read Alex's reasoning. Told you it was a while ago.
Last edited by redsnail; 4th Jun 2004 at 10:53.
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The rhumb line distance from A at 57N 12*30' W to B at 57N 167*30E is :
a. 3690 nm
b. 5882 nm <--- answer given
c. 10800 nm
d. 9057 nm
Looking at the longitude it is exactly 180 degrees apart which means you could travel a line of constant direction (a rhumb line) North up to the N pole and then South to 57N. That would then employ GC techniques eg. 1 degree = 60 NM
90 degrees - 57 degrees = 33 degrees x 2 = 66 degrees
66 degrees x 60 = 3960 NM.
Otherwise it is Departure formula
Departure in NM = dlong (in min) x cos mlat
Departure in NM = 180 x 60 x cos 57
Departure in NM = 180 x 60 x 0.545
Departure in NM = 5882
I would query this question, the answer of hose given to chose is therefore 5882. But can you see something correct in what I write about over the pole as an option?
a. 3690 nm
b. 5882 nm <--- answer given
c. 10800 nm
d. 9057 nm
Looking at the longitude it is exactly 180 degrees apart which means you could travel a line of constant direction (a rhumb line) North up to the N pole and then South to 57N. That would then employ GC techniques eg. 1 degree = 60 NM
90 degrees - 57 degrees = 33 degrees x 2 = 66 degrees
66 degrees x 60 = 3960 NM.
Otherwise it is Departure formula
Departure in NM = dlong (in min) x cos mlat
Departure in NM = 180 x 60 x cos 57
Departure in NM = 180 x 60 x 0.545
Departure in NM = 5882
I would query this question, the answer of hose given to chose is therefore 5882. But can you see something correct in what I write about over the pole as an option?
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The great circle bearing of position F longitude 24E from position 27N 8W measured at longitude 8E is 084T
The great circle bearing of E from F, measured at F is 272t
What is the bearing of F from E measured at E and what latitude
is F??
Convergency has changed by 8 degrees in the last third of dlong from 8E to 16E. So going back towards E decrease the GC track by two lots of 8 degrees from 264 (this number came from the reciprocal of 084).
264 - 8 - 8 = 248.
This is your Initial GC track.
Convergency = dlong x sin mlat
Convergency (IGC plus or minus FGC) = dlong x sin mlat
272 - 248 = 24 x sin mlat
24 = 24 x sin mlat
1 = sin mlat
Since sin 90 = 1, the mean lat is the pole!
for that to happen the other lat must be 27 N also
What a question! Not in the exam I hope.
Someonelse check my logic as well please
The great circle bearing of E from F, measured at F is 272t
What is the bearing of F from E measured at E and what latitude
is F??
Convergency has changed by 8 degrees in the last third of dlong from 8E to 16E. So going back towards E decrease the GC track by two lots of 8 degrees from 264 (this number came from the reciprocal of 084).
264 - 8 - 8 = 248.
This is your Initial GC track.
Convergency = dlong x sin mlat
Convergency (IGC plus or minus FGC) = dlong x sin mlat
272 - 248 = 24 x sin mlat
24 = 24 x sin mlat
1 = sin mlat
Since sin 90 = 1, the mean lat is the pole!
for that to happen the other lat must be 27 N also
What a question! Not in the exam I hope.
Someonelse check my logic as well please
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Chintito
The second one has, effectively, been answered between Alex and Aim High, i.e. Aim high provided 2 answers, Alex said why it is the second.
The first question is not like any I expect you to find in the exam, although it is interesting. The key is to understand that the whole reason we ever bother with Earth Convergence is that it tells us how the bearing of a great circle changes, depending on where it is measured. Therefore the change in great-circle bearing is equal to Earth Convergence.
So the change in GC bearing from 008°E (bearing 084°T) to F (at 024°E, bearing 092°, reciprocal of 272°) is 8°. Therefore the convergence is 8°, and ch. long. x sin mean lat = 8°. Since ch. long. is 16°, mean lat must be inverse sin 1/2, which is 30°. From this point approximation is the only option, unless you want maths way beyond ATPL level. Assume the same change of latitude between 'E' and 008°E as between 008°E and 'F', you're at 30°N 3/4 of the way up the track from 27°N, so, F is at 31°N.
The rest is simple. Earth Convergence between E and F is ch. long. sin mean lat = 32 x sin 29° = 15.5°. The question shows that tracks are decreasing westward (a diagram confirms this) so since F from E measured at F is 092° the same line measured at E is 092° - 15.5° = 76.5°.
That is a scratch calculation for an unusual question after 3 pints, so I will not be offended if anyone has anything to add.
Note to all students: this is beyond ATPL General Navigation requirements. I just enjoy a challenge. Sad, I know
The second one has, effectively, been answered between Alex and Aim High, i.e. Aim high provided 2 answers, Alex said why it is the second.
The first question is not like any I expect you to find in the exam, although it is interesting. The key is to understand that the whole reason we ever bother with Earth Convergence is that it tells us how the bearing of a great circle changes, depending on where it is measured. Therefore the change in great-circle bearing is equal to Earth Convergence.
So the change in GC bearing from 008°E (bearing 084°T) to F (at 024°E, bearing 092°, reciprocal of 272°) is 8°. Therefore the convergence is 8°, and ch. long. x sin mean lat = 8°. Since ch. long. is 16°, mean lat must be inverse sin 1/2, which is 30°. From this point approximation is the only option, unless you want maths way beyond ATPL level. Assume the same change of latitude between 'E' and 008°E as between 008°E and 'F', you're at 30°N 3/4 of the way up the track from 27°N, so, F is at 31°N.
The rest is simple. Earth Convergence between E and F is ch. long. sin mean lat = 32 x sin 29° = 15.5°. The question shows that tracks are decreasing westward (a diagram confirms this) so since F from E measured at F is 092° the same line measured at E is 092° - 15.5° = 76.5°.
That is a scratch calculation for an unusual question after 3 pints, so I will not be offended if anyone has anything to add.
Note to all students: this is beyond ATPL General Navigation requirements. I just enjoy a challenge. Sad, I know
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Send Clowns' method is correct although he has made a small calculation error, no doubt explained by the third pint of beer. The lat has changed from 27 N to 30 N in half the ch long, not 3/4, so it should be at 33N at the other end. This gives a GC bearing of 076 at E.
Last edited by oxford blue; 4th Jun 2004 at 19:37.
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Send Clowns is correct in all respects, despite his beer intake! The chlong between 8W and 8E is 16º, the same as the chlong between 8E and 24E. So we can divide the track into four equal parts – 27N 8W to midlat1; mid lat1 to 8E; 8E to midlat2; and midlat2 to 24E. As midlat2 is 30N, it follows that the latitude at 24E (F) is 31N.
As to the second question, Alex is quite correct in saying that "over the pole" would not be a rhumb line because of the change in direction at the pole. We must therefore use the departure formula, correctly quoted by Aim High, to get the answer of 5882nm.
As to the second question, Alex is quite correct in saying that "over the pole" would not be a rhumb line because of the change in direction at the pole. We must therefore use the departure formula, correctly quoted by Aim High, to get the answer of 5882nm.
