G Nav Questions
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G Nav Questions
Hi could anyone help me with a few questions?
I need help with the "method" used to get the answers !
Think thats the main problem !
The shortest distance in kilometres between 17 27'N 019 30'W and 38 00'S 019 30'W is ...
a) 1,233
b) 2,285
c) 3,327
d) 6,160
The rhumb line distance in nautical miles between 26 35'S
054 07'E is ...
a) 1,566
b) 1,751
c) 4,286
d) 4,793
And the last one ...
The great circle distance in nautical miles between 56 12'N 057 00'W and 64 00'N 123 00'E is ...
a) 1,328
b) 1,863
c) 3,588
d) 4,729
Thanks a lot
Appreciate any help here
Seabass
I need help with the "method" used to get the answers !
Think thats the main problem !
The shortest distance in kilometres between 17 27'N 019 30'W and 38 00'S 019 30'W is ...
a) 1,233
b) 2,285
c) 3,327
d) 6,160
The rhumb line distance in nautical miles between 26 35'S
054 07'E is ...
a) 1,566
b) 1,751
c) 4,286
d) 4,793
And the last one ...
The great circle distance in nautical miles between 56 12'N 057 00'W and 64 00'N 123 00'E is ...
a) 1,328
b) 1,863
c) 3,588
d) 4,729
Thanks a lot
Appreciate any help here
Seabass
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first question, it s D.
55 degree 27*60=3300 miles(*2.2)=over 6000km
second question:???need more datas and I do not think the JAA ask for rhumb line distances, the formula is too long.
third questions:
Difference of long.=180 degree
180*60 miles*cos 60=5400 miles
latitude:total distance 8 *60=480 miles
distance square=480 square+5400 square=5421 miles
no right given answer!!!??? closest one is D.(correction:wrong answer)
(correction: fly over the pole, it s faster!!!)
55 degree 27*60=3300 miles(*2.2)=over 6000km
second question:???need more datas and I do not think the JAA ask for rhumb line distances, the formula is too long.
third questions:
Difference of long.=180 degree
180*60 miles*cos 60=5400 miles
latitude:total distance 8 *60=480 miles
distance square=480 square+5400 square=5421 miles
no right given answer!!!??? closest one is D.(correction:wrong answer)
(correction: fly over the pole, it s faster!!!)
Last edited by Hulk; 22nd Aug 2003 at 22:09.
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C - there is no right answer
The formula is
Departure = ch long (in mins) x cos mean latitude.
ch long is 180 x 60 = 10800
Mean latitude = (64-56.2)/2 + 56.2 = 60.1
10800 x cos 60.1 = 5383.67 nm.
There is no correct answer given.
The formula is
Departure = ch long (in mins) x cos mean latitude.
ch long is 180 x 60 = 10800
Mean latitude = (64-56.2)/2 + 56.2 = 60.1
10800 x cos 60.1 = 5383.67 nm.
There is no correct answer given.
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I haven't got the conversion figures in front of me but the first question is asking for the distance in km's not nm's. I'd suggest that 'b' looks pretty good.
Classic gen nav question...... you've got to be REALLY careful in the exam that you are answering in the correct units.
Classic gen nav question...... you've got to be REALLY careful in the exam that you are answering in the correct units.
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The answer to (1) is (d). The change of latitude is 55 deg 27 min along a meridian of longitude. This comes to 3327 nm, which is 6161.6 km.
Not possible to answer (2) without the full question. Where is the second position?
You're all wrong on (3). You're trying to treat it as a departure question. It is not. The 2 positions are on a meridian and its anti-meridian. Therefore the shortest way is the Great Circle joining them, which is over the North Pole. So you take the co-latitude of 5612N (which is 33 degree 48 min) and the co-latitude of 6400N (which is 26 degrees), add them together to get 59 degrees and 48 min, which is 3588 nm along the arc of a Great Circle track.
I that part of your problem, Seabass, is that you're looking for 'a method'. There isn't any one method. You have to look at each individual case. Some are change of latitude along the same meridian in the same hemisphere, some are change of latitude along the same meridian in opposite hemispheres, some are meridian and anti-meridian, and some are departure problems. If you draw a little sketch of each problem, it all becomes much clearer.
Not possible to answer (2) without the full question. Where is the second position?
You're all wrong on (3). You're trying to treat it as a departure question. It is not. The 2 positions are on a meridian and its anti-meridian. Therefore the shortest way is the Great Circle joining them, which is over the North Pole. So you take the co-latitude of 5612N (which is 33 degree 48 min) and the co-latitude of 6400N (which is 26 degrees), add them together to get 59 degrees and 48 min, which is 3588 nm along the arc of a Great Circle track.
I that part of your problem, Seabass, is that you're looking for 'a method'. There isn't any one method. You have to look at each individual case. Some are change of latitude along the same meridian in the same hemisphere, some are change of latitude along the same meridian in opposite hemispheres, some are meridian and anti-meridian, and some are departure problems. If you draw a little sketch of each problem, it all becomes much clearer.
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Haaaa, there you go, that proves my point !! You have to be really careful.
Even knowing the right method it so easy to be a chimp and divide instead of multiply. 110% concentration is required.
Even knowing the right method it so easy to be a chimp and divide instead of multiply. 110% concentration is required.
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ha ha oxford blue , well seen, I d not have any problems to resove problems, but I go to fast instead to take a deep breath and look carefully!
BTW, do you know lot of aircrafts flying over the poles??
BTW, do you know lot of aircrafts flying over the poles??
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Yes, it is becoming more and more common with long range aircraft with a good ETOPS capability, such as the 777. The wide availability of IRS/FMS and GPS and the end of the Cold War has now meant that the polar routes are a far more feasible option than they were, say, 20 years ago. The shortest way between Europe and Japan, or USA and China is over the pole, and that is where more and more airlines are now routing.
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I'm having probs with the second one (that no one has answered yet).
Here's the complete question with the destination point:
The rhumb line distance in nautical miles between 26 35'S
024 56'E and 26 35S 054 07E is ...
a) 1,566
b) 1,751
c) 4,286
d) 4,793
Now i can work out that the answer is a) 1566 by (erroneously) taking it to be on a great circle meridian so distance = 1751nm. Obv. wrong but I'd guess it must be less than that being a small circle, so must be 1566. Checking with an online calculator confirms this. However, this is obviously not the way to do it!
How do you work out a rhumb line distance like this?
Here's the complete question with the destination point:
The rhumb line distance in nautical miles between 26 35'S
024 56'E and 26 35S 054 07E is ...
a) 1,566
b) 1,751
c) 4,286
d) 4,793
Now i can work out that the answer is a) 1566 by (erroneously) taking it to be on a great circle meridian so distance = 1751nm. Obv. wrong but I'd guess it must be less than that being a small circle, so must be 1566. Checking with an online calculator confirms this. However, this is obviously not the way to do it!
How do you work out a rhumb line distance like this?
