Great circles!!
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Great circles!!
I can generally do most great circle questions by drawing out a diagram depending on what hemisphere were in, drawing the straight line as a rhumb line and the great circle towards the nearest pole. However i seem to be having trouble with great circle bearings, and i dont know if you need to draw a different diagram to help.
Most great circle questions involve convergency, which i can never figure out to add or take away, that's the reason for me drawing the diagram as described above, but do i need to draw a different type of diagram for different questions?
There are 2 questions i am struggling with at the minute, which i can't figure out at all.
1) The great circle bearing of position B from position A in the northern hemisphere is 040 degrees. If Conversion Angle is 4 degrees, what is the great circle bearing of A from B?
For this question i take the reciprical of 040, which is 220 degrees, i know CA is 4 degrees so i add it on to give 224 degrees but the answer states that it is actually 228 degrees, can anyone explain how they get this?
2) The great circle bearing from A (70S 030W) to B (70S 060E) is approximately?
For this question i figure out that the rhumb line track is going to be 090 and the conversion angle is 42 degrees, but looking at my drawing it tells me to take the conversion angle away, but the answer is 132 degrees implying i have to add it on, so i am getting very confused. Can anyone explain the error in my ways?
Thank You
Most great circle questions involve convergency, which i can never figure out to add or take away, that's the reason for me drawing the diagram as described above, but do i need to draw a different type of diagram for different questions?
There are 2 questions i am struggling with at the minute, which i can't figure out at all.
1) The great circle bearing of position B from position A in the northern hemisphere is 040 degrees. If Conversion Angle is 4 degrees, what is the great circle bearing of A from B?
For this question i take the reciprical of 040, which is 220 degrees, i know CA is 4 degrees so i add it on to give 224 degrees but the answer states that it is actually 228 degrees, can anyone explain how they get this?
2) The great circle bearing from A (70S 030W) to B (70S 060E) is approximately?
For this question i figure out that the rhumb line track is going to be 090 and the conversion angle is 42 degrees, but looking at my drawing it tells me to take the conversion angle away, but the answer is 132 degrees implying i have to add it on, so i am getting very confused. Can anyone explain the error in my ways?
Thank You
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I draw the diagram a little differently, as if it were a Lambert chart, so in the northern hemisphere draw the meridians converging at the top. GCs are straight lines a to b and rhumb lines are concave to the pole. From this diagram you can see exactly what should be happening to your GC track (increasing or decreasing measure relative to the local meridian).
1) Your GC changes by Earth Convergence between A and B, not by conversion angle. Remember that CA = EC/2. So to go from A to B in the northern hemisphere you start at A on 040T and arrive at B on 048T since going east in NH you add the angle change. 042T would be your track midway between A & B. The reciprocal to go back B to A is therefore 228T.
2)CA is EC/2 ; EC= chlong sine Lat. Remember that for this question in the middle the GC track is the same as the rhumb line track of 090. CA converts between GC & RL track at the middle, not between GC tracks at your start point and destination. So in the middle of the trip GC track = 090T and to go back the the start add the CA (42 degs) since you are going back to the west in the southern hemisphere. Between A and B the GC track changes by earth convergence of 84 degrees, so the track at B is 090-42 = 048T. Subtract going east in Southern Hemisphere.
Good Luck!
1) Your GC changes by Earth Convergence between A and B, not by conversion angle. Remember that CA = EC/2. So to go from A to B in the northern hemisphere you start at A on 040T and arrive at B on 048T since going east in NH you add the angle change. 042T would be your track midway between A & B. The reciprocal to go back B to A is therefore 228T.
2)CA is EC/2 ; EC= chlong sine Lat. Remember that for this question in the middle the GC track is the same as the rhumb line track of 090. CA converts between GC & RL track at the middle, not between GC tracks at your start point and destination. So in the middle of the trip GC track = 090T and to go back the the start add the CA (42 degs) since you are going back to the west in the southern hemisphere. Between A and B the GC track changes by earth convergence of 84 degrees, so the track at B is 090-42 = 048T. Subtract going east in Southern Hemisphere.
Good Luck!
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Think i understand the first one now, so you have to multiply the 4 by 2 to get the convergency, which is what great circles cahnge by and then apply the rules of adding and subtracting. I managed to work everything else out, just not why it was 228 instead of 224 but i understand that now, i think i have been confusing intial and fianl great circle tracks with bearings, that's my problem.
Still not sure on the second one, going to read it again, takes me a few times for it all to sink in.
Thanks for the help
Still not sure on the second one, going to read it again, takes me a few times for it all to sink in.
Thanks for the help
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Neil
1. “The great circle bearing of position B from position A in the northern hemisphere is 040 degrees. If Conversion Angle is 4 degrees, what is the great circle bearing of A from B?”
If the GC bearing from A to B is 040, and the CA is 4, the RL bearing A to B is 044. As the RL direction is constant, the RL bearing from B to A is the reciprocal of 044, that is 224. To find the GC bearing B to A apply CA again, that is 224 + 4 = 228. A sketch (best drawn with the GC as a straight line as Sinbad suggests) should help decide the ‘plus or minus’ question:
2. “The great circle bearing from A (70S 030W) to B (70S 060E) is approximately?”
As the two points are at the same latitude and B is East of A, the RL bearing A to B must be 090. The CA is ˝ chlong x sin lat = 42. The GC bearing is thus 090 ± 42. Again, a sketch will resolve the ‘plus or minus’ question – in this case it is +, so the answer is 132. I suspect you might not have drawn a southern hemisphere diagram.
PM me with your e-mail address if you want more.
Bob
1. “The great circle bearing of position B from position A in the northern hemisphere is 040 degrees. If Conversion Angle is 4 degrees, what is the great circle bearing of A from B?”
If the GC bearing from A to B is 040, and the CA is 4, the RL bearing A to B is 044. As the RL direction is constant, the RL bearing from B to A is the reciprocal of 044, that is 224. To find the GC bearing B to A apply CA again, that is 224 + 4 = 228. A sketch (best drawn with the GC as a straight line as Sinbad suggests) should help decide the ‘plus or minus’ question:
2. “The great circle bearing from A (70S 030W) to B (70S 060E) is approximately?”
As the two points are at the same latitude and B is East of A, the RL bearing A to B must be 090. The CA is ˝ chlong x sin lat = 42. The GC bearing is thus 090 ± 42. Again, a sketch will resolve the ‘plus or minus’ question – in this case it is +, so the answer is 132. I suspect you might not have drawn a southern hemisphere diagram.
PM me with your e-mail address if you want more.
Bob
