a general navigation question...

[air_cowboy]

....which of the following differences in latitude will give the biggest difference in the initial great circle track and the mean great circle trackbetween 2 points separated by 10 degrees change of longitude?

a) 60N and 60S
b) 60N and 55N
c) 30S and 30N
d) 30S and 25S

....and y???

thanks...

[air_cowboy]

this was probably not he best forum for this post... but since it is here...any answers please!!!

[bfd777]

60N and 55N:8:O

[airborneforever]

South Asia and the Far East News and views on the fast growing and changing aviation scene on the planet.


that is the name of the forum.
how does your post fit in anywhere here?
try and act a little bit more mature, ok?

[A330AV8R]

airborneforever

I don't think maturity has anything to do with someone wanting to find out the answer to a question ? and why not this is an aviation forum isin't it granted its not in tech but so what .... give the guy a break


:=

[divdby0]

use the formula (dlong * cos lat)

dlong in this case is 10. so in case (a) and (c), the answer is the same coz of same latitude(cos 60 = 1/2 is same for north and south and cos 30 = sqrt(3)/2 = for N and S lats). So its either (b) or (d). Will let you know the answer soon as its 4.30 am in my corner of the world and I dont know where my scientific calculator is.

And Monsieur airborneforever,

If you dont know the answer, you can rather shut up. If you cant help an other person with your knowledge, that knowledge is useless.

Cheers,

Divdby0

[air_cowboy]

hi ...

the formula for convergence is actually...

convergence = ch.long * sin (mean lat.)

so the answer would be (b) coz sin (mean lat) between 55N and 60N would give the highest value. therefore the greatest change of track.

[divdby0]

true, it is sin mean lat. But the answer cannot be obtained with the formula alone. Sin (mean lat) would give max value for 60 (8.66) compared to 57.5 (8.43). Has gotten me thinking.

[divdby0]

ok...now once you get Convergency for a particular lat, you can find the gc track. Along the same latitude, r/l track from A to B and B to A is 90 and 270 respectively. And now that you have C, you can get CA and hence GC track A to B and B to A for any particular lat. That way, you can find the mean lat and hence the difference.

Note : the Gc track will curve in case of same lat in the direction of the hemisphere unlike r/l which curve in the direction opposite to the Hemisphere of the lat. Will prove to be important coz different direction means higher/lower gc tracks.

Hopefully I didnt embarrass myself again giving solution for distances rather than GC:ugh:

[aztecdriver]

Plug it into GPS and test, bah...:cool:

[CAPTAIN WOOBLAH]

Perhaps a navigation example would help.
1) Find track and distance from position 30deg 20' S / 20deg 05' W to position 32deg 55' S / 18deg 50' W
d.long = 1deg 15' East = 75'E
d.lat = 2deg 35' South = 155'S
Mean Lat = 32deg 55' + 30deg 20'
___________________ = 31deg 37 1/2'
2
Mid Lat = 31deg 37 1/2 ' + 1 = 31deg 38 1/2'
Dep = 75 Cos 31deg 38 1/2' = 63.85 miles
Log 75 = 1.87506
L Cos 31deg 38 1/2' = 1.93011
L Dep = 1.80517
Log 63.86 = 1.80517
Log 155 = 2.19033
________
1.61484
Tan Tr = 63,85
_____
155
Thus Track = S 22deg 23 1/2' E
= approx 157 1/2 T
Distance:
= 63.85 Cosec 22deg 23 1/2'
=167.65 miles (nautical)
=167 1/2'
Log 63.85 =1.80517
L Cosec 22deg 23 1/2' = 0.41915
_________
2.22432