Help please re: Gen_Nav longitude distance calcs
Thread Starter
Join Date: Jul 2010
Location: CA
Posts: 5
Likes: 0
Received 0 Likes
on
0 Posts
Help please re: Gen_Nav longitude distance calcs
Hi all,
First post here, hoping for someone's help as I can't seem to figure this out.
Currently studying Gen_nav, converting from FAA to JAA, and I'm having some troubles with calculating distances:
I know how to get great circle distances and everything, and am able to use the departure and convergency formulas, but here's what I don't know how to tackle:
Obviously I know 60' / 1° of lat. equals 60 nautical miles, all clear there, but wanting to calculate distance east / west, i.e. change in longitude, this can not be applied, as 60' equals 60nm only at the equator, right.
Could anybody please explain how to calculate, i.e. the distance from 20W to 40W at 70N? Or what longitude I get when moving 60NM west from, say 20W at 70N?
Thanks a lot for the help in advance.
First post here, hoping for someone's help as I can't seem to figure this out.
Currently studying Gen_nav, converting from FAA to JAA, and I'm having some troubles with calculating distances:
I know how to get great circle distances and everything, and am able to use the departure and convergency formulas, but here's what I don't know how to tackle:
Obviously I know 60' / 1° of lat. equals 60 nautical miles, all clear there, but wanting to calculate distance east / west, i.e. change in longitude, this can not be applied, as 60' equals 60nm only at the equator, right.
Could anybody please explain how to calculate, i.e. the distance from 20W to 40W at 70N? Or what longitude I get when moving 60NM west from, say 20W at 70N?
Thanks a lot for the help in advance.
Join Date: Aug 2001
Location: Dorset
Posts: 775
Likes: 0
Received 0 Likes
on
0 Posts
Departure is the distance between two meridians along a speciied parallel of latitude.
Departure in nm = Change in longitude in minutes x Cosine of latitude.
From 20W to 40 W is 20 degrees x 60 minutes/degree = 1200 minutes
So departure = 1200 x Cos 70 = 410.424 nm.
Departure in nm = Change in longitude in minutes x Cosine of latitude.
Rearranging this gives
Change of longitude in minutes = Departure / Cos latitude
Inserting 60 nm at 70N gives
Change of longitude in minutes = 60 / Cos 70 = 175.428 minutes
Dividing by 60 gives 2 degrees 55.428 minutes.
Adding this to the initial longitude of 20W gives 22 degrees 55.428 minutes West.
All of the above relates to Rhumb line distance along lines of latitude.
If you want great circle distance you must use the great circle distance method.
Departure in nm = Change in longitude in minutes x Cosine of latitude.
Could anybody please explain how to calculate, i.e. the distance from 20W to 40W at 70N?
So departure = 1200 x Cos 70 = 410.424 nm.
Or what longitude I get when moving 60NM west from, say 20W at 70N?
Rearranging this gives
Change of longitude in minutes = Departure / Cos latitude
Inserting 60 nm at 70N gives
Change of longitude in minutes = 60 / Cos 70 = 175.428 minutes
Dividing by 60 gives 2 degrees 55.428 minutes.
Adding this to the initial longitude of 20W gives 22 degrees 55.428 minutes West.
All of the above relates to Rhumb line distance along lines of latitude.
If you want great circle distance you must use the great circle distance method.
Thread Starter
Join Date: Jul 2010
Location: CA
Posts: 5
Likes: 0
Received 0 Likes
on
0 Posts
Doh, I guess that was really something that made me look stupid!
Still can't figure out why I thought I need to solve such problems without using departure
Thanks a bunch for clearing my head Keith.
Still can't figure out why I thought I need to solve such problems without using departure
Thanks a bunch for clearing my head Keith.
Join Date: Dec 2004
Location: A place where something is or could be located; a site.
Posts: 455
Likes: 0
Received 0 Likes
on
0 Posts
Hi Keith,
It been a while since my exams so could be wrong, but is that just 1 degree of arc = 60 nm?
Or is there a formula to go with that.
I'm at work and my BGS books are at home, not being lazy!
Thanks,
EK
If you want great circle distance you must use the great circle distance method.
Or is there a formula to go with that.
I'm at work and my BGS books are at home, not being lazy!
Thanks,
EK
Join Date: Aug 2001
Location: Dorset
Posts: 775
Likes: 0
Received 0 Likes
on
0 Posts
The JAR syllabus is limited to special cases in which the 1 degree of arc = 60 nm. These cases are; two points on the same meridian, two points on a meridain and anti-meridian, or two points on the equator.
But even with these we need to be careful in reading the question. Going north/south along a meridian does not change longitude, unless we pass over/under one of the poles. As we go over/under the poles there is of course an abrubt change of longitude. This change is not proportional to the distance flown.
For other cases such as two points on different meridans and different latitudes, the calulations require spherical geometry which is (thankfully) beyond JAR ATPL.
But even with these we need to be careful in reading the question. Going north/south along a meridian does not change longitude, unless we pass over/under one of the poles. As we go over/under the poles there is of course an abrubt change of longitude. This change is not proportional to the distance flown.
For other cases such as two points on different meridans and different latitudes, the calulations require spherical geometry which is (thankfully) beyond JAR ATPL.
