Help please re: Gen_Nav longitude distance calcs

[SPar.ca]

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.

[Keith.Williams.]

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.

Could anybody please explain how to calculate, i.e. the distance from 20W to 40W at 70N?

From 20W to 40 W is 20 degrees x 60 minutes/degree = 1200 minutes

So departure = 1200 x Cos 70 = 410.424 nm.


Or what longitude I get when moving 60NM west from, say 20W at 70N?

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.

[SPar.ca]

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 :ouch:

Thanks a bunch for clearing my head Keith. :zzz:

[EK4457]

Hi Keith,

If you want great circle distance you must use the great circle distance method.

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

[Keith.Williams.]

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.