JAA ATPL: GEN NAV Q

[frogone]

Hi Guys,

Here’s a quick question for any of you guys studying the Gen., Nav ATPL syllabus. How do you find the great circle distance in nautical miles between: e.g.

56˚12’N 057˚00W and 64˚00N 123˚00E :confused:

I’ve been looking through notes, etc but my brain is fried at this stage. It’s probably a simple example but I’m stumped.

Cheers

EIDW

[Paul Hickley]

The 2 longitudes are meridian and anti-meridian. The shortest way is the Great Circle track via the North Pole.

As you say, you probably just needed a break.

All the best,

Paul

[Alex Whittingham]

The thing to notice is that the two longitudes add up to 180º. This means that the shortest route between the two points is over the pole. Running up or down a meridian one minute of arc is 1nm. 56º12'N to the pole is 33º48'of arc or 2028' or 2028nm. Similarly, from the pole to 64ºN is 26º, 1560' or 1560nm. 2028 + 1560 = 3588nm.

[frogone]

Thanks for the replys guys! Alex, if both longitudes add up to 180 does this always imply the quickest way is over the poles? Also its making sence now :) , do you, with these questions, go up and then across?? Not across at a diagonal as that's what I was thinking!

Cheers

EIDW

[Unusual Attitude]

Give Paul a bit of string and a Globe and all will become clear !

Just a quick note to say thanks to Paul and all the other OATS ground instructors for the excellent instruction.

If they can get me through all 14 first time there's hope for anyone !

Regards

UA

[Alex Whittingham]

Distance questions should only be asked about

1. Two points on the same longitude, in which case the change of latitude in minutes is the distance in nm.

2. Two points on the same latitude, in which case the departure formula is used. Distance (departure) = change of long in minutes x cos lat

3. Two points with the longitudes 180º out, in which case the shortest distance is found by going over or under the nearest pole. What I did above was use the change of latitude in minutes to find the distance from one point to the pole then the same method to find the distance from the pole to the second point.

Distance questions 'at an angle' require 3D trig over on a spherical surface and are not asked in the ATPL exams. The closest you get to this is finding the amount of drift, in nm, on an IRS. As this is always a short distance we assume the earth is flat and find the distance 'at an angle' by first finding the east/west distance, then the north/south distance, then using pythagoras. You will meet this problem later on when you study the IRS/INS so don't worry about it now.

[Chirpy Pilot]

Is big Pete still teaching Navigation at Oxford. Top bloke!

[Paul Hickley]

Yes, he is and yes, you're right. Top bloke.

Paul

[Send Clowns]

Paul and Alex, do you have nothing better to do ??? :D I'm the unemployed General Nav instructor, should have more time on my hands than you two. Yet both post within 10 minutes of the query!

All the best
SC