Could someone give me the formula for this problem?

Find the orthodromic (great circle) distance between A and B

A= 59° 34,1'N 008° 08,4'E
B= 30° 25,9'N 171° 51,6'W

the answer is 5400NM

Try here Spherical Trigonometry (http://www.boeing-727.com/Data/fly%20odds/distance.html#sertrig)

The shortcut here is to spot that the two points are 180 deg. of longitude apart.

So, the GC will inevitably be a N-S rhumb line via one of the poles.

As both points are in the N hemisphere, you can add the distances of each point from the N pole (i.e. 90-latitude * 60 NMiles).

In this case the GC included angle is 90 degrees.

90*60=5400NM

As per Whopity's post, but condensed to just the equation:


a = sin²(Δlat/2) + cos(lat1).cos(lat2).sin²(Δlong/2)

c = 2.atan2(√a, √(1−a))

d = R.c


R = earth’s radius (mean radius = 6,371km);

Pass all the angles in radians!

Or in Javascript, if that floats your boat:


var R = 6371; // km

var dLat = (lat2-lat1).toRad();

var dLon = (lon2-lon1).toRad();

var lat1 = lat1.toRad();

var lat2 = lat2.toRad();

var a = Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.sin(dLon/2) * Math.sin(dLon/2) * Math.cos(lat1) * Math.cos(lat2);

var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));

var d = R * c;



From here (http://www.movable-type.co.uk/scripts/latlong.html) which I've used before for several mapping programming projects. Works a treat :)

Is this a question appearing on a PPL exam? Sounds ridiculous to me.

Heck, even for a CPL/ATPL, having to do great circle calculations by hand sounds a bit over the top. I can understand having to know the principles behind it, but the actual calculations...? Even in a simple corner case like this.

AIUI, the whole point is that no triangulation is required. Indeed a typical exam question, and I doubt if the point would be clear to most candidates. It wasn't to me, until Mark1 made things clear!

Thank you!