If you are at a known location (you know your precise latitude and longitude for example) and have an unobstructed view of another known location you can:
A: Take a precise visual bearing to the other location (for example with a compass and correcting for compass variation) – This is the Line-of-Sight bearing.
B: Calculate the Initial Heading of a Great Circle Path between the two points using a known formula. In each case you arrive at a true bearing.
My Question: Are these two equal?

At My location magnetic variation (from current aviation map) is approximately 14 ½ degrees East. Converting from observed magnetic bearing to true (astronomical) bearing you add the easterly variation.

From: https://www.movable-type.co.uk/scripts/latlong.html

where φ1,λ1 is the start point, φ2,λ2 the end point (Δλ is the difference in longitude)

(all angles  in radians)
var y = Math.sin(λ2-λ1) * Math.cos(φ2);
var x = Math.cos(φ1)*Math.sin(φ2) -
var brng = Math.atan2(y, x).toDegrees();

My reasoning for why I believe they are equal:

A Great Circle path between two points lies in a plane which passes through each point and the center of the earth. A straight line between the two points (the notional line-of-sight) also lies in this plane.
Therefore the line-of-sight is directly above the great circle path, and
the line of sight bearing is equal to the great circle initial bearing.

Is this reasoning correct?


For a spherical representation of the Earth, yes, your reasoning is correct. A great circle is defined as being the intersection of the sphere and a plane that passes through the sphere's center point. A line of sight, for most practical purposes, can be considered a straight line, and of course, lies on the same plane as its corresponding great circle plane. The initial bearings are angles with respect to North and are measured on the sphere's surface from the projection of the line along the normal. The angles are equal for both line of sight and great circle, and for all altitudes as well.

In reality, the Earth being slightly squashed, it is actually closer to an ellipsoid of revolution. Bearings and distances on an ellipsoid are computed on a geodesic. A geodesic on the ellipsoid does not lie on a plane (except for the equator and meridians), and the line of sight would not necessarily have the same start and end bearings as the geodesic, but we are talking about very small differences here, especially for short distances, negligible in most practical cases.

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No, they are not. Take two points at $80^\circ N$, one at $0^\circ$ and one at $179^\circ E$. The bearing from the first to the second is due East, but the great circle is almost over the North pole so the initial heading is almost North.

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  • $\begingroup$ I'm asking about line-of-sight bearing - in order to measure it you must be able to see one point from the other. Your example would need a great elevation for each point but, if you had that line-of-sight bearing would also be almost due North just like the Great Circle. Thanks for your response. $\endgroup$ – Andy Fawcett Oct 14 '18 at 22:52
  • $\begingroup$ No, it is just an extreme example. Even if you have two points close together on the $80^{th}$ parallel the great circle will go North of East even though the bearing is due East. The line of latitude is not a great circle, so it cannot be the shortest route. $\endgroup$ – Ross Millikan Oct 14 '18 at 23:09
  • $\begingroup$ Hi Ross"...even though the bearing is due east" - by which I assume you mean the line-of-sight bearing is due east. This is only approximately true for small differences in longitude. Line-of-Sight bearing also deviates increasingly North with larger difference in longitude. $\endgroup$ – Andy Fawcett Oct 14 '18 at 23:19
  • $\begingroup$ While your conclusion may still be correct your example does not make that case. I would like to make a minor adjustment to your example thus: Take two points at 89∘ 59' N, one at 0∘ and one at 179∘ E. The change is in scale and keeps the sense of the original while allowing for Line-of-Sight. Now imagine looking down on the N pole you will see the 2 points each approximately 1 nautical mile from the pole, one on the prime meridian, the other almost diametrically opposite. The LOS passes very close to the pole so the bearing is close to due N (not East as in your answer) $\endgroup$ – Andy Fawcett Oct 15 '18 at 22:31
  • $\begingroup$ The bearing from the first to the second is due East - You are referring to a line of constant bearing (rhumb line, or loxodrome) which is very different from a line of sight or a great circle. The OP was referring to initial bearings of a line of sight and a great circle. $\endgroup$ – FSimardGIS Oct 16 '18 at 2:36

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