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    <p>As BenjiSmith said, there are potentially several paths that connect any A &amp; B on the globe, but the two most popular (by far!) are the "great circle" and "rhumb line" paths. </p> <p>A great circle gives the shortest distance (by constructing a plane from the two points &amp; the center of the earth &amp; following a circular arc in that plane). </p> <p>A rhumb line maintains a constant heading, trading some distance (can be extreme at high latitudes) for ease of use. That is, in a boat or plane, you simply point at the desired heading and go until you arrive at your destination (whereas with a great circle the heading changes continuously). In mid latitudes the distance penalty isn't too severe.</p> <p>Be warned, both path types have discontinuities involving the poles and ambiguities when dealing with antipodal points (pts opposite each other on the sphere).</p> <hr> <p>To build a great circle, you'll want to convert the points to 3D cartesian coordinates (I'll leave this step out but it's trivial for a spherical earth &amp; found iteratively for an oblate earth model a la WGS-84).</p> <blockquote> <p>Let <strong>a</strong> be the unit vector pointing at the start point from the center of the earth. </p> <p>Let <strong>b</strong> be the unit vector pointing at the end point from the center of the earth.</p> <p>Let <em>r</em> be the radius of the earth.</p> <p>Let <em>d</em> be the (given) distance traveled.</p> </blockquote> <p>Construct the unit vector normal to the G.C. plane by taking the cross product of the unit vectors <strong>a</strong> and <strong>b</strong>. That is, let <strong>n = a x b</strong>.</p> <p>The (given) distance traveled is the length of the arc formed by sweeping the vector *r***a** around <strong>n</strong> by some angle <em>theta</em>. Recalling that the circumference of the full great circle is 2 * pi * <em>r</em>, we find <em>theta</em> = <em>d</em>/<em>r</em>.</p> <p>The cartesian point corresponding to the new location is thus found by rotating *r***a** around <strong>n</strong> by <em>theta</em> radians. Convert that cartesian point to lat/long &amp; you're done.</p> <p>I won't derive the rhumb line math here, but will say that the Mercator map projection has the property that rhumb lines are straight. You can easily construct a rhumb line with the mercator projection formula, but you'll have to define some error tolerance so you can break the path up into short, straight segments.</p> <p>Good luck!</p>
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    1. POIf I am here and heading towards there and I've covered this much ground, where am I?
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    1. COi updated my question taking into account your answer.
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