For all the math geniuses out there...given two lat/lon points (point A and point B), what is the simplest way to find the vector needed to reach point B from point A. For example, the vector needed to reach 36, -85 from 36, -82 would be 270°. I am not entirely sure if "vector" is the technically correct term so if there is a more accurate term, comment and let me know. The fact that the earth is a sphere is throwing me off...I feel as if I could tackle this problem if given a plane but the sphere complicates it too much for me. A chunk of code would be ideal but a simple algorithm that I can translate into code myself is perfectly fine also.
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PolyGeo
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Brandon Schmidt
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You're looking for a great circle distance calculation. Take a look at this question or this question. – om_henners Dec 05 '13 at 22:52
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The Earth isn't a sphere -- it's a spheroid, which will mess up the accuracy of spherical trig. The real solution is a partial differential equation, only solvable through iterative means. – Vince Dec 05 '13 at 23:03
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1The direction (or "bearing") from (36,-85) to (36,-82) is not 270 degrees! It is actually a bit greater than that. – whuber Dec 05 '13 at 23:26
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If they're lat-lon coords, the bearing is just under 90°. If they're lon-lat coords, it is exactly 0°. – Martin F Dec 06 '13 at 01:03
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What you're looking for is the initial bearing (or forward azimuth), which if followed in a straight line along a great-circle arc will take you from the start point to the end point.
Here is some simple JavaScript from this link:
var y = Math.sin(dLon) * Math.cos(lat2);
var x = Math.cos(lat1)*Math.sin(lat2) -
Math.sin(lat1)*Math.cos(lat2)*Math.cos(dLon);
var brng = Math.atan2(y, x).toDeg();
The above link has a wealth of useful information beyond this for related calculations.
As your question states, this is the simplest method - since the Earth is not a true sphere this calculation will not be 100% accurate, but it is a close approximation.
Radar
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If this is accurate within a few degrees that is fine. In the final product, this will be indiscernible due to the affects of pixel rounding. I'll try this out...looks promising! – Brandon Schmidt Dec 05 '13 at 23:13
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If your tolerance is really "a few degrees" (though i suspect it's a bit tighter) then the spherical earth assumption will be plenty good enough for you. – Martin F Dec 06 '13 at 01:05