[PATCH] Simplify distance calculations

[Robert C. Helling]

On 13 Jun 2015, at 08:53, Linus Torvalds <torvalds at linux-foundation.org> wrote:

> On Fri, Jun 12, 2015 at 2:34 PM, Anton Lundin <glance at acc.umu.se> wrote:
>> 
>> Sorry, One _degree_ in latitude is about 111.5 km where i live, and
>> thats the basis for the convertion that 10 udeg is about a meter.
>> 
>>> +     return sqrt(lat1.udeg*lat2.udeg + lon1.udeg*lon2.udeg) / 10;
> 
> Ugh. I think that's a bit *too* simplified.
> 
> You do want to have that multiplication with "cos(latitude)" because
> longitude lines are much further apart around the equator than in
> temperate areas..
> 
> And don't you want to square the *differences* (with that latitude
> correction) rather than just multiply the lat/long pairwise together?
> That just looks insane.The distance from 0,0 to any other point is
> always zero? That's an interesting distance function.
> 
> I guess there are few divesites where we need to worry about
> wraparound (much less 20 m apart), so those kinds of simplifications I
> can see, but you seem to have gone a bit overboard.
> 

A somewhat simplified formula that saves a bit of trigonometry but deals correctly with wrap around and the fact that meridians have a distance proportional to cos(latitude) would be

sin(lon1-lon2)^2 * cos(lat1)^2 + sin(lat1-lat2)^2 < distance^2 / radius_of_earth^2

This should be fine for distances much smaller than the radius of the earth. The right hand side is a constant known at compile time, so it should be figured out by the compiler already.

Best
Robert

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Robert C. Helling     Elite Master Course Theoretical and Mathematical Physics  
                      Scientific Coordinator                                   
                      Ludwig Maximilians Universitaet Muenchen, Dept. Physik    
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