GPS formulas

If the point is fixed, you can simplify your calculations by pre- calculating the sin and cosine of its latitude and longitude.

If processing time is at all a concern, you can also assume that the sine and cosine of your own position are also the same as the sine and cosine of the point. There will be an error--that will disappear as you approach the point.

fmod(atan2(sin(dLONG)*cos(LAT2),cos(LAT1)*sin(LAT2)-sin(LAT1)*cos(LAT2)*cos(dLONG)),

If you want to avoid excessive trig, first calculate the distance as sqrt( mNorth *mNorth + mEast * mEast) where the north and east distances are in meters. You can get these numbers with very simple plane trignometric calculations based on latitude, longitude and the radius of the earth.

When you have meters north and east to the point, the determination of bearing is a simple atan2() calculation.

These assumptions work nicely if you only need bearing to the nearest degree and the distance is less than 40 or 50 kilometers. If the distance is much larger, you are probably stuck with the more complex spherical trignometric equations.

Mark Borgerson

fmod(atan2(sin(dLONG)*cos(LAT2),cos(LAT1)*sin(LAT2)-sin(LAT1)*cos(LAT2)*cos(dLONG)),

The great circle formulae get increasingly inaccurate when you get near the waypoint. This is because the great circle distance is calculated as the sine or cosine of the route seen from Earth's center (6400 kilometers / 4000 miles away). The accuracy of the trigonometric functions will play a significant role in short distances.

The shortest route (great circle route) will have a changing heading (direction) unless the route is north-south or along the equator. There is a little longer route with a constant heading called the rhumbline.

One reference is . There are plenty of others, Google for 'great circle navigation'.

HTH

--

Tauno Voipio, CPL(A), aeronautic navigation instructor
tauno voipio (at) iki fi

Pretty much guaranteed to be wrong at this point already.

Differences between longitudes don't have much of a useful meaning. First because longitude is circular, and difference across a wrap-around will yield wild results. Second because the meaning of a longitude difference changes with latitude. The best way to generate such formulae is usually to not use longitude and latitude at all, but

3D cartesian coordinates of points either on the unit sphere, or on the actual earth surface --- those will even be easier to extract from GPS raw signals, as an extra bonus.

The distance between two such points along the earth's surface is then roughly

arccos(dotproduct(point1, point2))/R

the length of an arc along the great circle through the two points. I say "roughly" because earth isn't really a sphere.

Generally, two points on the sphere, together with the center, define a plane in space. The intersection of that plane with the sphere is a great-circle, and the shortest path between the two points is part of it.

--
Hans-Bernhard Broeker (broeker@physik.rwth-aachen.de)
Even if all the snow were burnt, ashes would remain.

Have a look at

--
Peter Bennett, VE7CEI  
peterbb4 (at) interchange.ubc.ca  
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca

The above formula will cause all sorts of grief for short distances. Due to (in)accuracy of the acos() function for small angles, the computed result will be quite erratic for two points that are very close to each other.

[...]

That grief is due to the problem being numerically tricky, though, not because of the formula being wrong. The formula cited by the OP will have very similar problems: differences among longitudes and latitudes of nearby points will lose up to 5 decimal digits to cancellation (GPS resolution vs. range), even before any trigonomtric function is called.

--
Hans-Bernhard Broeker (broeker@physik.rwth-aachen.de)
Even if all the snow were burnt, ashes would remain.

That'a why a plain plane trigonometry formula with latitude correction gives the best results on short tracks. Take one degree to correspond 111.11 km and multiply the east-west (longitude) difference with the cosine of the avreage latitude of the track, then use Pythagoras' theorem for the rest.

--

Tauno Voipio
tauno voipio (at) iki fi

I've used a similar approach: calculate the km/degree of latitude factor for the average latitude of the two points, and then use Pythagoras.

This approach is simple, gives increasingly accurate results with shorter distances, but doesn't work too well near the poles or over very large distances. Polar explorers and astronauts probably shouldn't use it.

Steve