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**posted on**

- Richard Sloan

May 29, 2005, 2:54 am

I have been looking for formulas to help me create a special GPS device, I

need to know distance from current position to a point (LAT,LONG) and also I

need a function to let me know when I am passing by a point and if its to

the left, right, ahead, or behind me, and it would be nice to know by how

much.

Below are formula I found on the web and thought would be at least part of

my answer, these were referred to as great circle equations.

I am not a math wiz, I am very good at math, but not to the point I could

come up with these equations myself :-)

Can anyone help with whats required?

Thanks!

Richard.

Now for the distance I have used successfully:

dLAT = LAT1 - LAT2;

dLONG = LONG1 - LONG2;

R = 6731000.0

double dist2(void) {

return R

cos(LAT1)

}

And for nearing I have used rather unsuccessfully:

double bearing(void) {

C =

fmod(atan2(sin(dLONG)

2*PI);

C = ((C*180)/PI)

return C;

}

need to know distance from current position to a point (LAT,LONG) and also I

need a function to let me know when I am passing by a point and if its to

the left, right, ahead, or behind me, and it would be nice to know by how

much.

Below are formula I found on the web and thought would be at least part of

my answer, these were referred to as great circle equations.

I am not a math wiz, I am very good at math, but not to the point I could

come up with these equations myself :-)

Can anyone help with whats required?

Thanks!

Richard.

Now for the distance I have used successfully:

dLAT = LAT1 - LAT2;

dLONG = LONG1 - LONG2;

R = 6731000.0

double dist2(void) {

return R

***2***asin(sqrt((sin((dLAT)/2))*(sin((dLAT)/2)) +cos(LAT1)

***cos(LAT2)***(sin((dLONG)/2))*(sin((dLONG)/2))));}

And for nearing I have used rather unsuccessfully:

double bearing(void) {

C =

fmod(atan2(sin(dLONG)

***cos(LAT2),cos(LAT1)***sin(LAT2)-sin(LAT1)***cos(LAT2)***cos(dLONG)),2*PI);

C = ((C*180)/PI)

return C;

}

Re: 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

Re: GPS formulas

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 <http://williams.best.vwh.net/avform.htm .

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

HTH

--

Tauno Voipio, CPL(A), aeronautic navigation instructor

tauno voipio (at) iki fi

Tauno Voipio, CPL(A), aeronautic navigation instructor

tauno voipio (at) iki fi

Re: GPS formulas

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 ( snipped-for-privacy@physik.rwth-aachen.de)

Even if all the snow were burnt, ashes would remain.

Hans-Bernhard Broeker ( snipped-for-privacy@physik.rwth-aachen.de)

Even if all the snow were burnt, ashes would remain.

Re: GPS formulas

[...]

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 ( snipped-for-privacy@physik.rwth-aachen.de)

Even if all the snow were burnt, ashes would remain.

Hans-Bernhard Broeker ( snipped-for-privacy@physik.rwth-aachen.de)

Even if all the snow were burnt, ashes would remain.

Re: GPS formulas

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

Tauno Voipio

tauno voipio (at) iki fi

Re: GPS formulas

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

http://www.fivetrees.com

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