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### GPS Formula

```Date: 11/27/2002 at 02:12:36
From: Arfay
Subject: GPS formula

How can I calculate the position from 3 points (or more) in latitude
longitude format? This is very likely the formula that is used in GPS
```

```
Date: 11/27/2002 at 09:23:03
From: Doctor Tom
Subject: Re: GPS formula

Hello Arfay,

I do not understand exactly. GPS requires four times, not distances.
It receives the time and locations of four satellites and works out
the location on earth based on this. It can make a guess based on
three, but it has to assume that the elevation is the same as the last
time it did the calculation, which may or may not be true.

The time measurements have to be extremely accurate since the radio
signal from the satellites travels at about one meter every 3
nanoseconds. An error of one microsecond (one millionth of a second)
represents an error of 300 meters. GPS is accurate to about 10 meters.

The GPS formula is NOT simple because such accuracy is required.
Einstein's theory of general relativity tells us that clocks run
slower in a "gravity well," so the clocks on earth run slower than the
clocks in the satellites that are about 30,000 kilometers above the
earth. Thus a relativistic correction must be made for the times

- Doctor Tom, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 11/29/2002 at 07:27:50
From: Arfay
Subject: GPS formula

Dear Dr. Tom,

Thank you very much for your answer, and sorry that the question was
confusing. GPS formula is not simple, as you mentioned, and I agree.
All I would like to know is, in case we can get the latitude and
longtitude (not the pure signal) from three satellites (or
more), how can we calculate our location?

The process of distance and measurement is quite hard to understand
so I just would like to ask only the position calculation part. How
can it be done by using sin cos tan?
```

```
Date: 11/29/2002 at 08:32:50
From: Doctor Tom
Subject: Re: GPS formula

Keep in mind that what follows is NOT exactly correct, because I
assume that we live in a Euclidean space, and Einstein's theory of
general relativity tells us that we do not. In other words, in the
real world the three angles of a triangle do not add to 180 degrees,
the Pythagorean theorem does not hold, et cetera. They do
approximately, but if you make those approximations, you will have
an error that is larger than the GPS error.

If you know the distances to three satellites and the positions of
those three satellites, you can calculate your position.

It is easy to think of this first in lower dimensions. Suppose you
live on a line (one dimensional space). If the satellite is on the
line and you know the distance to it, there are only two places you
could be on the line, right? If you know roughly where you are, it is
usually easy to eliminate one of the two possibilities.

Now imagine that you live on a plane (two dimensions). If you know the
distances to two satellites (that are on the plane) you are at a
distance of d1 from satellite 1 and d2 from satellite 2. Thus you are
somewhere on a circle of radius d1 centered at satellite 1 and on a
circle of radius d2 about satellite 2. You are therefore at the
intersection of the two circles. Generally, there are two
intersections, so you are at one of two points in the plane. If you
know roughly where you are, it is easy (usually) to eliminate one
possibility.

In three dimensions, you need three distances to three satellites. You
are on the surface of three spheres centered at each of the
satellites. Three spheres intersect at 2 points in space, and you can
usually eliminate one.

I would solve this with cartesian coordinates with the origin at the
center of the earth. I'd write my position as (x, y, z) and use the
Pythagorean theorem to write the equation of the distance from me to
each of the satellites as a quadratic equation in x, y, and z. You
will have three equations and three unknowns so you can solve them,
and you will obtain the two solutions. Then use the standard
conversion of cartesian to spherical coordiantes to convert your
location to latitude/longitude/altitude.

But remember that your solution is not quite correct because the
Pythagorean theorem is not quite true in the real world.

Note also that to obtain the distances to three satellites you need to
measure the time delay to four of them (since you don't know the
correct time, accurate to a fraction of a millionth of a second).
Also, these distances will not be what you would get with Newtonian
physics because of what I said in the previous note about living in a
gravity well relative to the satellites, so the apparent speed of
light changes because space-time is more bent near the surface of the
earth than out near the satellites.

- Doctor Tom, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 12/01/2002 at 22:26:24
From: Arfay
Subject: Thank you (GPS formula)

Thank you very much Dr. Tom, it's very helpful!
Arfay
```
Associated Topics:
College Higher-Dimensional Geometry
High School Higher-Dimensional Geometry

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