**IN WHICH** you learn the basic theoretical framework of GPS position finding, and practice using a GPS receiver tocollect computer-readable data.

## OVERVIEW

### How’d They Do That?

**By now you have determined that** GPS really works. That little gadget can actually tell you where you are! How?!

**The fundamentals of the system** are not hard to comprehend. But mis-conceptions abound, and it is amazing how many people don’t understand the principles. In a few minutes, assuming you keep reading, you will not be among them.

**We look first at a two-dimensional analogy**. You are the captain of an oceangoing ship off the western coast of some body of land. You wish to know your position. You have aboard:

**• an accurate** timepiece;

**• the ability to pick up distant sound signals** (a megaphone with the narrow end at your ear, perhaps?);

**• a map showing the coast and the** locations of any soundhouses (a soundhouse is like a lighthouse, but it emits noise instead); and

**• the knowledge that sound travels about** 750 miles per hour, which is about 20 kilometers (km) per minute, or 1/3 km per second.

**Suppose:**

**Figure 2-1. Measuring distance by measuring time.**

**1.** There is a soundhouse located at “S1″ on the diagram in Figure 2—1.

**2.** Each minute, exactly on the minute, the soundhouse horn emits a blast.

**With these elements,** you can determine the distance of your vessel, “V,” from the soundhouse. To do so, you note when your clock marks an exact minute. Then you listen for the sound signal. When it comes, you again note the second hand of the clock. You then may calculate “d” from

where “d” is the distance in km from the soundhouse and “s” is the number of seconds it took for the sound to reach you.

**Suppose it took 30** seconds for the signal to arrive. You would know your ship was 10 km from sound house S1.

**In geometric terms,** you know only that your ship is located somewhere on the surface of a sphere that has a radius of 10 km. You can reduce this uncertainty considerably since your ship is floating on the ocean, so you know your altitude is fairly close to mean sea level (MSL). Thus you could consider yourself to be on a circle with a radius of 10 km.

**That, however**, does not pinpoint your location. And, by looking at the diagram, you can see that, if you are moving, you might be in a lot of trouble, since contact with the ground is not recommended for ships. How could you determine your position more exactly? By finding your distance from a second soundhouse.

**Suppose you use the same technique as above,** listening to soundhouse “S2.” You find it is 15 km away. Then the question is: where is (are) the point(s) that are 10 km away from “S1″ and 15 km away from “S2″? If you solve this problem graphically, by drawing two circles, you find they intersect in two places. One of these locations you can pretty well eliminate by noting that your ship is not sitting on a prairie or in a forest. Based on the measurement from the second soundhouse, you know your position as accurately as your measuring devices and map will let you know it.

**It is obvious that you** can find your position by drawing circles. You can also find your position by purely mathematical means. (You would determine the coordinates of the intersection of the two circles from their formulas by solving two “simultaneous” equations. Then you would consider lengths of the sides of the triangle formed by the two soundhouses and the ship. This process is sometimes called “trilateration.” (A triangle is a “trilateral,” as a four-sided figure is a “quadrilateral”)

**As with the soundhouses,** distances—determined by the measurement of time—form the foundation of GPS. (Such distances are referred to as ranges.) Recall the basis for the last three letters of the acronym NAVSTAR: Time And Ranging.

**Figure 2-2. Earth and a point to be found.**

### How It Works: Measuring Distance By Measuring Time

How do the concepts illustrated above allow us to know our position on or near the earth’s surface? The length of that answer can vary from fairly short and simple to far more complicated than you are (or I am) interested in. At its most detailed, GPS is rocket science, brain surgery, nuclear physics, and the theory of relativity all rolled into one.

**Basically,** distances to several satellites (four are needed for a good 3-D spatial fix) are calculated from measurements of the times it takes for radio waves to reach from the satellites, whose positions are known precisely, to the receiver antenna.

**To illustrate,** let’s look at these basic ideas by seeing how the “ship” example differs from NAVSTAR.

**• NAVSTAR gives us 3-D locations:** Unless we are in fact on the sea, in which case we know our altitude, the problem of finding our location is three-dimensional. GPS can provide our position on or above the earth’s surface. (“Below” is tricky because of the radio wave line-of-sight requirement.) But the method translates from two to three dimensions beautifully.

**• The “soundhouses” are satellites:** Rather than being situated in concrete on a coast, the device that emits the signal is a satellite, zipping along in space, in an almost-exactly-circular orbit, at more than two miles per second. It is important to note, however, that at any given instant the satellites (space vehicles, or SVs) are each at one particular location.

**• NAVSTAR uses radio waves instead of sound:** The waves that are used to measure the distance are electromagnetic radiation (EM). They move faster than sound—a lot faster. Regardless of frequency, in a vacuum EM moves at about 299,792.5 km/sec, which is roughly 186,282 statute miles per second.

**Figure 2-3. The Earth and GPS satellites.**

**Let’s now look at the configuration of GPS** with a drawing of true but extremely small scale, starting with two dimensions. Suppose we represent the earth not as a sphere but as a disk (like a coin), with a radius of approximately 1 unit. (One unit represents about 4000 statute miles.) We are living on the edge (pun intended) so we are interested in points on, or just outside (e.g., airplanes and orbiting spacecraft) the edge of the coin. We indicate one such point by “x” on the diagram of Figure 2—2. We want to find out where “x” is.

**Suppose now that we have** two points on the drawing of Figure 2—3, called “a” and “b.” They represent two of the NAVSTAR satellites, such that we can draw straight lines from “a” to “x” and from “b” to “x.” We require that the lines not pass through our “Earth.” These points “a” and “b” are each about four units from the center of the coin (thus three units from the edge of the coin).

**Figure 2-4. Distances from “x” to the GPS satellites.**

**Measure the distance,** or length, from “a” to “x” (call it “La”) and from “b” to “x” (call it “Lb”). The lines “La” and “Lb” represent the unobstructed lines of sight from the satellites to our receiver antenna. If we know the positions of “a” and “b” and the lengths of “La” and “Lb” we can calculate the position of “x” through the mathematical process of trilateration1 shown in Figure 2—4.

**It is intuitively obvious that** we can locate any “x,” given knowledge of the positions of “a” and “b,” and of the lengths “La” and “Lb”: “x” must lie on a circle centered at “a” with radius “La” and also must lie on another circle centered at “b” with radius “Lb.”

**To formalize a bit what** we learned from the example with the ship and the soundhouses, two such circles either (1) do not intersect, (2) touch at a single point, (3) touch everywhere, or (4) intersect at two points. Possibility (1) is clearly out, because we have constructed our diagram so that the circles intersect at “x.” Possibility (2) seems unlikely, given the geometry of the situation, and given that the line segments connecting the points with x must not pass through the disk. Possibility (3) is out: the circles would have to have the same centers and radii. We are left then with possibility (4), shown as Figure 2—5: there are two points that are at distance “La” from “a” and “Lb” from “b.” One of these is “x.” The other is a long way from the disk, is therefore uninteresting (we’ll call it point “u”), and is certainly not “x.”

**The points “a” and “b,”** of course, represent two of the GPS satellites at a precise instant of time, and point “x” represents the position of the GPS receiver antenna at that same precise moment. (The 0.1 second that it takes for the signals from the satellites to reach the receiver is compensated for in the design of the system.)

**This is fine for two dimensions**. What happens when we move to three? From a conceptual point of view, the coin becomes a sphere, “x” resides on the sphere (or very slightly outside it), and we need to add a point (satellite) “c,” somewhere else in space (not in the plane formed by “a,” “b,” and “x”) in order to be able to locate the position of “x.” The problem we solve here is finding the intersection of three spheres (instead of two circles).

While the process of finding this intersection is more difficult mathematically, and harder to visualize, it turns out, again, that there are only two points, just one of which might reasonably be “x.”

**You may deduce this as follows**. Consider initially the intersection of the surfaces of two spheres. First, forget about the special case possibility that they don’t intersect at all (their centers are separated by more than the sum of their radii, or one sphere completely contains the other), and the case in which they touch everywhere (same centers, same radii). Now they must either touch at a single point (unlikely) or intersect forming a circle. So our problem of visualization is reduced to finding the intersection of this circle and the surface of the third sphere. If you again discard some special, inapplicable cases, you see that the intersection can be only two points, where the circle pierces the sphere. Only one of these points can be “x.”

**In theory,** one should need only three satellites to get a good, three-dimensional (3-D) fix. You may recall that when you took the receiver into the field, however, you needed four satellites before the unit would calculate a position. Why? Briefly, the reason has to do with the fact that the clock in a receiver is not nearly as good as the four $50,000 clocks in each satellite, so the receiver must depend on the satellite clocks to set itself correctly. So, in a sense, the fourth satellite sets the receiver’s clock. In actuality, however, all four satellites contribute to finding a point in four-dimensional space.

**Figure 2-5. Finding “x” by the intersection of circles.**

You now know the theoretical basis for GPS. It is not unlikely that you have several questions. I will defer the answers to a few that we have anticipated so that we can take up an issue that bears more directly on the fieldwork you are about to do.

### Factors Affecting When and How to Collect Data

Neither GPS nor any other method can tell you exactly where an object on the earth’s surface is. For one thing, an object, no matter how small, cannot be considered to be in exactly one place, if by “place” we mean a zero-dimensional point specified by coordinates. Any object occupies an infinite number of zero-dimensional points. Further, there are always errors in any measuring system or device.

**Now that** we have thrown out the idea of “exact” location, the issue becomes: what kind of approximation are you willing to accept (and pay for)? There are two philosophies you might consider:

**• good enough **

**• the best that is reasonable**

**You might use “good enough**” when you know for certain what “good enough” is. For example, if you are bringing a ship into a harbor, “good enough” might be locational values guaranteed to keep your ship out of contact with the harbor bottom, or between buoys.

**You might, on the other hand,** use “the best that is reasonable” when you don’t really know what “good enough” is. For example, if you are building a database of city block outlines, you might not be able to foretell the uses to which it might be put. Your immediate needs might suggest that one level of accuracy would be appropriate, but several months later you (or someone else) might want a higher level of accuracy for another use. So it might be worth expending the extra resources to collect data at the highest level of accuracy that your budget and the state of the art will allow.

**The major factors that relate to the accuracy of GPS measurements are:**

**• satellite clock errors **

**• ephemeris errors (satellite position errors) **

**• receiver errors **

**• ionosphere errors (upper atmosphere errors) **

**• troposphere errors (lower atmosphere errors) **

**• multipath errors (errors from bounced signals)**

**There are several 50-cent** words above that haven’t been defined. We will defer discussion of most of the sources of error until later, primarily because, at the moment, there is little or nothing you can do about them besides knowing of their existence and understanding how they affect your results. In the long run you can do a lot about these errors by a method called “differential correction.” But first we will consider an issue called “Dilution of Precision” (DOP) because:

**• High (poor)** DOP values can magnify the other errors,

• DOP values can be monitored during data collection and the data logger can mask out data with excessive DOP values,

**• DOP values,** which can be predicted considerably in advance for any given location, can perhaps be reduced by selecting appropriate times to collect data, and

• Differential correction cannot eradicate errors created by inappropriate DOP values.