260
`
`MATHEMATICS MAGAZINE
`
`Global Positioning System:
`The Mathematics of GPS Receivers
`RICHARD B. THOMPSON
`University of Arizona
`Tucson, AZ 85721
`
`Introduction
`
`GPS satellite navigation, with small hand-held receivers, is widely used by military
`units, surveyors, sailors, utility companies, hikers, and pilots. Such units are even
`available in many rental cars. We will consider the mathematical aspects of three
`questions concerning satellite navigation.
`
`How does a GPS receiver use satellite information to determine our position?
`
`Why does the determined position change with each new computation, even though
`we are not moving?
`
`What is done to improve the accuracy of these varying positions?
`
`We will see that receivers use very simple mathematics, but that they use it in
`highly ingenious ways.
`Being able to locate our position on the surface of the earth has always been
`important for commercial, scientific, and military reasons. The development of
`navigational methods has provided many mathematical challenges, which have been
`met and overcome by some of the best mathematicians of all time.
`Navigation by means of celestial observation, spherical trigonometry, and hand
`computation had almost reached its present form by the time of Captain James Cook's
`1779 voyage to the Hawaiian Islands. For the next 150 years these methods were used
`to determine our location on land or sea. In the 1940s electronic navigation began
`with the use of fixed, land-based, radio transmitters. The present-day LOng RAnge
`Navigation (LORAN-C) system uses sequenced chains of such transmitters.
`The use of satellites in navigation became common in the 1970s, with the introduc-
`tion of the Navy Navigation Satellite System (NAVSAT or TRANSIT). This system
`uses the Doppler shift in radio frequencies to determine lines of position and
`locations.
`
`The Satellites
`
`Almost all satellite navigation now uses the Global Positioning System (GPS). This
`system, operated by the United States Department of Defense, was developed in the
`1980s and became fully operational in 1995. The system uses a constellation of
`satellites transmitting on radio frequencies, 1227.60 mHz and 1575.42 mHz.
`The original design of the system provided for eighteen satellites, with three
`satellites in each of six orbits. Currently, there are four satellites in each orbit. In the
`basic plan, the six orbits are evenly spaced every 600 around the Earth, in planes that
`
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`261
`
`FIGURE 1
`The System of Satellites.
`
`are inclined at 550 from the Equator. Orbits are circular, at a rather high altitude of
`20,200 kilometers above the surface of the Earth, with periods of twelve hours.
`FIGURE 1 displays one configuration of the basic eighteen satellites. Although not
`drawn to scale, it gives the correct feeling that we are living inside a cage of orbiting
`satellites, several of which are "visible" from any point on the surface of the Earth at
`any given time.
`
`Receivers
`
`Current GPS receivers are electronic marvels. They are hand-held, run on small
`batteries, weigh as little as nine ounces, and can cost under $150. We can turn on a
`receiver at any point on or above the surface of the Earth and, within a few minutes,
`see a display showing our latitude, longitude, and altitude. The indicated surface
`position is usually accurate to within 100 meters, and the altitude is usually in error by
`no more than 160 meters.
`How does a small radio receiver listen to a group of satellites, and then compute our
`position, with great accuracy? We start by noting exactly what sort of information is
`received from the satellites. Each satellite sends signals, on both of its frequencies,
`giving (i) its position and (ii) the exact times at which the signals were transmitted.
`The receiver also picks up time signals from the satellites, and uses them to
`maintain its own clock. When a signal comes in from a satellite, the receiver records
`the difference, At, in the time at which the signal was transmitted and the time at
`which it was received. If the Earth had no atmosphere, the receiver could use the
`speed, c, of radio waves in a vacuum to compute our distance d = c At from the
`known position of the satellite. This information would suffice to show that we are
`located at some point on a huge sphere of radius d, centered at the point from which
`the satellite transmitted. However, the layer of gasses surrounding the Earth slows
`down radio waves and, therefore, distorts the measurement of distance. Receivers can
`partially correct for this by allowing for the effect of mean atmospheric density and
`thickness.
`Information
`from several satellites
`the
`is combined
`to give
`longitude, and altitude-of our position in any selected refer-
`coordinates-latitude,
`ence system.
`Several factors restrict the accuracy of this process, including: (i) errors in the
`determined positions of the satellites; (ii) poor satellite positioning; (iii) limitations on
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`the precision with which times and distances can be measured; and (iv) the vaxying
`density of Earth's atmosphere and the angles at which the radio signals pass through
`the atmosphere. Some of these difficulties are overcome by the use of an ingenious
`plan that provides the key to GPS technology. It is rather complicated to explore this
`method in the actual setting of positioning on the Earth: The distances are large, the
`time differences are small, and the geometry is all in three dimensions. Fortunately,
`we can capture most of the salient features of GPS receiver operation in a simple
`two-dimensional model.
`
`A Simple Model
`
`Suppose that you are standing somewhere in a circular lot, with a radius of 100 ft. The
`lot is paved, except for an irregularly-shaped gravel plot that surrounds you. The mean
`distance from your position to the edge of the gravel is 20 ft. Cars circle the lot on a
`road. To determine your position, messengers leave from cars on the road and walk
`straight toward you. When such a messenger arrives, he tells you where and at what
`time he left the road. You have a watch and know that all messengers walk at a rate of
`5 ft/sec on pavement but slow down to 4 ft/sec on gravel. Our model is shown in
`FiGURE 2.
`
`|
`
`\
`
`f C~~~~ravrel
`]
`
`Pavement
`
`10 f
`
`FIGURE 2
`The Model.
`
`Consider a rectangular coordinate system with its origin at the center of the lot.
`Distances will be measured to tenths of a foot, and time will be measured to tenths of
`a second. The location of a point on the road will be described by its angular distance
`from due north, measured in a clockwise direction.
`At noon a messenger leaves a position 45? from north. When he arrives, your watch
`shows that it is 20.2 seconds after noon. Since you have no way to know the exact
`distance that he walked on the gravel, you assume that he covered the mean distance
`of 20 ft. At 4 ft/sec, this took him 5 sec. For the remaining 15.2 sec he walked on
`pavement, covering 5 ft x 15.2 sec = 76.0 ft. Allowing for the assumed distance of 20
`ft on the gravel, you know that you are located at some point on a circle of radius 96.0
`ft, centered at the starting location of the messenger.
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`263
`A second messenger leaves the road at a point 1350 from north at 12:01 pm and
`walks to your position. On his arrival, your watch shows that it is 29.5 sec after he
`started. Assuming that he took 5 sec to walk 20 ft on the gravel, he walked 5 ft/sec
`24.5 sec = 122.5 ft on the pavement. Hence, you are on a circle of radius 142.5 ft,
`centered at this messenger's point of departure.
`The coordinates of the departure points for the two messengers are P1 = (100
`sin 450, 100 i cos 450) and P2 = (100 sin 1350, 100 cos 1350), respectively. Using our
`precision of one tenth of a foot, these are rounded to (70.7 70.7) and (70.7,- 70.7).
`Thus, the coordinates (x0, yo), of your position satisfy
`
`{ (xO - 70.7)2 + (yo - 70.7)2
`142.52J
`70.7)2
`(xo - 70.7)2 +(yo+
`The system has two solutions, ( - 20.0, 39.2) and (161.4,39.2), rounded to tenths.
`Since the latter point is outside of the lot, you can conclude that you are located 20.0
`ft west and 39.2 ft north of the center of the lot. The situation is shown in FIGURE 3.
`
`96.02
`
`PI
`
`(xo, yo)
`
`pi
`
`P.,
`
`P3
`
`FIGURE 3
`Twvo Messengers.
`
`FIGURE 4
`Three Messengers.
`
`So far so good. Suppose that, just to be careful, you decide to check your position
`by having a third messenger leave the road at a point 1800 from north and walk to
`your location. He leaves at 12:02 pm and, according to your watch, arrives 32.2 sec
`later. As before, you compute your distance from this departure point P3. FIGURE 4
`shows the result of adding information from the third messenger to your picture.
`What has happened? The mnost likely problem is that your watch does not agree
`with the times used at the departure points on the road. Suppose that your watch runs
`steadily but has a fixed error of e seconds, where a positive e means that your watch
`is ahead of the road times and a negative e means that your watch is behind the road
`times. If we let At be the time difference between departure and arrival, as shown on
`your watch, then the estimate for the distance traveled is
`d(At, e) = 20 ft + (At sec - e sec - 5 sec)5 ft
`sec
`Thus, the radius of each circle is in error by the same amount, -5 e ft, and there
`must be a value of 8 for which the three circles have a common point. FIGURE 5 shows
`the effect of various watch errors.
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`P.
`e= -1lsec
`
`2
`
`P
`e-1sec
`
`p2
`
`P3
`E= 3 sec
`
`PI
`
`P3
`
`E= 4 sec
`
`P. p.,
`E=5sec
`
`2
`
`p(
`
`P*3
`E 6sec
`
`P
`/2
`E= 7sec
`
`p
`
`p2
`
`E=9sec
`
`FIGURE 5
`Effect of Watch Error.
`
`It appears that your watch has an error of approximately 5 sec. The error and the
`coordinates of your position are a solution for the following system of equations:
`
`( xo - 70.7)2 + ( y0 - 70.7)2 = d(20.2, 8)2
`- 70.7)2+ ( y0 + 70.7)2 = d(29.5, 8)2
`(x
`)
`( xo - 0.0)2 + ( yo + 100.0)2 = d(32.2, )2 )
`
`The system can be solved numerically, starting with seed values of 0 for e and
`estimated coordinates of your position for x0 and yo. There is only one solution giving
`a location inside of our lot. Rounding this to our level of precision yields (x0, yo, 8) =
`(10.9,31.2,4.9). You conclude that you are 10.9 ft east and 31.2 ft north of the center
`of the lot, and that your watch is 4.9 sec fast. You note the coordinates of your
`position, and discard the watch error, which is of no further interest to you.
`As this example of our GPS model shows, you can use time difference information
`from three messengers to determine your position, relative to a coordinate system in
`the lot. The only tools needed for this effort are a steady, but not necessarily accurate,
`watch and the ability to approximate the solution of a system of three equations in
`three unknowns.
`
`Back to the Satellites
`
`Our "lot" is now the region inside of the satellite orbits (including the Earth), "cars on
`the road" are satellites, "messengers" are radio waves, and "gravel" is the Earth's
`atmosphere. We take the center of the Earth as the origin in our coordinate system.
`Working in three dimensions, we need information from four satellites. Call these S1,
`S2, S3, and S4; and suppose that Si is located at (Xi, Yi, Zi) when it transmits a signal
`at time Ti. If the signals are received at times Ti', according to the clock in our
`receiver, we let A ti = Ti' - Ti, and let e represent any error in our clock's time. The
`receiver allows for the mean effects of passage through the Earth's atmosphere and
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`265
`computes distances d(Ati, 8) that indicate how far we are from each of the satellites.
`Our position (x0, yo, z0) is located on each of four huge spheres. In most situations,
`there will be only one sensible value of e that allows the spheres to have a point in
`common. Our location is determined by solving a system equations.
`
`(xo -
`XI2
`yX2
`(
`
`+ ( yo
`((zo
`
`+
`
`)
`
`Z2
`
`= d(Atl,
`
`8)2
`
`_
`
`( XO - X2 )2 + ( Yo
`y2)2 + ( Zo - Z2 )2 = d( At2, 8)2
`l( xo - X3 )2 + ( YCO - Y3 )2 + ( ZO _ Z3 )2 = d( At3, 8)2
`-X4) 2 + ( yO - Y4)2 + ( zO - Z4) = d(At4, 8
`J
`(x
`
`When a numerical solution is found, the rectangular coordinates (x0, Yo' Z0) are
`converted into the essentially spherical coordinates of latitude, longitude, and altitude
`above sea level.
`As a practical matter, there are times and locations when a GPS receiver can
`receive usable data from only three satellites. In such cases, a position at sea level canl
`still be found. The receiver simply substitutes the surface of the Earth for the missing
`fourth sphere.
`To summarize our results so far, the receiver is expected to (i) receive time and
`position information from the satellites, (ii) maintain a steady (but not necessarily
`accurate clock), (iii) select four satellites with a good range of positions, (iv) find an
`approximate numerical solution for a system of four equations, and (v) make a
`transformation of coordinates. Given the current state of electronics, these are easy
`tasks for a small hand-held instrument.
`
`Variability of Positions
`
`Our second question about GPS positioning causes a great deal of discussion and
`confusion among those who use the system. If a person stands in one fixed location
`and determines repeated positions with a receiver, the coordinates of these positions
`tvill vary over time. Since the observer's location has not changed, the changing
`positions are often attributed to alteration of the satellite signals by the Department of
`Defense. The Department does, at times, degrade the satellite data and cause a loss of
`GPS accuracy. This Selective Availability (SA) will be phased out within the next few
`years. (It is stated that SA is used for reasons of national security.) However,
`manipulation of the signals explains very little of the variation in positioning. The
`variation is primarily caused by random errors in measurement, the selection of
`different satellites, and by the effects of the atmosphere. We will illustrate these
`problems by returning to our simple 2-dimensional model.
`In our example, you determined that the coordinates of your position were
`(10.9,31.2) and that your watch was 4.9 seconds fast. Suppose that these values are
`exactl? correct. After a couple of minutes you again use three messengers to
`determine your location. This time the first messenger leaves from the road at a point
`that is 47.2? from north. Rounding to your level of precision, you record the departure
`point as P1 = (73.4,67.9). In this case we will assume that your information on the
`location of the departure point is not quite correct, and that the messenger actually
`left from Q, = (74.1340,67.3568). This is only a 1.0% error in the first coordinate and
`a 0.8% error in the second coordinate. Suppose also that the messenger actually
`encountered 25.9 ft of gravel on his way to your position.
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`Keeping track of 6 places, the distance from Q( to your location is 72.841286 ft.
`Covering the 25.9 ft of gravel at 4 ft/sec took the messenger 6.475000 sec, and
`covering the 46.941286 ft of pavement at 5 ft/sec took 9.388257 sec. The actual
`walking time was 15.863287 sec, which with your watch error of 4.9 sec, is 20.763257
`sec. Using the allowed one place of precision, you would note At = 20.8 sec. Recall
`that, as you stand in the lot, you have no way of knowing the amount of gravel over
`which a messenger has walked. Hence, you always assume the mean distance of 20 ft.
`Under this assumption, the messenger would take 5 sec to cover the gravel, leaving
`15.8 sec to walk on the pavement. At 5 ft/sec he would cover 79.0 ft. You conclude
`that the messenger has traveled 99.0 ft, and that you are at that distance from P1.
`To find your position, messengers leave from the road at points 138.5? and 8.10
`from north. You record these departure points as P2 = (66.3, - 74.9) and P3 =
`(14.1,99.0). Now suppose that your information is slightly incorrect, and that the
`departure points are actually Q2 = (66.8404, - 75.6490) and Q3 = (13,9731,98.0100).
`In addition, assume that the second messenger walked over 22.1 ft of gravel and that
`the third messenger walked over 12.0 ft of gravel. Working in the same way as you did
`for the first messenger, you record time differences of 30.1 sec and 18.9 sec for the
`second and third messengers, and solve the following system of equations.
`
`(xo - 73.4)2 + ( yo - 67.9)2 = d(20.8, 8)2
`( xo - 66.3)2 + ( yo + 74.9)2 = d(30.1, 8)2
`(xO - 14.1)2 + ( yo _ 99.0)2 = d(18.9, 8)2)
`
`The solution, when rounded, gives your location as (xO, Yo) = (5.4,32.3) and your
`watch error as 4.4 sec. Small errors in the location of the departure points, variation in
`the amount of gravel covered, and the rounding of numbers to one-place have
`produced a "position" that is 5.61 ft from your actual location of (10.9,31.2).
`We can let a computer simulate what happens if you stay in your fixed location and
`make repeated computations of your position. Each determination of a position is
`made with the following assumptions. (i) Three points of departure for messengers are
`picked at random, assuming that the angle between any two points of departure is at
`least 30?, but not more than 150?. (ii) The distance over which a messenger must walk
`on gravel is a normal random variable with a mean of 20 ft and a standard deviation of
`5 ft. (iii) The relative error in each coordinate of the point of departure is a normal
`random variable, with a mean of 0 and a standard deviation of 0.3%.
`It is common to discuss accuracy of positioning in terms of circular errors of
`probability (c.e.p.). The n% c.e.p. is the distance, d 1, such that the probability of an
`error that is less than or equal to dn is n%. A set of 1,000 simulated positions allowed
`us to estimate c.e.p.'s for our model. We found d50 = 2.11 ft and d95 = 7.69 ft. (It is
`interesting to note that one simulated position was 16.9 ft from the correct location.)
`The positions computed in a run of 50 simulations are plotted on the left side of
`FIGURE 6, along with circles of radii d50 and d95. Our probabilistic model yields
`results that agree quite well with plots of successive positions found with an actual
`GPS receiverfrom a fixed location.
`Commercially available GPS units operate under what is called the Standard
`Positioning Service (SPS), measuring distances using satellites' 1575.42 mHz fre-
`quency. Under the best circumstances, the 50% c.e.p. for the SPS is 40 meters. As we
`mentioned, selective availability adds a small amount of random error into the SPS. At
`almost all times the 50% c.e.p. is no more than 100 meters, with a common value
`being around 50 meters. Under these conditions, the 95% c.e.p. for SPS is
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`267
`
`i
`
`\
`
`x
`
`/
`
`Double
`Messengers
`
`Single
`Messengers 0 2 4 6 8 10
`Feet
`FIGURE 6
`Simulations.
`
`approximately 100 meters. As in our model, almost all of the variability in SPS
`positions comes from random errors that are inherent in the components of the
`system.
`
`PPS and Differential GPS
`
`What is done to remove somne of the random errorsfrom GPS positions? At the present
`time, there are two common methods of improving GPS accuracy. One of these is the
`Precise Positioning Service (PPS), which is available for governmental use only. This
`uses signals transmitted on both of the GPS frequencies to eliminate much of the
`variability caused by the Earth's atmosphere. Just as with various colors of light in
`the visible spectrum, the reduction in the speed of a radio wave as it passes through
`the atmosphere depends upon its frequency. Hence, measurements of the arrival
`times of two signals of different frequencies can be used to greatly improve the
`accuracy of our distance estimates.
`As before, the situation is most easily understood in terms of our simple two-dimen-
`sional model. To model the PPS we will suppose that each messenger is accompanied
`by an assistant, who also walks at 5 ft/sec over pavement. However, while the
`messenger walks at 4 ft/sec over gravel, the assistant is slowed to 3 ft/sec when
`walking on gravel. We will return to our first example of the variability of positions
`and see what improvement in accuracy results from knowledge gained with assistant
`messengers.
`Recall that your location in the lot has coordinates (10.9, 31.2), and that your watch
`error is 4.9 sec. The first messenger departed from Q, = (74.1340,67.3568) and
`walked over 25.9 ft of gravel while covering the 72.841286 ft to your position. With
`your watch error, you recorded a time difference of At = 20.8 sec.
`The assistant messenger will require 8.633333 sec to cover the 25.9 ft of gravel at 3
`ft/sec and 9.388257 sec to cover the paved part of the route at 5 ft/sec. Hence, his
`total walking time will be 18.021591 sec. Due to the error in your watch and the
`allowed level of precision, you record an elapsed time of As= 22.9 sec for the
`assistant messenger. The time differences for the messenger and the assistant messen-
`ger give you enough information to estimate the amount of gravel that lies between
`you and the point of departure on the road, and to estimate the total distance from
`your location to the point of departure. If we let G be the number of feet of gravel
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`and let D be the total distance, in feet, then we have the following system of linear
`equations:
`
`{20.8= D-G + G
`4
`5
`
`22.9= D-G
`5
`
`+ G
`3
`
`Solving the system at your level of precision, you conclude that the messenger and
`his assistant crossed 25.2 ft of gravel, and came 97.7 ft from their point of departure.
`Thus, as best you can tell, you are at some point on a circle of radius 97.7 ft, centered
`at the nominal point of departure P1 = (73.4,67.9).
`Similarly, suppose the second and third messengers also have assistants. Computa-
`tions similar to those above show that the second messenger traveled 144.8 ft,
`including 22.8 ft over gravel, and that the third messenger traveled 91.5 ft, including
`12.0 ft over gravel. (These values include any possible watch error.) Your estimate for
`the actual distance that a messenger has traveled is now a function of the distance, G,
`of gravel covered; the time difference, At; and your watch error, 8.
`
`d(G, At, 8) = G ft + At sec- 8sec-
`
`G ft
`ft
`
`ft
`5 sec
`
`This distance formula and the three points of departure lead to a system of
`equations whose solution (x0, yo, s) gives an estimate of the coordinates for your
`position and for the error of your watch.
`
`t (x0 - 73.4)2 + ( yo - 67.9)2= d(25.2, 20.8, 8)2
`( x0 - 66.3)2 + ( y0 + 74.9)2 = d(22.8, 30.1, 8)2
`( x0 - 14.1)2 + ( yo
`
`_ 99.0)2 = d(12.0, 18.9, 8)2)
`
`.
`
`Solving this system, at your level of precision, yields a position of (x0, yo)=
`(9.2, 31.6) and a watch error of 4.8 sec. Your current estimate is only 1.75 ft from your
`correct location of (10.9,31.2). This compares with an error of 5.61 ft found by using
`single messengers.
`A computer-generated set of 1,000 simulations for positions computed with messen-
`gers and assistants gave estimates of 0.50 ft and 1.90 ft for the c.e.p.'s d50 and d95,
`respectively. The maximum distance of a computed position from the actual location
`was 3.41 ft. These simulations were based upon the same conditions that we used for
`our model of the SPS. The positions computed in a run of 50 simulations for our
`model of PPS are plotted on the right side of FIGURE 6, along with circles of radii d50
`and d95. Comparison of the two sides of FIGURE 6 shows that there is a considerable
`gain in accuracy when most of the variation due to distance walked over gravel is
`eliminated.
`In the real world of satellites and positions on the Earth, the use of two radio
`frequencies in the PPS produces considerably more accuracy than can be obtained
`with the single-frequency SPS. It is believed that the PPS has a 50% c.e.p. of
`approximately 16 meters.
`A second method for improving the accuracy of the usual SPS locations is coming
`into use at airports and major harbors. This is called the Differential Global Position-
`ing System (DGPS). Most of the error in a GPS position is due to random variables in
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`the atmosphere and the satellite system. Hence, within a small geographical area, the
`error at any instant tends to be independent of the exact location of the receiver.
`DGPS exploits this situation by establishing a fixed base station, whose exact location
`is already known. Equipment at the base station computes its current "GPS position,"
`compares this with its known location, and continuously broadcasts a correction term.
`A DGPS receiver in the area receives its own satellite information and computes its
`position. Simultaneously it receives the current correction from its base station, and
`applies this to its computed position. The result is a very accurate determination of the
`receiver's position; 50% c.e.p.'s for GDPS run close to 9 meters.
`
`Conclusions
`
`The very ingenious idea of leaving clock error as a variable allows a GPS receiver to
`display our position on the Earth at any location and at any time, using nothing more
`than simple algebra. The variations in computed positions are almost entirely due to
`inherent limitations on precision within the system. A second clever plan allows the
`use of two radio frequencies to eliminate much of the variability caused by the passage
`of signals through the Earth's atmosphere.
`
`REFERENCES
`
`1. Nathaniel Bowditch, American Practical Navigator, Defense Mapping Agency Hydrographic/Topo-
`graphic Center, 1984.
`2. GPS Global Positioning System, a supplemental program produced by the United States Power
`Squadrons, 1995.
`3. Product information, West Marine, Inc., 1997.
`4. United States Coast Guard navigation web site, http:
`
`//www.navcen.uscg.mil/
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