PHYSICS UPDATE The global positioning system Alan J Walton and Richard J Black University of Cambridge,† Department of Physics,‡ Cavendish Laboratory, Madingley† Road, Cambridge CB3 0HE, UK University of Glasgow, Computing Science Department, 17‡ Lilybank Gardens, Glasgow G12 8RZ, UK A hand-held global positioning system receiver displays the operator’s latitude, longitude and velocity. Knowledge of GCSE-level physics will allow the basic principles of the system to be understood; knowledge of A-level physics will allow many important aspects of their implementation to be comprehended. A discussion of the system provides many simple numerical calculations relevant to school and first-year undergraduate syllabuses. For a little over 100 it is now possible to buy a hand-held global positioning system (GPS) receiver which displays the user’s current latitude, longitude, altitude and velocity (speed and compass bearing). Figure 1 shows one such receiver; its operation relies on receiving information from a set of satellites deployed by the US Department of Defense. Typically, the displayed horizontal position is accurate to 100 m and the altitude to 140 m; the system is inherently capable of greater precision, but the current US policy is to impose ‘selective availability’ (SA) which degrades its performance to these limits. Most hand-held receivers will have Figure 1. A typical hand-held GPS receiver (Garmin model 12). many additional features such as automatically displaying the route being taken and allowing waypoints to be entered. Particular models may be optimized for different applications such as hiking, On being shown a GPS receiver, students boating or flying. A GPS receiver and a mobile commonly ask ‘how does it work?’ The phone should perhaps be made compulsory safety purpose of this article is to show that the basic equipment for every school expedition. principles can be understood using only GCSE Phys. Educ. 34(1) January 1999 37 PHYSICS UPDATE Figure 2. A schematic (two-dimensional) diagram showing that a signal transmitted at time t on the satellite clock is received at a later time t + 1t on the receiver’s clock. physics. To understand how these principles satellite transmits a signal at, say, 42.0000 s on are implemented calls for A-level physics (and its clock and this signal is received 80 ms later at beyond). time 42.0800 s on the set’s clock, it follows that the receiver is at a distance (a range)ofc1t from the satellite, where c is the speed of light, i.e. at a 8 1 3 7 The basics range of 3 10 ms− 80 10− s 2.4 10 m. To simplify× the discussion× × a little,= × we will To begin with, we will consider a single satellite consider the two-dimensional case shown in in orbit around the Earth. On board is a clock figure 2. If we draw a circle of radius c1t which (for now) we will suppose is identical to— about the satellite’s position when it transmitted and synchronized with—a clock in the receiver. the signal, then we know that the receiver must By this we mean that if the two clocks are placed be somewhere on this circle. To find out where next to one another they will forever keep identical we must perform a similar timing measurement time. As the satellite orbits the Earth, it transmits using a second satellite as shown in figure 3. its current time and position. In practice the The receiver is clearly located where the two position is transmitted in the form of the current circles intercept (there are, of course, two such ‘ephemeris constants’; these specify the equation points, but the one ‘up in the sky’ can be readily of motion of the satellite and allow its current discounted). Rather than deducing the receiver position to be deduced from its current time. To position by means of a compass and scale diagram, put it simply, the satellite is a speaking clock we can (see figure 3) use Pythagoras’ theorem which says ‘My current time is t and my current to interrelate the signalling positions (x1,y1) and position is x,y,z’. This message is received by (x2,y2)of the two satellites, their measured ranges c1t and c1t and the receiver position (X, Y ). a (stationary) GPS set located at some unknown 1 2 This immediately gives position X, Y, Z not at time t—and this is the key 2 2 2 point—but at a later time t 1t. Thus if the (x1 X) (y1 Y) (c1t1) + − + − = 38 Phys. Educ. 34(1) January 1999 PHYSICS UPDATE Figure 3. A schematic (two-dimensional) diagram showing that the transit times 1t 1 and 1t 2 of the signals received from two satellites at known positions (x1, y1) and (x2, y2), respectively, will allow the set’s location to be determined from the intersection of the circles of radii c1t 1 and c1t 2, respectively. 2 2 2 (x2 X) (y2 Y) (c1t2) . on the receiver. In practice the satellites are − + − = These simultaneous equations with their two equipped with highly-stable rubidium and caesium unknowns X and Y can be solved numerically to (‘atomic’) clocks while the receiver makes do with find (X, Y ), the set’s position. a less expensive (and lighter!) quartz clock. This It is not difficult to show that three- means that the clock on the receiver may run fast dimensional navigation demands a minimum of or slow. If fast by δt (this is called the clock bias three satellites (again assuming perfect synchro- error), we must subtract δt from each of the so- nization between the satellite and receiver clocks). called pseudo transit times 1t1, 1t2 and 1t3 to If their signalling positions are (x1,y1,z1), obtain the true transit times required in equations (x2,y2,z2) and (x3,y3,z3) and the measured sig- (1), (2) and (3). This gives nal transit times are 1t1, 1t2 and 1t3, respec- 2 2 2 2 2 tively, then (x1 X) (y1 Y) (z1 Z) c (1t1 δt) − + − + − = − (4) 2 2 2 2 (x1 X) (y1 Y) (z1 Z) (c1t1) (1) 2 2 2 2 2 − + − + − = (x2 X) (y2 Y) (z2 Z) c (1t2 δt) 2 2 2 2 − + − + − = − (x2 X) (y2 Y) (z2 Z) (c1t2) (2) (5) − + − + − = 2 2 2 2 2 2 2 2 2 (x3 X) (y3 Y) (z3 Z) c (1t3 δt) . (x3 X) (y3 Y) (z3 Z) (c1t3) (3) − + − + − = − + − + − = − (6) which can be solved numerically to give the Since δt is unknown there are now four unknowns receiver’s position (X,Y,Z). and only three equations. The problem is resolved by utilizing a fourth satellite at (x4,y4,z4) with a Clock synchronization problems pseudo transit time 1t4, giving 2 2 2 2 2 So far the discussion has assumed that there are (x4 X) (y4 Y) (z4 Z) c (1t4 δt) . identical synchronized clocks on the satellites and − + − + − = − (7) Phys. Educ. 34(1) January 1999 39 PHYSICS UPDATE Solving equations (4) to (7) numerically gives run slow’) and for the (opposing) time dilation the receiver position (X,Y,Z). Likewise, it takes attributable to the change in gravitational potential at least three (rather than two) satellites to give between the Earth’s surface and the satellite’s orbit a two-dimensional fix. When the number of as described by general relativity. Their combined satellites in view drops from four to three it is effect, which yields a net increase in clock speed, the altitude which is sacrificed in a receiver (of is allowed for by offsetting the satellite’s clocks no consequence in marine receivers which usually prior to launch. The GPS provides a rare example suppress the ‘altitude’ anyway). It is implicit in of special and general relativity at work in the equations (4) to (7) that c is the same for the mechanical world. signals coming from all four satellites. This is not warranted, since the signals reaching the receiver The signal structure from satellites of different elevation will have to pass through different regions of the ionosphere As has been emphasized, the measurement of and troposphere. The resulting variation in c transit times lies at the heart of GPS. To is allowed for in the receiver in a mathematical measure a range to an accuracy of, say, 30 m model of the ionosphere and troposphere. therefore demands a timing accuracy of 30 m/(3 It is perhaps worth mentioning that the 8 1 7 × 10 ms−) 10− s, or better. The way principles behind GPS—but applied in reverse— that this is accomplished= is explained in figure 4. were in operation in the First World War. By All satellites transmit at the same frequency, but timing the arrival of the shock wave produced as a each has its own unique binary code known as shell left a gun barrel at three microphones located the coarse/acquisition (or C/A) code, consisting at known coordinates, it was possible to deduce the of a pseudo-random sequence of 0s and 1s. gun position (Bragg 1921). The three unknowns The C/A code is 1023 bits long and repeats here—which demand three microphones—are, of every millisecond. Thus each bit (or ‘chip’) has course, the gun two-dimensional coordinates and 3 7 a duration of 10− s/1023 9.77 10− s, its firing time. Using this technique, it was which is some ten times greater= than× the timing possible to locate a gun position to about 45 m. accuracy we would require for 30 m resolution. The receiver generates the same C/A code as that The satellites transmitted by the satellite, but because of the clock bias δt it will not be synchronized with that The American Navstar (Navigational Satellite being produced in the satellite (see figure 4).