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I am currently working on a GPS project, so I am doing some research about the subject. I understand that the GPS receiver knows its location by calculating its distances from 3 satellites, and by eliminating one of the two points of the intersection of the 3 spheres.

  • Why then do we need the fourth satellite?
  • How accurate will the position b without the fourth satellite?
  • What do "clock accuracy" and "clock bias" mean?
  • Does time synchronization between the receiver and the satellite mean that the receiver clock will get the time very precisely?

I've found that some receivers use this method described below to compute the time of flight between the satellite and the device. Does it mean we don't need the fourth satellite to correct the clock bias?

Coarse/Acquisition (C/A) Code

A pseudorandom noise code (PRN) modulated onto a L1 signal which helps the GPS receiver to compute the distance from each satellite. Specifically, the difference between the pseudorandom number code generated by the GPS rover software and the pseudorandom number code coming in from the satellite is used to quickly compute the distance to a satellite and therefore calculate your position.

AncientSwordRage
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Yahya Aouled Amer
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8 Answers8

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Why then do we need the fourth satellite ?

Because there is a fourth unknown: time. GPS works one-way, so the receiver's clock needs to be aligned with the senders' (the GPS satellites) clocks in order to compute the time-of-flight. Think about it this way: if the GPS satellite tells you "I sent this message at time X", and you receive it at time Y, you can compute time-of-flight only if your clocks are aligned. Typically, the clocks in GPS receivers is "bad", at least compared with the atomic clocks on the satellites, so the difference between the receiver clock and the GPS clock needs to be determined.

How much will be the position accuracy without the fourth satellite ?

It will be "useless", in te sense of (much) reduced accuracy. With 3 satellites you could do some attempt with the assumption that you are on a sphere, or using knowledge of where you were earlier, or using other inputs (e.g. altimeter), but it will not be as accurate.

What does it mean by "clock accuracy" and "clock bias" ?

See above.

Does time synchronization btw the receiver and the satellite mean that the receiver clock will get the time very precisely ?

Not "get", "compute". Basically the receiver determines a best guess at the correct time, "best" in the least-square-error sense.

Ludo
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    Does it mean that the atomic clocks are synchronized on all the satellites very precisely ? – Yahya Aouled Amer Feb 14 '22 at 12:12
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    and if I use an atomic clock on my receiver, does it mean I don't need a fourth satellite ? – Yahya Aouled Amer Feb 14 '22 at 12:12
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    @YahyaAouledAmer yes and maybe. Just an atomic clock doesn't help - it needs to be in sync with those on the GPS satellites also. You can probably do it with some effort, but not sure what the added value would be. – Ludo Feb 14 '22 at 12:25
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    I vaguely remember something from an early GPS system I had where if it could only find 3 satellites it could narrow your position down to one of two places on the surface of the planet within a few hundred yards, and couldn't determine altitude. So if you had a rough idea where you were already in order to eliminate the extra possibility and knew you were on the surface it would get you roughly in the right couple of football fields. – Perkins Feb 14 '22 at 21:06
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    @Ludo It's impossible to keep them in sync because time flows at different rates on the ground and in orbit. – J... Feb 14 '22 at 21:51
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    @J... but known different rates, surely? So one could account for those differences, and remain synchronised? – Tim Feb 14 '22 at 22:24
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    Real GPS receivers that I have seen will give a position with three sat fixes, by assuming you're near where you were before, and near the surface of the Earth. Current phones will have an altimeter and use "augments" to supplement the GPS. – JDługosz Feb 14 '22 at 23:07
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    @JDługosz Current phones can get the time component using 3G/4G networks. Augments is Ephemeris data + local atmospheric corrections. – Aron Feb 15 '22 at 02:57
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    the satellites clocks run slower on orbit than on the ground due to special relativity, and faster on orbit than on the ground due to general relativity. Both these effects (overall, the GR is about twice the opposite of SR, so they sum to about the negative of the SR term) are taken into account in designing and building the satellite clocks. the tricky part left over is the additional clock rate skewing from the Doppler shift, which depends on the projection of the satellite orbital velocity onto the line of sight from the receiver, which i describe in my answer. – Ryan C Feb 15 '22 at 05:21
  • @RyanC: Also, both the GR and SR terms vary based on where a satellite is in its orbit and on whether the satellite is in its "morning" orbit or its "evening" orbit (GPS satellites use semisynchronous orbits, making two orbits per sidereal day), as the earth's gravitational field is imperfectly-spherically-symmetric, causing the gravitational force felt by the satellite (and, thus, the GR term) to vary as it passes over higher-or-lower-gravity regions of the earth, which, in turn, causes variations in the satellite's altitude and orbital speed (and, thus, the SR term) which are... (1/2) – Vikki Feb 15 '22 at 08:19
  • @RyanC: (2/2) ...90 degrees out of phase with the variations in gravitational force (the variations in altitude themselves affect the gravitational force felt by the satellite [and, thus, the GR term], by altering its distance from the earth's center, which produces further knock-on effects on orbital shape and speed and the GR and SR terms). – Vikki Feb 15 '22 at 08:25
  • @JDługosz "current phones will have an altimeter" isn't strictly false because some do, but many don't. Plenty of dedicated GPS receivers do too, but they're not much help unless that's been calibrated. Phones usually use GPS altitude, but some apps use GPS coordinates to place you on a map, then get altitude from the map. – Chris H Feb 15 '22 at 13:56
  • @Aron if they've got a data signal, of course. The assumptions JDługosz mentions work without data, which is why you often have poor accuracy when you first turn on GPS in the middle of nowhere. If your interface displays that of course - my preferred phone app does (and I often use it in aeroplane mode to save battery) and when I use my handheld GPS it's an old Mk1 etrex – Chris H Feb 15 '22 at 13:58
  • I've heard that the 3 satellite version works very good if you know your altitude (trivial at sea). – Joshua Feb 15 '22 at 16:57
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    @ChrisH No. That isn't correct. 4G timing signals have nothing to do with "data". NTP isn't nearly accurate enough for GPS signals. NTP on a good ethernet connection can be accurate to the ms, which is to say, a few hundred kms. 4G timing signals need to coordinate user access to the spectrum, and hence is insanely accurate. – Aron Feb 16 '22 at 03:47
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    @Aron if that was true, network location would often be better than GPS, and it varies from worse to dreadful. It's also far too sensitive to distance from the towers for that to hold. Spectrum sharing doesn't need absolute timing at the 10s of ns you'd need for location, which is just as well as propagation in many places is far too variable for that. – Chris H Feb 16 '22 at 09:15
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Your GPS cannot directly determine the distance from any satellite, it has to go indirectly. It gets a signal from the first satellite, say "it was exactly 10:30:25.123456789 seconds according to my extremely precise clock when this signal was sent", and it gets a signal from the second satellite, say "it was exactly 10:30:25.123556789 seconds according to my extremely precise clock when this signal was sent". The clocks are 0.0001 seconds apart. So the signal from the first satellite travelled 0.0001 seconds longer. At 299,792,458 meter/sec, that is 29,979.2458 meters difference. So you are 29,979 meters closer to the second satellite than to the first. And your GPS also knows the exact location of the satellites.

With the third satellite, you also learn how much closer or further away you are to the third satellite compared to the first and the second. You can turn that into three rather complicated equations, and try to solve those equations, but there is not just one solution: There is a whole curve of solutions.

Now if three satellites is all you've got, your GPS can make a guess: It guesses that you are located on the surface of the earth. Your GPS has likely a map of roads in your area, but it also has a map of elevations. So it guesses first that you are at height zero and calculates where that curve intersects with the earth surface at height zero. That might be off a bit because you are in a hilly area, 1000 meters above sea level. But the GPS knows your approximate location, so it guesses you are about 1,000 meters above sea level, recalculates where you are, and that location might be 980 meters above sea level, and then the next calculation gives you your precise location. But only if you are on the surface. If you are at the top of a church tower, your location will be guessed wrong. If you are on an airplane, with a window seat so your GPS gets a signal, it will be quite imprecise, maybe kilometres off if you are 10,000 meters above ground.

With a fourth satellite, there are four ways to take three satellites and calculate the curve where you should be, so you get four curves. And then the GPS picks the point that it is closest to all four curves. That gives you your location quite precisely, and at the same time, if the curves don't meet exactly in one point but are maybe ten meters apart, then you also know the precision of your location.

(Some smartphones nowadays have a barometer. That could also be used to estimate your height above sea level, not very precise, because air pressure also depends on the weather, and help you get your location if you are high above ground. I don't think anyone does that. )

If you had a very precise atomic clock, you'd need only three satellites. But atomic clocks are big and expensive, so there isn't one in your mobile phone.

PS. If you think that it's kind of unfair that you need four satellites to get three coordinates, you are actually getting four. You also get the time with very high precision (maybe 100ns). Annoyingly no phone that I have seen uses this ability to set its clock. Actually, just a single satellite gives you the time with less than 100ms error: The fact alone that you can receive the satellite gives you your location with a ridiculous error of thousands and thousands of miles - but if you divide this error by the speed of light, then you get a better approximation for the time than your wristwatch will give you.

PPS. With four satellites your position is already overspecified. There are four curves, and with infinite precision they would intersect exactly at the point where the GPS is. But we don't have infinite precision, so we take the point that is closest to them. Five or six satellites would work exactly the same, except you have more curves.

gnasher729
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    I find this answer the easiest to understand, but have I got this right? 3 satellites give a curve in space that (hopefully) intersects the Earth in one (or two) places where the satellites are visible, so some combination of assumptions and barometric pressure can provide a likely position. The fourth satellite narrows the fix to a single point on that curve, thus producing an assumption-less fix? – uhoh Feb 14 '22 at 23:33
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    The problem with "three satellites plus surface of the Earth" is that it tends to produce large errors. If you're standing on a second-floor balcony, the equivalent "surface of the Earth" location might be hundreds of meters away. – Mark Feb 15 '22 at 00:49
  • @uhoh that's a bit too simple. each pair of satellites gives you a surface, with a thickness dependent on the uncertainty (take the pancakes from the bistatic radar post and stretch them into other shapes). due to measurement error, most likely there is NO point at which ALL the surfaces intersect, but you can solve for which point is the closest to all of them, averaged in a way that involves weighting by the estimated uncertainties. more detail coming in my answer the the OP. – Ryan C Feb 15 '22 at 05:32
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    @RyanC "most likely there is NO point at which ALL the surfaces intersect" I do not think that that applies to either three satellites plus the Earth surface + altitude solution or the four satellite without surface assumption solution that are being talked about here for the OP's question "Why does GPS need the fourth satellite?". I could be wrong, but I don't think those are over-specified scenarios. It's only when you have more information than that where the problem becomes over-specified and you have to start thinking about how to handle it. – uhoh Feb 15 '22 at 05:50
  • @uhoh no, that's not the issue. the problem is measurement error means you are trying to solve the wrong equations. if you could measure perfectly with zero error, then all the surfaces would coincide; but since the things you are intersecting are not quite the right surfaces, they in general do not all intersect anywhere. – Ryan C Feb 15 '22 at 06:20
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    @RyanC I certainly understand error, but when a problem is underspecified or exactly specified as discussed here, it is not the same thing as when the problem is over-specified. A GPS unit is working with three or four satellites as discussed here, it does not have to reconcile with non-intersecting surfaces. Certainly the final number on the screen that the user sees is a best fit taking account the thicknesses of the surfaces but the non-intersection you mentioned only happens when there is more information available (e.g. 4 satellites plus Earth surface assumption, or 5+ satellites etc. – uhoh Feb 15 '22 at 06:26
  • @RyanC I'm just talking about the phrase most likely there is NO point at which ALL the surfaces intersect not applying to the comment directly above it and "Why does GPS need the fourth satellite?" and not the general problem where you have 4 or more surfaces where non-intersection sets in. – uhoh Feb 15 '22 at 06:27
  • @uhoh with sufficiently large error, there can be cases where the three spheres do not intersect at all. – Ryan C Feb 15 '22 at 06:33
  • @RyanC If the error is of order of the radius of the Earth, then perhaps. But at that point you have a profound malfunction and worse things to worry about. But with three satellites and Earth's surface and errors of hundreds of meters (less than a microsecond, not thousands of kilometers) for each satellite's signal I don't believe there can be any possibility of non-intersection and just repeating there can be won't convince me. Can you construct some non-intersection scenario with real numbers? – uhoh Feb 15 '22 at 09:34
  • I didn't understand why you said "but there is not just one solution: There is a whole curve of solutions." The intersection of three spheres gives a two points not a curve, isn't ? https://upload.wikimedia.org/wikipedia/commons/thumb/c/c7/Sphere3-intersect.svg/1280px-Sphere3-intersect.svg.png – Yahya Aouled Amer Feb 15 '22 at 09:58
  • Don't assume the receiver has a map, or at least not one with precise ground altitudes. Most don't (it's not baked in to phones, for example, though apps may support it) – Chris H Feb 15 '22 at 14:05
  • Yahya, you don't have spheres (or other surfaces) to start with. You don't know the distance to any satellite, you know for example that you are 30km closer to satellite A than to satellite B. You could be 500km from A and 530km from B (one surface), or 510km from A and 540 km from B (another surface), and so on, and anything in between. You get one surface for every possible distance from A. – gnasher729 Feb 15 '22 at 23:29
  • @gnasher729, Re, "You don't have spheres (or other surfaces)..." Actually, there are relevant "other surfaces." They are the surfaces of constant time difference between any two satellite signals. They are hyperboloid surfaces. (see https://space.stackexchange.com/a/58294/11958). – Solomon Slow Nov 12 '22 at 23:50
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The receiver has no very precise time, the fourth satellite is needed to calculate a 3D position without knowing the precise time.

After knowing the position, the receiver may calculate the GPS signal delay between a satellite and the receiver. Using the delay the the precise time for the receiver may be calculated from the time of the satellite.

The receiver clock frequency may be a little bit too small or too large, that is the clock accuracy. The time of the receiver may be a little bit too early or too late, that is the clock bias.

Uwe
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  • If I use the time provided by the receiver clock, does it mean I will get the position but with very low accuracy, and if use the fourth satellite I will get good accuracy ? – Yahya Aouled Amer Feb 14 '22 at 12:08
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    @YahyaAouledAmer Generally yes, but most standalone receivers have at best mediocre clocks. With something like a smartphone with a solid network link you can generally have an accurate enough estimate of the ‘exact’ time to not need the fourth satellite though, but if you’ve got such a device you can also triangulate off the cell network to get a generic location which you can use to narrow things down with just three satellites (or fewer possibly), or even use an internal altimeter/barometer to provide that extra bit of data. – Austin Hemmelgarn Feb 15 '22 at 02:30
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    @YahyaAouledAmer: At the speed of light, even a microsecond difference is a 300 meter difference. That's why GPS requires 4 atomic clocks: they can achieve nanosecond accuracy. The smartphones bit in the comment above is mostly nonsense - smartphones use the mobile network to download the satellite orbits, so they can acquire the 4th signal faster. Even 5G doesn't have atomic clocks in the network. Knowing your 5G tower gets you the same ~300 meter error and therefore doesn't really help. – MSalters Feb 15 '22 at 12:43
  • what exactly does the satellite orbits information contain ? – Yahya Aouled Amer Feb 15 '22 at 12:54
  • @MSalters yes, I doubt mobile networks can provide a time accuracy much better than NTP, which is ~ms. But tower triangulation can be surprisingly good in ideal circumstances - or (as I found the other day) over a mile out; then I picked up the satellites and saw the dot jump to where I knew I was, and the error circle vanish – Chris H Feb 15 '22 at 14:02
  • @MSalters are you sure about that? This company specifically mentions cellphone systems as a use case for their Rubidium clocks. I guess they're probably only used as frequency references and not kept synchronised perfectly to be a good time reference, but I think it's not true that atomic clocks aren't used in cell networks https://freqelec.com/low_g_oscillators/ – llama Feb 15 '22 at 21:16
  • @llama GPS needs rubidium clocks with better stability than that used for cellphone systems. About 3E-13 per day, the clocks of your link 2E-11 per day. – Uwe Feb 15 '22 at 22:17
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GPS is much easier to understand in the one-dimensional case.

You are somewhere on the road between Springfield and Newtown. The road is straight and the two towns are six kilometers apart. Around noon, you hear the church bells of Springfield. Ten seconds later, you also hear the bells of Newtown.

  • Where exactly are you?
  • What exact time is it?

The speed of sound is 300m/s. You heard the bells of Springfield 10s earlier, so you must be 3km closer to Springfield than to Newtown. The distance is 6km in total, which means you are 1.5km from Springfield and 4.5km from Newtown. The bells rang at noon and sound needs 15s to cover the 4.5km, so it's now 12:00:15 when you hear the bells of Newtown.

In the example above, we needed two satellites to solve for two variables - position and time. If you hear only the bells of Springfield, you cannot determine your position at all. That is, unless you have an accurate clock to measure the time of arrival directly instead of calculating backwards from time difference and total distance. In the three-dimensional case, you need four satellites to solve for the three position variables and for time.

To answer your questions one by one:

Why then do we need the fourth satellite?

Because we cannot measure the time of arrival accurately, only the differences. We thus cannot use the three-spheres method. We have to work backwards from time differences and total distances, as in the example of Springfield and Newtown. This requires one extra satellite, but as a bonus it gives you time as a calculated variable.

How much will be the position accuracy without the fourth satellite?
What does it mean by "clock accuracy" and "clock bias"?

With the three-spheres method, positioning error is proportional to clock error of the receiver clock. The proportionality constant is the speed of light, which is 300m/μs. If the clock of your receiver device differs from global atomic time by merely 10μs, your position will be off by 3km. Good luck with that.

Does time synchronization between the receiver and the satellite mean that the receiver clock will get the time very precisely?

Yes, a GPS receiver is by far the most accurate clock you can buy. It receives its signals directly from atomic clocks aboard the satellites and corrects them for time of flight, using the known speed of light and receiver position. Traditional radio time standards like DCF77 or WWVB don't correct for time of flight, which makes them greatly inferior to GPS.

Rainer P.
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A GPS satellite sends out a constant stream of "I transmitted this at time x" messages. There are two ways of turning this into a location fix:

The three spheres method you describe requires a fourth parameter: time. If you have an atomic clock synchronized to the clocks of the GPS satellites, you can compute your distance from each satellite, find the intersection of three spheres, and determine your location.

Most GPS receivers don't include a synchronized atomic clock, though. In this case, you can use the time information to compute the difference in distances between two satellites. This tells you that you are somewhere on the surface of a hyperboloid of revolution; three satellites gives you three pairs, three hyperboloids, and, unfortunately, many intersection points. You need the fourth satellite to boost the number of pairs to six, giving you six hyperboloids and hopefully a single intersection point.

Mark
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  • Why is the shape of all the places that have some delta D of the distances to points A and B a hyperboloid? (I get the of revolution part) – Eugene Feb 15 '22 at 00:31
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    @Eugene, because that's the definition of a hyperbola/hyperboloid: the set of all points such that the difference between the distances to the two foci is constant. – Mark Feb 15 '22 at 00:42
  • I mean: given the shape of a hyperboloid, where would the point directly on the line between A and B be? – Eugene Feb 15 '22 at 00:44
  • @Eugene, there are two points directly on the line from A to B: one close to A, and one close to B. – Mark Feb 15 '22 at 00:46
  • Ah I see, wrote the second comment before seeing your reply, sorry. – Eugene Feb 15 '22 at 00:53
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    I've been thinking about this, and I think there's something missing. You don't have just the delta, you have accurate readings of A and B, which are synchronized with atomic clocks, then you can tell which "half" of the hyperbola you're on: if A<B, you are on the one that's closer to A and visa versa. Three of these half hyperboloids have only a single intersect. – Eugene Feb 15 '22 at 08:09
  • I didn't exactly why we are talking here about hyperboloids, and in other documents about spheres intersection? Does it mean that there are two methods of computing the position, one by using spheres calculation and one by using hyperboloids from differences between satellites transmitting time ? if you have drawings or illustrations it will be fantastic – Yahya Aouled Amer Feb 15 '22 at 10:08
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    @YahyaAouledAmer It's easier to grok in 2D 1st. If you have an accurate local time, then you can measure the "time of flight" and therefore the exact distance to a satellite A, the shape of where you could be with relation to A is a circle. If you don't have an accurate local time, but you can measure the difference between A and B times, the shape of where you could be with relation to A and B is a hyperbola. If you rotate those shapes to get a 3d object, you get a sphere and a hyperboloid respectively. – Eugene Feb 15 '22 at 18:02
  • to get an "accurate local time" I need a fourth satellite, so it means I don't need that satellite if I will use the difference between two satellites ? – Yahya Aouled Amer Feb 16 '22 at 08:08
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    @YahyaAouledAmer There are no spheres with GPS. Anyone talking about spheres in the context of GPS is oversimplifying it in an attempt to explain to laypeople who have never heard of a hyperboloid but can imagine a sphere. – TooTea Feb 16 '22 at 09:45
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The "spheres" illustration in regard to GPS functioning is simple, but slightly misleading.

One could get meaningful spheres if the time when the clock pulses were transmitted by the satellites was exactly known in the frame of reference of the receiver .

This time is not known with any meaningful precision because of number of reasons, including, but not limited to, relativistic differences between the clock rates and the limited long-term precision of the receiver's clock.

In reality, one gets a rotational hyperboloid of possible positions from the time difference between two satellites and a complex curve of equally possible positions from 3 satellites. When you have 4 satellites, you get a point.

In the case of 3 satellites, the curve in question may happen to be near-vertical near the Earth surface and, if lucky, to cross the surface in only one place near the satellites you hear. In this case the receiver may assume you are sufficiently near the surface and give you a position estimate based on this assumption (with a gross estimated inaccuracy of hundreds of meters or even kilometers).

A barometric altimeter, if properly calibrated, may help in this case by providing a better estimate for your altitude than the basic assumption that you are near the surface.

fraxinus
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I remember working on software that tracked the GPS satellite orbits and could therefore predict the accuracy of a fix at a given location in the future based on the position of the satellites at that time. In short: You need at least to receive signals from at least 3 satellites to get a location fix. With 4 satellites, you can tell if one satellite was giving out incorrect information such as it's location, as the accuracy of your fix would be compromised. Some more expensive GPS receivers can display the precision of your location based on data received. With 5+ satellites you can determine which data is inaccurate and safely ignore it. Obviously being to receive data from more satellites greatly improves the accuracy of your location.

user3177081
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    You've got to be remembering wrong. Three satellites alone can't give you a location fix. You need three satellites plus time, or four satellites. And virtually every GPS receiver will give you an accuracy estimate -- it's not just a feature of the expensive ones. – Mark Feb 15 '22 at 04:09
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    This type of software is well-known in the aviation community, where it's called RAIM Prediction. Your numbers are incorrect for pure GPS: 5 satellites are required to detect a fault and 6 are required to determine which satellite to exclude. However if the receiver can use barometric or other augmentation, then you are correct with 4 and 5. – Steve V. Feb 15 '22 at 06:29
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All other answer contain a lot of fluff and unrelated tangentials, unnecessarily complicating the subject. Meanwhile, the actual "meat" needed to address your question is quite simple.

GPS works based on time dilation, which means that accurate time measurements are absolutely essential. The problem is, the clock installed in most GPS devices is worth less than 1$, while the clock installed on the fourth satellite is worth millions of dollars. The clock on the fourth satellite is used to periodically synchronize the cheap clock in GPS device to overcome this problem.

user46610
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    If we can synchronize the receiver clock to the satellite clock, does it mean we can can get our position to a centimeter accuracy? Then why in my smartphone I get an accuracy of 5-10 meters ? – Yahya Aouled Amer Feb 15 '22 at 13:50
  • and without the fourth satellite, does it mean that we will get inaccuracy of kilometers ?? – Yahya Aouled Amer Feb 15 '22 at 13:56
  • @YahyaAouledAmer Even with perfect synchronization, you still cannot get a centimeter accuracy because the signal gets distorted a bit by reflecting off buildings, terrain, cars, etc. and also, the signal could get distorted before it gets to Earth by solar flares, etc. Without fourth satellite, the cheap clock in your device would quickly drift away and the inaccuracy will keep decreasing and decreasing. And yes, you would get inaccuracy of kilometers, then as the cheap clock drifts away more it will be tens of kilometers, hundreds of kilometers, etc. and the GPS reading would get useless. – user46610 Feb 15 '22 at 20:57
  • You can't "synchronize to the fourth satellite" because you don't know how far away it is. To overcome this, GPS does a combined "position and time" calculation that requires at least four satellites, and uses all of them equally. – Mark Feb 15 '22 at 23:00
  • user46610, time dilation doesn't matter (except we have to take it into account obviously). We start with no knowledge. When we receive the first signal, that gives us the time within 100ms. Knowing the time within 100ms, we can figure out where the satellites are within a few kilometres. With that, we get a quite precise time, maybe within microseconds. That gives us the satellite locations within metres, and that finally gives us the exact time and the exact location of the satellites. There is no need for a clock in the GPS receiver at all. – gnasher729 Feb 16 '22 at 00:01
  • @gnasher729 It looks like I was misled by a pop-science channel material about why that 4th satellite is needed, and I believed that the operation of GPS was a lot simpler than it really is. Where do I click to delete my answer? – user46610 Feb 16 '22 at 20:46