The Hyperbola depiction of how the GPS equations are solved:

From several psuedoranges measured at the same time, we can calculate the (absolute) difference in the distances between us and these GPS satellites. 2 satellites gives us 1 known difference in distance, and we can plot our position along the surface of a hyperboloid where all points at the surface have the same (absolute) diff in distance to the 2 satellites. From there we can add an other or several other additional satellites to make more hyperboloids which will create an intersection point at our (receiver) position. We will need 3 hyperboloids to make one point of intersection.

Why will 3 satellites give us only 2 hyperboloids as opposed to 3? (hyperboloids between satellites 1 + 2, 2 + 3, AND 1 + 3?)

And why will 4 satellites give us only 3 hyperboloids as opposed to 6? (hyperboloids between satellites 1 + 2, 2 + 3, AND 3 + 4 AND 1 + 3, 2 + 4, 1 + 4 ?)

Formula taken from one of the answers under: k satellites in general position do determine k(k-1)/2 hyperboloids

So, why do we need the 4th satellite, when it is not needed to make 3 hyperboloids?

  • 1
    Please note that your questions do not contradict anything in the quotation, which only asserts that (distance or time) signals from two satellites determine one hyperboloid. The quotation is counting hyperboloids, not satellites. So in fact k satellites in general position do determine k(k-1)/2 hyperboloids. – whuber Sep 23 '15 at 14:22
  • Yes, but then why would we need 4 satellites to determine 3 hyperboloids to intersect to a point? In your (correct) formula 4 sats gives 6 hyperboloids, and we would only need 3 sats to create 3 hyperboloids. Why do we need the 4th? ( Time error will be solved by solving the position and then knowing the exact difference between psuedorange and real range ) – wacgyver Sep 24 '15 at 4:16
  • Just thinking about Joon Ho Cho's answer, I suspect that means that if there are 3 satellites called A, B and C: 1) The constant TDOF (time difference of flight) surface between satellites A & B, A & C, and B & C give three hyperboloid of revolution surfaces. 2) Intersecting any two surfaces gives a line of infinite possible positions. 3) This is the one I'm not sure about: the intersection line sits on the third hyperboloid surface. So adding the third surface to the mix doesn't produce a unique intersection, it just confirms the line you already have. It doesn't add any new information. That – Rory Sep 25 '18 at 23:11

EDIT: I answered this question originally with the assumption that there was only one way of understanding the way GPS works. There's a good few paragraphs on Wikipedia: https://en.wikipedia.org/wiki/Global_Positioning_System#Geometric_interpretation

I don't understand why you can't have more hyperboloids. I wonder if there is no value at adding more hyperboloids, i.e., including more than 3 'differences' when you have 4 satellites doesn't increase the amount of information?

--old post follows

I thought having two satellites to measure your distance from would give you a circle (the intersection of two spheres. And that three satellites gives you two points, one near the Earth's surface and one up in space somewhere. I don't think that you do get hyperboloids so I am keen to see a reference.

Also, note that in order to measure distances, your receiver needs to solve for time too, so you need at least four satellites. (You also need the almanac that tells you where the satellite is at a particular time, and that's transmitted in the signal from the satellites, or used to be, which is why it took ages to get a first fix on old hardware).

This seems like a pretty good resource for reading up: https://www.maptoaster.com/maptoaster-topo-nz/articles/how-gps-works/how-gps-works.html


When you switch the GPS on, the time-to-first-fix varies depending on how long it is since you last used the GPS. To get a fix, the GPS receiver needs a valid almanac, initial location, time, and ephemeris data.

The terms "cold/warm/hot start" indicate how many of these pieces of data the GPS receiver already has. The terms mean different things to different GPS manufacturers.

Cold start - if the GPS not been used for a long time and/or has moved several hundred kilometres it will take some time to get the first fix. In this state, the GPS receiver does not have a current almanac, ephemeris, initial position or time. Older GPS units may take up to an hour to search for satellites, download the almanac and ephemeris data and obtain an initial position, though newer GPS units may require much less than this.

If the GPS receiver has moved several hundred kilometres, its assumptions about which satellites to use will be incorrect and it will have to search for them. Most units will let you enter an approximate location to speed the process.

satellite positioning example from http://www.nptel.ac.in/courses/105102015/Flash/gps3.jpg

  • 1
    Three gives a location (intersection of 3 spheres) 4th and subsequent only refine the location / measure accuracy... obviously the solution that's not on/near the earths' surface is ignored. You would need to take that into account if you're using GPS to navigate a trans-atmospheric vessel (which is not something I'd plan on doing). – Michael Stimson Sep 23 '15 at 4:57
  • 2
    You don't get spheres, because there's no way to measure the distance to a satellite - all you can do is look at the difference in timestamps you're receiving from various satellites, and use that to work out which are nearer or further away (and by how much). If you compare the timestamps from satellites A and B, and that shows you you are X miles closer to A than B, then the set of points that are X miles closer to A than B forms a hyperboloid. – psmears Sep 23 '15 at 9:14
  • Hi @psmears, have you got a source? I was taught that you are measuring a distance from the satellite to you. At least in consumer GPS gear, it's a pseudo random code that's generated on your device and on the satellite at the same TIME, and matching them together you get the TIME delay between generating the code and receiving the code from the satellite, which with the speed of light, gives you a distance. Since it's distance not direction, it's a sphere. – Alex Leith Sep 23 '15 at 23:11
  • You can not measure distance due to inaccurate clock in the receiver. Due to the high speed of light you will be 300m out 1 second after correcting your clock to an atomic clock as exact as the satellites clocks. But since you measure several satellites range at the same time, the DIFF between the psuedo ranges will be real. So you only know the difference in range to several satellites, not the actual range from you to any satellite. Therefore the sphere model does not work well to explain gps. en.wikipedia.org/wiki/Pseudorange – wacgyver Sep 24 '15 at 4:06
  • 1
    Yes, but then why would we need 4 satellites to determine 3 hyperboloids to intersect to a point? If we would only need 3 sats to create 3 hyperboloids. Why do we need the 4th? ( Time error will be solved by solving the position and then knowing the exact difference between psuedorange and real range ) Put as a formula, k satellites can be used to determine k(k-1)/2 hyperboloids, i.e. 3 sats is 3(3-1) /2 = 3*2 /2 = 6/2 = 3 hyperboloids – wacgyver Sep 27 '15 at 8:07

3 Satellites indeed provide 3 equations, not for hyperboras but for hyperboloids. Whenever 2 hyperboloids intersect, there generates a curve. Thus, it appears that 3 satellites are enough that generate 2 curves that meet at a point. However, there are special cases where the 2 curves are identical. That is why we need one more satellite to resolve the uncertainty.

An example is as follows. Consider the case where 3 satellites are located at three verteces of a regular triangle and the receiver is at the cercumcenter. Then, the intersection of 3 hyperboloids becomes not a single point in the three dimensional space but a straight line perpendicular to the triangle plane and intersecting it at the circumcenter.


You can construct the hyperboloid with combination of your choice. However, if you have n satellite only n-1 of them will actually carry information. All the other would be half correlated with two of the n-1 hyperboloid and so will not provide additional information. In general when you do combine observation, the combination have to be linear independent to carry additional information.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.