I read this thread Trilateration using 3 latitude and longitude points, and 3 distances explaining about trilateration using 3 latitude and longitude with measured distance.

The problem is I want to find out an unknown target location with only 3 knowns latitude and longitude co-ordinates without knowing each distance point. For example, I have 3 following longitudes and latitudes:

lat="-6.28530175" lon="106.9004975375"
lat="-6.28955287" lon="106.89573839"
lat="-6.28388865789474" lon="106.908087643421"

Thank you very much :)

Based on @D.E.Wright answer I am googling to find out more and I get this following forum http://www.coderanch.com/t/453432/Programming/implement-cell-triangulation-mobile-phones, it gives 3 steps to triangulation. The problem I couldn't understand what he means about "Calculate distance from the first tower based on speed which gives a radius value".

And with @Kirk Kuykendall suggestion I decide to put my 3 lat/long to google map and I get this following image :

Google Maps

Where :
A, B, C : Cell Towers location
A : lat -6.28955287, lon 106.89573839 with cell tower Radius 6000m
B : lat -6.28530175, lon 106.9004975375 with cell tower Radius 6000m, and
C : lat -6.28388865789474, lon 106.908087643421 with cell tower Radius 6000m

Now, with these data can I doing a triangulation to get my real position? And how is the mathematical method? I've try to modify phyton code to remove the distance variable from @wwnick in this thread Trilateration using 3 latitude and longitude points, and 3 distances but until now I couldn't find the answer.

  • Do you mean you want to find the centroid of a triangle on a sphere? May 27 '11 at 16:08
  • Is the 'Unknown Target' the intersecting point of all three lat/lngs?
    – Mapperz
    May 27 '11 at 16:12
  • 3
    This question cannot be answered with the information given. What is the relationship between the three points and the unknown target location?
    – whuber
    May 27 '11 at 16:51
  • @Kirk Kuykendall & @whuber : Let's say I have information (including longitude, latitude and radius) about my current cell tower and its neighboring (2 neighboring cell towers) from my cell phone, then I want to measure my location with these information. If I use my current cell tower information its accuracy are about 5km, so if I use these 3 cell tower information can I get my accurate position? May 27 '11 at 17:30
  • 1
    Yes please edit your question to include signal strength. I believe this could be used to help determine location. I suppose the power rating (watts) of each cell tower transmitter would also be a factor, if you have that, please post that too. Jun 6 '11 at 2:40

Yes, I agree that I don't see how to get exact location without a distance. You would be in this situation:

enter image description here

So you would only be able to tell within a certain area where you were. And that's with only a 1000m buffer (radius). Your 6000m buffer has such an overlap that the potential location would be huge.

The solutions would be to:

  1. Get some sort of distance measurement (even if it's approximate)
  2. Use more cell towers
  3. Use better placed cell towers

By better placed cell towers I mean that these three form a very flat triangle. A more equilateral triangle would give smaller areas of overlap, meaning a more accurate position.

  • Mark; when we worked with EMS for these soltutions the answer was not a 100% certain; but your graphic does illustrate what is the industry answer; being the point where the 3 radi cross.
    – D.E.Wright
    May 31 '11 at 16:42
  • Thank you for your illustration. When I request cell neighboring I got more than 3 cell towers, about 4 until 5 cell tower and for each cell tower I also got signal strength level. If it's not possible to calculate without the distance, do you have any resource that related to calculate user position with each cell tower with signal strength given? Jun 5 '11 at 16:30

Your answer will have you doing concentric circles radiating out from each point; when all three intersect you will have our point. So you may have one radius of 15 miles, one of 10 miles and one of 25 miles, but at a point in space all three will intersect of have a mean point to intersect that will represent you originating point.

In EMS we use this triangulation method from Cell Towers a lot; but its programatic from our side since a cell signal doesn't give your range or bearing you need to use triangulation.

  • Is that mean we can doing triangulation if we know each cell tower radius? May 27 '11 at 17:31
  • If you know the towers locations then yes, you need to have atleast 3 for any degree of accuracy, the more you have the more precise it works. This is how EMS used to track 911 calls from Cells before the mandatory GPS signaling.
    – D.E.Wright
    May 27 '11 at 18:12
  • How do you do triangulation without distance?
    – Sean
    May 27 '11 at 19:39
  • You simply need to expand from your point outwards until you have your circles intersect. Each point in the Cell example is a radio transmitter that can't always give you a distance. So you are relying on the other points and there offset radi to intersect with eachother to tell you where the target point is.
    – D.E.Wright
    May 27 '11 at 19:44
  • @DE Makes sense, but this approach assumes you are equidistant from all three points. That assumption doesn't sound tenable. One could create polygons around each cell tower showing where it can be "heard" and intersect all three polygons to narrow down a location, but going further than that requires more data.
    – whuber
    May 27 '11 at 20:20

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