# Trangulation method to find the location

I am trying to finding the location of the user using mobile tower signal strengths. Scenario is given in the picture

Here, I know the coordinates (latitude and longitude, like x,y) of Tower 1, 2 and 3. Also I know the signal strength (can I consider it as radius? But for nearest towers radius will be high, and for far towers it will be lower, basically the opposite of radius)

If I have these parameters, how could I calculate my Location?

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See Wikipedia on path loss –  Henry Oct 30 '13 at 8:30
Signal strength won't be the radius for sure. But you can investigate empirically if the signal strenght is proportional to $\frac{1}{r^k}$ for some $k$ or something else (Numerical Calculus for the win). –  Victor Chaves Oct 30 '13 at 8:47
Thank you for your reply. I will try to find out a proportion like that. –  Hari Oct 30 '13 at 10:50

You can consider the signal strength as a function (decreasing) of the distance.

Let $T_1$, $T_2$ and $T_3$ be the positions of the towers. The signal strength from a point $P$ to the tower j is

$$ST_j(P)=g(|P-T_j|)$$ where $g$ is a decreasing function.

Note that if $g$ is decreasing, it is one to one. So knowing the signal strength is equivalent to knowing the distance from $P$ to the towers.

Once you have determined the distances, you just have to solve the system: $$|P-T_j|=d_j,\,\;\; j=1,2,3.$$

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Thank you for your answer. I'm sorry. I am not that much good at mathematics, so if I am wrong please correct me. First of all, I want to find out the Point where I am standing (I suppose you named it as P, which is unknown to me). Also if I got the three distances, can I take them as the radius and find out the point of intersection? –  Hari Oct 30 '13 at 10:42
Yes, you are right. Note that in general the system has no solution. So you might need to find an approximate solution. –  Pocho la pantera Oct 30 '13 at 14:15
Thank you, but how to calculate g(P-Tj) if P is unknown? –  Hari Oct 31 '13 at 4:01
You need to know the function $g$. Say for example that $g(t)=\frac{1}{t}$, then if $ST_j(P)=g(|P-T_j|)=g_j$ (your measurements), you conclude that $|P-T_j|=\frac{1}{g_j}$. –  Pocho la pantera Oct 31 '13 at 11:44

In open space the signal strength is a function of distance that is a function of $\frac{1}{Distance^2}$. If you are not in open space there are too many variables and you practically can not determine accurately location for such cases.

The idea of the GPS was to support exactly the need for location using "open space" and even this solution may suffer from in accuracies.

If you insist, having a distance you solve equations of the three circles (one for each tower) intersection - you really use only two that are more accurate and the third will be used to determine which of the two solution is the better one.

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Thank you very much Moti –  Hari Mar 11 at 6:21