# locate a audio source by 3 microphones, same plane, by using the volume

Let have 3 microphones MIC1: @ mic1x,mic1y MIC2: @ mic2x,mic2y MIC3: @ mic3x,mic3y. MIC1-MIC2 separated 2500mm at 90degr MIC2-MIC3 separated 2500mm at 90degr

If a sound produced by source S reach MIC1 with a peak of let's say 75, MIC2 of 20 and MIC3 of 10. How could I locate the audio source?

-
I might be wrong, but I think this might belong to physics.stackexchange. Basically, you need to check the rate of volume dissipation (should be something quadratic, I think...), and then use that to calculate the distance to each of the three points, and that to find the location. Well, assuming it's an empty room, that is... – tomasz Sep 2 '12 at 18:37
A microphone in fornt of me will "hear" more of what I say than one in the same distance behind me (well, that may here other sounds better). I assume that timing differences might be more accurate. – Hagen von Eitzen Sep 2 '12 at 18:49
Also, the microphones themselves may not hear equally well in all directions. They can be designed explicitly to approximate this, but if you just pick some random mics off the shelf in the electronics store for your location system, you can't be sure either way. – Henning Makholm Sep 2 '12 at 20:55

Assuming no reflection or absorption of sound, the intensity of sound from a point source follows an inverse-square law: if the source is at point $s$, the intensity $I(x)$ at point $x$ is $C/\|s-x\|^2$ for some constant $C$, where $\|\cdot\|$ is Euclidean distance. Suppose $s$ and the three microphones $m_1$, $m_2$, $m_3$ are all in the plane, with $m_i$ known. Then the ratios of intensities give you two equations $I(m_1) \|s - m_1\|^2 = I(m_2) \|s - m_2\|^2$ and $I(m_1) \|s - m_1\|^2 = I(m_3) \|s - m_3\|^2$. Each of these defines a circle in the plane (or a straight line if the intensities happen to be equal). Usually, if they intersect the circles will intersect at two points, so there are two possible locations for the source.
EDIT: In your example let $m_1 = (0,0)$, $m_2 = (2500,0)$, $m_3 = (2500,1)$ with (assuming the numbers you give refer to intensities) $I(m_1)=75, I(m_2)=20, I(m_3) = 10$. Then if $s = (x,y)$, $75 (x^2 + y^2) = 20 ((x - 2500)^2 + y^2)$ and $75 (x^2 + y^2) = 10 (x^2 + (y - 2500)^2)$ become
\eqalign{(x + 10000/11)^2 + y^2 &= 375000000/121\cr x^2 + (y + 5000/13)^2 &= 187500000/169\cr} which correspond to a circle of radius $\sqrt{375000000/121} \approx 1760.446975$ centred at $(-10000/11, 0)$ and a circle of radius $\sqrt{187500000/169} \approx 1053.312611$ centred at $(0, - 5000/13)$. These intersect at approximately $(831.8160827, 261.5652866)$ and $(115.4863013, -1431.577834)$.