Approximate (but as accurate as it can) location of sound

I quess that this is relatively easy question, but I have been struggling this for a two days now (basicly investigating different formulas) and couldn't find a solution.

So, let's think this case: Three (or more) internet nerds from same area (like in 10 kilometer radius) listens for a thunder. When each one hears that lightning strike, they tell to server that 'Wow now I heard it!' (clicks a button). Then server somehow calculates that where that lightning was. Let's assume that there is zero lag between what user hears and server got the info (network, human reflexes).

Things we know:

• We know listeners Latitude and Longitude.
• Also we know speed of sound in that horrible thunder.
• Time differences between each person. (For example Listener 1 hears it first, then Listener 2 after 1.2 seconds and then Listener 3 after 2.3 seconds)

I know it can be made easily when you know how far that was (by hearing and seeing), and my awesome brain simulation tells me that it can be done with only hearing OR seeing.

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2 Answers

A "simple" way to do it would be solve the following equations. First, WLOG, assume that the person who heard the first sound is at the location $(0,0,0)$ and the second is $(x_{P_2},y_{P_2},z_{P_2})$ and the third is $(x_{P_3},y_{P_3},z_{P_3})$. Let $P_2$ have heard it $t_1$ seconds later, and $P_3$ have heard it $t_2$ seconds later. Then, you have the equalities

Note: speed of sound is 340 m/s and the coordinates are in meters.

$$340t_1 + \sqrt{x^2+y^2+z^2} = \sqrt{(x-x_{P_2})^2+(y-y_{P_2})^2+(z-z_{P_2})^2}$$ $$340t_2 + \sqrt{x^2+y^2+z^2} = \sqrt{(x-x_{P_3})^2+(y-y_{P_3})^2+(z-z_{P_3})^2}$$

However, note that you don't have enough info with just this.. you need another person. (because 2 equations and 3 unknowns)

EDIT: An simplifying criteria is $z=0$, which is valid if you assume that the sound originated from the ground. Note, this assumes that $z_{P_2}=z_{P_3}=0$.

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You can consider things to happen in a plane, so set all the $z$'s to $0$. Then you have two equations in two unknowns. Otherwise you need a fourth observation. – Ross Millikan Jul 31 '12 at 22:14
@Ross, I don't see why that assumption is valid. The OP mentioned Latitude and Longitude. My guess is that that implies 3 dimentions. In addition, the sound should be assumed to have come from the air (but I am not certain of this fact). – picakhu Jul 31 '12 at 22:18
As long as you are on a 2D surface, you can compute distances based on 2 coordinates. This includes the surface of a sphere. True, z=0 is the planar approximation, which simplifies things. If you want to consider the height of the strike above the ground you need another data point. The same thing happens with GPS-if you assume you are on the surface of the earth you need 3 satellites, if you worry about altitude you need 4. – Ross Millikan Jul 31 '12 at 22:22
Ahh, I get it now.. – picakhu Jul 31 '12 at 22:24

If you have two listeners who see the lightning, you can calculate the distance of the strike from each one. Then just swing a circle of that radius around each one. The circles will intersect in two points (unless they are tangent) and you know the location up to that ambiguity.

If you have three listeners you have just enough data to resolve it, but I don't know an easy way. Let the listeners be at points $A,B,C$ and the time of day that they hear the thunder be $a,b,c$. Then if the strike is at $D,d$ and the speed of sound is $s$ you have $|D-A|=s(a-d)$ and two more equations like it. The three unknowns are the two coordinates of $D$ and the time $d$. Three equations in three unknowns. You generally will not have an ambiguity. If the data (positions and times) are perfect the solution will be exact.

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How on earth can I calculate distance of lightning by knowing that friend of mine heard that 3 seconds after me? What about if we hear it at the same time. It can be right between us, or it can be 10 kilometers away. – user1018237 Jul 31 '12 at 17:49
@user1018237 Assume that the speed of light is instantaneous, since it's not far enough away to notice the difference. From the moment you see the lightning, time until you hear the thunder. If you know the speed of sound, you can get an estimate for the distance. – Robert Mastragostino Jul 31 '12 at 17:52
Ok, maybe I should highlight more that I can only listen it. – user1018237 Jul 31 '12 at 17:54
@user1018237: I recognize that you can only listen to it, which is why I took $d$ as a variable. As you say, from two observations 3 seconds apart, you only know that the strike was 1020 meters closer to the first than the second. That gives a hyperbola in the plane. The third observation can be combined with one of the first two to give another hyperbola. The point where they intersect is where the strike was. Nobody needs to see it if you have three observations. – Ross Millikan Jul 31 '12 at 17:58
If, however, you insist on using the speed of light and the ability to "see" the strike, you have an overdefined system, and you can obtain instead a least-squares solution. – Emily Jul 31 '12 at 18:33