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I have the three receivers set up: $A (0.5, 0)$ $ B (-0.5, 0)$ $ C (0, 1)$

I know that the signal arrives at $C$ first. It then arrives at $B$ $2.63464*10^{-4}$ later. It finally arrives at $A$ last $7.07023 * 10^{-4}$ seconds after it arrives at $C$.

How would I use this information to find the point of emission? I know it has to do with trilateration. (Note, please do not link a Wikipedia page. I've already tried that as well as intensive googling.)

Martin Sleziak
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the fool
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  • The exponents will display correctly using 10^{-4} rather than 10^-4. I believe you need to know the speed of the signal to solve this. Also wikipedia says trilateration uses absolute time measurements, while multilateration uses differences, so I guess this is actually to do with the latter. – stewbasic Sep 15 '16 at 04:52

1 Answers1

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I think you can only approximate the position of the emitter

Let's call the emitter E and the speed of the signal s

then we have $ EB-EC = 2.6364 *10^{-4} s$

and $EA-EC= 7.07023 * 10^{-4} s$

and s is the same in each formula

I (using geometers sketchpad) played around with it a bit and found that E is around (-1 , 1.3) but be aware this is just a very rough guess where E is (it can be much further away).

the proper way is to plot some hyperbolics and then E on the intersection of some of them and so but that is abit beyond my knowledge.

still a wikipedia link (sorry :)

https://en.wikipedia.org/wiki/Hyperbolic_navigation

https://en.wikipedia.org/wiki/Multilateration

sadly did not find a good geometry reference for this

Martin Sleziak
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Willemien
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