# How many boats does it take to find an acoustic buoy by Doppler shift?

Inspired by this question on the Doppler shift, suppose there is buoy somewhere on the surface of the ocean emitting a pure frequency.

You get to place some boats wherever you want on the surface of the ocean, moving whatever direction you want with whatever speed you want.

The boats listen to the pure frequency, which will in general be Doppler shifted due to the relative motion between buoy and boat. If you do not know ahead of time the frequency the buoy is emitting, is it possible to deduce the buoy's location simply by observing the frequency measured by the boats? What is the minimum number of boats necessary, how should they be positioned, and what velocities should they have?

This is a toy problem with the mathematics of the Doppler effect, so let's leave out attenuation of sound over distance and assume the surface of the ocean is a plane. Also, if the boats could continuously monitor the Doppler shift, they could collect extra information based on their own changes in position and velocity, so imagine locating the buoy based solely on a single frequency measurement from each boat.

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Is the buoy fixed to ground? BTW: I love "pure" frequency as much as Ted Bunn loves pure energy :) physics.stackexchange.com/questions/9731/… –  Georg May 14 '11 at 10:17
Yes, the buoy is fixed. I think you can deal with your qualms about pure frequency if you give it a little effort. Obviously it makes no real difference from a practical point of view. –  Mark Eichenlaub May 14 '11 at 15:35
unlike the "pure energy" misnomer, "pure frequency" actually refers to a mathematical concept, albeit strictly unphysical, still is applicable in many approximations –  lurscher May 16 '11 at 22:27
What is wrong with a pure frequency? This is perfectly well-defined ($\exists T\neq 0\colon p(\mathbf{x},t)=p(\mathbf{x},t+nT)\ \forall n\in\mathbb{Z},\mathbf{x},t$) and actually quite easy to achieve experimentally. –  leftaroundabout Jun 23 '11 at 11:12

Lemme make sure I have the assumptions correct:

1) The buoy is at rest in a known inertial frame in a known plane; the boats are in the same plane.

2) Each boat may make a "single measurement" of the frequency. I will assume this means that the time of measurement is long enough that the boat can resolve the frequency to arbitrary precision, but the boat's velocity is constant over the measurement time and the change in the boat's position (relative to the buoy) is negligible. I will assume the angle of origin of the signal cannot be determined, only the frequency. The velocity and positions of the boats are known.

A three-boat solution:

First boat: at rest in the inertial frame. Measures the frequency.

Second boat: nonzero velocity in the inertial frame.

From the Doppler shift measured by the second boat, you can determine the absolute value of the angle of the buoy with respect to the velocity vector of the second boat (but can't distinguish "right" from "left").

Third boat: nonzero velocity, nonparallel to the second boat. Different location than the second boat.

From the Doppler shift measured by the third boat, you can determine the absolute value of the angle of the buoy with respect to the velocity vector of the third boat.

Draw lines from the 2nd and 3rd boats along all the two possible angles from each. Where they intersect is the location of the buoy. That's the "straightforward" solution. Is it possible to do it with two boats if you get tricky?

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This looks right to. With only two boats, you get just a single number - the frequency difference between the boats. A single number shouldn't be able to give you a two-dimensional vector, so three ought to be the minimum. –  Mark Eichenlaub May 17 '11 at 11:45
I'm pretty sure that for any particular geometry you propose I can find a place to put the buoy that develops at least one ambiguity. For instance let the moving boats diverge along the x and y axis. There are places behind each moving boat where I can place the buoy and get two (or even four) solutions. This will work in most cases, but not every case. –  dmckee May 17 '11 at 14:56
dmckee: I don't understand your objection, but perhaps that's because I'm not envisioning the same case you are (I don't know what it means for boats to "diverge"). Could you give a specific example, such as the x and y coordinates of the two boats, their velocity vectors, and the location of the buoy? –  Anonymous Coward May 17 '11 at 17:41
@Anon: Boats at $(1,0)$ and $(0,1)$ each moving in the direction of increasing coordinate. If I put the buoy at $(0,0.057)$ the boat at $(0,1)$ knows it is directly behind, and the boar at $(1,0)$ gets an angle of $\pm 1/10$ radian from the rear, making for two solutions: ${ (0,0.57), (0,-0.57) }$. I think I can do that for any geometry you choose for you two moving boats. If I put the buoy at $(\pm 0.057, \pm 0.057)$ you end up with a four fold ambiguity. –  dmckee May 21 '11 at 17:47
'Course, the real world solution to that issue is to pick a geometry where all the ambiguous solutions are either withing buoy-sighting distance of at least on boat or all within buoy sighting distance of each other. And this may be possible. –  dmckee May 21 '11 at 18:11

Really idealized approach.

Assuming the buoy is on the surface.

Three boats. They move at the same non-zero velocity much less than the speed of sound (but enough for the instruments to get read a change) perpendicular to the distance between them. Their separation is small compared to the length scales of the problem, but nearly twice the "spotting something on the surface" distance. At a given time they take a reading. Each boat gets a angle between it's velocity and the buoy, leading to a left-right ambiguity for each boat as well as the range ambiguity. Further, while the pairs should be able to resolve the range issue, there remains the possibly of several solution ambiguity for each pair. The third boat should resolve them.

Pathological case: the buoy lies on the line between the boats when they take their reading.

Solution to the above: Turn the formation through a significant angle and try again. Or just wait a while. Or if we insist on only one reading, have the center boat lead or trail the others by a significant distance.

Problems with this: The big one seems to be precision. Each boat does not get a perfectly defined opening angle. They get a rough value. So in principle resolving the ambiguities could depend on the spacing of the boats and the relative location of the buoy. BUut I'm not going to do any math tonight.

To make it harder: Allow that the buoy might be sunken. Now the pairwise ambiguities are the intersections of cones instead of discrete points.

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So the three boats are in a line, right? With a single reading, a single boat gets no information because the boats don't know the true frequency of the buoy. –  Mark Eichenlaub May 14 '11 at 19:15
Ah...I missed that. We can add a boat riding at anchor to solve that. –  dmckee May 14 '11 at 21:33
If all the boats reverse direction and follow their path back, the true frequency can be read as the mean of forward/back at any point of the path. –  Georg May 15 '11 at 11:15
@Georg I agree, but the problem statement was for each boat to make a single reading. It's a bit contrived, to be sure. I just thought someone might have fun working it out within those constraints. –  Mark Eichenlaub May 15 '11 at 11:54
Ahh, I did not see that single reading. Of course that makes rubbish from my answer. –  Georg May 15 '11 at 12:25

Three boats is the minimum. And I don't think even that is enough in the genral case, if one of them is not moving (i.e., dedicated to directly solving bouy frequency).

Using just two boats with doppler lines will result in two lines of position for each boat. The intersection of these lines from two boats will result in 4 possible positions. You need a third boat to resolve the ambiguity. Even using a third boat, there are special cases where its LOPs would intersect 2 of the intersections of LOPs from the first two boats; but it would not be likely.

I would use 3 boats, which have traveled some distance along rays separated by 120 degrees and radiating from the same point. Doppler shifts could be used to qualitatively isolte the bouy to the 1/3 of the ocean directly away from the boat with lowest received frequency. The actual bouy frequency could also be bracketed to a very narrow frequency. From there it is fairly simple to solve for the actual frequency. Proceed from there to draw the LOPs from each boat. The two boats whose paths bound the wedge of water containing the bouy would have LOPs defining 1 possible location for the bouy.

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I disagree that 3 boats is not sufficient for the general case. I think we agree that 2 boats (with a 3rd to measure frequency) will give two "lines of position" extending out each boat. But these lines will only intersect at one point, not 4. They are not infinite lines, but only "half-infinite" (with one end at the boat of origin, the other at infinity). –  Anonymous Coward May 16 '11 at 21:45
@Anon: Only one intersection is the best case---indeed in most cases if you choose your geometry properly. But you have to find a geometry where that is true for every possible buoy position. –  dmckee May 21 '11 at 17:50