# Calling All Math Geeks...

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I've been searching around the web in vain for a worked solution of a 2D trilateration problem, which I need to help me solve for a cache location. It's been a while since my math classes, so I'm a bit hazy. I'll copy down what I was able to figure out, and hope someone can help me continue it on!

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The basic concept is fairly straightforward:

Given three points on a plane,

P1(x1,y1) , P2(x2,y2), P3(x3,y3)

as well as three vectors

/d1, /d2, /d3

and the fact that

(P1 + /d1) = (P2 + /d2) = (P3 + /d3) = P(x,y)

Determine the location of P(x,y).

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So the general idea is to trace a circle around each of the points....

i) a circle of radius |/d1| around P1,

ii) a circle of radius |/d2| around P2,

iii) a circle of radius |/d3| around P3

And the one location where all three circles intersect will be P(x,y).

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The general equation for a circle with origin C(a, and radius r is:

(x - a)^2 + (y - ^2 = r^2

Applying this equation to the circles described above, the following three equations for the circles around P1, P2 and P3 can be written out:

(x - x1)^2 + (y - y1)^2 = |/d1|^2

(x - x2)^2 + (y - y2)^2 = |/d2|^2

(x - x3)^2 + (y - y3)^2 = |/d3|^2

but this is as far as I got... so I put it out to the forums! How can I contrinue from here to solve for P(x,y)?

Hmm I wonder if we don't have enough info. I think (and we're talking very rusty math skills here) you've taken it to a point where we need some correlation between the points and or vector lengths. Is there any more info provided? What cache is it by the way.

There is a crib sheet on this cache.

Determine the location of P(x,y).

Lift the seat before you start and you won't have to be so accurate, but more important, put it back down when you're done. I've been searching around the web in vain for a worked solution of a 2D trilateration problem, which I need to help me solve for a cache location.

The problem as you have stated it is overdetermined; that is, there is too much information there for a solution in general. However, for those cases in which there is a solution, take a look at this document, which contains the closed-form answer.

Well, plot the points.

Then, using the law of cosines.

Notation: a2=a squared

a2=b2+c2-2bc(cosA)

therefore, a squared is equal to b squared plus c squared minus two times b times c times cosine of angle A.

You will find the measure of one angle.

Then, using that angle use the next eqn:

sinA/a=sinB/b=sinC/c

therefore, sine of angle B divided by side b is equal to .....

You will find the measure of each angle.

Divide each angle by two( to find the middle)

Then, go out to the site, using a compass, and do it manually from there.

Draw it in AuctoCAD and ID the point of intersection. It's easy to solve this problem graphically, and you don't even need a map to determine the coordinates of the unknown waypoint. Simply use a sheet of grid paper and scale the X and Y axes to include all the given points. In this example each UTM grid square is 1km by 1km with 100-meter subdivisions. You could also add 10-meter subdivisions if your scale is fine enough. Most GPS receivers will resolve UTM coordinates down to 1 meter. You could also solve this algebraically, but I think this is just as good considering the inherent inaccuracies. Hope this helps.

Cheers ...

Edited by Rich in NEPA

I'm guessing that you want a formulaic approach (though the graphic one certainly works), and that instead of vectors, you meant distances (otherwise the vectors would specific direction as well as distance, and you've already got the solution three different ways).

Let the three known points be P1, P2, and P3. Their x,y coordinates are, respectively, x1,y1; x2,y2; and x3,y3. The three radii are R1, R2, and R3. The coordinates of the solution point S (the only unknown) are xs and xy.

So:

(Xs - X1)**2 + (Ys - Y1)**2 = R1**2

(Xs - X2)**2 + (Ys - Y2)**2 = R2**2

(Xs - X3)**2 + (Ys - Y3)**2 = R3**2

or

Xs**2 - 2X1Xs + X1**2 + Ys**2 - 2Y1Ys + Y1**2 = R1**2

Xs**2 - 2X2Xs + X2**2 + Ys**2 - 2Y2Ys + Y2**2 = R2**2

Xs**2 - 2X3Xs + X3**2 + Ys**2 - 2Y3Ys + Y3**2 = R3**2

or

Xs**2 + Ys**2 = R1**2 + 2X1Xs + 2Y1Ys - X1**2 - Y1**2

Xs**2 + Ys**2 = R2**2 + 2X2Xs + 2Y2Ys - X2**2 - Y2**2

Xs**2 + Ys**2 = R3**2 + 2X3Xs + 2Y3Ys - X3**2 - Y3**2

So, the right hand sides of the three equations are all equvalent. Set one equal to one of the others, and solve for Xs in terms of Xy (the only squares are of known values), then plug in that value of Xs into the third, and at worst you might have to use something like the quadratic equation, but it's no worse than an algebraic equation of the second degree in one unknown. Wow, I feel REALLY dumb after looking at this thread!!!

I do not know if you ever got an answer to your question or not but I have several pages on the subject out of one of my Surveying books that may be either helpfull or overwhelming. It would be easier to mail to you. So if you like you could PM me a mailing address and they will go out the next day. They are too much to scan and keep legible for email or posting.

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