**Triangulating your position**

Triangulating your position is a method of determining your location based on your position relative to other, established locations or signal sources by taking bearings. Often, you will have very limited information to work with. There are three common situations which are each handled differently. Case 3 is properly called trilateration, not triangulation, but I will cover it here for completeness.

- You know the
**angle and distance** between you and **one** established point.
- You know the
**angles** but not the distances between you and **two** established points.
- You know the
**distances** but not the angles between you and **three** established points.

**Graphical Solution:**

In all cases, you can simply draw on a map if you have a protractor and a ruler. For Case 1, just draw the line, scaled appropriately for length, at the required angle. For Case 2, draw lines from your known points at the measured angles and your position is at the intersection of the two lines. For Case 3, you need a circle compass. Draw circles around your known points at the appropriate distances (radius) and your location is at the intersection of the three circles.

**Mathematical Solution:**

This is inelegant however, and occasionally not an option. Your map might not be to scale, or you might not have the appropriate drawing tools. If you're programming a computer to perform these functions, it's obviously not going to be drawing on a map either. For whatever reason, you might need to do this mathematically.

**1 - Angle and Distance Known**

If you know both the angle and the distance to an established point, the problem is almost trivial. Simple trigonometry will give you your position. This might happen if you have a compass and a laser rangefinder, or if you have traveled a known distance (d) in a straight line at a known angle (α) from an established point.

P1
|\
| \
| \d
| \
|___α\
You

Given P1 = (x,y), your position is: **( x+cosα*d , y+sinα*d )**

**Example:**

A friendly artillery battery is located at grid coordinates +5 miles W, +10 miles N. Your compass reads 338.2° to the artillery battery. The time delay between the muzzle flash and the boom is 5.1 seconds, allowing you to calculate your distance as 1.08 miles. Determine your position.

Be sure your calculator is set to use degrees, not radians.

( x+cosα*d , y+sinα*d )

( 5+cos(338.2)*1.08 , 10+sin(338.2)*1.08 )

( 5 + 1.00 , 10 + -0.40 )

**(6,9.6)**

**2 - Angles to Two Points Known**

If you know the angles to two established points, the solution requires finding the equations of the lines between you and the known points. Your position is the intersection of these lines. This is by far the most common situation when hiking or otherwise navigating on land. All you need is a compass and two landmarks that would be on your map, such as water towers, street intersections, hilltops, or any other major terrain feature.

__________P2
| /
| /
P1_____________| /
\ | /
\ | / Note:
\ | / up is North
\ | /
\ | /
\ α|β/
\|/
You

Step one is to determine the equations of the two lines. We have enough information to use the point-slope method, understanding that the cotangent of the measured angle is the slope. Since all we need is the ratio of the rise to the run to find the slope of the line, the actual distances we have do not matter. By invoking the law of similar triangles, we can assume the hypotenuse (our unknown distance) is 1, for the purposes of getting this ratio. This makes our rise **cosθ** and our run **sinθ** -- therefore our slope is **cotθ.**

Point-slope format is the familiar **y=mx+b**, where m is the slope and b is the y-intercept. We have determined our slope to be **cotθ**, so we still need to find the y-intercept. This is accomplished by plugging our known coordinate into x and y and solving for b (**b = y - mx**). So our line equation, in terms of our known values, is:

y = cotθx + y_{Pn} - cotθx_{Pn}

Once we have our two line equations, all we have to do is find their point of intersection. There are numerous methods to do this, for example using linear algebra. Feel free to use whatever method you are most familiar or comfortable with.

**Example:**

You are hiking in rough terrain when your partner stumbles and hurts himself very badly, he cannot walk and it is too far to carry him back over the uneven ground. After administering first aid, you call for help on your radio, and the rescue team needs to know your position. You have a compass and a map, and are able to spot a hilltop and a water tower. The hilltop is at grid coordinates (1,5) and the water tower is at (9,7). The compass readings from your position to the landmarks are 312.27° and 19.65° respectively. Where are you?

First find the equations of the lines from you to the terrain features.

**P**_{1}: y = cot(312.27)*x + 5 - cot(312.27)*1 ⇒ y = -0.91x + 5.91

**P**_{2}: y = cot(19.65)*x + 7 - cot(19.65) * 9 ⇒ y = 2.80x - 18.21

Then find the point of intersection. This is only one of many ways to do it:

P_{1} = P_{2}

-0.91x + 5.91 = 2.80x - 18.21

5.91 + 18.21 = 2.80x + 0.91x

24.12 = 3.71x

**6.5 = x**

y = -0.91(6.5) + 5.91

**y = 0**

**Your position is (6.5,0).**

**3 - Three Distances Known**

Seismologists find the epicenter of Earthquakes by a third method of triangulation (actually trilateration). Earthquakes send out two kinds of shockwaves that travel at different speeds. By measuring the time delay between the first shockwave and the second shockwave, they can determine how far away the Earthquake occurred.

Just knowing the distance defines a circle around the station where the measurement took place. The Earthquake could have taken place anywhere along the perimeter of this circle. If two stations in different locations take measurements, it narrows the possible locations down to two -- the points of intersection where the two circles overlap. A third measurement at another station will pinpoint which of the two points is the true epicenter.

Extended to the three-dimensional case, this is how the Global Positioning System works. Every GPS satellite has a high-precision atomic clock in it. By measuring the speed-of-light delay of the transmissions between the GPS and your hand-held receiver (roughly 50 milliseconds), the satellite calculates how far away you are from it. This defines a sphere around the satellite, which is in a known position in orbit around the Earth. The intersection of two spheres is a circle, three narrows it down to two points, and four pinpoints it to a single point in three-dimensional space. Therefore, at least four GPS satellites are necessary to find your location. In practice, one of the two points defined by three satellites is above the satellite orbit, and the GPS system disregards it.

I will only explain the two-dimensional case here though, requiring three distance measurements. Given the coordinates (x,y) of the three stations and the distances (r) from the stations to the location you are looking for, you can easily find the equations of the circles defined. For each station:

(x-x_{n})^{2} + (y-y_{n})^{2} = r_{n}^{2}

If you know how to solve systems of simultaneous, non-linear equations, more power to you. Just solve the three simultaneous circle equations and get your answer. The rest of us will need a simpler method. If you take two of the three equations for the circles and subtract one from the other, the result will be the equation of the line passing through the two points of intersection of the two circles. The big advantage being that the result is therefore a *linear* equation, which is much easier to deal with.

Next, subtract any other two of the three circle equations to get a second line equation. The intersection of these two lines is the location you are looking for. Both line equations will pass through two points of intersection between two circles, and one of these points is the point you are looking for. This is where they will cross.

**Example:**

Seismology research station A is located at mile marker (100,100), B at (160,120), and C at (70,150). Analysis of the S and P shockwaves indicates that a particular Earthquake was 50.00 miles from station A, 36.06 miles from station B, and 60.83 miles from station C. Locate the epicenter of the Earthquake.

First find the equations for the three circles, using the distances from the epicenter as the radii.

**A:** (x-100)^{2} + (y-100)^{2} = 50.00^{2}

**B:** (x-160)^{2} + (y-120)^{2} = 36.06^{2}

**C:** (x-70)^{2} + (y-150)^{2} = 60.83^{2}

These equations expand to:

**A:** x^{2} - 200x + 10000 + y^{2} - 200y + 10000 = 2500

**B:** x^{2} - 320x + 25600 + y^{2} - 240y + 14400 = 1300

**C:** x^{2} - 140x + 4900 + y^{2} - 300y + 22500 = 3700

Now subtract two pairs of equations. There are three unique combinations but you only need two. Notice that the x^{2} and y^{2} terms cancel out, leaving us with linear equations.

**A - B:** 120x - 15600 + 40y - 4400 = 1200 ⇒ y = -3x + 530

**B - C:** -180x + 20700 + 60y - 8100 = -2400 ⇒ y = 3x - 250

Finally, find the intersection point between the two linear equations we found. This is our answer.

A - B = B - C

-3x + 530 = 3x - 250

780 = 6x

**130 = x**

y = 3(130) - 250

**y = 140**

**The epicenter of the Earthquake is (130,140)**

All examples have been checked with AutoCAD and verified as correct. Special thanks to Swap who helped me with the solution to Case 3, and everyone in the Chatterbox who pitched in. Thanks also to The Custodian, whose Global Positioning System writeup corrected several misconceptions I had. This node would not have been possible without the sense of community we have at Everything2.