Trilateration

[+WiiRhyme]

Hi,

 

We were looking for a program or simple method of doing trilateration. We have three sets of coordinates and their distances from the coordinates we are looking for.

 

If anyone has any programs or knows of any, could you please write back?

 

Thank you in advance for your help.

[+fehrgo]

You're actually looking at finding the points of intersection of 3 separate circles. First a word of warning, the precision used in the coordinates you have and the distance you have will likely mean that you can't solve for this, and will need to calculate it differently, if it can indeed be calculated at all.

 

In a perfect world, with perfect coordinates and measurements:

 

Take the formula for the three circles, and make them equal.

 

You are trying to solve for x and y (the intersection point), where:

 

(x+a1)^2 + (y-b1)^2 = r1^2

(x+a2)^2 + (y-b2)^2 = r2^2

(x+a3)^2 + (y-b3)^2 = r3^2

 

a1, a2, a3, and b1, b2, b3 are the center points of the circles, or rather the coordinates that you already have. r1, r2, r3 are the distances.

 

Let's re-state those formulas slightly differently:

 

(x+a1)^2 + (y-b1)^2 - r1^2 = 0

(x+a2)^2 + (y-b2)^2 - r2^2 = 0

(x+a3)^2 + (y-b3)^2 - r3^2 = 0

 

Since they all now equal 0, logic extends:

 

(x+a1)^2 + (y-b1)^2 - r1^2 + (x+a2)^2 + (y-b2)^2 - r2^2 + (x+a3)^2 + (y-b3)^2 - r3^2 = 0

 

or

 

(x+a1)^2 + (y-b1)^2 + (x+a2)^2 + (y-b2)^2 + (x+a3)^2 + (y-b3)^2 = r1^2 + r2^2 + r3^2

 

Of course, this assumes a flat map projection with coordinates in a unit rather than degrees, and the distance being in that same unit. IE, UTM projection with distances in meters.

 

If you plug your numbers in there, you'll be left with a formula for x and y that is much easier to solve for.

 

Another option is to use a paper map, compass, and pencil

 

http://gc.kls2.com/ also will show you ranges from coordinates, but does not have the resolution you are after. If any online map supports multiple range circles and lets you zoom, that would do it to.

 

Back to the non perfect world, do it as three sets of intersection:

 

(x1+a1)^2 + (y1-b1)^2 - r1^2 + (x1+a2)^2 + (y1-b2)^2 - r2^2 = 0

(x2+a1)^2 + (y2-b1)^2 - r1^2 + (x2+a3)^2 + (y2-b3)^2 - r3^2 = 0

(x3+a2)^2 + (y3-b2)^2 - r2^2 + (x3+a3)^2 + (y3-b3)^2 - r3^2 = 0

 

You will end up with three sets of x,y coordinates that will be very close to each other, assuming that the circles only touch and do not overlap. If they overlap or don't quite touch, you will need to fiddle with the radius slightly until the values work out for you, or work with six sets of coordinates instead, which are the intersection points of the individual arcs.

 

I'm willing to bet the owner is just after people drawing circles on maps.

 

Good luck!

[+fehrgo]

It also shouldn't take long for someone to bang something together using Google Mapplets.

[+fehrgo]

As suspected, go to http://maps.google.com/preview and use "circle filter". Playing with that should get you 99% of the way there.

[+Shakedown.dave]

You're actually looking at finding the points of intersection of 3 separate circles. First a word of warning, the precision used in the coordinates you have and the distance you have will likely mean that you can't solve for this, and will need to calculate it differently, if it can indeed be calculated at all.

 

In a perfect world, with perfect coordinates and measurements:

 

Take the formula for the three circles, and make them equal.

 

You are trying to solve for x and y (the intersection point), where:

 

(x+a1)^2 + (y-b1)^2 = r1^2

(x+a2)^2 + (y-b2)^2 = r2^2

(x+a3)^2 + (y-b3)^2 = r3^2

 

a1, a2, a3, and b1, b2, b3 are the center points of the circles, or rather the coordinates that you already have. r1, r2, r3 are the distances.

 

Let's re-state those formulas slightly differently:

 

(x+a1)^2 + (y-b1)^2 - r1^2 = 0

(x+a2)^2 + (y-b2)^2 - r2^2 = 0

(x+a3)^2 + (y-b3)^2 - r3^2 = 0

 

Since they all now equal 0, logic extends:

 

(x+a1)^2 + (y-b1)^2 - r1^2 + (x+a2)^2 + (y-b2)^2 - r2^2 + (x+a3)^2 + (y-b3)^2 - r3^2 = 0

 

or

 

(x+a1)^2 + (y-b1)^2 + (x+a2)^2 + (y-b2)^2 + (x+a3)^2 + (y-b3)^2 = r1^2 + r2^2 + r3^2

 

Of course, this assumes a flat map projection with coordinates in a unit rather than degrees, and the distance being in that same unit. IE, UTM projection with distances in meters.

 

If you plug your numbers in there, you'll be left with a formula for x and y that is much easier to solve for.

 

Another option is to use a paper map, compass, and pencil

 

http://gc.kls2.com/ also will show you ranges from coordinates, but does not have the resolution you are after. If any online map supports multiple range circles and lets you zoom, that would do it to.

 

Back to the non perfect world, do it as three sets of intersection:

 

(x1+a1)^2 + (y1-b1)^2 - r1^2 + (x1+a2)^2 + (y1-b2)^2 - r2^2 = 0

(x2+a1)^2 + (y2-b1)^2 - r1^2 + (x2+a3)^2 + (y2-b3)^2 - r3^2 = 0

(x3+a2)^2 + (y3-b2)^2 - r2^2 + (x3+a3)^2 + (y3-b3)^2 - r3^2 = 0

 

You will end up with three sets of x,y coordinates that will be very close to each other, assuming that the circles only touch and do not overlap. If they overlap or don't quite touch, you will need to fiddle with the radius slightly until the values work out for you, or work with six sets of coordinates instead, which are the intersection points of the individual arcs.

 

I'm willing to bet the owner is just after people drawing circles on maps.

 

Good luck!

 

My head hurts.

[Mag Magician]

The answer to your problem lies not in programming, but in basic map reading, and the ability to use a compass. As stated above, the hider of the particular cache probably requires the seeker to have such knowledge. Do not forget to include the variable of magnetic variation into your equation if using a compass, or your GPS database.

If one is adept at map reading and compass work, any co-ordinate, anywhere, can be reached within 30 feet without the aid of a GPSr. Triangulation is merely a method of putting two lines on a map, and walking to where the two lines meet.

 

edit, on second thought:

Topographical maps are helpful.

Edited July 30, 2007 by Team Magic

[+EScout]

Do your preliminary work on an map program and find the cache point by using the map program's distance function from the 3 triangle points you are given. It will only take a few minutes. Enter this position and the three triangle points in your GPSr. Go to this position, which should be close to the actual cache. Then, use your closest waypoint function on your GPSr and scroll between the three waypoints until the distances match what you are given.

 

I have set up a cache in a park, and the three tags you find give bearings to the cache. Using only the closest waypoint function on your GPSr, you scroll until the reciprocal bearings match. It can be done fairly quickly and is very precise.

[nicksmkp]

I have a similar problem. Only I'm sure trilateration is needed for a solution. Circles only work in an x,y plane. I am also interested in the z coordinates. I'm quite sure I either need a program to solve trilateration, or a program to solve the intersection of three or more spheres. Which I think are one in the same.

 

The problem

We are conduction a survey on a marine slipway. With tight specs. The rail needs to be straight to 1/4 inch over 600'. 90% of rail is underwater.

 

possible solution

Use a pole with a known constant length. Lean it in all which directions and locate coordinates in the x,y,z plane. To solve for the end of the pole (which is put on the center of the rail by a diver) use all collected coordinates and intersect them with the known rod height.

[+TomToad]

Hi,

 

We were looking for a program or simple method of doing trilateration. We have three sets of coordinates and their distances from the coordinates we are looking for.

 

If anyone has any programs or knows of any, could you please write back?

 

Thank you in advance for your help.

If you have an iPhone, iPod Touch, or iPad, there is a program called Geocaching Toolkit. One function is Circle/Circle intersect. Put in two of the coordinates and their distance. You will get two answers, write them down. Then replace one of the coordinates with the third. You will get another two coordinates. The ones that match up is the correct one.

[AZcachemeister]

Do your preliminary work on an map program and find the cache point by using the map program's distance function from the 3 triangle points you are given. It will only take a few minutes. Enter this position and the three triangle points in your GPSr. Go to this position, which should be close to the actual cache. Then, use your closest waypoint function on your GPSr and scroll between the three waypoints until the distances match what you are given.

 

I have set up a cache in a park, and the three tags you find give bearings to the cache. Using only the closest waypoint function on your GPSr, you scroll until the reciprocal bearings match. It can be done fairly quickly and is very precise.

 

Exactly what I have done, and yes it works.

[+Markwell]

There is a spreadsheet out there for something called "You Are the GPS" - I still have it. You enter in three points and the distance, and it calculates the intersection. Heavy-duty math.

 

But I would use Microsoft Streets and Trips and can probably get pretty close. I will draw radius circles from points, and while it may not give you the mathematical precision that the calculation would do, it would get you pretty darn close - probably as close as a GPS.

[+Markwell]

http://forums.Groundspeak.com/GC/index.php?showtopic=55888&view=findpost&p=673732

 

http://www.geocaching.com/seek/cache_details.aspx?ID=16760

[+The Jester]

But I would use Microsoft Streets and Trips and can probably get pretty close. I will draw radius circles from points, and while it may not give you the mathematical precision that the calculation would do, it would get you pretty darn close - probably as close as a GPS.

Some GPSr's have Proximity waypoint function - you can set a circle around the waypoint at a given distance (you may have to convert the distance to the units used = feet to decimal miles or such), generally as a way to avoid something. You can then look on the GPSr map and see where the three interesect and navigate to there. This way you can solve such a problem in the field.

[+fizzymagic]

I have written my own that uses true ellipsoidal distances. I use an iterative minimization method with a absolute-value objective function. It works great, but it is overkill for you unless the distances involved are very large.

 

Your best bet for solving a single cache problem would be to draw circles on a paper map, or else draw circles on a Google Map. From each waypoint, draw a circle with radius equal to the distance from it to the cache. The intersection of the three circles is the cache.

[+fizzymagic]

If one is adept at map reading and compass work, any co-ordinate, anywhere, can be reached within 30 feet without the aid of a GPSr.

How special for you that you are so gifted. I, personally, bow in awe at your expertise.

 

Unfortunately, giving a response that had nothing to do with the question doesn't enhance that reputation all that much.

 

Triangulation is merely a method of putting two lines on a map, and walking to where the two lines meet.

Too bad the OP was asking about trilateration, which is not the same as triangulation and cannot be done simply in the field with map and compass (unless by "compass" you mean one of those things that draws circles.)

[+Markwell]

Yea - I was going to let the trilateration thing slip.