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triangulation


CiCiS

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we are working on finding a cache that uses triangulation to find the final set of cords. we were given 3 landmarks to find, and with each landmark there is a set distance away from said landmark; where those 3 distances intersect is the final stage. we went out into the field and took cord readings at each landmark. then shot projections from each landmark using the specific distance published on the cache page. we then took cord readings at each of the projections. we found a city map, used a compass and drew the 3 circles on the map. the 3 circles intesect over quite an area and have not pinpointed gz for the final stage. does anyone know of a better way to come up with the final stage?

thanks

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http://www.freemaptools.com/radius-around-point.htm

 

First, convert gps coords to decimal with this:

 

http://boulter.com/gps/

 

Then put in the radius in miles that you want and then put in the coords you want and hit "draw radius"

 

Do that for all three! Remember to put in the new radius BEFORE putting in the new coords and hitting "Draw Radius"

 

The CO probably expected you to do the math...but hey, the internet works great too!

 

This is my cache that is like this:

 

http://www.geocaching.com/seek/cache_details.aspx?wp=GC2YBPR

Edited by jacob501
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we are working on finding a cache that uses triangulation to find the final set of cords. we were given 3 landmarks to find, and with each landmark there is a set distance away from said landmark; where those 3 distances intersect is the final stage. we went out into the field and took cord readings at each landmark. then shot projections from each landmark using the specific distance published on the cache page. we then took cord readings at each of the projections. we found a city map, used a compass and drew the 3 circles on the map. the 3 circles intesect over quite an area and have not pinpointed gz for the final stage. does anyone know of a better way to come up with the final stage?

thanks

ok. thanks for that quick reply. we will give that a try and see what kind of results that brings in! appreciate the promptness. cache on!

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we are working on finding a cache that uses triangulation to find the final set of cords. we were given 3 landmarks to find, and with each landmark there is a set distance away from said landmark; where those 3 distances intersect is the final stage. we went out into the field and took cord readings at each landmark. then shot projections from each landmark using the specific distance published on the cache page. we then took cord readings at each of the projections. we found a city map, used a compass and drew the 3 circles on the map. the 3 circles intesect over quite an area and have not pinpointed gz for the final stage. does anyone know of a better way to come up with the final stage?

thanks

ok. thanks for that quick reply. we will give that a try and see what kind of results that brings in! appreciate the promptness. cache on!

You're welcome! Hope it works!

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There must be something wrong with your calculations somewhere. Three circles on a plane can only intersect at a single point. I have used Mapsource (which came with my Garmin maps). Plot the points. Select each one and tell it to draw a radius and then look where the three intersect. A lot simpler than GE.

 

I would have not bothered with shooting projections. If you have three points and a distance from each one you can solve it.

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There must be something wrong with your calculations somewhere. Three circles on a plane can only intersect at a single point.

 

Um, no.

 

Two circles in a plane can intersect in zero, one, two, or an infinitely large number of points.

 

In the usual trilateration example, they'd intersect in two points.

 

A third circle may touch none, one or both of those two points.

 

Given precise and exact center points and radii, the solution would be a single point. But with only the slightest inaccuracy or imprecision, there would be no single point that touches all three circles. That's what you get when you use real-world data. A good approximation is the best you can get.

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