Coords of Intersection of Two Circles

[+LibDocAbq]

Hello,

I'm working on a problem where I have two known distances from two waypoints. I need to know the intersection coordinates or coordinate of these two circles. There must a simple way, I just can't figure it out.

 

Thanks

[+the pooks]

Google gave me this

 

I once did it by trial and error in the field. Used the projection function of the GPS. Project from both points and adjust angles until they coincide. One does get close witha few iterations.

[+DeepButi]

Use Google Earth. Create the two waypoints and by trial&error with a couple of routes you should get the correct coordinates close enough!

[+redsox_mark]

Yes, Google Earth. With this Circle Generator for Google Earth it will draw circles from your 2 points, you can see where they intersect.

 

I found a nice puzzle cache that way (it gave 3 points and the distances from each of those 3 points).

[+Walts Hunting]

Mapsource. Add both waypoints. When you add them uncheck the proximity box and put in the distance in miles. You might have to do a little division go get the decimal equilavent. This will show a circle around each waypoint. Problem solved

[+sbell111]

The third post of this thread includes a link to a zipped excel document that is supposed to solve this problem.

Edited November 27, 2009 by sbell111

[+fizzymagic]

I'm working on a problem where I have two known distances from two waypoints. I need to know the intersection coordinates or coordinate of these two circles. There must a simple way, I just can't figure it out.

Heh. I think I know which one you're working on.

 

Be aware that for two circles and two distances, there are going to be two points that lie on the intersections of the circles. Three circles and three distances give a unique solution. Any more is just gravy. []

[+DENelson83]

It's called trilateration.

[+jmd65]

There's a nice little iPhone/iPod Touch app called Geocaching Toolkit that does circle intersections and much more. I’ve used it to solve quite a few puzzle caches.

[John E Cache]

Hello,

I'm working on a problem where I have two known distances from two waypoints. I need to know the intersection coordinates or coordinate of these two circles. There must a simple way, I just can't figure it out.

 

Thanks

Not simple, but you can get the angle using "The Law of Cosines".

 

Where the sides of the triangle are:

 a = the distance between the waypoints

 bc = the the two distances to the cache at angle A.

 

Law of Cosines

[+Cardinal Red]

Hello,

I'm working on a problem where I have two known distances from two waypoints. I need to know the intersection coordinates or coordinate of these two circles. There must a simple way, I just can't figure it out.

 

Thanks

Not simple, but you can get the angle using "The Law of Cosines".

 

Where the sides of the triangle are:

 a = the distance between the waypoints

 bc = the the two distances to the cache at angle A.

 

Law of Cosines

 

The Law Of Cosines will solve an oblique triangle. As I read the problem I did NOT envision a triangle here. Did I read it wrong?

[John E Cache]

Hello,

I'm working on a problem where I have two known distances from two waypoints. I need to know the intersection coordinates or coordinate of these two circles. There must a simple way, I just can't figure it out.

 

Thanks

Not simple, but you can get the angle using "The Law of Cosines".

 

Where the sides of the triangle are:

 a = the distance between the waypoints

 bc = the the two distances to the cache at angle A.

 

Law of Cosines

 

The Law Of Cosines will solve an oblique triangle. As I read the problem I did NOT envision a triangle here. Did I read it wrong?

You have three points (2 waypoints and the cache). I was thinking 3 points define a triangle, but I missed the oblique part.

[+BigFurryMonster]

I'm working on a problem where I have two known distances from two waypoints. I need to know the intersection coordinates or coordinate of these two circles. There must a simple way, I just can't figure it out.

 

If you figure it out, you can do my cache next

 

... From each of the waypoints A, B, C and D, all located in the park, 4 pictures were taken (and given). The cache is located at the given distances from these waypoints.

 

http://www.geocaching.com/seek/cache_details.aspx?wp=GC17HVF

[+Dread_Pirate_Bruce]

With two waypoints and two distances, there can be either one or two intersection points. There are two intersection points if the two distances, when added together exceed the straight line distance between the two waypoints. There is one intersection point if the two distances, when added together exactly equal the straight line distance between the two.

 

It seems to me that for a puzzle cache, there must be a unique solution. Therefore, if the cache is at the intersection or is located somewhere based on the intersection, there should be a single intersection point. And, that is on the straight line between the two waypoints, exactly the distance from the one as given by the radius of its associated circle.

[+Markwell]

With two waypoints and two distances, there can be either one or two intersection points. There are two intersection points if the two distances, when added together exceed the straight line distance between the two waypoints. There is one intersection point if the two distances, when added together exactly equal the straight line distance between the two.

If we're going to get nitpicky, the intersection of two circles with different center points can create AT MOST 2 points. You can have two or one, as you said, but you can also have none if the straight line is less than the two given distances. In that case there would be no solution.

It seems to me that for a puzzle cache, there must be a unique solution. Therefore, if the cache is at the intersection or is located somewhere based on the intersection, there should be a single intersection point. And, that is on the straight line between the two waypoints, exactly the distance from the one as given by the radius of its associated circle.

That is, if the person setting up the puzzle realized that there would indeed be likely two solutions. That's a big assumption of knowledge on the part of the puzzle creator. I would hazard that the cache owner just took two figuring that it would be enough not knowing that a third would be necessary for the unique point.

 

However, we have a ready analogy that can help: GPS units working in three dimensions require four known distances from points to accurately determine your location. Two spheres intersect and create a circle. That circle can intersect a third sphere in two points. The fourth sphere (known distance) confirms which of the two points is the correct one. However, your GPS can guess your location with only three satellites by disregarding one of the two, which will be shown to be traveling at an obscene velocity over time. Likewise, if you have two points and one of them is ridiculous (in the middle of a lake, sitting in someone's backyard) and the second point is in the middle of the forest, I'd try the forest point first.

 

Real life example:

If I tell you the point is 5 kilometers from the southern support of the Golden Gate Bridge, and 5 kilometers from the intersection of Locust and Bonita in Sausalito, you should find that the two points are either in San Francisco Bay, or...

[+mountainman38]

Yes, Google Earth. With this Circle Generator for Google Earth it will draw circles from your 2 points, you can see where they intersect.

 

I found a nice puzzle cache that way (it gave 3 points and the distances from each of those 3 points).

 

I used that program for a cache with 4 points and known distances from the cache, but they didn't come close to meeting at a point. The result was a roughly quadrilateral area about 1/2 mile across, which is way too big to start guessing coordinates for.

 

I'm not sure if the KML generator doesn't use enough precision in it's calculations, or if the distances given weren't precise enough to use this method.

[+fizzymagic]

I used that program for a cache with 4 points and known distances from the cache, but they didn't come close to meeting at a point. The result was a roughly quadrilateral area about 1/2 mile across, which is way too big to start guessing coordinates for.

 

You don't say how far apart the original points are, but if it is more than a few miles the difference between a spherical approximation and the WGS-84 ellipsoidal distance can become significant.

 

Also, if you are using cache locations from the Geocaching kml Google Maps, be aware that the coordinates are intentionally fuzzed.

[+RIclimber]

Hello,

I'm working on a problem where I have two known distances from two waypoints. I need to know the intersection coordinates or coordinate of these two circles. There must a simple way, I just can't figure it out.

 

Thanks

If you gave a Garmin 60c, you can use the built in Proximity Alarm, it will put circles around the waypoints on the GPS screen.

[+The Jester]

Hello,

I'm working on a problem where I have two known distances from two waypoints. I need to know the intersection coordinates or coordinate of these two circles. There must a simple way, I just can't figure it out.

 

Thanks

If you gave a Garmin 60c, you can use the built in Proximity Alarm, it will put circles around the waypoints on the GPS screen.

More models than just 60 will do this, I've done it on the GPS 12, and Rino series.

[+mountainman38]

I used that program for a cache with 4 points and known distances from the cache, but they didn't come close to meeting at a point. The result was a roughly quadrilateral area about 1/2 mile across, which is way too big to start guessing coordinates for.

 

You don't say how far apart the original points are, but if it is more than a few miles the difference between a spherical approximation and the WGS-84 ellipsoidal distance can become significant.

 

Also, if you are using cache locations from the Geocaching kml Google Maps, be aware that the coordinates are intentionally fuzzed.

 

Fizzy,

 

I'm working on this cache. I'm not looking for spoilers -- I really want to solve this myself -- but so far I'm getting some pretty poor results.

 

Distance are between 120 and 160 NM, and I've used both the KML generator for Google Earth, as well as MapSource with no joy.

[+fizzymagic]

I'm working on this cache. I'm not looking for spoilers -- I really want to solve this myself -- but so far I'm getting some pretty poor results.

 

Believe it or not, I had already solved this one while testing my trilateration algorithm. I had to use great circle distances instead of ellipsoidal distances to get a good match. That means that the cache owner used a spherical model of the Earth in creating the puzzle.

 

If you are not getting good results with Google Earth, then the problem may be with it using the ellipsoidal estimate.

 

I minimize the sum of squared distances between the target and actual distances, which is arguably not the ideal metric, but which works well for me. For this problem, the difference between my ellipsoidal and spherical solutions was about 600 feet.

 

There may be someplace online that does spherical trilateration for you; I am not familiar with any, but that doesn't mean they don't exist.

 

The other option is to do it yourself using a combination of your method now and FizzyCalc. Here is one method for manual iteration that should work pretty well.

 

First, find an approximate solution where the distances are all right to within a quarter mile or so. Then calculate the distance between one of the reference points and your guess with FizzyCalc in spherical mode. Save the reverse azimuth! Now project the distance you were off along that reverse azimuth from your guessed point. Now do the same for the next reference point, using your new guess. Continue this process with the rest of the reference points. When you finish continue with the first one until the distance you are moving is very small. This process should converge on the correct answer.

 

Here's an example. I am using spherical distances:

Reference Point A: N 47 15.700, W 105 18.500 Distance = 603.994 nm

Reference Point B: N 48 01.000, W 122 20.600 Distance = 558.828 nm

Reference Point C: N 38 10.050, W 112 14.560 Distance = 186.123 nm

 

OK, so let's say I map this and get a guess of N 40 12.341, W 115 15.827

 

Now, from A to the guess is 603.652 nm, rev az = 42.1907 degrees.

So since the target distance is greater than the actual distance, I project from the -0.342 nm at 42.1907 degrees, giving a new estimate of N 40 12.088, W 115 16.128 (yes, FizzyCalc will let you project a negative distance!)

 

From B to this new estimate is 558.703 nm, rev az = 329.393 degrees.

Again the target distance is larger so I project -0.327 nm at 329.393 degrees, giving a new estimate of N 40 11.807, W 115 15.910

 

From C to this new estimate is 185.946 nm rev az = 129.934 degrees.

Target distance is still greater, so I project -0.177 nm at 129.934 degrees, which gives me a new estimate of N 40 11.921, W 115 16.088

 

I start again with Point A. Remember to be careful of the sign of the projection distance; in this case, it will be positive because the distance is now smaller than the target. I won't go through the details, but here are the points you should get by continuing:

 

N 40 11.997, W 115 15.997

N 40 12.001, W 115 16.000

N 40 12.001, W 115 16.000

N 40 12.000, W 115 16.002

N 40 11.999, W 115 16.001

 

As you can see, we have converged. We started out off by 0.366 nm, and in less than 10 iterations we have the correct answer.

 

Try it and let me know how it goes!

[+redsox_mark]

Yes, Google Earth. With this Circle Generator for Google Earth it will draw circles from your 2 points, you can see where they intersect.

 

I found a nice puzzle cache that way (it gave 3 points and the distances from each of those 3 points).

 

I used that program for a cache with 4 points and known distances from the cache, but they didn't come close to meeting at a point. The result was a roughly quadrilateral area about 1/2 mile across, which is way too big to start guessing coordinates for.

 

I'm not sure if the KML generator doesn't use enough precision in it's calculations, or if the distances given weren't precise enough to use this method.

 

In my case, the 3 circles converged on a single point. I was amazed that I was able to determine the exact coordinates (in standard HH.MMM format). The cache page had a geochecker link which told me my coordinates were exactly correct.

 

The cache was Circular Confusion. Note for this one to make sense you need to determine what "THOP", "CFCNCA", and "BB" are.

[+mountainman38]

I'm working on this cache. I'm not looking for spoilers -- I really want to solve this myself -- but so far I'm getting some pretty poor results.

 

Believe it or not, I had already solved this one while testing my trilateration algorithm. I had to use great circle distances instead of ellipsoidal distances to get a good match. That means that the cache owner used a spherical model of the Earth in creating the puzzle.

 

If you are not getting good results with Google Earth, then the problem may be with it using the ellipsoidal estimate.

 

I minimize the sum of squared distances between the target and actual distances, which is arguably not the ideal metric, but which works well for me. For this problem, the difference between my ellipsoidal and spherical solutions was about 600 feet.

 

There may be someplace online that does spherical trilateration for you; I am not familiar with any, but that doesn't mean they don't exist.

 

The other option is to do it yourself using a combination of your method now and FizzyCalc. Here is one method for manual iteration that should work pretty well.

 

First, find an approximate solution where the distances are all right to within a quarter mile or so. Then calculate the distance between one of the reference points and your guess with FizzyCalc in spherical mode. Save the reverse azimuth! Now project the distance you were off along that reverse azimuth from your guessed point. Now do the same for the next reference point, using your new guess. Continue this process with the rest of the reference points. When you finish continue with the first one until the distance you are moving is very small. This process should converge on the correct answer.

 

Here's an example. I am using spherical distances:

Reference Point A: N 47 15.700, W 105 18.500 Distance = 603.994 nm

Reference Point B: N 48 01.000, W 122 20.600 Distance = 558.828 nm

Reference Point C: N 38 10.050, W 112 14.560 Distance = 186.123 nm

 

OK, so let's say I map this and get a guess of N 40 12.341, W 115 15.827

 

Now, from A to the guess is 603.652 nm, rev az = 42.1907 degrees.

So since the target distance is greater than the actual distance, I project from the -0.342 nm at 42.1907 degrees, giving a new estimate of N 40 12.088, W 115 16.128 (yes, FizzyCalc will let you project a negative distance!)

 

From B to this new estimate is 558.703 nm, rev az = 329.393 degrees.

Again the target distance is larger so I project -0.327 nm at 329.393 degrees, giving a new estimate of N 40 11.807, W 115 15.910

 

From C to this new estimate is 185.946 nm rev az = 129.934 degrees.

Target distance is still greater, so I project -0.177 nm at 129.934 degrees, which gives me a new estimate of N 40 11.921, W 115 16.088

 

I start again with Point A. Remember to be careful of the sign of the projection distance; in this case, it will be positive because the distance is now smaller than the target. I won't go through the details, but here are the points you should get by continuing:

 

N 40 11.997, W 115 15.997

N 40 12.001, W 115 16.000

N 40 12.001, W 115 16.000

N 40 12.000, W 115 16.002

N 40 11.999, W 115 16.001

 

As you can see, we have converged. We started out off by 0.366 nm, and in less than 10 iterations we have the correct answer.

 

Try it and let me know how it goes!

 

Hey Fizzy,

 

Thanks for the help! That looks like a lot of work, and reminds me of my surveying days. When I have time I will certainly follow the steps you suggested, and let you know how it turns out.

[Andronicus]

There is a windows mobile app called GateCacher that solves these types of problems. I used it on a puzzle cache and it just automaticaly spitts out the two waypoints. Worked super!

[+fizzymagic]

That looks like a lot of work, and reminds me of my surveying days. When I have time I will certainly follow the steps you suggested, and let you know how it turns out.

 

Although it may look complicated, that's because I tried to be complete in my description and give intermediate results you could check yourself. In reality, doing the process only took me about 10 minutes.

 

It's a good method to solve these kinds of problems.

[+rosebud55112]

With two waypoints and two distances, there can be either one or two intersection points. There are two intersection points if the two distances, when added together exceed the straight line distance between the two waypoints. There is one intersection point if the two distances, when added together exactly equal the straight line distance between the two.

 

Another nitpick: You will have exactly one point of intersection if the sum of the two radii equals the distance between the waypoints OR if the difference of the two radii equals the inter-waypoint difference. If the difference between the radii is greater than the inter-waypoint distance, you will have no intersections. And of course, if you have two waypoints which are the same, and two radii which are the same, you have an infinite number of intersections, and a poor puzzle.

[+Renegade Knight]

...The Law Of Cosines will solve an oblique triangle. As I read the problem I did NOT envision a triangle here. Did I read it wrong?

You read it right. However the puzzle breaks down into triangles and triangles can be solved fairly simply. Especially if you convert the coords involved to UTM.

 

Consider a line between the two known points and a line from each of those points to the unknown point. That's a triagle.

 

Using some coordinate geomotry you can solve the trianges (both that one and right triangles that you sketch in to help.)

 

It's a use for Pythragriams theorem. I can't spell that right...dang it.

[+Renegade Knight]

I used that program for a cache with 4 points and known distances from the cache, but they didn't come close to meeting at a point. The result was a roughly quadrilateral area about 1/2 mile across, which is way too big to start guessing coordinates for.

 

You don't say how far apart the original points are, but if it is more than a few miles the difference between a spherical approximation and the WGS-84 ellipsoidal distance can become significant....

 

Good point. I've only solved these for a mile or two at the most. I've set up larger using UTM which assumes flat but I can see where if someone doesn't assume flat they can come up with a different answer.

[John E Cache]

...The Law Of Cosines will solve an oblique triangle. As I read the problem I did NOT envision a triangle here. Did I read it wrong?

You read it right. However the puzzle breaks down into triangles and triangles can be solved fairly simply. Especially if you convert the coords involved to UTM.

 

Consider a line between the two known points and a line from each of those points to the unknown point. That's a triagle.

 

Using some coordinate geomotry you can solve the trianges (both that one and right triangles that you sketch in to help.)

 

It's a use for Pythragriams theorem. I can't spell that right...dang it.

The Law of Cosines for one triangle is derived by making the two right triangles so you don't have to draw the right triangles. All you need is the three lengths. Since the lengths are geo lengths(not straight lines), I wonder how accurate the computed angle is? Edited December 6, 2009 by John E Cache

[+edscott]

Hello,

I'm working on a problem where I have two known distances from two waypoints. I need to know the intersection coordinates or coordinate of these two circles. There must a simple way, I just can't figure it out.

 

Thanks

 

The math guys have it right, but if you want it 10th grade geometry "simple" just use a good satellite image of a known scale and a drawing compass. There will be two intersections and the cache is at one of them. With some luck one will look like a better location than the other so try it first. You can get the coordinates by clicking on that spot with most map programs or just use the map to navigate to the spot.

 

worked for this one

[+sbell111]

...The Law Of Cosines will solve an oblique triangle. As I read the problem I did NOT envision a triangle here. Did I read it wrong?

You read it right. However the puzzle breaks down into triangles and triangles can be solved fairly simply. Especially if you convert the coords involved to UTM.

 

Consider a line between the two known points and a line from each of those points to the unknown point. That's a triagle.

 

Using some coordinate geomotry you can solve the trianges (both that one and right triangles that you sketch in to help.)

 

It's a use for Pythragriams theorem. I can't spell that right...dang it.

The Law of Cosines for one triangle is derived by making the two right triangles so you don't have to draw the right triangles. All you need is the three lengths. Since the lengths are geo lengths(not straight lines), I wonder how accurate the computed angle is?

I'm sure Fizzy could give you the 'right' answer, but I would imagine that over short distances that the error would be smaller than the normal search area.

[+sbell111]

[+fizzymagic]

Dude, that is awesome!