I am pulling my hair out of my head. Can someone help me?!?!

[ThomasFamily102]

Since we started geocaching we have mastered alot of the puzzles around here. Can someone please explain how to do the triangular logic where you have to find the center point from three given coords. An example is

 

Cache is equidistant.

A. 3880271709

B. 7337678624

C. 6839267149

 

Please go easy when you explain it as I am only 19 and didn't take that class in high school before graduating. Thanks everyone for helping.

[knowschad]

I'd think that Google should be able to help you out: http://www.google.com/search?hl=en&sou...&aqi=g2g-m6

[+fizzymagic]

Since we started geocaching we have mastered alot of the puzzles around here. Can someone please explain how to do the triangular logic where you have to find the center point from three given coords. An example is

 

Cache is equidistant.

A. 3880271709

B. 7337678624

C. 6839267149

 

Please go easy when you explain it as I am only 19 and didn't take that class in high school before graduating. Thanks everyone for helping.

 

There is always one point equidistant from any 3 points on the Earth's surface. Solving for that point can be done in several ways.

 

But the values you gave were not points on the Earth's surface, so I really don't know how to help you. You need 2 numbers to define a point.

[+reveritt]

Since we started geocaching we have mastered alot of the puzzles around here. Can someone please explain how to do the triangular logic where you have to find the center point from three given coords. An example is

 

Cache is equidistant.

A. 3880271709

B. 7337678624

C. 6839267149

 

Please go easy when you explain it as I am only 19 and didn't take that class in high school before graduating. Thanks everyone for helping.

Well, first you need to interpret the coordinates. I would start by assuming that each point contains both longitude and latitude in degrees, as follows:

 

A: N38.802 W71.709

 

Then I would average the three longitudes to get a resulting longitude, and do the same with the latitudes. If you have transcribed the coordinates correctly, and I didn't make a mistake (big assumption there), and my assumptions about format and N/W are correct, then the cache is somewhere in the northern Ungava peninsula, not far south of the Arctic Circle. If that sounds like an unlikely location, then reexamine my assumptions and your data.

[+baloo&bd]

Since we started geocaching we have mastered alot of the puzzles around here. Can someone please explain how to do the triangular logic where you have to find the center point from three given coords. An example is

 

Cache is equidistant.

A. 3880271709

B. 7337678624

C. 6839267149

 

Please go easy when you explain it as I am only 19 and didn't take that class in high school before graduating. Thanks everyone for helping.

 

Cache number?

[+DeepButi]

...

Then I would average the three longitudes to get a resulting longitude, and do the same with the latitudes. If you have transcribed the coordinates correctly, and I didn't make a mistake (big assumption there), and my assumptions about format and N/W are correct, then the cache is somewhere in the northern Ungava peninsula, not far south of the Arctic Circle. If that sounds like an unlikely location, then reexamine my assumptions and your data.

This will have no relationship at all with an equidistant point!!!

 

The true point is "near" the Great Lakes, a more accessible location

[Clan Riffster]

I had a puzzle cache where triangulation was the key. As a twist, I made the three reference points the finals for three other puzzle caches, owned by three other cachers. You had to solve those three first, before you could attempt mine. Rather than make mine equidistant, I gave distance in feet for each point. I.e: 56,204' from point A, 71, 663' from point B, 201, 769' from point C. The solution lay in creating a radius around each point which was the described size, then locating the point where those three circles converged. It was an imperfect puzzle, because it didn't take into account the curvature of the earth, but folks who practiced flat earth solutions were able to get a final within just a couple feet of ground zero.

 

Not sure that this would work with your puzzle though.

[+bflentje]

I solved one like this once in Northfield, MN by simply plotting the points and drawing circles on paper to precision scale, using a precision ruler and fine point instrument and was able to locate the cache within a 100' circle. The container was just a small but I managed to locate it. I guess I got lucky somewhat.

[+reveritt]

...

This will have no relationship at all with an equidistant point!!!

...

You're quite right--I didn't notice the inherent contradiction in the original post, which mentions "center".

[+reveritt]

I solved one like this once in Northfield, MN by simply plotting the points and drawing circles on paper to precision scale, using a precision ruler and fine point instrument and was able to locate the cache within a 100' circle. The container was just a small but I managed to locate it. I guess I got lucky somewhat.

A graphical method like that would be easy if the points were closer together, but these are quite far apart, and the distortion of any map projection is bound to introduce error. I think a mathematical method is needed.

 

I sure would like to read the cache listing for myself.

[+DeepButi]

For a flat surface or small distances (100/200 miles) it should be easy to solve:

 

(LatX - Lat1)^2 + (lonX - lon1)^2 = (LatX - Lat2)^2 + (lonX - lon2)^2 = (LatX - Lat3)^2 + (lonX - lon3)^2

 

but in this case with aprox 2,000miles earth curvature will have a significant impact.

 

Maybe, only maybe, you can solve it for flat, plot it in Google Earth and play a little bit until the right spot is found.

 

There must be a formula out there (Google it) for the spheric distance, but just now I'm myself mad at a fractal image with the hint HINT and not finding it on my trigonometric tool bag .

 

Edited: the formula ... not easy but it can be done applying it three times ...

Cos(a) = Cos(b ) × Cos(c ) + Sin(b ) × Sin(c ) × Cos(A)

 

where

 

* A is an angle measured in degrees. It is the difference in Longitude between points B and C.

* The great circle arc, b, is 90° minus the Latitude of C. This is called the Polar Distance.

* The great circle arc, c, is 90° minus the Latitude of B.

 

Edited January 19, 2010 by DeepButi

[+reveritt]

For a flat surface or small distances (100/200 miles) it should be easy to solve:

 

(LatX - Lat1)^2 + (lonX - lon1)^2 = (LatX - Lat2)^2 + (lonX - lon2)^2 = (LatX - Lat3)^2 + (lonX - lon3)^2

 

...

Even over short distances, this will not work, because the scale (miles per degree) of longitude depends on the latitude.

 

Here is a formula for distance that gives a very good approximation over hundreds or even thousands of miles:

 

Distance (in NM) between points A and B:

 

1) average LatA andLatB; and call the result Q.

 

2) D = 60* SQRT((LatA - LatB)^2 + (cos(Q)*((LongA - LongB)^2)))

 

But knowing how to figure the distance is not much help in solving for the single point that is equidistant from three other points.

[+DeepButi]

reveritt,

you're absolutely right. I wrote without thinking too much (the fractal image is making me really mad )

[+reveritt]

If you are running a Windows machine, you might want to grab a free copy of FizzyCalc, which is a neat little utility that might be helpful in solving problems like this.

[+DeepButi]

I had to free my mind of the fractal image for a while or I will get (more) crazy. So, I used some of my spare time to this one ...

 

We have three "C" points and we will apply the distance formula three times:

 

Cos(a) = Cos(b ) × Cos(c1) + Sin(b ) × Sin(c1) × Cos(A)

Cos(a) = Cos(b ) × Cos(c2) + Sin(b ) × Sin(c2) × Cos(A+K2)

Cos(a) = Cos(b ) × Cos(c3) + Sin(b ) × Sin(c3) × Cos(A+K3)

 

As you can see ( ), I used A+K2 and A+K3 to designate the diferences in longitude.

 

Now let's equal the formula for c1/c3:

 

Cos(b ) × Cos(c3) + Sin(b ) × Sin(c3) × Cos(A+K3) = Cos(b ) × Cos(c1) + Sin(b ) × Sin(c1) × Cos(A)

 

divide by Cos(b )

 

Cos(c3) + tg(b ) × Sin(c3) × Cos(A+K3) = Cos(c1) + tg(b ) × Sin(c1) × Cos(A)

tg(b ) × (Sin(c3) × Cos(A+K3) - Sin(c1) × Cos(A)) = Cos(c1) - Cos(c3) ----> (1)

 

the same process for c1/c2 will give us

 

tg(b ) × (Sin(c2) × Cos(A+K2) - Sin(c1) × Cos(A)) = Cos(c1) - Cos(c2) ----> (2)

 

Now we divide (1) and (2) and use

Cos(x+y) = Cos(x)Cos(y)-Sin(x)Sin(y)

to get

 

(Cos(c1)-Cos(c2)) (Sin(c3) (Cos(A)Cos(K3)-Sin(A)Sin(K3)) – Sin(c1)Cos(A)) =

(Cos(c1)-Cos(c3)) (Sin(c2) (Cos(A)Cos(K2)-Sin(A)Sin(K2)) – Sin(c1)Cos(A))

On this formula there is only one unknown value: A.

All other values -c1, c2, c3, K2, K3- are known so we will end up with something similar to

 

Value1 * CosA = Value2 * SinA

 

so

 

tgA = some value

 

and we will get A ...

 

the remaining is easy ...

 

 

(I guess how many mistakes did I make!!!???).

Edited January 20, 2010 by DeepButi

[ThomasFamily102]

Holy cow....I SAID GO EASY LOL. The cache number is GC12W45.

[ThomasFamily102]

...

Then I would average the three longitudes to get a resulting longitude, and do the same with the latitudes. If you have transcribed the coordinates correctly, and I didn't make a mistake (big assumption there), and my assumptions about format and N/W are correct, then the cache is somewhere in the northern Ungava peninsula, not far south of the Arctic Circle. If that sounds like an unlikely location, then reexamine my assumptions and your data.

This will have no relationship at all with an equidistant point!!!

 

The true point is "near" the Great Lakes, a more accessible location

 

Nice guess we live in Auburn Indiana.

[+Renegade Knight]

Since we started geocaching we have mastered alot of the puzzles around here. Can someone please explain how to do the triangular logic where you have to find the center point from three given coords. An example is

 

Cache is equidistant.

A. 3880271709

B. 7337678624

C. 6839267149

 

Please go easy when you explain it as I am only 19 and didn't take that class in high school before graduating. Thanks everyone for helping.

 

Step one may be figuring out how those numbers relat to a location since they don't fit any standard way of displaying a location.

 

That said, if A, B, and C are known points and the numbers are distances (in mm?) ...then you have something you can solve.

[+Renegade Knight]

...

Well, first you need to interpret the coordinates. I would start by assuming that each point contains both longitude and latitude in degrees, as follows:

 

A: N38.802 W71.709...

 

Good thought.

[+reveritt]

OK, the solution to the puzzle is the CIRCUMCENTER of the triangle formed by the three points, A, B, and C.

 

For any 3 points that are not all on the same line, there is exactly one circle that passes through all the points. The center of that circle will, by definition, be equidistant from each of the 3 points. That center point is the circumcenter of the triangle formed by the 3 points.

 

I will try to look at this some more in the near future. I suggest you look for methods for finding the circumcenter of a triangle, given the coordinates of the 3 vertices.

[+PirateKatz]

Maybe I just haven't had enough coffee today but...

 

Is it even possible to find the center of three points while using geographic coordinates, without getting into some very complex math to correct for varying X/Y scales and not working on a Cartesian plane? Geographic coordinates are incredibly cumbersome for determining length & area.

 

The cartographer in me says to project the three points into an equidistant projection appropriate for the location of the cache and then solve from there using basic geometry. Though, I can't believe any cache would require this... :-)

[+DeepButi]

Look at the hint: Earth is flat

 

No need to use spheric trigonometry? Neither "real" distances?

 

Then it should be a lot easier, my first set of formulas probably would suffice (the "wrong" ones).

[+fizzymagic]

For any 3 points that are not all on the same line, there is exactly one circle that passes through all the points.

 

Here's the cool thing: in planar geometry, the above is true, but for spherical geometry, there is always a point equidistant from the three points. In fact, if the points lie on a line (which would be a great circle) then there are 2 points equidistant from them.

 

There is actually a closed-form solution for the circumcenter problem on a sphere, but it's pretty nasty. I've built a general-purpose solution to these problems by getting a good estimate and then iterating to the exact solution. The advantage of my approach is that it works with the real (ellipsoidal) distances just as well as it does with the spherical approximation.

 

I haven't made my general solution app publicly available because I don't want to ruin these kind of puzzles for the creators, but recently I have friends trying to get me to sell it. I dunno.

[+fizzymagic]

...

Well, first you need to interpret the coordinates. I would start by assuming that each point contains both longitude and latitude in degrees, as follows:

 

A: N38.802 W71.709...

 

Good thought.

Unfortunately, that gives a solution about 1140 miles from the posted coordinates. So I think that interpretation is likely to be wrong.

 

ETA: Solved it using Vincenty distances, which gives a solution good enough to validate. Here's a small hint that I don't think gives anything away: the points are all with 15 miles of the posted coords.

Edited January 20, 2010 by fizzymagic

[+reveritt]

...

Unfortunately, that gives a solution about 1140 miles from the posted coordinates. So I think that interpretation is likely to be wrong.

Yeah, that was my impression, as well.

ETA: Solved it using Vincenty distances, which gives a solution good enough to validate. Here's a small hint that I don't think gives anything away: the points are all with 15 miles of the posted coords.

Well, I'm not going to bust my brain any more trying to figure out how to interpret what is posted in the cache listing. The math is the fun part.

[knowschad]

I'd think that Google should be able to help you out: http://www.google.com/search?hl=en&sou...&aqi=g2g-m6

 

 

I guess I oversimplified this a mite, huh?

[ThomasFamily102]

Since we started geocaching we have mastered alot of the puzzles around here. Can someone please explain how to do the triangular logic where you have to find the center point from three given coords. An example is

 

Cache is equidistant.

A. 3880271709

B. 7337678624

C. 6839267149

 

Please go easy when you explain it as I am only 19 and didn't take that class in high school before graduating. Thanks everyone for helping.

Well, first you need to interpret the coordinates. I would start by assuming that each point contains both longitude and latitude in degrees, as follows:

 

A: N38.802 W71.709

 

Then I would average the three longitudes to get a resulting longitude, and do the same with the latitudes. If you have transcribed the coordinates correctly, and I didn't make a mistake (big assumption there), and my assumptions about format and N/W are correct, then the cache is somewhere in the northern Ungava peninsula, not far south of the Arctic Circle. If that sounds like an unlikely location, then reexamine my assumptions and your data.

 

How can West be 71.709? Isn't there only 60 degrees allowed?

[ThomasFamily102]

Wow. Not asking for a handout answer for the coordinates but I graduated with just Algebra 2 knowledge. None of these classses you are talking about lol. You guys are giving me some helpful hints tho. I just gotta work out how to plug all these numbers into it so it can be solved!! Everything is easier when you can just plug the original equation into the computer and there you have it. Like suduko puzzles. Thats how I got through most classes in high school.

[+bittsen]

Want to know how to solve a puzzle like that?

 

Ignore it.

 

 

 

Not to be mean but if you look at a puzzle and can't solve it, ignore it. Why stress yourself out over a smiley face on a cache map?

[ThomasFamily102]

Want to know how to solve a puzzle like that?

 

Ignore it.

 

 

 

Not to be mean but if you look at a puzzle and can't solve it, ignore it. Why stress yourself out over a smiley face on a cache map?

 

Kinda just wanna learn how to do them because there are about twenty of them this way.

[+reveritt]

...

How can West be 71.709? Isn't there only 60 degrees allowed?

Both longitude and latitude coordinates can be up to 90 degrees. I personally live at 71 degrees, 18.5 minutes west longitude, and you apparently live well west of that.

 

That said, however, Fizzymagic pointed out above that my interpretation of the data is probably incorrect.

[+DeepButi]

...

How can West be 71.709? Isn't there only 60 degrees allowed?

Both longitude and latitude coordinates can be up to 90 degrees. I personally live at 71 degrees, 18.5 minutes west longitude, and you apparently live well west of that.

 

That said, however, Fizzymagic pointed out above that my interpretation of the data is probably incorrect.

ehem ...

 

Latitude goes from 0º (equator) to 90ºN (North pole) or 90ºS (South pole)

longitude goes from 0º (Greenwich Meridian) to 180ºW or 180ºE

 

[+reveritt]

...

Latitude goes from 0º (equator) to 90ºN (North pole) or 90ºS (South pole)

longitude goes from 0º (Greenwich Meridian) to 180ºW or 180ºE

 

Quite right. Perhaps I should have another coffee.

 

I'm still looking for some suggestions on how to interpret the coordinates in the cache listing. If my initial idea was incorrect, then what do the numbers mean? They don't seem to be UTM, or MGS

[+reveritt]

Perhaps they are UTM offsets to be applied to the posted coordinates. This problem would be much easier to solve with UTM coordinates and plane geometry ("Earth is flat", says the hint).

 

Perhaps Point A (3880271709) is two offsets, 38802 and 71709. Yeah--I like that approach.

[ThomasFamily102]

If it helps I live at N 41 and W 085

[knowschad]

Want to know how to solve a puzzle like that?

 

Ignore it.

You didn't wait for us to answer!! I would also ignore it (as I have been pretty much doing with this thread) but some people like the challenge. Remember that ice cream thing? Edited January 21, 2010 by knowschad

[+reveritt]

I have the solution as a UTM coordinate set, but when I use FizzyMagic's Fizzycalc utility to convert UTM to Lat/Lon, I get a bogus latitude.

 

My steps:

1. divide each of the three datapoints given in the cache listing into two 5-digit numbers, which are assumed to be offsets (easting and northing, respectively) to the posted cache UTM coordinates.

 

2. Add the offsets to the cache UTM coordinates to get UTM coordinates for 3 new points, which are assumed to be equidistant from the actual cache location.

 

3. Using the UTM coordinates, solve for the circumcenter of the triangle formed by the 3 points. I used the formula in this Wikipedia article (the Cartesian coordinates solution)

 

4. Check the distances between the solution point and each of the three triangle points, using Pythagorean theorem. The three distances are all 18227.31 meters (about 11.3 miles).

 

5. Check solution with Evince, using link on cache listing. Solution is incorrect.

 

6. Jump off tall building.

Edited January 21, 2010 by reveritt

[ThomasFamily102]

GRRRRRRRRRRRRRR!!!!! Someone just create a cache puzzle solver for ones you can't solve on your own!!!

[+reveritt]

GRRRRRRRRRRRRRR!!!!! Someone just create a cache puzzle solver for ones you can't solve on your own!!!

Hey TF102, calm down; some of us are doing our best to help.

[ThomasFamily102]

My bad!

[+reveritt]

I tried subtracting the assumed UTM offsets from the Cache coordinates (step 2 in my post above), which landed me in the middle of a cornfield with no joy.

[uperdooper]

[+reveritt]

Hey TF102,

 

I am running out of ideas. At this point, if I were trying to find this cache, I would ask the cache owner to review this thread and offer a hint. However, it is your thread, and you are the one looking to find the cache, so that is your decision. I am happy to keep working on the puzzle, but I would like to know if I am getting warm, or if I am completely out in left field. (hey--may that's where the cache is hidden--out in left field!)

[ThomasFamily102]

We do have alot of baseball fields around here. I will shoot him an email and see if he wants to give some insight

[7rxc]

Holy cow....I SAID GO EASY LOL. The cache number is GC12W45.

First off... Thanks for providing the Cache number... that was really a help.

Second... Thanks for providing an interesting puzzle... I'm a bit limited in my math skills myself.

Not just lackadaisical, but much out of practice... been years since I did this sort of thing.

Point being, I left the math to the knowledgable, and went to see if there was another way...

 

Short answer is... There is, I tried methods outlined in the thread, and got similar weird results.

I did sort it out, and my advice to ALL... I misread one number on the first run and it haunted me till today.

I also didn't read the numbers accurately I entered (fine print) on my choice of processor. That also cost me...

Details, Details, Details. Be accurate... I likely would have had it back at the begining, but for that.

Of course I was more interested in confirming my ideas on what was represented than accuracy... or finding the cache, since I'm way across and out of country too.

But I verified this a.m. so there is a different approach that works. it's still math as in geometry etc. but not a fancy formula.

 

Good luck Doug 7rxc

Edited January 23, 2010 by 7rxc

[4wheelin_fool]

You could always use the redneck approach and enter all three points in your gps and then go to a likely spot on the map. Next hit "go to" in sequence for each one until they are equidistant.

 

You could also use a electric razor to shave your head which would be preferable to forcefully pulling on it.

[ThomasFamily102]

You could always use the redneck approach and enter all three points in your gps and then go to a likely spot on the map. Next hit "go to" in sequence for each one until they are equidistant.

 

You could also use a electric razor to shave your head which would be preferable to forcefully pulling on it.

 

Hm...using an electric razor on my head???LOL I am a girl. Might look a little funny

[+thedeadpirate]

You could always use the redneck approach and enter all three points in your gps and then go to a likely spot on the map. Next hit "go to" in sequence for each one until they are equidistant.

 

You could also use a electric razor to shave your head which would be preferable to forcefully pulling on it.

 

Hm...using an electric razor on my head???LOL I am a girl. Might look a little funny

 

Depends on how much you shave off.

 

[ThomasFamily102]

HAHAHAHAHAHAHAA I think I will leave my hair alone lol! Thank you very much!

[+mountainman38]

GRRRRRRRRRRRRRR!!!!! Someone just create a cache puzzle solver for ones you can't solve on your own!!!

 

What's the point, then? There are lots of easier caches to go out and find. For me, the harder the cache is to find, the more rewarding.

 

Having a program do all the work would mean you didn't really solve it, and that defeats all the work the CO did to make the puzzle.