Help! Spherical Trigonometry To Find A Cache...

January 26, 2006

If a point in the sky was located with an azimuth of 272.2089 degrees and an elevation of 59.3436 degrees and was viewed from an observation tower at 49.3742 degrees from the North celestial pole and 75.3708 degrees west of the Prime Meridian, how would you go about finding the coordinates for the point on the globe that lies directly beneath this point in the sky? Is this enough information to even solve the equation?

The final coordinates should be within a 1/2 mile of 35.1037 degrees N and 113.8879 degrees W...

This cache has been plaguing me for almost 2 months. No one has found it.

TIA,

Dan (AKA BRUZR's Daddy)

Edited January 26, 2006 by BRUZRs_Daddy & Sugar-Pie

January 26, 2006

OUCH!

That sound you just heard was my brain exploding!

Seriously, it has been so long since I had Trig I couldn't begin to tell you what to do. That wasn't one of my favorite classes, it just kept me out of study hall...

Sorry!

January 26, 2006

I can see it in my head, but I don't know trig, much less spherical trig...

The "observation tower" is a point in Nova Scotia (I think) and the final coords are in Arizona. So, that's the azimuth line, right?

January 26, 2006

As far as I can tell the puzzler has given you a location with the degrees from the celestial pole and prime meridian. These are also geographic coordinates. You will need to find the the location. From there I can't see as there is enough information to located a point. You have an azimuth(direction) and an elevation. Unless these directions point to a specific star or building or mountain.

The coordinates that I figured fall on an interestingly named town.

Email me to discuss more

24.7k · January 26, 2006

It's not spherical trig. it's a simple right triangle to solve the equation. When you know the point's coord in the sky, that's the same coord as on the ground.

You really only have information about 2 points and those can be related to each other via the right triangles.

The key thing is to turn your coordinates to UTM which are much simpler unit's to work with on these types of problems.

Plan B. Borrow some astronomy software or find it on the web.

Plan C. Talk to your friendly neighborhood surveyor. Some of them like these problems and work with COGO all the time. (Coordinate Geometry)

I agree with bobbarley. Something is missing to solve the problem.

Edited January 26, 2006 by Renegade Knight

3.8k · January 26, 2006

It seems to me that there needs to be a distance measurement included somewhere.

Plan D: Ask Fizzy.

3.8k · January 26, 2006

Wouldn't you need the length of at least one of the sides of the triangle?

January 26, 2006

use law of cosines

1.3k · January 26, 2006

How high is the tower?

January 26, 2006

use law of cosines

That's what I found on a website on S.T. But it was all greek to me.

January 26, 2006

Here's the waypoint #

GCRJY0

January 26, 2006

Here's an exmple using easier coordinates to get you started:

The north star, Polaris, is always viewed directly overhead(90 degrees from the horizon) at the north pole (latitude, 90 degrees). If you were to move one degree south, the star would appear to have moved one degree toward the horizon. This continues until you get to the equator, at which point polaris touches the horizon. Thus: Your latitude is always equal to Polaris' altitude in the sky.

Let's take an example using the north star as our "point in the sky" and my location, roughly n42 w76.

If you are standing at n42 w76 and spy an object at 42 degrees elevation at 0 degrees azimuth, where is it located overhead?

Well, obviously we are talking about the north star, because the azimuth is 0 degrees (which means north) and the elevation is equal to our latitude. These questions are easy when the azimuth is 0 degrees. Ask yourself: If the elevation is 42 degrees, how far do I have to travel until the elevation is 90 degrees (overhead). The answer is obviously 48 degrees, because 42 is 48 less than 90.

Here's one step harder:

You are standing at n42 w 76 and observe an star at azimuth 0 and elevation 80 degrees. Where is the point on the globe where that star is overhead?

Well, since the object is 10 degrees away from being overhead, we need to walk 10 degrees towards it's azimuth to get under it. So we walk north to n52 w76 and the star is overhead (actually, we would have to teleport because stars move, but you get the idea)

Now, the real hard part comes in when the azimuth is not north. Now we are moving diagonally, so it's not as easy to count degrees. I don't know how to do this math, but hopefully I have helped yo see the problem in a clearer light.

January 26, 2006

that site they give for help explains it....I'm going to try it out even though I will probably never make it our that way. I know trig pretty well but there are a lot more variables involved....thanks for showing me the challenge!

January 26, 2006

that site they give for help explains it....I'm going to try it out even though I will probably never make it our that way. I know trig pretty well but there are a lot more variables involved....thanks for showing me the challenge!

Just remember me when you solve it.

You do the math and I'll hoof it up the mountain for the find... Long distance teamwork!!!

I'll log it on your behalf... Got a TB you want to mail me to put in it?

Edited January 26, 2006 by BRUZRs_Daddy & Sugar-Pie

January 26, 2006

Wouldn't you need the length of at least one of the sides of the triangle?

Actually you have this

The elevation angle converts handily to a distance (length). Subtract the elevation angle from 90, then multiply it by 60 nautical miles per degree and you have the distance in nautical miles. From there, divide by 6000 to get distance in feet and so on to get it in whatever units you want.

Once Upon a time, Tiffany's Head Slave was a ship's navigator (with sextant) and had to learn all of this stuff. If you have more questions, i can try to look it up in some reference books that are at home.

January 26, 2006

Wouldn't you need the length of at least one of the sides of the triangle?

Actually you have this

The elevation angle converts handily to a distance (length). Subtract the elevation angle from 90, then multiply it by 60 nautical miles per degree and you have the distance in nautical miles. From there, divide by 6000 to get distance in feet and so on to get it in whatever units you want.

Once Upon a time, Tiffany's Head Slave was a ship's navigator (with sextant) and had to learn all of this stuff. If you have more questions, i can try to look it up in some reference books that are at home.

So, where does that put the coords to the cache I'm trying to find?!!

January 26, 2006

I'm kinda surprised that no one has pointed out that it is generally considered bad form to get help on a puzzle cache here in the forums. :lol:

5.2k · January 26, 2006

I'm kinda surprised that no one has pointed out that it is generally considered bad form to get help on a puzzle cache here in the forums.

There's already a Godwin's Law but now we need a name for the likelihood of someone interrupting a discussion about a particular puzzle to point out that its bad form to have a discussion about a particular puzzle :lol:

3.8k · January 26, 2006

Godwin's Law has nothing to do with this; it may be considered bad form to solve, or provide solutions to puzzles in these forums, but I personally don't believe it is in poor taste, or bad form, to discuss or solicit help regarding puzzle caches. :lol: :lol:

January 26, 2006

I'm kinda surprised that no one has pointed out that it is generally considered bad form to get help on a puzzle cache here in the forums.

It's called compromise and teamwork.

I offered to do the literal legwork of going to get the cache and giving the mathematician due credit for his/her help.

What's wrong with that?

I have strengths and weaknesses... So do you. If people work together, utilizing each others' strengths to compensate for the weaknesses, I can't see what's wrong with that...

Edited January 26, 2006 by BRUZRs_Daddy & Sugar-Pie

24.7k · January 26, 2006

Wouldn't you need the length of at least one of the sides of the triangle?

Actually you have this

The elevation angle converts handily to a distance (length). Subtract the elevation angle from 90, then multiply it by 60 nautical miles per degree and you have the distance in nautical miles. From there, divide by 6000 to get distance in feet and so on to get it in whatever units you want.

Once Upon a time, Tiffany's Head Slave was a ship's navigator (with sextant) and had to learn all of this stuff. If you have more questions, i can try to look it up in some reference books that are at home.

Don't you need an elevation of either the observed point, or the point you are observing from?

In navigationg a ship I would assume that you can asume your elevation to be more or less Sea Level?

8k · January 26, 2006

Spherical triangles have 3 sides and 3 vertices. Unlike the triangles flat-earthers are familiar with, both the sides and the vertices are measured in angular measurement. Lengths of sides are irrelevent (as they are measured in multiples of the radius of the sphere called radians). If you know any 3 three of the angles you can calculate the others using the law of sines or the law of cosines. The cache page provides enough information to solve the problem. Of course if you'd ask fizzymagic, he would tell you that to get the right answer you would not only have to use the WGS-84 ellipsoid but in fact need to use the geoid if 'directly overhead' has the common meaning. However, it is clear that, in this case, cache owner wants you to assume a spherical earth.

Edited January 26, 2006 by tozainamboku

5.3k · January 26, 2006

Oops. Why does it say Ringbone?

Edited January 26, 2006 by fizzymagic

5.3k · January 26, 2006

There is a distance involved, but it's not made explicit. You have to assume that the star is very far away compared to the radius of the Earth (the lines of sight to the star from anywhere on the Earth are parallel). Also, you have to assume that the Earth is a sphere and that there is no atmospheric refraction. Given those, it is solvable with the information given.

In fact, if you know the radius of the Earth (6366707 meters), you can solve it using GeoCalc to project the waypoint.

January 26, 2006

Wouldn't you need the length of at least one of the sides of the triangle?

Actually you have this

The elevation angle converts handily to a distance (length). Subtract the elevation angle from 90, then multiply it by 60 nautical miles per degree and you have the distance in nautical miles. From there, divide by 6000 to get distance in feet and so on to get it in whatever units you want.

Once Upon a time, Tiffany's Head Slave was a ship's navigator (with sextant) and had to learn all of this stuff. If you have more questions, i can try to look it up in some reference books that are at home.

Don't you need an elevation of either the observed point, or the point you are observing from?

In navigationg a ship I would assume that you can asume your elevation to be more or less Sea Level?

For the purpose of this type of problem (spherical trig solution of celestial nav), you assume obersver is at 0 and the observed object is at infinity. When doing celestial nav via the more traditional sight reduction techniques, you will always make allowance for the "height of eye" which is deck height plus the height of the sextant above it.

5.3k · January 26, 2006

For those of you who find this amusing, here is a related question:

What is the furthest distance this star can be from the Earth? How does that compare to the distance from the Sun to Pluto?

Edited January 26, 2006 by fizzymagic

1k · January 26, 2006

Wouldn't you need the length of at least one of the sides of the triangle?

Actually you have this

The elevation angle converts handily to a distance (length). Subtract the elevation angle from 90, then multiply it by 60 nautical miles per degree and you have the distance in nautical miles. From there, divide by 6000 to get distance in feet and so on to get it in whatever units you want.

Once Upon a time, Tiffany's Head Slave was a ship's navigator (with sextant) and had to learn all of this stuff. If you have more questions, i can try to look it up in some reference books that are at home.

I concur, but use 6080 , not 6000.

edited to correct number.

Edited January 26, 2006 by reveritt

January 26, 2006

There is a distance involved, but it's not made explicit. You have to assume that the star is very far away compared to the radius of the Earth (the lines of sight to the star from anywhere on the Earth are parallel). Also, you have to assume that the Earth is a sphere and that there is no atmospheric refraction. Given those, it is solvable with the information given.

In fact, if you know the radius of the Earth (6366707 meters), you can solve it using GeoCalc to project the waypoint.

Where do I find the correct GeoCalc? I googled it and found a slew of different makes...

5.3k · January 27, 2006

Where do I find the correct GeoCalc? I googled it and found a slew of different makes...

Sorry. It's here. Figure out the arc length in radians, multiply by the radius of the Earth, and you have the distance to project.

January 27, 2006

Where do I find the correct GeoCalc? I googled it and found a slew of different makes...

Sorry. It's here. Figure out the arc length in radians, multiply by the radius of the Earth, and you have the distance to project.

...Annnnnnnnnnnd how do I find the arc length in radians?

5.3k · January 27, 2006

...Annnnnnnnnnnd how do I find the arc length in radians?

That's what the puzzle is about. It wouldn't be appropriate for me to solve it for you. I think that doing a Web search of celestial navigation will help you find the information you are looking for. GeoCalc will eliminate all the trigonometric calculations for you.

January 27, 2006

...Annnnnnnnnnnd how do I find the arc length in radians?

That's what the puzzle is about. It wouldn't be appropriate for me to solve it for you. I think that doing a Web search of celestial navigation will help you find the information you are looking for. GeoCalc will eliminate all the trigonometric calculations for you.

Is this page what I'm looking for?

http://www.themathpage.com/aTrig/arc-length.htm

I don't want you to do the work for me, just hold my hand... :lol:

Edited January 27, 2006 by BRUZRs_Daddy & Sugar-Pie

5.2k · January 27, 2006

Oops. Why does it say Ringbone?

Man, I gotta remember not to drink soda while reading the forums :lol: Now I have to clean my monitor :lol:

January 27, 2006

Oops. Why does it say Ringbone?

Man, I gotta remember not to drink soda while reading the forums Now I have to clean my monitor

I feel like a dork... I actually just looked up ringbone...

I'm still lost. :lol:

11.3k · January 27, 2006

This is exactly why I don't "order" Puzzle caches in my Pocket Queries. :lol:

Puzzles don't even exist in my world.

January 27, 2006

Here's how I approached the problem.

The elevation given can be directly converted to distance from the cache.

We have a bearing from our location to the cache.

We know our coordinates. (be careful these are tricky)

Now all that's necessary is to project a waypoint from our location, to the given bearing and distance.

The answer I came up with was just less than .3 miles from the posted coords. I've emailed the cache owner to see if I'm right.

Nice puzzle, I wish I lived closer.

January 27, 2006

Here's how I approached the problem.

The elevation given can be directly converted to distance from the cache.

We have a bearing from our location to the cache.

We know our coordinates. (be careful these are tricky)

Now all that's necessary is to project a waypoint from our location, to the given bearing and distance.

The answer I came up with was just less than .3 miles from the posted coords. I've emailed the cache owner to see if I'm right.

Nice puzzle, I wish I lived closer.

Is the observation tower in Nova Scotia?

I took the coords as literal:

49deg 22' 27" N

75deg 22' 15" W

Am I wrong?

January 27, 2006

I took the coords as literal:

Did you?

Read his description of the tower location again.

Anyway, I used a different value. I haven't heard back. I may be wrong.

January 27, 2006

I took the coords as literal:

Did you?

Read his description of the tower location again.

Anyway, I used a different value. I haven't heard back. I may be wrong.

Wouldn't the North pole and Prime Meridian each be 0?

January 27, 2006

Wouldn't the North pole and Prime Meridian each be 0?

The Prime Meridian is 0.

January 27, 2006

Wouldn't the North pole and Prime Meridian each be 0?

The Prime Meridian is 0.

North Pole is 90... DUH

January 27, 2006

I think I'm doing the HAPPY DANCE!!!

N ## ##.###, W ### ##.###

Is that what you got?

Edited January 27, 2006 by BRUZRs_Daddy & Sugar-Pie

January 27, 2006

It's very close, I'll email you exactly what I got and how I got there so you can compare.

Thanks for starting this topic. I enjoyed working on the puzzle.

January 27, 2006

Cool... I'm going to edit my post with the coords above...

Edited January 27, 2006 by BRUZRs_Daddy & Sugar-Pie

January 27, 2006

So now I AM doing the HAPPY DANCE!!!

I got an email from the cache owner this morning...

The coords I emailed him for verification are within 12 feet of his actual placement!!!

Thanks to everyone for their help... Especially Fizzy and Tiffany's Slaves!!! Without GeoCalc, I would have never gotten it.

Oh yeah, thanks to VegasCacheHounds, who was basically bagging on me for asking for help in here... It spurred me on. I hate confrontation, but don't know why, as it always inspires me...

Mtngoat50 and I were both working on it last night and I posted my findings, but have since edited the post so that those who took to the challenge of solving this puzzle could still do so...

Edited January 27, 2006 by BRUZRs_Daddy & Sugar-Pie

January 27, 2006

So Tiffany's Slaves get none of the love here.

May your caches be infested with ticks and tupperware eating animals!

January 27, 2006

I said thanks to everyone...

Aren't you part of the "everyone that helped"?

Wait a minute... I'll edit my "Thank You" post...

January 27, 2006

I got the right answer....you email me your answer and I'll tell you if I'm right...........

If the answer comes up to Nova Scotia, I'm willing to be a stunt cacher

January 27, 2006

I was wrong to begin with, when I said Nova Scotia... Someone else made the comment about the coords for the tower location being tricky...

Bottom line is: The tower location is NOT in Nova Scotia. Not to mention that was the starting point for the whole equation. The final coords are in AZ, but that's all I'm gonna say.

Canningclan, did you verify your final coordinates with the cache owner?

Edited January 27, 2006 by BRUZRs_Daddy & Sugar-Pie

February 5, 2006

So now I AM doing the HAPPY DANCE!!!

I got an email from the cache owner this morning...

The coords I emailed him for verification are within 12 feet of his actual placement!!!

Thanks to everyone for their help... Especially Fizzy and Tiffany's Slaves!!! Without GeoCalc, I would have never gotten it.

Oh yeah, thanks to VegasCacheHounds, who was basically bagging on me for asking for help in here... It spurred me on. I hate confrontation, but don't know why, as it always inspires me...

Mtngoat50 and I were both working on it last night and I posted my findings, but have since edited the post so that those who took to the challenge of solving this puzzle could still do so...

Glad to be of service.

BTW, did I ever say that I had a problem? Where did I bag on you, or confront you? All I said was that I was surprised that no one had said that yet. Every other time I've seen someone ask a question like that, people jump all over it, I was just shocked no one had yet.

But hey, if being angry at me for making an observaion spurred you on, then cool, at least I could help.

Help! Spherical Trigonometry To Find A Cache...

Recommended Posts

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment

Link to comment