calculating waypoints

[+Maasoz]

Hello,

 

With one geocache I have the following problem. I have to calculate a waypoint with starting point a previous waypoint.

The distance to that waypoint is known by me and the the latitude How can I create the new waypoint with my garmin GPSmap 60CSx?

[+Bear and Ragged]

It's awhile since I used mine, but Project Waypoint is what you need to look for...

 

Or.

If you have the coordinates, http://www.fizzymagic.net/Geocaching/FizzyCalc/ is a handy little programme on the computer!

[AZcachemeister]

I don't think waypoint projection will 'get us there' on this problem.

We need to find the point on a line that is a specific distance from another specified point.

 

I'm sure there are mathematical ways to do this, but I think I would use some mapping software (on my PC) to 'draw a line' on the map.

You could also use the 'battleship' method on your GPSr by marking waypoints along the specified latitude line until you start getting close to the specified distance from the given point. This would be easiest if you are standing at the specified waypoint. Or you could make a GOTO for the specified point and start walking the latitude line until the distance matches.

[+cx1]

Complain to the cache owner that this is a poor puzzle concept

 

Projecting waypoints can be terribly complicated. Some projection models take into account the not perfectly round shape of the Earth and some don't. So depending on how it is done and where any rounding occurs you can get two different data points using different tools, yet both would be 'correct'.

 

I discovered this while attempting to create a puzzle cache of my own. Seemed simple enough, start with a given coordinate and then give a list of bearings and distances to get to the final location. But bearings and distances used to get to the correct final location that would work using one formula would start to drift in another formula and the final location would be slightly different. And since my puzzle solutions must be exact (I hate iffy solutions) I gave up on the idea.

 

My serious advice...

Use Google Earth. Plot and pin your previous waypoint on the map.

Use the ruler tool and place one end on your previous waypoint and stretch the ruler line to the distance you were given.

Maintain that stretched distance while moving the free end of the tool to the given latitude.

Make note of the longitude provided by the nice folks at Google.

Sorry that this does not make direct use of the Garmin device you listed. The method AZcachemeister suggested would work with your device as you asked but requires you be out in the field. I post this only as an alternative to all the extra walking.

[+RobDJr]

Projecting waypoints can be terribly complicated. Some projection models take into account the not perfectly round shape of the Earth and some don't. So depending on how it is done and where any rounding occurs you can get two different data points using different tools, yet both would be 'correct'.

 

Doesn't that depend heavily on the distance involved? Projecting long range might have that issue, but within a reasonably short distance, not so much?

 

I'm no math whiz, but this strikes me as a problem that might have a purely mathematical approach, especially if in short distance. Maybe using UTM to change it to a rectangular system as a simplistic approach?

 

If the distance involved isn't too much, my starter approach would be to plot a circle with a radius of the distance and centerpoint of the point given, and see where it intersects the latitude line given. That brings me to one thing I dislike about this kind of thing, that there will be two solutions. Hopefully one can be eliminated as impossible.

[+cx1]

 

Doesn't that depend heavily on the distance involved? Projecting long range might have that issue, but within a reasonably short distance, not so much?

 

You know, that is exactly what I thought too when I came up with the puzzle idea.

But it didn't take much.

The longest distance involved was I believe 1700 feet.

The puzzle was to project 9 waypoints in sequence based on the previous waypoint.

I knew from experience that some fellow cachers use an iphone app called geocaching toolkit(iGCT) for projecting waypoints along with many other useful functions. I myself use it while doing field puzzle to make sure a solution I get is not going to take me somewhere obviously 'wrong' because it can quickly map the new waypoint.

When I am at home I often use (and know other cachers also use) the tools at GPSVisualizer for waypoints, converting formats etc.

 

So to use a fresh example:

Given coordinate N 39 27.943, W 087 21.237

Distance 1700 feet at a bearing of 75 degrees.

 

iGTC solution N 39 28.015, W 087 20.887

GPSVisualizer N 39 28.015, W 087 20.888

 

Not much difference granted, but add another distance and bearing to those solutions and then again and again for a total of nine times and the difference was I believe 4 digits on the north and 3 digits on the west. Occasionally they would give the same result but more often they would not. And since I could not force a finder to use a particular provider of waypoint projection and could not even mention the problem since part of the puzzle was to figure out what the numbers(the bearings and distances) were for it seemed like a lost cause for a puzzle idea.

[AZcachemeister]

I think I would use some mapping software (on my PC) to 'draw a line' on the map.

 

 

My serious advice...

Use Google Earth. Plot and pin your previous waypoint on the map.

...

I post this only as an alternative to all the extra walking.

 

I think I suggested that approach!

 

If I was 50/100 miles from home without my laptop (an unlikely scenario), doing a bit of walking might be the quicker solution.

 

 

Doesn't that depend heavily on the distance involved? Projecting long range might have that issue, but within a reasonably short distance, not so much?

 

...4 digits on the north and 3 digits on the west...

 

Of course you realize 0.001 minute is only about 5.28 feet, so the approximate error (on the part of the projections) would only be about 20 feet or so. Unless your hide was an evil 'needle-in-a-haystack', I would consider that 'close enough'.

I certainly do applaud your desire for accuracy, though.

[+cx1]

 

Of course you realize 0.001 minute is only about 5.28 feet, so the approximate error (on the part of the projections) would only be about 20 feet or so. Unless your hide was an evil 'needle-in-a-haystack', I would consider that 'close enough'.

I certainly do applaud your desire for accuracy, though.

 

Well it is true that 20 feet isn't much...

But even though I am very careful determining my coordinates for my puzzles by using 2 or 3 gps units and taking several readings from each and calculating a best average out the data I feel I still need to allow for my coordinates to be off by 15 feet.

And considering typical finders GPS units may also be off by 15 feet or more when they arrive at a GZ that gives the potential now with the incorrect data from the puzzle solution of where they think they are to where I think they should be at over 50 feet apart.

After solving one of my puzzles that much variance would be mean to do.

[+fizzymagic]

Projecting waypoints can be terribly complicated. Some projection models take into account the not perfectly round shape of the Earth and some don't. So depending on how it is done and where any rounding occurs you can get two different data points using different tools, yet both would be 'correct'.

 

No, only one would be correct. That would be the one that uses the correct nmodel.

 

The geodesic distance uses the correct model and gives the correct answers, although people often forget to include the elevation in such puzzles, so they get slightly inaccurate answers. But in no way is there more than one "correct" answer. Using a spherical model of the Earth is incorrect. Period.

[+fizzymagic]

Not much difference granted, but add another distance and bearing to those solutions and then again and again for a total of nine times and the difference was I believe 4 digits on the north and 3 digits on the west.

 

And you thought this was problem with the projection algorithms and not your method? Good thing you didn't do the puzzle.

 

The problem here is that you rounded intermediate results, not anything wrong with the projections. Rounding to a 5-foot precision 9 times will give an expected error of about 15 feet.

 

If I were you, I would refrain from giving other people advice about projections and what is and is not a good puzzle for the time being.

[+cx1]

Not much difference granted, but add another distance and bearing to those solutions and then again and again for a total of nine times and the difference was I believe 4 digits on the north and 3 digits on the west.

 

And you thought this was problem with the projection algorithms and not your method? Good thing you didn't do the puzzle.

 

The problem here is that you rounded intermediate results, not anything wrong with the projections. Rounding to a 5-foot precision 9 times will give an expected error of about 15 feet.

 

If I were you, I would refrain from giving other people advice about projections and what is and is not a good puzzle for the time being.

 

Snobby, but accurate perhaps.

I guess you missed the laughing smiley in my first post. One might take it that I was joking about the cache owner and relating an humorous but frustrating experience I personally have had with using projection techniques. Clearly you did not.

I believe I also mentioned possible issues with rounding, but hey thanks for pointing that back out to me.

I also explained how because I knew how other people would try and attempt to solve my puzzle which involves an app that only goes to D M.M in precision and comparing the results from that to an interface that allows D.DDDDDDDDDD I was getting varying results. They were both 'correct' in that they were 'close enough' for a geocache.

 

If I were you, I would refrain from.....nah I'm not going to be snooty like that

[+fizzymagic]

I guess you missed the laughing smiley in my first post. One might take it that I was joking about the cache owner and relating an humorous but frustrating experience I personally have had with using projection techniques. Clearly you did not.

 

I did notice that you said waypoint projections make bad puzzles because waypoint projection is so hard.Was that a joke? Sure didn't look like it.

 

I believe I also mentioned possible issues with rounding, but hey thanks for pointing that back out to me.

 

It wasn't a possible issue, it was the issue.

 

I didn't notice you saying that puzzles subject to cumulative rounding errors were bad. You said, instead, that projection puzzles were bad.

 

Sorry, but I tend to get snippy when people say incorrect things about math and geodesy. There is too much misinformation out there already, and since the correct information is quite accessible, I tend to point out the errors.

 

Projection puzzles are just fine, and your advice to the OP to tell the CO that the puzzle is a bad one was inappropriate.

[+cx1]

 

Projection puzzles are just fine, and your advice to the OP to tell the CO that the puzzle is a bad one was inappropriate.

Well at least I tried to help the OP.

I did offer advice how to roughly accomplish what he was wanting.

Where was your help to the OP?

Oh wait, why be helpful when you can be 'snippy'.

[+rosebud55112]

Isn't another issue with the situation as described by the OP the fact that two points legitimately meet the criteria?

 

"The distance to that waypoint is known by me and the the latitude"

 

So if I'm at N 45 00.000 W 90 00.000 and I am trying to find "the" point 800 feet away on the latitude N 45 00.050, then there is a point at that latitude with a longitude of W 89 59.something and another at W 90 00.somethingelse that also works. One would be NE of the starting point and the other would be NW of it.

 

If you know the distance and azimuth, then you have a unique point.

[AZcachemeister]

Isn't another issue with the situation as described by the OP the fact that two points legitimately meet the criteria?

 

GOOD CATCH!

 

Indeed you are correct, there are TWO POINTS that are XXX distance from a given waypoint along a given latitude line.

 

This is going to take a lot more walking than I thought...

[+edscott]

...

My serious advice...

Use Google Earth. Plot and pin your previous waypoint on the map.

Use the ruler tool and place one end on your previous waypoint and stretch the ruler line to the distance you were given.

Maintain that stretched distance while moving the free end of the tool to the given latitude.

Make note of the longitude provided by the nice folks at Google.

Sorry that this does not make direct use of the Garmin device you listed. The method AZcachemeister suggested would work with your device as you asked but requires you be out in the field. I post this only as an alternative to all the extra walking.

 

Yes this is the easy way. It is a bit haphazard to do in the field but a piece of cake on a desktop. If you are trying it on the fly for a multi be sure to preprint a map that is large enough to cover the area where you expect all the stages to be. If you guess wrong you are looking at finishing it on another trip.

[+jellis]

It's awhile since I used mine, but Project Waypoint is what you need to look for...

 

Or.

If you have the coordinates, http://www.fizzymagic.net/Geocaching/FizzyCalc/ is a handy little programme on the computer!

Yes I use the Fizzycalc a lot. Saved me many times.

I use a Garmin 60CSX and some other Garmin models have a tool called Sight n Go which is good too.

Edited December 29, 2011 by jellis

[+rjb43nh]

AZcachemeister-"Indeed you are correct, there are TWO POINTS that are XXX distance from a given waypoint along a given latitude line."

Not totally correct. If the longitude of the point you're looking for at XXX distance is the same as the original waypoint's longitude, the latitude line you draw will be tangent to the circle and there will be a unique point. The 2 cases this will happen is when the points are either directly north (0°) or directly south (180°)of the original.

[+RobDJr]

Isn't another issue with the situation as described by the OP the fact that two points legitimately meet the criteria?

 

GOOD CATCH!

 

Indeed you are correct, there are TWO POINTS that are XXX distance from a given waypoint along a given latitude line.

 

This is going to take a lot more walking than I thought...

Hey hey HEY! I said it first!

 

I was aware that there could be a unique solution, but I discounted that for a projection puzzle because it would simply be a solution of "walk this many feet north or south".

 

As long as the location of the final wasn't a needle in a haystack type rock field or something similar, I think would be fairly comfortable with the error introduced by simply projecting waypoints in the field using my unit. I can imagine, for hiding, one would want to be more accurate.

 

I've found that, for projecting up to a thousand feet, in my experience the error is negligible as far as searching is considered. So at what point would the error go beyond what would be considered reasonable?

[AZcachemeister]

AZcachemeister-"Indeed you are correct, there are TWO POINTS that are XXX distance from a given waypoint along a given latitude line."

Not totally correct. If the longitude of the point you're looking for at XXX distance is the same as the original waypoint's longitude, the latitude line you draw will be tangent to the circle and there will be a unique point. The 2 cases this will happen is when the points are either directly north (0°) or directly south (180°)of the original.

 

A possibility I hadn't considered, but certainly could be the case.

Perhaps this is how the CO got around the 'bad puzzle design' problem?

 

 

Hey hey HEY! I said it first!

 

I was aware that there could be a unique solution, but I discounted that for a projection puzzle because it would simply be a solution of "walk this many feet north or south".

 

As long as the location of the final wasn't a needle in a haystack type rock field or something similar, I think would be fairly comfortable with the error introduced by simply projecting waypoints in the field using my unit. I can imagine, for hiding, one would want to be more accurate.

 

I've found that, for projecting up to a thousand feet, in my experience the error is negligible as far as searching is considered. So at what point would the error go beyond what would be considered reasonable?

 

Oops! How did I miss seeing that? My apologies!

 

I've not seen a local projection that was more than a couple tenths of a mile...walking distance.

I do recall reading about a cache that required a projection of 20-30 miles. If I recall correctly it was somewhere in So Cal, the final was up in the Sierras...not a lot of finders.

[+niraD]

I've not seen a local projection that was more than a couple tenths of a mile...walking distance.

I do recall reading about a cache that required a projection of 20-30 miles. If I recall correctly it was somewhere in So Cal, the final was up in the Sierras...not a lot of finders.

There used to be a nearby mystery/puzzle cache that required you to find the point where two projected lines intersected. The reference points for the projections were a few miles away, which was far enough that you really needed to use an accurate projection algorithm.