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Dolphin has an evil idea


Harry Dolphin

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So, we were walking along the Hudson River waterfront in Edgewater, NJ, and noticed that we could see no fewer than eight intersection stations! From Jeffrey Hook Lighthouse to the Empire State building. So, if we compute the angles to each (Okay. Done that already), and post them on a cache page, how hard would this be for a geocacher to find? Hee hee hee. (Have not set out the container yet.)

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Depends on whether the geocachers know about FORWARD or an equivalent program. If they try to do the determination on their handheld using whatever "project" funciton it offers, they will have trouble entering a precise enough bearing (maybe whole degrees) to pinpoint the spot.

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Yes, I like it! But will you use linear or great circle math?

 

I've seen something like it done. It looked like an ordinary cache by a rather clever (some might / have said evil) local geocacher by the name of OtisPug. At the location, there was a bearing and distance. Both were out to about eight decimal places, as I recall. the distance was loooong. I think 500+ miles. I did the projection for a friend going that direction (Ranboze), using several different methods, one of which (Forward) tuned out to be correct (and very close). Cache was found.

 

Here is the cache. . And yes, I still have the numbers, somewhere.... Maybe someday I'll get up there.

Edited by Klemmer & TeddyBearMama
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Only three known intersection stations are needed to compute a position using a procedure known as "resection". Using this method the angle between the left and center intersection is measured and also the angle between the center and right intersection stations. The output is the position of the unknown point where the angles were measured. This was used for many years to position hydrographic survey vessels prior to using electronic methods. For the hydrographic surveys, sextants were used to measure the angles, but the intersection stations were normally quite close. For greater distances, a surveying instrument may be required.

 

For the resection equations, see: http://www.sli.unimelb.edu.au/planesurvey/...c/top10-01.html .

 

For an on-line resection program, see: http://www.civl.port.ac.uk/structures/java...ion/default.htm .

 

If one has more angles to more intersection station, these could provide a check.

 

GeorgeL

NGS

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For the resection equations, see: http://www.sli.unimelb.edu.au/planesurvey/...c/top10-01.html .

 

For an on-line resection program, see: http://www.civl.port.ac.uk/structures/java...ion/default.htm .

 

These probably work for points a short distance away from the unknown point, but what about when the points are so far away that you can't assume a flat earth? or even a spherical earth?

 

Does anyone know of a WGS84 ellipsoid version of this method?

 

Also, the discussion here gives me ideas for a geocache puzzle. I've found the FORWARD and INVERSE programs at the NGS website, but does anyone know of any programs that will do any of the following on a WGS84 ellipsoid:

 

1) compute a point given a bearing from one reference point, and another bearing from another reference point?

 

2) compute a point given a bearing from one reference point and a distance from another reference point?

 

3) compute a point given distances from 2 reference points (actually you would get 2 possible points from this, add a distance or bearing from a third reference point to eliminate ambiguity)?

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For an on-line resection program, see: http://www.civl.port.ac.uk/structures/java...ion/default.htm .

 

If one has more angles to more intersection station, these could provide a check.

 

GeorgeL

NGS

 

It would be nice if this applet actually worked. The resulting network sketch and X coordinate are correct but the Y coordinate isn't even close. I've done the calculations in a stand-alone program so the input data I am using is correct. First email contact on page is non-deliverable, will see if I get a response from second.

Too bad, I had an idea for the geocache using this applet.

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Arrrgh.

 

Went to try out the given formula with a set of simply arranged data points, which I had arranged in a perfect square (for sake of simplicity).

The formula blows up before you can even get started, because computing K1 gives you 1/0 ...

 

You picked the worst possible arrangement, one that is guaranteed to be indeterminate. There is one rule that must be followed:

 

Draw a circle through your three control points. The unknown point cannot lie on that circle. If it does, the solution is indeterminate. Or restated in terms of the following diagram, if (x + y) = 180°, then there is no solution.

 

resection.gif

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You picked the worst possible arrangement, one that is guaranteed to be indeterminate.

 

Thanks for your help on Tienstra's formula.

 

I was actually able to work through an example problem and get the right answer, so I made a cache out if it.

I guess I stole Harry Dolphin's Evil Idea. :P

See GC1B0Q9.

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No one has ever figured it out yet but I have a whole bunch of geocaches that deal with the measurements of the PLSS and some other things I have been playing with all these years.

A

It all began here.

GCD28A THE STONE

 

Joseph C. Brown. GCD3BB

Prospeckt K. Robbins. GCD3B9

 

Roaring River Township SE. GC92D4

 

Now that we have come NORTH and go through the GEOID.

GEOID A GCD065

GEOID B GCD067

 

The GEOID CENTER has a different name.

Columbia Memorial GCCDD0

 

Now the GEOID TOP is Benchmark

CASSVILLE GF0908

 

This is just some of it.

Hope you enjoy the math and the path.

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Klemmer's "Evil One":

Can You Help Me Find My cache?

 

I will be happy to confirm or deny coords from anyone by email (use my GC.com profile). Please don't post them here.

 

I'm right on it!

 

Since you gave bearings to the triangulation stations instead of angles between them, it shouldn't be too hard to solve just using normal trig and algebra..

 

And don't be too certain somebody won't solve it by drawing lines on a map.

 

The FTF of mine did just that, and my tri-stations are much farther away from the cache than yours (one is more than 18 kilometers). I was disappointed...

Edited by Mesa Mike
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Yeah, I was sort of trying to discourage that idea. Depending on how well you do it, it can work. But the math is more fun. I've been working thru the resection equations (on Excel)..... got an answer late last night. OK Lat, but not Long. Must have a typo somewhere.... back to it again tonight (maybe).

 

I'd be interested to see if "standard" trig could work. I just used the azimuth differences and did (am doing...) the resection. The azimuths on the cache page came from NGS INVERSE.

 

One interesting thing I learned last night: INVERSE gives you a Back Azimuth for a good reason - it's not just the complement (180° reverse) of the Forward Azimuth. I'm guessing that is because the ellipsoid is not a perfect sphere. Surveyors: right?

Edited by Klemmer & TeddyBearMama
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The surveyors may have better answers, but one reason the azimuth and back azimuth are rarely the same is because the meridians converge, even on a perfect sphere. Or another way of looking at it is: if you plot a great circle on a Mercator map (or any other map that has parallel meridians), the line is not straight, and so the azimuth and back azimuths are not complements.

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Holo:

 

Riiiighttt.... I agree a Great Circle Route (GCR) is not straight, but on a perfect sphere, wouldn't they be the same curved line? In other words, a GCR from Seattle to Oslo would be the same flight path as from Oslo to Seattle (curved when viewed on a flat paper map). [Edit: Yes, I was a pilot, but that was a good while ago, and anyway , in the planes I flew, we didn't fly far enough over water / remote areas to worry about it]

 

So - an Azimuth being by definition (?) a straight line (line of sight, right?), if it is a GCR, it shouldn't be straight on paper.... Ummm.... I'm getting confused here.... Are the azimuths from INVERSE GCR's or do they go THROUGH the ellipsoid? Or is this all mute at short ranges anyway. The couple miles or so I was dealing with, the differences were in the several minutes of azimuth, at most.

 

Time out for some research during lunch......

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I'll post a map later when I have time, but here is a verbal explanation, if you understand the difference between a rhumb line (a line of constant bearing) versus a great circle route (i.e. the shortest route or line of sight). A rhumb line crosses each meridian at the same angle, so the back azimuth would be exactly 180 from the forward azimuth. A Mercator map is specifically designed to show rhumb lines as straight lines.

 

A great circle is not a rhumb line, because it does not cross each meridian at the same angle. Thus, the back azimuth and forward azimuths won't be 180 degrees apart, even on a perfect sphere. A great circle, or line of sight, on a Mercator map looks something like a sine curve.

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Holo:

 

No need for a map (unless you really feel like it). Between some lunch reading and your post, I'm all refreshed (or as refreshed as I'll get...). I understand now.

 

I'm so used to working with Mercator maps & rhumb lines, I got brain locked on those.

 

Thanks.

 

Klem

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Just for completeness, and because I wanted to see how much difference it makes, here is a map showing the difference between the great circle azimuths (as calculated by INVERSE) versus the rhumb line azimuth, otherwise known as the true course. I calculated the lines between Omaha, Nebraska and Denver, Colorado. These two cities are far enough apart that the difference between the great circle and rhumb lines are just perceptible.

 

For anyone else who has been following this thread, a rhumb line is the path that an airplane would follow if it navigated on a constant bearing, and is the path that most navigators choose for simplicity. If a pilot wanted to fly on a great circle, he would have to constantly adjust his bearing. The map below is projected in a Mercator projection, which was designed for navigation and it shows rhumb lines as straight lines. Although a great circle appears to be curved on this map projection, in reality it is the straightest and shortest distance between two points on the globe.

 

The red line is the great circle, and the blue line is the rhumb. It shows that difference in forward and back geodetic azimuths (i.e. great circle azimuths) is not 180°, but the difference in the forward and back rhumb line azimuths is exactly 180°.

 

azimuths.gif

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Now that is ground measure correct?

 

The distance would be different than if you were to fly above the curvature of the Earth.

Since the Great Circle is a curve then wouldn't it be shorter to fly the Great Circle route?

 

Angular Measurement

 

The measure of the central angle of a circle is defined as the ratio of the subtended arc of the circle divided by the radius, that is, a ratio of two lengths. Thus, this measure is dimensionless but is assigned a special name of radians. Additionally one may express the angle in degrees by noting that an angle of 1 radian equals about 57.3°. The fact that both radian measure and degree measure are dimensionless means that the numerical value of an angle does not change from one system of units to another.

 

The RADIAN

To deep for me.

 

I should have had a v.jpg

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So the shortest distance between two points is a straight line only if the two points are on a truly two-dimensional surface.

An interesting side-note is that the Great Circle would (perhaps obviously?) be south of the rhumb line in the southern hemisphere.

The shortest distance between two points in any situation (called a Geodesic) is usually constrained by the situation. For example, since airplanes cant fly through the earth and thus follow a straight line (in the 3 dimensional sense), the next best thing it can do is fly along the surface of the earth. Then the shortest route is the great circle. So when we say "What's the the shortest route between Omaha and Denver?" we obviously mean "What's the the shortest route between Omaha and Denver without traveling through the earth?"

 

The straight and curved lines on holograph's diagram are purely the result of the projections used. Clearly (again in a 3 dimensional sense) both paths are curved since they follow the surface of the earth. If you picked another projection, you could have the great circles appear straight and the rumb lines appear curved.

 

If you want to extend the concept of "straight line" to curved surfaces, you might say a great circle is a "straight line" when you're constrained to the surface of the earth. Most of us would just say let's just fuggedaboutit.

 

It gets even harder to visualize when you go to 3 and 4 dimensions (as in the case of gravity) but we won't go there. :mad:

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The straight and curved lines on holograph's diagram are purely the result of the projections used.

Indeed. Here is a different map projection that is designed to show all great circle routes to or from Omaha as straight lines. Great circle routes through points other than Omaha would not necessarily be straight lines in this projection. The same two lines are shown on this projection -- the red is the great circle, the blue is the rhumb. However, unless you lay a straight edge against your monitor, your eyes could trick you into thinking either or neither of the lines is straight.

 

azimuths2.gif

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And lets not even talk about the different altitudes aircraft must fly (for a variety of reasons), even while flying a route with a ground track that is a great circle route.

 

Back about 10 posts or so:

Klemmer's "Evil One":

Can You Help Me Find My cache?

 

I will be happy to confirm or deny coords from anyone by email (use my GC.com profile). Please don't post them here.

Two local cachers & three not-local benchmark hunters from this forum solved Klemmer's "Evil Cache". In the case of several, they were close enough to have probably found it. Only one was really exact. Turned out not to be so hard afterall. Oh well. Lots of smart folks out there.

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Nope. Turns out that it's not that evil or difficult.

Triangulation

Published at noon, logged by 3:30. But it still makes for a fun geocache! :sad:

 

Intersection requires only 2 known stations, and is fairly easy.

Resection requires 3 stations, and unless you know Tienstra's Formula, is not so easy.

 

Have any of you read this paper?

 

Make your puzzle harder by not specifying the bearings from your cache to the reference points, but the angles between them instead.

 

As for my puzzle, Help! I've Lost My Cache! the first finder spent many hours with a map and protractor trying to zero in on a point that would give the proper angles. He drove a hundred miles and then searched for an hour before he found it. Lucky for him his final best guess wasn't too far off. The second finder got the coordinates exact after he wrote a computer program to basically automatically do what the FTF did manually. His computer program ran for a while before it decided it had found the best answer.

 

By the way, my encrypted hint is "Dolphin has an evil idea". Let's see if anyone has the presence of mind to search Google to see if they can get any insight on what that means...

Edited by Mesa Mike
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