+Team On Posted January 6, 2019 Share Posted January 6, 2019 Just out of curiosity. Apparently it is easy to type coordinates into a field, choose distance and bearing, press a button and then get the result for a waypoint projection. But how is that mathematically done? What is that probably super long formula? It must be more than rather simple geometry for several reasons ? Quote Link to comment
+arisoft Posted January 6, 2019 Share Posted January 6, 2019 (edited) The formula is "Vincenty's formulae" You can read all dirty details from here https://en.wikipedia.org/wiki/Vincenty's_formulae This is the correct formula because we are using WGS84 coordinates with geocaching and this formula especially is capable of calculating correct coordinates on the surface of a WGS84 spheroid. Edited January 7, 2019 by arisoft 1 Quote Link to comment
medoug Posted January 8, 2019 Share Posted January 8, 2019 (edited) For short distances at a given latitude, you can get really close if you know the change in N coordinates for a given number of feet in the north direction and change in W coordinates for a given number of feet in the west direction. Then you can use a couple very simple trigonometry formulas to convert a distance and heading to the projected coordinates. This makes the assumption that the earth is flat and on a rectangular grid which is close to true for a small area. Edited January 9, 2019 by medoug Quote Link to comment
+icezebra11 Posted January 8, 2019 Share Posted January 8, 2019 43 minutes ago, medoug said: For short distances at a given latitude, you can get really close if you know the change in N coordinates for a given number of feet in the north direction and change in W coordinates for a given number of feet in the west direction. Then you can use a single very simple trigonometry formula to convert a distance and heading to the projected coordinates. This makes the assumption that the earth is flat and on a rectangular grid which is close to true for a small area. Even easier if you use UTMs and trig because you don't need to find the ft/degree factors. But most cachers don't have a clue as to what UTM coordinates are. Quote Link to comment
+fizzymagic Posted January 8, 2019 Share Posted January 8, 2019 2 hours ago, icezebra11 said: Even easier if you use UTMs and trig because you don't need to find the ft/degree factors. But most cachers don't have a clue as to what UTM coordinates are. Including you, apparently. UTM northing is not the same as north, for example. And projection using azimuths in UTM does not work, in general. In short: UTM is not suitable for projections. Vincenty's formula is good to a few centimeters over hundreds of kilometers. I recommend it. I have a Windows app that you can use for it called FizzyCalc. 2 Quote Link to comment
+JohnCNA Posted January 8, 2019 Share Posted January 8, 2019 9 hours ago, fizzymagic said: Including you, apparently. UTM northing is not the same as north, for example. And projection using azimuths in UTM does not work, in general. In short: UTM is not suitable for projections. Vincenty's formula is good to a few centimeters over hundreds of kilometers. I recommend it. I have a Windows app that you can use for it called FizzyCalc. FizzyCalc is a wonderful app on Windows. On Android, I highly recommend 'GCC - Geocache Calculator' for when you need to do projection (and many other) calculations in the field. Quote Link to comment
+icezebra11 Posted January 8, 2019 Share Posted January 8, 2019 (edited) 18 hours ago, fizzymagic said: Including you, apparently. UTM northing is not the same as north, for example. And projection using azimuths in UTM does not work, in general. In short: UTM is not suitable for projections. Vincenty's formula is good to a few centimeters over hundreds of kilometers. I recommend it. I have a Windows app that you can use for it called FizzyCalc. I am well aware of UTMs and that their axes are not parallel with Lat/Long but I was responding to medoug's post that specifically said "short distances" and "small area." So let's do the math. How about a 1/4 mile projection (I'd call that a short distance) at 45 degrees from the coordinates N40 00.000, W105 00.000 or 500000mE, 4427757mN. A quarter mile equals 402.34 meters. 402.34sin45 = 284.5 meters therefore the projected coordinates are: 500284.5mE, 4428041.5mN A 402.34 meter projection at 45 degrees, done on my Oregon 600 set to UTMs and from a starting point of 500000mE, 4427757mN results in coordinates: 500284mE, 4428041mN I'd call that close enough for caching purposes. Edited January 8, 2019 by icezebra11 Typos Quote Link to comment
+Joe_L Posted January 8, 2019 Share Posted January 8, 2019 3 hours ago, icezebra11 said: 500184mE, 442804mN Please check this. Quote Link to comment
+icezebra11 Posted January 8, 2019 Share Posted January 8, 2019 1 hour ago, Joe_L said: Please check this. Thanks for catching the typos Joe_L. I've corrected them in my post. Quote Link to comment
Tahosa and Sons Posted January 17, 2019 Share Posted January 17, 2019 I use UTM's on many of my back country caches. They work so well for projections. And if UTM's are useless then why does Geocaching.Com put them on the cache page. Quote Link to comment
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