+Go Play Outside Posted April 9, 2005 Share Posted April 9, 2005 what would this be in decimal form 303 degrees 30 minutes 39.34059 seconds I got 303.555059 am I close??? Quote Link to comment
+Hemlock Posted April 9, 2005 Share Posted April 9, 2005 (edited) First off, the 303 degrees doesn't make sense, from a geographical point of view. Latitude should be in the range -90 to 90, and longitude in the range -180 to 180. To convert to decimal we need to go in two steps. First, convert the seconds 39.34059 to minutes by dividing by 60 to get 0.6556765 Add that to the minutes to get 30.6556765 Now convert the minutes to degrees by dividing by 60 to get 0.510928 Add that to the degrees and you get 303.510928 Edited April 9, 2005 by Hemlock Quote Link to comment
+Go Play Outside Posted April 9, 2005 Author Share Posted April 9, 2005 Thanks for the advice. This is actually a bearing given in a geocache. GCNBJ9 It was given in an odd maner and I though that was sort of the way it was done but I had not added the seconds to the minutes before converting the minutes again. There was also a typo in the answer I posted above but thanks again. Quote Link to comment
+ZoomZoom Posted April 9, 2005 Share Posted April 9, 2005 (edited) 303 degrees 30 minutes 39.34059 seconds. Divide 39.34059 by 60=6556765. Now divide oncemore 30.6556765 by 60= 303.510927941. Ooops! Should have read more carefully that the answer was already given. Edited April 9, 2005 by ZoomZoom Quote Link to comment
+ZoomZoom Posted April 9, 2005 Share Posted April 9, 2005 Strange... I thought these type of caches were forbidden since awhile back. Quote Link to comment
+coast2coast2coast Posted April 9, 2005 Share Posted April 9, 2005 It's the first one of these I've seen sence all the KWFB ones Quote Link to comment
+fizzymagic Posted April 9, 2005 Share Posted April 9, 2005 Answer is different by about .33 km depending on whether you use a spherical approximation or the real geodesic distance. That's a long ways. It is especially funny given the absurd accuracy of the projection distance (to 1 cm). Doing projections over such a long distance is a tricky thing unless you are very specific about what model parameters you are using for the Earth. Quote Link to comment
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