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there are several categories in the utm section of esri crs projected. go to the projected coordinate systems. then to UTM, Then look at the nad83 BLM (US Feet). That should work in both autodesk and esri. (EPSG) 32165


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Writing my comment as an answer: It looks to me like there was an issue with the geometry of the shapefile, so running Repair Geometry should (and did) solve this problem.


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I can tell you from experience using GPS enabled cameras, and doing research on handheld GPS accuracy, there is a portion of your data that will be completely wrong (I had photos collected in Qatar showing up in Brazil). Additionally, the accuracy of the photo will be dependent on the surrounding (forest cover, buildings, etc.). Consumer grade cameras are ...


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Try .. 70°28'41.94"N 67°47'58.24"E From Here


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Have you tried making a custom projection? Here is a walkthrough from esri knowledgebase.


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I'm not sure who you consulted, but this doesn't seem like great advice. The problem seems to be not one of negative values but more the order of x and y co-ordinates. A latitude (y) can only be in the range of -90 to +90. Longitude (x) can be -180 to +180, broadly speaking. These co-ordinates are from the equator and the Greenwich meridian, respectively. ...


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When you say that you have imported your CAD drawings into ArcGIS, I am assuming that you now have either file geodatabase feature classes or shapefiles. Consequently you should be able to perform a spatial adjustment transformation on them: Transformations move or shift data within a coordinate system. They are often used to convert data from unknown ...


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Haversine is used for the great circle distance (shortest distance following the sphere curvature). So it does not take the height into account. For straight line distance, I would first convert your lat/long/radius+alt triplets to XYZ in a 3D cartesian system (see here). Then you can apply your equation. Because your points are very close from each ...


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Those are not not latitude and longitudes. It's (likely) going to be some Projected Coordinate System. The easiest way to figure out which would be to look in the metadata (if it came with any) or contact the agency. They'll know what coordinate system it is. Edit: If for whatever reason you can't get this information from them, I'd start your search ...


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I guess the bogus coordinates are set to what they are because HTC is a smartphone made in Taiwan, and the coordinates they are set with are either those at their factory, or from the store from which the phone was sold (the map did show two "HTC"s close by). To prove that something like that is the case, I checked another photo which I know was taken in ...


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For a line, it is a plunge and not a dip (for 3D planes) It is an elementary problem in analytic geometry: The distance = SQRT((x2 –x1)2+(y2 –y1)2+(z2 –z1)2) The plunge = arcsin ((z2 – z1) / distance) The azimuth = arctan((x2 –x1)/(y2 –y1)) (always in two dimensions) The value θ returned will be in the range of ±90° and must be corrected to give ...


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Another way to determine what coordinate system it is in (if you have some reasonable guesses) would be to load some existing data with the projection defined and then load the data with the coordinates and change the coordinate system of the data frame (right click the "Layers" data frame) to what you think it might be. Reasonable guesses would be the ...


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For Openstreetmap, the parameters are quite simple: http://www.openstreetmap.org/#map=10/47.1911/2.4884 http://www.openstreetmap.org/#map= : base url 10/ : zoom level 47.1911/ : latitude of center, North positive 2.4884 : longitude of center, East positive ...


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Here you have some additional information on the location of the 2 new ones they found. Good luck and please come back with your findings" http://siberiantimes.com/science/casestudy/news/now-two-new-large-holes-appear-in-siberia/


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You should modify the table, by adding 4 fields: x1, y1, x2, y2 or similar. Now do table update: x1=objectnodex(obj,1,1) y1=objectnodey(obj,1,1) For the end it is a little trickier: x2=objectnodex(obj,1,objectinfo(obj,22)) y2=objectnodey(obj,1,objectinfo(obj,22)) The above only works with polylines, not lines. Since the default coordsys is wgs84, you ...


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If you are working with anything but points, you need to convert your shapes to WKT (well known text) columns, and then save that as csv. There is an expression in qgis 2.2 field calculator, it is in a geometry branch, named - geomtoWKT. Haven't tried it. If that doesn't work for you, try searching for other ways to get WKT columns. Just saving layer as csv ...


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Short answer The containerPoint methods date from a feature request back in 2012, and today, they're a bit confusing. The best answer is Leaflet maintainer Vladimir Agafonkin's description: "layerPoint is actually a point relative to the map layer (the div which contains tiles and markers), not the outer map container. What you need is ...



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