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2

You know the location of point A and point B so you can use trigonometry to calculate the direction of travel. You know the times at point A and point B so you can calculate the speed of travel. Create a Euclidean Distance surface around point B such that each cell size represents one time unit of measure (one cell = one second of travel?). Create a ...


1

coordinates(new_sp_grd) returns the center coordinates of each cell. Here is a simple example: library("sp") grid <- SpatialGrid(GridTopology(c(0, 0), c(1, 1), c(5, 4))) # get coordinates of center coordinates of each cell grid_ctr <- coordinates(grid) plot(grid, axes=TRUE) points(grid_ctr, col="red", pch=16)


0

Isn't this enough? x <- 1:(cd[1,1])*cc['x'] y <- 1:(cd[1,2])*cc['y'] grid_cells_centre <- data.frame(x=rep(x, length(y)), y=rep(y, each=length(x)))


1

Your source data is in GDA94, supposed to be Vicgrid94 covering Philip Island. gdalwarp transforms correctly to a target CRS of epsg:3857, but gdal_translate will not do that. -a_srs has a different purpose: just assign a CRS, no reprojection. So you have to use: gdalwarp -of GTiff -s_srs epsg:3111 -a_srs epsg:3857 test.vrt 25k.tif BTW I used this ...


0

Your coordinates look as though they are in the format x,y,z and are space separated. This is already pretty close to GeoJSON format. Presuming your Excel sheet looks like this: A B C D |-------|-------|-----------------------------|---------| 1 | data1 | data2 | coordinates | ...


2

Just in case you want the distance along the parallel of latitude (and not the shortest distance) between two points at a given latitude, distance = r * (longitude 2 - longitude 1) with longitude in radians and where r, the radius of the parallel of latitude, is r = R * cosine latitude where R is Earth radius.


2

If you aren't too concerned with a precise location, this is pretty close to the SW corner of the highlighted part of the image you provided: 33.061973, -117.202343 This was found by google map search for "Fortuna Ranch Rd, Encinitas" and then zoom into the property location. I chose the SW corner because of the fence line crossing is a little more ...


5

I'm not really sure but maybe check if you meant EPSG 4326 instead of 3857.


0

In the python code, replace in the function getDestinationLatLong d = distance/1000 with d = float(distance)/1000 Or d will be rounded ....


2

Use the field calculator. If you have a field (like start_x), use the field calculator with the expression !SHAPE!.firstPoint.X, and for a field like end_x you would use !SHAPE!.lastPoint.X. for your Y coordinates, you would use !SHAPE!.firstPoint.Y and !SHAPE!.lastPoint.Y Check out the arcpy geometry documentation(http://desktop.arcgis.com/en/arcmap/10.3/...


5

I don't believe you can calculate a position from these values. These look to be bearings and distances - a direction around a circle from a known location. The known location could be anywhere in the world, and the boundary lines are drawn in the direction and distance listed from that known location. These are used to create lot boundaries. Basically ...


-2

I take it that looks like standard latitude and longitude, in degrees - minutes - seconds (N87 degrees, 40 minutes and 20 seconds). Apparently this can be converted to decimal using the following: Decimal Degrees = Degrees + minutes/60 + seconds/3600 Though I'm unsure if this is correct in your particular case (and also because I haven't done that in a ...


2

I had the same problem (Arcmap would not import my csv file with headers, instead it would show Field1, Field2, Field3 etc). When I examined the file headers in Excel, I found...decimals numbers, spaces, percent signs etc. After replacing all of that with underscores or text, I had no problems. I dragged the .csv file into arcmap, opened the attribute ...


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I've just found out what's going on. You have to switch interval units from "millimeters" or "centimeters" to "units of map". Then the software will use the shp coordinate system to generate the grid.


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First, when you load the data using shp2gpsql-gui, ensure that you've set the SRID on the data correctly: you have determined your data are EPSG:3763, so use that number in the SRID field of the GUI. Now that your data is correctly loaded, you can get back geographic coordinate by using the ST_Transform function to convert them from their local system into ...


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I wrote this method which does the trick using JTS: public LineString refineLineString(LineString ls, double maximalDistance){ // list to store coordinates for resulting line string ArrayList<Coordinate> resultLineStringCoordinates = new ArrayList(); // list of LineSegments from input LineString ArrayList<LineSegment> segmentList ...


0

I just found this page for matlab: http://uk.mathworks.com/help/aeroblks/ecefpositiontolla.html I hope it can help.


1

In your csv, create a new 'size' field that counts the number of decimals in the lat/long fields. Use that as a sort of key to buffer the points at variable distances (and hence, create your polygons of different sizes). EDIT for less vagueness: In Excel or similar, extract the number of decimal places into a new column (let's call it 'size' since you are ...


0

You may be right with the central meridian, but 43°N is not your origin latitude, because the parallel of origin should be a straight line. In most cases, the origin is at the equator. Since you do not know the exact projection, I suggest to georeference the map to the imprinted degree grid. You can use QGIS for such tasks, grabbing as much grid ...


2

For javascript you can use proj4.js. Its conversion function is called proj4.


0

For those interested I've managed to get this working, I'm converting from MGA meters to a local map grid projection, also in meters. private Esri.ArcGISRuntime.Geometry.MapPoint ConvertToLocalMapGridCoords(Esri.ArcGISRuntime.Geometry.MapPoint sourceMapPoint, string projectionFile) { var wkTextDest = System.IO.File.ReadAllText(projectionFile); var ...



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