I've created a new file of intersecting polygons (20 metre buffer of points). I would like to combine the polygons using QGIS, but not to combine them into one layer, with one single attribute table entry if you see what I mean - in order to reduce the number of single polygons in the layer. I've created an attribute field in the table for each polygon as ROWID. I would then like to output the centroid co-ordinates of each combined polygon (eastings & northings) using QGIS.
I have an encoded polyline, this is the content:
kbwyFyycpDiEhA~QnIvYTlSvk#pK_CjaBte#fxApKah#mmBqIoh#aEgEot#iS_To[mSoEaLu[mv#fQuKoRiEkc#aOkYgr#NiVlQySlDsHwIez#g]aU`J`AnO{InBsEuQ~G_Z_ScKwO_`#uScMyXhEuUre#c]fOuQfl#kDsCwAoRaj#v_#_FdNuB|YwZ|[wDnSeJvJ}MwHuRiu#cW}`#oByXgJ{Gc\fW_S_G_Qx#yGtMy[fGaEjL_ViG}|#g|ArZom#dKgCE}FeSoTcWuM}v#cSqb#cb#gP_IwX_l#aJcd#qTqUip#sM_ClAbBrEwEj#mE{FvE`GfEk#uAoEhByNDvKfr#xO|RlTrI~b#xYzm#zO`Hbb#ra#zmAx`#|TlWYlEuIxAiF`IuSvg#tZdb#tVjk#|_#|TkBrH_E_E`AfOjFrLqGAmLtXuR`QyAhIt#lJhHfJ}ChDrDtLwEzRtUtw#cPdNwD~Lj#zXjQbU~QjJjCpYrHbEj#dJnTtGnG~Fn#`IhaAqDv_#ai#dJgYjWeIz[ud#z]_QjAqEyEu[|H_Pna#k_#v{AhLr~A}GhE`HnH~m#jJzSve#p]re#jO~y#oScIg#iBiNvEqIuN{`#rEsNl`#iJmTyIxD[
How can I find the longitude and latitude values for all the points in the polyline?
I used Interactive Polyline Encoder Utility but with this I get the values point by point.
Using the Interactive Polyline Encoder Utility, you can already get the latitude and longitude values using the "Decode Polyline" button . Under the "Locations List" box you should see all the coordinates for the polyline. Those points in the Locations List box are already the latitude and longitude values that you need.
I have a list of coordinates of lat/lon values consisting of cities around the world. I have put together a SVG map of the US which I would like to display the major cities of the world as pins on top of the SVG map. So far I've figured out that the map projection that I am displaying is a Mercator projection of the US so the next step is how can I get the X/Y coordinates for that map of the map for each city? Once I get the mercator projection from the lat/lon then how is that converted to X/Y values and then to relative X/Y values based on the size of my map?
The Wikipedia article on the Mercator projection provides the equations you need to implement. The Mercator projection transforms lat/long to x/y, that's what map projections do. All you have to do once you have the x/y values is translate them into (in your case) pixels or whatever measure you use.
I.e. all objects that contain this coordinates (for polygons) or cross them (for lines)?
You can collapse the bounding box you query with to a point.
Given this grid ( http://i.stack.imgur.com/Nz39I.jpg is a trapezium/trapezoid, not a square), how do you find the point clicked by the user? I.e. When the user clicks a point in the grid, it should return the coordinates like A1 or D5.
I am trying to write pseudo code for this and I am stuck. Can anyone help me? Thanks!
EDIT: I am still stuck... Does anyone know of any way to find the height of the grid?
If it is a true perspective projection, you can run the click-point through the inverse projection to find it's X,Z coordinates in the 3D world. That grid has regular spacing and you can use simple math to get the A1,D5,etc.
If it's just something you drew, then you'll have to compare the Y coordinates to the positions of the horizontal lines to figure out which row. Then you'll need to check its position (left/right) relative to the angled lines to get the column - for that, you'll need either coordinates of the end-points, or equations for the lines.
Yet another option is to store an identical image where each "square" is flood-filled with a different color. You then check the color of the pixel where the user clicked but in this alternate image. This method assumes that it's a fixed image and is the least flexible.
If you have the coordinates of end points of the grid lines then
Try using the inside-outside test for each grid line and find the position
Since this grid is just a 3D view of a 2D grid plane, there is a projective transform that transforms the coordinates on the grid into coordinates on the 2D plane. To find this transform, it is sufficient to mark 4 different points on the plane (say, the edges), assign them coordinates on the 2D plane and solve the resulting linear equation system.