polygon

mongo 2.6 I have some amount of stored polygons. And I have a point. I what to know if this point fits any of stored polygons document example { ..., "polygons" : [ [ 17.60083012593064, 78.18557739257812 ], [ 17.16834652544664, 78.19381713867188 ], [ 17.17490690610013, 78.739013671875 ], [ 17.613919673106714, 78.73489379882812 ] ], ... } There is nearly the same question already Mongodb : Check if a point is inside a stored polygon . But it is not working for me - this query has to give at least one result(the one in example) - but it does not. db.areas.find( { polygons : { $geoIntersects : {

I working on a project at the moment, where I have a point feature -- the point feature includes a 142 points -- and multiple polygon (around 10). I want to calculate the distance between every single point and the nearest polygon feature in R. My current approach is tedious and a bit long winded. I am currently planning to calculate the distance between every single point and every single polygon. For example, I would calculate the distance between the 142 points and Polygon A, the distance between the 142 points and Polygon B, the distance between 142 points and Polygon C, etc. Here is a

I am looking for an accurate algorithm or a service to calculate surface area on earth where points are calculated on the basis of GPS Coordinates. I am using Google Map Api version 3 and are drawing polygons based on recorded coordinates but I don't think the standard ways of calculating area of the polygon will take into account of slopes(hills). Do I need to work on contours for such thing? Are there any third party services may be ArcGis or some other that takes into account of slopes as well. Yes, this is definitely possible. There is a quick tutorial with example code here: https://web

For a polygon defined as a sequence of (x,y) points, how can I detect whether it is complex or not? A complex polygon has intersections with itself, as shown: Is there a better solution than checking every pair which would have a time complexity of O(N 2 )? There are sweep methods which can determine this much faster than a brute force approach. In addition, they can be used to break a non-simple polygon into multiple simple polygons. For details, see this article , in particular, this code to test for a simple polygon . See Bentley Ottmann Algorithm for a sweep based O((N + I)log N) method

I am playing with a spatial data set (mostly polygons over an area of a city) and I would like to produce different views, based on different levels of zoom. Everything is fine when I have the plot boundary box larger than the area that contains the polygons. But on a zoom in, some polygons get parts that are outside the boundary box and the output for those polygons is problematic : the points of the polygon outside the boundary box are not displayed, resulting in polygons that are not cropped at boundary box limits (as would be expected) but rather are not considered. the polygons are

I want to calculate the overlapped area "THE GRAY REGION" between red and blue rectangles. Each rectangle is defined by its four corner coordinates. The resulted unit of the overlapped area is unit square. I could not imagine how can I do it? Any creative comments would be appreciated. This type of intersection is easily done by the "min of the maxes" and "max of the mins" idea. To write it out one needs a specific notion for the rectangle, and, just to make things clear I'll use a namedtuple: from collections import namedtuple Rectangle = namedtuple('Rectangle', 'xmin ymin xmax ymax') ra =

I am writing a program in which I need to draw polygons of an arbitrary number of sides, each one being translated by a given formula which changes dynamically. There is some rather interesting mathematics involved but I am stuck on this probelm. How can I calculate the coordinates of the vertices of a regular polygon (one in which all angles are equal), given only the number of sides , and ideally (but not neccessarily) having the origin at the centre? For example: a hexagon might have the following points (all are float s): ( 1.5 , 0.5 *Math.Sqrt(3) ) ( 0 , 1 *Math.Sqrt(3) ) (-1.5 , 0.5

It seems to be the standart that polygon are drawn in a convex shape. See stackoverflow.com/questions/15556929/open-gl-polygon However it is not for me if I choose it to be filled. How can I have my polygon filled while maintaining the shape I defined? void drawFloor(){ // White glPolygonMode(GL_FRONT_AND_BACK, GL_FILL); // GL_LINE works the way I want it too //glDisable(GL_POLYGON_SMOOTH); // has no effect :( glBegin(GL_POLYGON); glColor3f(1.0, 1.0, 1.0); glVertex3f(0, 0, 0); //0 glVertex3f(2*boxSize, 0, 0); //1 glVertex3f(2 * boxSize, -boxSize, 0); //2 glVertex3f(5 * boxSize, -boxSize, 0); /