If it is convex, a trivial way to check it is that the point is laying on the same side of all the segments (if traversed in the same order). You can check that easily with the dot product (as it is proportional to the cosine of the angle formed between the segment and the … Read more
This is a function i converted from another language into PHP: $vertices_x = array(37.628134, 37.629867, 37.62324, 37.622424); // x-coordinates of the vertices of the polygon $vertices_y = array(-77.458334,-77.449021,-77.445416,-77.457819); // y-coordinates of the vertices of the polygon $points_polygon = count($vertices_x) – 1; // number vertices – zero-based array $longitude_x = $_GET[“longitude”]; // x-coordinate of the point … Read more
You could use this collision algorithm: To be able to decide whether two convex polygons are intersecting (touching each other) we can use the Separating Axis Theorem. Essentially: If two convex polygons are not intersecting, there exists a line that passes between them. Such a line only exists if one of the sides of one … Read more
There are many algorithms for this problem. It is called “minimum bounding box“. You will find solutions too searching for “convex hull“, especially here. One way is to find the leftmost point and then repeat to search for a point where all other points are to the right of the line p(n-1)p(n).
The answer turns out to be quite simple: import numpy from PIL import Image, ImageDraw # polygon = [(x1,y1),(x2,y2),…] or [x1,y1,x2,y2,…] # width = ? # height = ? img = Image.new(‘L’, (width, height), 0) ImageDraw.Draw(img).polygon(polygon, outline=1, fill=1) mask = numpy.array(img)
One algorithm to solve this is ray-casting. See an explanation here. And you can find code implementing this for the Google Maps JS API V3 here. HTH.
This page: http://www.ecse.rpi.edu/Homepages/wrf/Research/Short_Notes/pnpoly.html shows how to do this for any polygon. I have a Java implementation of this, but it is too big to post here in its entirety. However, you should be able to work it out: class Boundary { private final Point[] points; // Points making up the boundary … /** * Return … Read more
The key to solving this problem is first making an observation: the center of the largest circle that will fit inside an arbitrary polygon is the point that is: Inside the polygon; and Furthest from any point on the edges of the polygon. Why? Because every point on the edge of a circle is equidistant … Read more
Matthew’s answer is a good solution. However, when using the Google Maps API v3, you might want to pass each point of the polygon to a LatLngBounds object through the extend() method, and then finally call the getCenter() method on the LatLngBounds object. Consider the following example: var bounds = new google.maps.LatLngBounds(); var i; // … Read more
Get the indices where the y value of two consecutive time steps have different sign. Use linear interpolation between these points to generate new x values where y is zero. First, a smaller example to make it easier to get a feeling for the linear interpolation and which points are added to the original data: … Read more