point

问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 6 years ago . I want to get angle between 3 points in JavaScript. If I have points A(x1,y1), B(x2, y2), C(x3, y3) I want to get angle that is formed with lines AB and BC. let A = {x:x1, y:y1}, B = {x:x2, y:y2}, C = {x:x3, y:y3} 回答1: Try this function : /* * Calculates the angle ABC (in radians) * * A first point, ex: {x: 0, y

问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 6 years ago . I want to get angle between 3 points in JavaScript. If I have points A(x1,y1), B(x2, y2), C(x3, y3) I want to get angle that is formed with lines AB and BC. let A = {x:x1, y:y1}, B = {x:x2, y:y2}, C = {x:x3, y:y3} 回答1: Try this function : /* * Calculates the angle ABC (in radians) * * A first point, ex: {x: 0, y

问题 If I have a series of points as longitude and latitude, how would I calculate the center of all of those points? 回答1: Geomidpoint covers 3 different methods for calculating this. 回答2: Several people have answered to take the mean of the latitudes and longitudes. This is sort of the right idea, but means are more complicated on the sphere. The latitude/longitude representation is essentially artificial and has discontinuities (at the poles, and opposite the prime meridian if you aren't careful

问题 I'm having trouble using the pointtype command on gnuplot. I've tried several ways such as: set pt 5 set pointtype 5 plot " " w pt 5 plot " " w pointtype 5 And for some reason nothing seems to work. I tried using the "help" feature, and apparently my version of gnuplot doesn't have "pt" or "pointtype" as an option. Is this perhaps listed as some other feature. I know the points are there, when I plot multiple sets of data, the point type automatically changes, but I have no personal control

问题 I am wondering how i could generate random numbers that appear in a circular distribution. I am able to generate random points in a rectangular distribution such that the points are generated within the square of (0 <= x < 1000, 0 <= y < 1000): How would i go upon to generate the points within a circle such that: (x−500)^2 + (y−500)^2 < 250000 ? 回答1: import random import math # radius of the circle circle_r = 10 # center of the circle (x, y) circle_x = 5 circle_y = 7 # random angle alpha = 2

问题 Well, I'm currently trying to move a list of points using its center as reference with a desired offset. But I'm having problems. I have implemented the current function: public static IEnumerable<Point> Translate(this IEnumerable<Vector2> points, float offset, Vector2 pivot) { foreach (Vector2 v in points) { float magnitude = (v - pivot).magnitude; Vector2 n = (v - pivot).normalized; Vector2 _v = n * (magnitude + offset) + pivot; yield return new Point(_v); } } Note: Point class is an own

问题 I am creating a Java app that will allow users to view images and to pan the image using their mouse. To implement the panning of the image I use a combination of mouseClicked and mouseDragged events using JViewports. The bulk of the code is in the mouseDragged method public void mouseDragged(MouseEvent e, WindowWrapper w) { final JViewport vp = someFieldViewPort; //Getting the point that the mouse is dragged to to Point cp = e.getPoint(); final Point vPoint = vp.getViewPosition(); //I found

问题 I am trying to use Shapely's within function to do a 'spatial join' of a LineString and a Point file (FYI, the point file was generated using the interpolate function on the LineString ). Problem is - nothing is being returned. # this condition is never satisfied if point.within(line): # here I write stuff to a file where: point = POINT (-9763788.9782693591000000 5488878.3678984242000000) line = LINESTRING (-9765787.998118492 5488940.974948905, -9748582.801636808 5488402.127570709) What am I

问题 Does anyone know of a way to fetch all polygons in a MySQL db within a given distance from a point? The actual distance is not that important since it's calculated for each found polygon later, but it would be a huge optimization to just do that calculation for the polygons that are "close". I've looked at the MBR and contains functions but the problem is that some of the polygons are not contained within a bounding box drawn around the point since they are very big, but some of their

问题 I have a need to convert Pixels to Points in C#. I've seen some complicated explanations about the topic, but can't seem to locate a simple formula. Let's assume a standard 96dpi, how do I calulate this conversion? 回答1: There are 72 points per inch; if it is sufficient to assume 96 pixels per inch, the formula is rather simple: points = pixels * 72 / 96 There is a way to get the configured pixels per inch of your display in Windows using GetDeviceCaps. Microsoft has a guide called "Developing