Anyone knows if I am able to user Google Maps Circle, Rectangle and Polygon classes in Node JS? In the frontend is easy with Google Maps Javascript SDK, but I can't figure out how to get a hold of this library within Node JS.
I need to be able to check if points are with bounds, something in the lines of:
const location = google.maps.LatLng(lat, lng);
const circle = new google.maps.Circle({
center: area.center,
radius: area.radius,
});
const doesContain = circle.getBounds().contains(location);
Thanks ahead!
Alright boys, after giving some thought I realized it's easier to create my own code for checking if a geometry contains a point than depend on Google Maps library to do so.
Although this does not offer and the functionality Google Maps SDK offers, it does solve the geometry problem.
For anyone else looking for other Google Maps SDK functionalities, checkout this Node.js Client for Google Maps Services. Though it does not include the geometry functions I was looking for.
Solution
Without further ado here is my code:
class Circle {
/**
* Circle constructor
* #param {array} center Center coordinate [lat, lng]
* #param {number} radius Radius of the circle in meters
*/
constructor(center, radius) {
this.name = "Circle";
this.center = center;
this.radius = radius;
}
/**
* Checks if a point is within the circle
* #param {array} point Coordinates of a point [lat,lng]
* #returns true if point is within, false otherwhise
*/
contains(point) {
const { center, radius } = this;
const distance = this.distance(center, point);
if (distance > radius) return false;
return true;
}
/**
* Calculate the distance between two points (in meters)
* #param {array} p1 [lat,lng] point 1
* #param {array} p2 p1 [lat,lng] point 2
* #returns Distance between the points in meters
*/
distance(p1, p2) {
var R = 6378.137; // Radius of earth in KM
var dLat = (p2[0] * Math.PI) / 180 - (p1[0] * Math.PI) / 180;
var dLon = (p2[1] * Math.PI) / 180 - (p1[1] * Math.PI) / 180;
var a =
Math.sin(dLat / 2) * Math.sin(dLat / 2) +
Math.cos((p1[0] * Math.PI) / 180) *
Math.cos((p2[0] * Math.PI) / 180) *
Math.sin(dLon / 2) *
Math.sin(dLon / 2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
var d = R * c;
return d * 1000; // meters
}
}
class Rectangle {
/**
* Rectangle constructor
* #param {arrar} sw South-west coorodinate of the rectangle [lat,lng]
* #param {array} ne North-east coordinate of the rectangle [lat, lng]
*/
constructor(sw, ne) {
this.name = "Rectangle";
this.sw = sw;
this.ne = ne;
}
/**
* Checks if a point is within the reactangle
* #param {array} point Coordinates of a point [lat,lng]
* #returns true if point is within, false otherwhise
*/
contains(point) {
const { sw, ne } = this;
const x = point[0];
const y = point[1];
if (x < sw[0] || x > ne[0] || y < sw[1] || y > ne[1]) return false;
return true;
}
}
class Polygon {
/**
* Polygon constructor
* #param {array} points Array of vertices/points of the polygon [lat,lng]
*/
constructor(points) {
this.name = "Polygon";
this.points = points;
}
/**
*
* #returns {obj} Returns the coordinate of the min/max bounds that surounds the polygon
* (south-west coordinate, north-east coordinage as in [lat,lng] format)
*/
getBounds() {
const { points } = this;
let arrX = [];
let arrY = [];
for (let i in points) {
arrX.push(points[i][0]);
arrY.push(points[i][1]);
}
return {
sw: [Math.min.apply(null, arrX), Math.min.apply(null, arrY)],
ne: [Math.max.apply(null, arrX), Math.max.apply(null, arrY)],
};
}
/**
* Checks if a point is within the polygon
* #param {array} point Coordinates of a point [lat,lng]
* #returns true if point is within, false otherwhise
*/
contains(point) {
const x = point[0];
const y = point[1];
const bounds = this.getBounds();
// Check if point P lies within the min/max boundary of our polygon
if (x < bounds.sw[0] || x > bounds.ne[0] || y < bounds.sw[1] || y > bounds.ne[1])
return false;
let intersect = 0;
const { points } = this;
// Geofencing method (aka Even–odd rule)
// See more at: https://en.wikipedia.org/wiki/Even%E2%80%93odd_rule
// Now for each path of our polygon we'll count how many times our imaginary
// line crosses our paths, if it crosses even number of times, our point P is
// outside of our polygon, odd number our point is within our polygon
for (let i = 0; i < points.length; i++) {
// Check if pont P lies on a vertices of our polygon
if (x === points[i][0] && y === points[i][1]) return true;
let j = i !== points.length - 1 ? i + 1 : 0;
// Check if Py (y-component of our point P) is with the y-boundary of our path
if (
(points[i][1] < points[j][1] && y >= points[i][1] && y <= points[j][1]) ||
(points[i][1] > points[j][1] && y >= points[j][1] && y <= points[i][1])
) {
// Check if Px (x-componet of our point P) crosses our path
let sx =
points[i][0] +
((points[j][0] - points[i][0]) * (y - points[i][1])) /
(points[j][1] - points[i][1]);
if (sx >= x) intersect += 1;
}
}
return intersect % 2 === 0 ? false : true;
}
}
module.exports = { Circle, Rectangle, Polygon };
Explanation
The Circle and Rectangle class is pretty straight forward, it's trivial to determine if a point lies within a boundary. The Polygon class is a bit more complicated because of obvious reasons.
The method used here to determine if a point P is within a polygon is called Geofencing (aka Even–odd rule), a common method in geospacial analysis.
Step 1
First we check if the point P falls within the max/min boundaries of the polygon (image 1), if it doesn't, we return false, problem solved.
Image 1 -- Polygon boundaries, P1 is within the polygon boundaries, P2 is not.
Step 2
Then we check if the point lies on a vertices (points) of the polygon, if it does, we return true, problem solved. (Image 2)
Image 2 -- Polygon boundaries, point P is on a vertices, return true.
Step 3
This next step is the most gratifying one, by now we know the point is with the polygon boundaries (from step 1) but we don't know if it's within it or not. The way to solve this we cast an imaginary line departing from the point to any direction, if it crosses the path of polygon even number of times, the point is outside of the polygon, if it crosses an odd number of times, the point is within the polygon. Like so:
Image 3 -- An imaginary line from P1 crosses the polygon paths an odd number of times (3 times), it's within the polygon boundaries. A imaginary line from P2 crosses an even number of times (4 times), it lies outside of the polygon.
Since we can pick any direction we want to cast the imaginary line from, we'll pick along the x-axis to simplify things, like so:
Image 4 -- Casting the imaginary line from point P parallel to the x-axis t0 simplify determining how many times it intersects our polygon.
To determine how many times the imaginary line intersects our polygon, we have to check each path of the polygon at a time. To do this, we break it down into two steps (see image 5 for references):
For each segment/path of the polygon we check if our point Py (y-component of our point P) is within the the boundaries of the path in question (Y1 and Y2). If it is not, we know our point is does not intersects that specific path and we can move on to the next one. If it is within the path's y-boundaries, then we have to check if it crosses our path in the x-direction (next step).
Assuming the step before is true, to check intersection in the x-direction we have calculate the equation for the path (using line equation: y2 - y1 = m(x2 - x1)) and plug in our Py component to solve for our intersection (in my code I call this Sx). Then we check if Sx is greater than Px, if so, then our imaginary line intersects the path in the x positive direction.
It's important to note that the imaginary line starts at our point P and we only count intersections in that direction we originally picked, in this case x-axis+. This is why Sx has to be grater than or equal to Px, otherwise the test fails.
Image 5 -- We break down each path of the polygon to determine the number of intersections.
Once this path is done we move to the next one and so on. In this case the line crosses 3 times our paths, and therefore we know it's within our polygon.
This is a very clever and simple way if you think about it, it works for any shape, it's truly amazing.
Read more
https://en.wikipedia.org/wiki/Even%E2%80%93odd_rule
Examples
Example 1 - Simple shapes
const p = new Polygon([
[-3, 3],
[-4, 1],
[-3, 0],
[-2, -1],
[0, 0],
[3, 2],
[0, 1],
[-1, 4],
]);
console.log("Contains: ", p.contains([-1, 1])); // returns true
JSFiddle 1
Example 2 - Complex shapes (overlapping areas)
This method works for more complex shapes, when the polygon coordinates creates overlappping areas and they cancel each other out.
const p = new Polygon([
[-2, 0],
[2, 0],
[2, 4],
[-2, 4],
[-2, 0],
[0, 2],
[2, 0],
[0, -2],
[-2, 0],
]);
console.log("Contains: ", p.contains([0, 1])); // returns false
JSFiddle 2
Side note
If you need to quickly plot points just to get a view of a shape/grid, this plotting tool helped a lot to get a visual of what's going on. Very often I thought my code had a bug when in fact my coordinates was skewed and code was correct.
https://www.desmos.com/calculator
I only wish it let you draw lines between points. Either way I found it helpful.
the drawing algorithm which I currently use:
a_max = Pi*2 (float)(num_segments - 1.0f)/(float)num_segments;
for (unsigned int i = 0; i<=num_segments;i++)
{
const float a = (float)i / (float)num_segments * a_max;
SetPixel(centre.x + cos(a) *radius, centre.y +sin(a) *radius);
}
Works fine, but it starts drawing at (centre.x+radius, centre.y). I would like to have it to start at the top , because I want to draw a compass and zero degree is at the top, not on the right, so that I don't have to make a hacky solution.
Try rotating 90 degrees to the left before you start drawing, this should solve it for you.
A compass not only starts at "north" instead of "east" but also goes clockwise instead of counter-clockwise.
For this case, just swap sin(a) and cos(a):
x = centre.x + sin(a) * radius
y = centre.y + cos(a) * radius
I am learning game programming with Phaser and I am currently building a simple breakout game.
When the ball hits the paddle I use the following code to determine the new x velocity of the ball:
if (ball.x < paddle.x)
{
// Ball is on the left-hand side of the paddle
diff = paddle.x - ball.x;
ball.body.velocity.x = (-10 * diff);
}
else if (ball.x > paddle.x)
{
// Ball is on the right-hand side of the paddle
diff = ball.x -paddle.x;
ball.body.velocity.x = (10 * diff);
}
else
{
// Ball is perfectly in the middle
// Add a little random X to stop it bouncing straight up!
ball.body.velocity.x = 2 + Math.random() * 8;
}
This code is originally taken I believe from stackoverflow, although I cannot remember from which post I'm afraid.
The problem with this is that when the ball goes left or right at an angle, it appears faster on screen than if it goes straight up. The more pronounced the angle, the faster it goes and appears.
Does anyone know how to solve this problem?
Regards
Crouz
The speed of the ball is given by a combination of the horizontal (x) speed and vertical (y) speed, as given by Pythagoras' theorem:
z2 = x2 + y2
where overall speed would be z.
If you increase x without decreasing y appropriately, then the overall speed will also increase.
If you want the speed to remain the same, then you need to adjust y as well. Example pseudo-code:
speed = sqrt(x^2 + y^2)
x = 2 + random()*8
y = sqrt(speed^2 - x^2)
You could incorporate this into your code by calculating speed before any adjustments are made, and then adjusting y afterwards:
// Calculate overall speed
var speed = Math.sqrt(
Math.pow(ball.body.velocity.x, 2) +
Math.pow(ball.body.velocity.y, 2)
);
if (ball.x < paddle.x)
{
// Ball is on the left-hand side of the paddle
diff = paddle.x - ball.x;
ball.body.velocity.x = (-10 * diff);
}
else if (ball.x > paddle.x)
{
// Ball is on the right-hand side of the paddle
diff = ball.x -paddle.x;
ball.body.velocity.x = (10 * diff);
}
else
{
// Ball is perfectly in the middle
// Add a little random X to stop it bouncing straight up!
ball.body.velocity.x = 2 + Math.random() * 8;
}
// Adjust y to maintain same overall speed
ball.body.velocity.y = Math.sqrt(
Math.pow(speed, 2) -
Math.pow(ball.body.velocity.x, 2)
);
This will always result in a positive y, so you might need to negate it depending on which direction (up or down) you want it to be moving.
At the moment I'm using the dot product of the mouse position and (0, 1) to generate radians to rotate an object, in three.js
Code below, works ok but the object 'jumps' because the radian angle skips from positive to negative when the clientX value goes between window.innerWidth / 2
onDocumentMouseMove : function(event) {
// rotate circle relative to current mouse pos
var oldPos = new THREE.Vector2(0, 1);
Template.Main.mouseCurrPos = new THREE.Vector2((event.clientX / window.innerWidth ) * 2 - 1, - (event.clientY / window.innerHeight) * 2 + 1);
Template.Main.mouseCurrPos.normalize();
//Template.Main.projector.unprojectVector(Template.Main.mouseCurrPos, Template.Main.scene);
var angle = oldPos.dot(Template.Main.mouseCurrPos);
Template.Main.mousePrevPos.x = event.clientX;
Template.Main.mousePrevPos.y = event.clientY;
if (event.clientX < window.innerWidth / 2) {
Template.Main.circle.rotation.z = -angle;
}
else {
Template.Main.circle.rotation.z = angle;
}
console.log(Template.Main.circle.rotation.z);
}
However if I add this to assign the value to oldPos:
if (event.clientX < window.innerWidth / 2) {
oldPos = new THREE.Vector2(0, -1);
}
else {
oldPos = new THREE.Vector2(0, 1);
}
Then the "jumping" goes but the effect of rotation is inverted when the mouse is on the left of the window.
I.e. mouse going up rotates anti-clockwise and vice-versa which is not desired.
It's frustrating.
Also if I keep the oldPos conditional assignment and leave out the conditional negation of the angle instead, the jumping comes back.
You can see a demo here: http://theworldmoves.me/rotation-demo/
Many thanks for any tips.
Why are you using the result of the dot product as the angle (radians)? The dot product gives you the cosine of the angle (times the magnitude of the vectors, but these are a unit vector and a normalized vector, so that doesn't matter).
You could change your angle computation to
var angle = Math.acos(oldPos.dot(Template.Main.mouseCurrPos));
However, you may get the wrong quadrant, since there can be two values of theta that satisfy cos(theta) = n. The usual way to get the angle of a vector (origin to mouse position) in the right quadrant is to use atan2():
var angle = Math.atan2(Template.Main.mouseCurrPos.y,
Template.Main.mouseCurrPos.x);
This should give the angle of the mouse position vector, going counterclockwise from (1, 0). A little experimentation can determine for sure where the zero angle is, and which direction is positive rotation.
Given a line made up of several points, how do I make the line smoother/ curvier/ softer through adding intermediate points -- while keeping the original points completely intact and unmoved?
To illustrate, I want to go from the above to the below in this illustration:
Note how in the above picture, if we start at the bottom there will be a sharper right turn. In the bottom image however, this sharp right turn is made a bit "softer" by adding an intermediate point which is positioned in the middle of the two points, and using averages of the angles of the other lines. (Differently put, imagine the lines a race car would drive, as it couldn't abruptly change direction.) Note how, however, none of the original points was "touched", I just added more points.
Thanks!! For what it's worth, I'm implementing this using JavaScript and Canvas.
with each edge (e1 & e2) adjacent to each 'middle' edge (me) do
let X = 0.5 x length(me)
find 2 cubic bezier control points by extending the adjacent edges by X (see algorithm below)
get midpoint of cubic bezier (by applying formula below)
insert new 'midpoint' between me's two coordinates.
FloatPoint ExtendLine(const FloatPoint A, const FloatPoint B, single distance)
{
FloatPoint newB;
float lenAB = sqrt((A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y));
newB.X = B.x - (B.x - A.x) / lenAB * distance;
newB.Y = B.Y - (B.Y - A.Y) / lenAB * distance;
return newB;
}
Edit: Formula for Bezier Curve midpoint: p(0.5) = 0.125(p0) + 0.375(p1) + 0.375(p2) + 0.125(p3)
The following code found elsewhere here does the job for me, in the specific context of JavaScript-Canvas which I'm using -- but please see Angus' answer for a more general approach:
var max = points.length;
context.beginPath();
var i = 0;
context.moveTo(points[i].x, points[i].y);
for (i = 1; i < max - 2; i++) {
var xc = (points[i].x + points[i + 1].x) * .5;
var yc = (points[i].y + points[i + 1].y) * .5;
context.quadraticCurveTo(points[i].x, points[i].y, xc, yc);
}
context.quadraticCurveTo(points[max - 2].x, points[max - 2].y, points[max - 1].x,points[max - 1].y);
context.closePath();
context.stroke();