Apply gradient to BufferGeometry vertices - three.js

I have an animation that uses a BufferGeometry to create a grid of particles which are then animated using Perlin noise. That all works perfectly but the final thing to do is to apply a gradient across the grid. I have tried everything I have found and nothing is working. I feel like using a ShaderMaterial is the best/easiest solution but the code I've found for gradients just isn't working so I'm asking what the best way to do this is and ideally an example of how to do it.
Here is a link to the codepen so you can see all of the code and the example working.
https://codepen.io/JJGerrish/pen/oNxyJXX?editors=0010
And here is an example of the what I want the grid to look like.
I've left my attempt at creating a gradient shader in so you are welcome to play around with that or come up with a better solution.

Your problem is that you are using uVu.y , but you don't have any uv coordinates so the value will always be 0.
Are you sure you don't want to be using the position x value?
gl_FragColor = vec4(mix(color1, color2, smoothstep(-10.0, 10.0, pos.x)), 1.0);
(demo in code below with a smoothstep, note sending the pos variable from the vertex to fragment shader).
Also, why not do the noise in the shader too rather than in the JS?
//noise library
/*
* A speed-improved perlin and simplex noise algorithms for 2D.
*
* Based on example code by Stefan Gustavson (stegu#itn.liu.se).
* Optimisations by Peter Eastman (peastman#drizzle.stanford.edu).
* Better rank ordering method by Stefan Gustavson in 2012.
* Converted to Javascript by Joseph Gentle.
*
* Version 2012-03-09
*
* This code was placed in the public domain by its original author,
* Stefan Gustavson. You may use it as you see fit, but
* attribution is appreciated.
*
*/
(function(global){
var module = global.noise = {};
function Grad(x, y, z) {
this.x = x; this.y = y; this.z = z;
}
Grad.prototype.dot2 = function(x, y) {
return this.x*x + this.y*y;
};
Grad.prototype.dot3 = function(x, y, z) {
return this.x*x + this.y*y + this.z*z;
};
var grad3 = [new Grad(1,1,0),new Grad(-1,1,0),new Grad(1,-1,0),new Grad(-1,-1,0),
new Grad(1,0,1),new Grad(-1,0,1),new Grad(1,0,-1),new Grad(-1,0,-1),
new Grad(0,1,1),new Grad(0,-1,1),new Grad(0,1,-1),new Grad(0,-1,-1)];
var p = [151,160,137,91,90,15,
131,13,201,95,96,53,194,233,7,225,140,36,103,30,69,142,8,99,37,240,21,10,23,
190, 6,148,247,120,234,75,0,26,197,62,94,252,219,203,117,35,11,32,57,177,33,
88,237,149,56,87,174,20,125,136,171,168, 68,175,74,165,71,134,139,48,27,166,
77,146,158,231,83,111,229,122,60,211,133,230,220,105,92,41,55,46,245,40,244,
102,143,54, 65,25,63,161, 1,216,80,73,209,76,132,187,208, 89,18,169,200,196,
135,130,116,188,159,86,164,100,109,198,173,186, 3,64,52,217,226,250,124,123,
5,202,38,147,118,126,255,82,85,212,207,206,59,227,47,16,58,17,182,189,28,42,
223,183,170,213,119,248,152, 2,44,154,163, 70,221,153,101,155,167, 43,172,9,
129,22,39,253, 19,98,108,110,79,113,224,232,178,185, 112,104,218,246,97,228,
251,34,242,193,238,210,144,12,191,179,162,241, 81,51,145,235,249,14,239,107,
49,192,214, 31,181,199,106,157,184, 84,204,176,115,121,50,45,127, 4,150,254,
138,236,205,93,222,114,67,29,24,72,243,141,128,195,78,66,215,61,156,180];
// To remove the need for index wrapping, double the permutation table length
var perm = new Array(512);
var gradP = new Array(512);
// This isn't a very good seeding function, but it works ok. It supports 2^16
// different seed values. Write something better if you need more seeds.
module.seed = function(seed) {
if(seed > 0 && seed < 1) {
// Scale the seed out
seed *= 65536;
}
seed = Math.floor(seed);
if(seed < 256) {
seed |= seed << 8;
}
for(var i = 0; i < 256; i++) {
var v;
if (i & 1) {
v = p[i] ^ (seed & 255);
} else {
v = p[i] ^ ((seed>>8) & 255);
}
perm[i] = perm[i + 256] = v;
gradP[i] = gradP[i + 256] = grad3[v % 12];
}
};
module.seed(0);
/*
for(var i=0; i<256; i++) {
perm[i] = perm[i + 256] = p[i];
gradP[i] = gradP[i + 256] = grad3[perm[i] % 12];
}*/
// Skewing and unskewing factors for 2, 3, and 4 dimensions
var F2 = 0.5*(Math.sqrt(3)-1);
var G2 = (3-Math.sqrt(3))/6;
var F3 = 1/3;
var G3 = 1/6;
// 2D simplex noise
module.simplex2 = function(xin, yin) {
var n0, n1, n2; // Noise contributions from the three corners
// Skew the input space to determine which simplex cell we're in
var s = (xin+yin)*F2; // Hairy factor for 2D
var i = Math.floor(xin+s);
var j = Math.floor(yin+s);
var t = (i+j)*G2;
var x0 = xin-i+t; // The x,y distances from the cell origin, unskewed.
var y0 = yin-j+t;
// For the 2D case, the simplex shape is an equilateral triangle.
// Determine which simplex we are in.
var i1, j1; // Offsets for second (middle) corner of simplex in (i,j) coords
if(x0>y0) { // lower triangle, XY order: (0,0)->(1,0)->(1,1)
i1=1; j1=0;
} else { // upper triangle, YX order: (0,0)->(0,1)->(1,1)
i1=0; j1=1;
}
// A step of (1,0) in (i,j) means a step of (1-c,-c) in (x,y), and
// a step of (0,1) in (i,j) means a step of (-c,1-c) in (x,y), where
// c = (3-sqrt(3))/6
var x1 = x0 - i1 + G2; // Offsets for middle corner in (x,y) unskewed coords
var y1 = y0 - j1 + G2;
var x2 = x0 - 1 + 2 * G2; // Offsets for last corner in (x,y) unskewed coords
var y2 = y0 - 1 + 2 * G2;
// Work out the hashed gradient indices of the three simplex corners
i &= 255;
j &= 255;
var gi0 = gradP[i+perm[j]];
var gi1 = gradP[i+i1+perm[j+j1]];
var gi2 = gradP[i+1+perm[j+1]];
// Calculate the contribution from the three corners
var t0 = 0.5 - x0*x0-y0*y0;
if(t0<0) {
n0 = 0;
} else {
t0 *= t0;
n0 = t0 * t0 * gi0.dot2(x0, y0); // (x,y) of grad3 used for 2D gradient
}
var t1 = 0.5 - x1*x1-y1*y1;
if(t1<0) {
n1 = 0;
} else {
t1 *= t1;
n1 = t1 * t1 * gi1.dot2(x1, y1);
}
var t2 = 0.5 - x2*x2-y2*y2;
if(t2<0) {
n2 = 0;
} else {
t2 *= t2;
n2 = t2 * t2 * gi2.dot2(x2, y2);
}
// Add contributions from each corner to get the final noise value.
// The result is scaled to return values in the interval [-1,1].
return 70 * (n0 + n1 + n2);
};
// 3D simplex noise
module.simplex3 = function(xin, yin, zin) {
var n0, n1, n2, n3; // Noise contributions from the four corners
// Skew the input space to determine which simplex cell we're in
var s = (xin+yin+zin)*F3; // Hairy factor for 2D
var i = Math.floor(xin+s);
var j = Math.floor(yin+s);
var k = Math.floor(zin+s);
var t = (i+j+k)*G3;
var x0 = xin-i+t; // The x,y distances from the cell origin, unskewed.
var y0 = yin-j+t;
var z0 = zin-k+t;
// For the 3D case, the simplex shape is a slightly irregular tetrahedron.
// Determine which simplex we are in.
var i1, j1, k1; // Offsets for second corner of simplex in (i,j,k) coords
var i2, j2, k2; // Offsets for third corner of simplex in (i,j,k) coords
if(x0 >= y0) {
if(y0 >= z0) { i1=1; j1=0; k1=0; i2=1; j2=1; k2=0; }
else if(x0 >= z0) { i1=1; j1=0; k1=0; i2=1; j2=0; k2=1; }
else { i1=0; j1=0; k1=1; i2=1; j2=0; k2=1; }
} else {
if(y0 < z0) { i1=0; j1=0; k1=1; i2=0; j2=1; k2=1; }
else if(x0 < z0) { i1=0; j1=1; k1=0; i2=0; j2=1; k2=1; }
else { i1=0; j1=1; k1=0; i2=1; j2=1; k2=0; }
}
// A step of (1,0,0) in (i,j,k) means a step of (1-c,-c,-c) in (x,y,z),
// a step of (0,1,0) in (i,j,k) means a step of (-c,1-c,-c) in (x,y,z), and
// a step of (0,0,1) in (i,j,k) means a step of (-c,-c,1-c) in (x,y,z), where
// c = 1/6.
var x1 = x0 - i1 + G3; // Offsets for second corner
var y1 = y0 - j1 + G3;
var z1 = z0 - k1 + G3;
var x2 = x0 - i2 + 2 * G3; // Offsets for third corner
var y2 = y0 - j2 + 2 * G3;
var z2 = z0 - k2 + 2 * G3;
var x3 = x0 - 1 + 3 * G3; // Offsets for fourth corner
var y3 = y0 - 1 + 3 * G3;
var z3 = z0 - 1 + 3 * G3;
// Work out the hashed gradient indices of the four simplex corners
i &= 255;
j &= 255;
k &= 255;
var gi0 = gradP[i+ perm[j+ perm[k ]]];
var gi1 = gradP[i+i1+perm[j+j1+perm[k+k1]]];
var gi2 = gradP[i+i2+perm[j+j2+perm[k+k2]]];
var gi3 = gradP[i+ 1+perm[j+ 1+perm[k+ 1]]];
// Calculate the contribution from the four corners
var t0 = 0.6 - x0*x0 - y0*y0 - z0*z0;
if(t0<0) {
n0 = 0;
} else {
t0 *= t0;
n0 = t0 * t0 * gi0.dot3(x0, y0, z0); // (x,y) of grad3 used for 2D gradient
}
var t1 = 0.6 - x1*x1 - y1*y1 - z1*z1;
if(t1<0) {
n1 = 0;
} else {
t1 *= t1;
n1 = t1 * t1 * gi1.dot3(x1, y1, z1);
}
var t2 = 0.6 - x2*x2 - y2*y2 - z2*z2;
if(t2<0) {
n2 = 0;
} else {
t2 *= t2;
n2 = t2 * t2 * gi2.dot3(x2, y2, z2);
}
var t3 = 0.6 - x3*x3 - y3*y3 - z3*z3;
if(t3<0) {
n3 = 0;
} else {
t3 *= t3;
n3 = t3 * t3 * gi3.dot3(x3, y3, z3);
}
// Add contributions from each corner to get the final noise value.
// The result is scaled to return values in the interval [-1,1].
return 32 * (n0 + n1 + n2 + n3);
};
// ##### Perlin noise stuff
function fade(t) {
return t*t*t*(t*(t*6-15)+10);
}
function lerp(a, b, t) {
return (1-t)*a + t*b;
}
// 2D Perlin Noise
module.perlin2 = function(x, y) {
// Find unit grid cell containing point
var X = Math.floor(x), Y = Math.floor(y);
// Get relative xy coordinates of point within that cell
x = x - X; y = y - Y;
// Wrap the integer cells at 255 (smaller integer period can be introduced here)
X = X & 255; Y = Y & 255;
// Calculate noise contributions from each of the four corners
var n00 = gradP[X+perm[Y]].dot2(x, y);
var n01 = gradP[X+perm[Y+1]].dot2(x, y-1);
var n10 = gradP[X+1+perm[Y]].dot2(x-1, y);
var n11 = gradP[X+1+perm[Y+1]].dot2(x-1, y-1);
// Compute the fade curve value for x
var u = fade(x);
// Interpolate the four results
return lerp(
lerp(n00, n10, u),
lerp(n01, n11, u),
fade(y));
};
// 3D Perlin Noise
module.perlin3 = function(x, y, z) {
// Find unit grid cell containing point
var X = Math.floor(x), Y = Math.floor(y), Z = Math.floor(z);
// Get relative xyz coordinates of point within that cell
x = x - X; y = y - Y; z = z - Z;
// Wrap the integer cells at 255 (smaller integer period can be introduced here)
X = X & 255; Y = Y & 255; Z = Z & 255;
// Calculate noise contributions from each of the eight corners
var n000 = gradP[X+ perm[Y+ perm[Z ]]].dot3(x, y, z);
var n001 = gradP[X+ perm[Y+ perm[Z+1]]].dot3(x, y, z-1);
var n010 = gradP[X+ perm[Y+1+perm[Z ]]].dot3(x, y-1, z);
var n011 = gradP[X+ perm[Y+1+perm[Z+1]]].dot3(x, y-1, z-1);
var n100 = gradP[X+1+perm[Y+ perm[Z ]]].dot3(x-1, y, z);
var n101 = gradP[X+1+perm[Y+ perm[Z+1]]].dot3(x-1, y, z-1);
var n110 = gradP[X+1+perm[Y+1+perm[Z ]]].dot3(x-1, y-1, z);
var n111 = gradP[X+1+perm[Y+1+perm[Z+1]]].dot3(x-1, y-1, z-1);
// Compute the fade curve value for x, y, z
var u = fade(x);
var v = fade(y);
var w = fade(z);
// Interpolate
return lerp(
lerp(
lerp(n000, n100, u),
lerp(n001, n101, u), w),
lerp(
lerp(n010, n110, u),
lerp(n011, n111, u), w),
v);
};
})(this);
//effective animation code
var wWidth = window.innerWidth;
var wHeight = window.innerHeight;
var scene = new THREE.Scene();
var camera = new THREE.PerspectiveCamera(75, wWidth / wHeight, 0.01, 1000);
camera.position.x = 0;
camera.position.y = 0; // 0
camera.position.z = 50; // 40
camera.lookAt(new THREE.Vector3(0, 0, 0));
var renderer = new THREE.WebGLRenderer({
alpha: true
});
renderer.setClearColor(0x000000, 0);
document.getElementById('sec-graphical-intro').appendChild(renderer.domElement);
//Animation parameters
var rows = 50;
var cols = 100;
var separation = 1;
var perlinScale = 0.025;
var waveSpeed = 0.1;
var waveHeight = 8;
var FPS = 45;
var startTime = new Date().getTime();
var particles = 0;
var count = 0;
noise.seed(Math.random());
function createGeometry() {
var numParticles = cols * rows;
var positions = new Float32Array( numParticles * 3 );
var i = 0
var j = 0;
for ( var ix = 0; ix < cols; ix ++ ) {
for ( var iy = 0; iy < rows; iy ++ ) {
positions[i] = ix * separation - ( ( cols * separation ) / 2 ); // x
positions[i + 1] = 0; // y
positions[i + 2] = iy * separation - ( ( rows * separation ) / 2 ); // z
i += 3;
j ++;
}
}
var geometry = new THREE.BufferGeometry();
geometry.addAttribute( 'position', new THREE.BufferAttribute( positions, 3 ) );
// geometry.dynamic = true;
// geometry.translate(-100, 0, -25);
return geometry;
}
var geo = createGeometry();
var material = new THREE.ShaderMaterial( {
uniforms: {
"color1": {
type : "c",
value: new THREE.Color(0x2753c9)
},
"color2": {
type : "c",
value: new THREE.Color(0x1dcdc0)
}
},
vertexShader: `
varying vec2 vUv;
varying vec4 pos;
void main() {
vUv = uv;
gl_PointSize = 4.0;
pos = projectionMatrix * modelViewMatrix * vec4(position,1.0);
gl_Position = pos;
}
`,
fragmentShader: `
uniform vec3 color1;
uniform vec3 color2;
varying vec2 vUv;
varying vec4 pos;
void main() {
if ( length( gl_PointCoord - vec2( 0.5, 0.5 ) ) > 0.475 ) discard;
gl_FragColor = vec4(mix(color1, color2, smoothstep(-10.0, 10.0, pos.x)), 1.0);
}
`
});
particles = new THREE.Points(geo, material);
scene.add(particles);
function perlinAnimate() {
var curTime = new Date().getTime();
var positions = particles.geometry.attributes.position.array;
var i = 0
var j = 0;
for ( var ix = 0; ix < cols; ix ++ ) {
for ( var iy = 0; iy < rows; iy ++ ) {
pX = (ix * perlinScale) + ((curTime - startTime) / 1000) * waveSpeed;
pZ = (iy * perlinScale) + ((curTime - startTime) / 1000) * waveSpeed;
positions[ i + 1 ] = (noise.simplex2(pX, pZ)) * waveHeight;
i += 3;
}
}
particles.geometry.attributes.position.needsUpdate = true;
count += 0.1;
}
function render() {
renderer.render(scene, camera);
}
function animate() {
perlinAnimate();
render();
window.setTimeout(function() {
requestAnimationFrame(animate);
}, 1000 / FPS);
}
function refreshCanvasState() {
wWidth = window.innerWidth;
wHeight = window.innerHeight;
camera.aspect = wWidth / wHeight;
camera.updateProjectionMatrix();
renderer.setSize(wWidth, wHeight);
}
//EVENTS && INTERACTIONS
window.addEventListener('resize', refreshCanvasState, false);
animate();
refreshCanvasState();
addEvent(document, "keypress", function(e) {
e = e || window.event;
// use e.keyCode
console.log(e.keyCode);
});
function addEvent(element, eventName, callback) {
if (element.addEventListener) {
element.addEventListener(eventName, callback, false);
} else if (element.attachEvent) {
element.attachEvent("on" + eventName, callback);
} else {
element["on" + eventName] = callback;
}
}
<head>
<script src="https://cdnjs.cloudflare.com/ajax/libs/three.js/84/three.min.js"></script>
</head>
<body>
<section id="sec-graphical-intro"></section>

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if(dist < r1){
ctx.moveTo(x2 + r2, y2);
ctx.arc(x2, y2, r2, 0, PI2, true);
}
ctx.fill();
ctx.stroke();
return;
}
// get the start end angles for outer then inner circles
const p = intercepts;
var startA1 = Math.atan2(p.y1 - y1, p.x1 - x1) + ang;
var endA1 = Math.atan2(p.y2 - y1, p.x2 - x1) + ang;
var startA2 = Math.atan2(p.y1 - y1, p.x1 - (x1 + dist)) + ang;
var endA2 = Math.atan2(p.y2 - y1, p.x2 - (x1 + dist)) + ang;
ctx.beginPath();
if(endA1 < startA1){
ctx.arc(x1, y1, r1, startA1, endA1);
ctx.arc(x2, y2, r2, endA2, startA2, true);
}else{
ctx.arc(x2, y2, r2, endA2, startA2);
ctx.arc(x1, y1, r1, startA1, endA1,true);
}
ctx.closePath();
ctx.fill();
ctx.stroke();
}
const outerRadius = 100;
const innerRadius = 80;
var w = canvas.width;
var h = canvas.height;
var cw = w / 2; // center
var ch = h / 2;
var globalTime;
ctx.font = "32px arial";
ctx.textAlign = "center";
ctx.lineJoin = "round";
ctx.lineWidth = 8;
ctx.strokeStyle = "#999";
// main update function
function mainLoop(timer){
globalTime = timer;
ctx.setTransform(1,0,0,1,0,0); // reset transform
ctx.globalAlpha = 1; // reset alpha
ctx.fillStyle = "black";
ctx.fillRect(0,0,w,h);
ctx.fillStyle = "white";
ctx.fillText("Move mouse over circle",cw,40);
drawCrescent(cw, ch-40, outerRadius, mouse.x, mouse.y, innerRadius);
requestAnimationFrame(mainLoop);
}
requestAnimationFrame(mainLoop);
canvas { border : 2px solid black; }
<canvas id="canvas"></canvas>
Solved it using globalCompositeOperation
https://codepen.io/matt3224/pen/oeXbdg?editors=1010

Dynamically change point color in Three.js PointClouod

I'm trying to render a matrix of points in Three.js but I need to treat each particle in the cloud as an individual "pixel" for which I can change the color of each on the fly. I figured out how to basically render the point cloud, and can set the initial color, but cannot figure out how to change the color of each point once it's set.
I'm generating the point cloud like this:
function generateRegularPointcloud( color, width, length ) {
var geometry = new THREE.Geometry();
var numPoints = width * length;
var colors = [];
var k = 0;
for( var i = 0; i < width; i++ ) {
for( var j = 0; j < length; j++ ) {
var u = i / width;
var v = j / length;
var x = u - 0.5;
var y = 0;
var z = v - 0.5;
var v = new THREE.Vector3( x,y,z );
var intensity = ( y + 0.1 ) * 7;
colors[ 3 * k ] = color.r * intensity;
colors[ 3 * k + 1 ] = color.g * intensity;
colors[ 3 * k + 2 ] = color.b * intensity;
geometry.vertices.push( v );
colors[ k ] = ( color.clone().multiplyScalar( intensity ) );
k++;
}
}
geometry.colors = colors;
geometry.computeBoundingBox();
var material = new THREE.PointCloudMaterial( { size: pointSize, vertexColors: THREE.VertexColors } );
var pointcloud = new THREE.PointCloud( geometry, material );
return pointcloud;
}
My basic code is here: http://jsfiddle.net/dg34sbsk/
Any idea how to change each point color separately and dynamically? (Data for the colors will be coming in from a web service).
You can directly change its's value pointclouds[0].geometry.colors=... and after that pointclouds[0].geometry.colorsNeedUpdate=true.
To set each point's color just set the colors's children's value like pointclouds[0].geometry.colors[22]=new THREE.Color("rgb(255,0,0)");.
see this:http://jsfiddle.net/aboutqx/dg34sbsk/2/ .click and you will see the color of one point changes.

How to calculate the coordinates of a arrowhead based on the arrow?

I have a line that is based on two (x,y) coordinates I know. This line has a starting and an end point. Now I want to add an arrowhead at the end point of the line.
I know that the arrow is an equilateral triangle, and therefore each angle has 60 degrees. Additionally, I know the length of one side, which will be 20. I also no one edge of the triangle (that is the end point of the line).
How can I calculate the other two points of the triangle? I know I should use some trigonometry but how?
P.s. The endpoint of the line should be the arrowhead's tip.
You don't need trig., just some vector arithmetic...
Say the line goes from A to B, with the front vertex of the arrowhead at B. The length of the arrowhead is h = 10(√3) and its half-width is w = 10. We'll denote the unit vector from A to B as U = (B - A)/|B - A| (i.e., the difference divided by the length of the difference), and the unit vector perpendicular to this as V = [-Uy, Ux].
From these quantities, you can calculate the two rear vertices of the arrowhead as B - hU ± wV.
In C++:
struct vec { float x, y; /* … */ };
void arrowhead(vec A, vec B, vec& v1, vec& v2) {
float h = 10*sqrtf(3), w = 10;
vec U = (B - A)/(B - A).length();
vec V = vec(-U.y, U.x);
v1 = B - h*U + w*V;
v2 = B - h*U - w*V;
}
If you want to specify different angles, then you will need some trig. to calculate different values of h and w. Assuming you want an arrowhead of length h and tip-angle θ, then w = h tan(θ/2). In practice, however, it's simplest to specify h and w directly.
Here's a sample LINQPad program that shows how to do that:
void Main()
{
const int imageWidth = 512;
Bitmap b = new Bitmap(imageWidth , imageWidth , PixelFormat.Format24bppRgb);
Random r = new Random();
for (int index = 0; index < 10; index++)
{
Point fromPoint = new Point(0, 0);
Point toPoint = new Point(0, 0);
// Ensure we actually have a line
while (fromPoint == toPoint)
{
fromPoint = new Point(r.Next(imageWidth ), r.Next(imageWidth ));
toPoint = new Point(r.Next(imageWidth ), r.Next(imageWidth ));
}
// dx,dy = arrow line vector
var dx = toPoint.X - fromPoint.X;
var dy = toPoint.Y - fromPoint.Y;
// normalize
var length = Math.Sqrt(dx * dx + dy * dy);
var unitDx = dx / length;
var unitDy = dy / length;
// increase this to get a larger arrow head
const int arrowHeadBoxSize = 10;
var arrowPoint1 = new Point(
Convert.ToInt32(toPoint.X - unitDx * arrowHeadBoxSize - unitDy * arrowHeadBoxSize),
Convert.ToInt32(toPoint.Y - unitDy * arrowHeadBoxSize + unitDx * arrowHeadBoxSize));
var arrowPoint2 = new Point(
Convert.ToInt32(toPoint.X - unitDx * arrowHeadBoxSize + unitDy * arrowHeadBoxSize),
Convert.ToInt32(toPoint.Y - unitDy * arrowHeadBoxSize - unitDx * arrowHeadBoxSize));
using (Graphics g = Graphics.FromImage(b))
{
if (index == 0)
g.Clear(Color.White);
g.DrawLine(Pens.Black, fromPoint, toPoint);
g.DrawLine(Pens.Black, toPoint, arrowPoint1);
g.DrawLine(Pens.Black, toPoint, arrowPoint2);
}
}
using (var stream = new MemoryStream())
{
b.Save(stream, ImageFormat.Png);
Util.Image(stream.ToArray()).Dump();
}
}
Basically, you:
Calculate the vector of the arrow line
Normalize the vector, ie. making its length 1
Calculate the ends of the arrow heads by going:
First back from the head a certain distance
Then perpendicular out from the line a certain distance
Note that if you want the arrow head lines to have a different angle than 45 degrees, you'll have to use a different method.
The program above will draw 10 random arrows each time, here's an example:
Let's your line is (x0,y0)-(x1,y1)
Backward direction vector (dx, dy) = (x0-x1, y0-y1)
It's norm Norm = Sqrt(dx*dx+dy*dy)
Normalize it: (udx, udy) = (dx/Norm, dy/Norm)
Rotate by angles Pi/6 and -Pi/6
ax = udx * Sqrt(3)/2 - udy * 1/2
ay = udx * 1/2 + udy * Sqrt(3)/2
bx = udx * Sqrt(3)/2 + udy * 1/2
by = - udx * 1/2 + udy * Sqrt(3)/2
Your points: (x1 + 20 * ax, y1 + 20 * ay) and (x1 + 20 * bx, y1 + 20 * by)
I want to contribute my answer in C# based on Marcelo Cantos' answer since the algorithm works really well. I wrote a program to calculate the centroid of a laser beam projected on the CCD array. After the centroid is found, the direction angle line is drawn and I need the arrow head pointing at that direction. Since the angle is calculated, the arrow head would have to follow the angle in any of the direction.
This code gives you the flexibility of changing the arrow head size as shown in the pictures.
First you need the vector struct with all the necessary operators overloading.
private struct vec
{
public float x;
public float y;
public vec(float x, float y)
{
this.x = x;
this.y = y;
}
public static vec operator -(vec v1, vec v2)
{
return new vec(v1.x - v2.x, v1.y - v2.y);
}
public static vec operator +(vec v1, vec v2)
{
return new vec(v1.x + v2.x, v1.y + v2.y);
}
public static vec operator /(vec v1, float number)
{
return new vec(v1.x / number, v1.y / number);
}
public static vec operator *(vec v1, float number)
{
return new vec(v1.x * number, v1.y * number);
}
public static vec operator *(float number, vec v1)
{
return new vec(v1.x * number, v1.y * number);
}
public float length()
{
double distance;
distance = (this.x * this.x) + (this.y * this.y);
return (float)Math.Sqrt(distance);
}
}
Then you can use the same code given by Marcelo Cantos, but I made the length and half_width of the arrow head variables so that you can define that when calling the function.
private void arrowhead(float length, float half_width,
vec A, vec B, ref vec v1, ref vec v2)
{
float h = length * (float)Math.Sqrt(3);
float w = half_width;
vec U = (B - A) / (B - A).length();
vec V = new vec(-U.y, U.x);
v1 = B - h * U + w * V;
v2 = B - h * U - w * V;
}
Now you can call the function like this:
vec leftArrowHead = new vec();
vec rightArrowHead = new vec();
arrowhead(20, 10, new vec(circle_center_x, circle_center_y),
new vec(x_centroid_pixel, y_centroid_pixel),
ref leftArrowHead, ref rightArrowHead);
In my code, the circle center is the first vector location (arrow butt), and the centroid_pixel is the second vector location (arrow head).
I draw the arrow head by storing the vector values in the points for graphics.DrawPolygon() function in the System.Drawings. Code is shown below:
Point[] ppts = new Point[3];
ppts[0] = new Point((int)leftArrowHead.x, (int)leftArrowHead.y);
ppts[1] = new Point(x_cm_pixel,y_cm_pixel);
ppts[2] = new Point((int)rightArrowHead.x, (int)rightArrowHead.y);
g2.DrawPolygon(p, ppts);
You can find angle of line.
Vector ox = Vector(1,0);
Vector line_direction = Vector(line_begin.x - line_end.x, line_begin.y - line_end.y);
line_direction.normalize();
float angle = acos(ox.x * line_direction.x + line_direction.y * ox.y);
Then use this function to all 3 points using found angle.
Point rotate(Point point, float angle)
{
Point rotated_point;
rotated_point.x = point.x * cos(angle) - point.y * sin(angle);
rotated_point.y = point.x * sin(angle) + point.y * cos(angle);
return rotated_point;
}
Assuming that upper point of arrow's head is line's end it will perfectly rotated and fit to line.
Didn't test it =(
For anyone that is interested, #TomP was wondering about a js version, so here is a javascript version that I made. It is based off of #Patratacus and #Marcelo Cantos answers. Javascript doesn't support operator overloading, so it isn't as clean looking as C++ or other languages. Feel free to offer improvements.
I am using Class.js to create classes.
Vector = Class.extend({
NAME: "Vector",
init: function(x, y)
{
this.x = x;
this.y = y;
},
subtract: function(v1)
{
return new Vector(this.x - v1.x, this.y - v1.y);
},
add: function(v1)
{
return new Vector(this.x + v1.x, this.y + v1.y);
},
divide: function(number)
{
return new Vector(this.x / number, this.y / number);
},
multiply: function(number)
{
return new Vector(this.x * number, this.y * number);
},
length: function()
{
var distance;
distance = (this.x * this.x) + (this.y * this.y);
return Math.sqrt(distance);
}
});
And then a function to do the logic:
var getArrowhead = function(A, B)
{
var h = 10 * Math.sqrt(3);
var w = 5;
var v1 = B.subtract(A);
var length = v1.length();
var U = v1.divide(length);
var V = new Vector(-U.y, U.x);
var r1 = B.subtract(U.multiply(h)).add(V.multiply(w));
var r2 = B.subtract(U.multiply(h)).subtract(V.multiply(w));
return [r1,r2];
}
And call the function like this:
var A = new Vector(start.x,start.y);
var B = new Vector(end.x,end.y);
var vec = getArrowhead(A,B);
console.log(vec[0]);
console.log(vec[1]);
I know the OP didn't ask for any specific language, but I came across this looking for a JS implementation, so I thought I would post the result.

Calculating the position of points in a circle

I'm having a bit of a mind blank on this at the moment.
I've got a problem where I need to calculate the position of points around a central point, assuming they're all equidistant from the center and from each other.
The number of points is variable so it's DrawCirclePoints(int x)
I'm sure there's a simple solution, but for the life of me, I just can't see it :)
Given a radius length r and an angle t in radians and a circle's center (h,k), you can calculate the coordinates of a point on the circumference as follows (this is pseudo-code, you'll have to adapt it to your language):
float x = r*cos(t) + h;
float y = r*sin(t) + k;
A point at angle theta on the circle whose centre is (x0,y0) and whose radius is r is (x0 + r cos theta, y0 + r sin theta). Now choose theta values evenly spaced between 0 and 2pi.
Here's a solution using C#:
void DrawCirclePoints(int points, double radius, Point center)
{
double slice = 2 * Math.PI / points;
for (int i = 0; i < points; i++)
{
double angle = slice * i;
int newX = (int)(center.X + radius * Math.Cos(angle));
int newY = (int)(center.Y + radius * Math.Sin(angle));
Point p = new Point(newX, newY);
Console.WriteLine(p);
}
}
Sample output from DrawCirclePoints(8, 10, new Point(0,0));:
{X=10,Y=0}
{X=7,Y=7}
{X=0,Y=10}
{X=-7,Y=7}
{X=-10,Y=0}
{X=-7,Y=-7}
{X=0,Y=-10}
{X=7,Y=-7}
Good luck!
Placing a number in a circular path
// variable
let number = 12; // how many number to be placed
let size = 260; // size of circle i.e. w = h = 260
let cx= size/2; // center of x(in a circle)
let cy = size/2; // center of y(in a circle)
let r = size/2; // radius of a circle
for(let i=1; i<=number; i++) {
let ang = i*(Math.PI/(number/2));
let left = cx + (r*Math.cos(ang));
let top = cy + (r*Math.sin(ang));
console.log("top: ", top, ", left: ", left);
}
Using one of the above answers as a base, here's the Java/Android example:
protected void onDraw(Canvas canvas) {
super.onDraw(canvas);
RectF bounds = new RectF(canvas.getClipBounds());
float centerX = bounds.centerX();
float centerY = bounds.centerY();
float angleDeg = 90f;
float radius = 20f
float xPos = radius * (float)Math.cos(Math.toRadians(angleDeg)) + centerX;
float yPos = radius * (float)Math.sin(Math.toRadians(angleDeg)) + centerY;
//draw my point at xPos/yPos
}
For the sake of completion, what you describe as "position of points around a central point(assuming they're all equidistant from the center)" is nothing but "Polar Coordinates". And you are asking for way to Convert between polar and Cartesian coordinates which is given as x = r*cos(t), y = r*sin(t).
PHP Solution:
class point{
private $x = 0;
private $y = 0;
public function setX($xpos){
$this->x = $xpos;
}
public function setY($ypos){
$this->y = $ypos;
}
public function getX(){
return $this->x;
}
public function getY(){
return $this->y;
}
public function printX(){
echo $this->x;
}
public function printY(){
echo $this->y;
}
}
function drawCirclePoints($points, $radius, &$center){
$pointarray = array();
$slice = (2*pi())/$points;
for($i=0;$i<$points;$i++){
$angle = $slice*$i;
$newx = (int)($center->getX() + ($radius * cos($angle)));
$newy = (int)($center->getY() + ($radius * sin($angle)));
$point = new point();
$point->setX($newx);
$point->setY($newy);
array_push($pointarray,$point);
}
return $pointarray;
}
Here is how I found out a point on a circle with javascript, calculating the angle (degree) from the top of the circle.
const centreX = 50; // centre x of circle
const centreY = 50; // centre y of circle
const r = 20; // radius
const angleDeg = 45; // degree in angle from top
const radians = angleDeg * (Math.PI/180);
const pointY = centreY - (Math.cos(radians) * r); // specific point y on the circle for the angle
const pointX = centreX + (Math.sin(radians) * r); // specific point x on the circle for the angle
I had to do this on the web, so here's a coffeescript version of #scottyab's answer above:
points = 8
radius = 10
center = {x: 0, y: 0}
drawCirclePoints = (points, radius, center) ->
slice = 2 * Math.PI / points
for i in [0...points]
angle = slice * i
newX = center.x + radius * Math.cos(angle)
newY = center.y + radius * Math.sin(angle)
point = {x: newX, y: newY}
console.log point
drawCirclePoints(points, radius, center)
Here is an R version based on the #Pirijan answer above.
points <- 8
radius <- 10
center_x <- 5
center_y <- 5
drawCirclePoints <- function(points, radius, center_x, center_y) {
slice <- 2 * pi / points
angle <- slice * seq(0, points, by = 1)
newX <- center_x + radius * cos(angle)
newY <- center_y + radius * sin(angle)
plot(newX, newY)
}
drawCirclePoints(points, radius, center_x, center_y)
The angle between each of your points is going to be 2Pi/x so you can say that for points n= 0 to x-1 the angle from a defined 0 point is 2nPi/x.
Assuming your first point is at (r,0) (where r is the distance from the centre point) then the positions relative to the central point will be:
rCos(2nPi/x),rSin(2nPi/x)
Working Solution in Java:
import java.awt.event.*;
import java.awt.Robot;
public class CircleMouse {
/* circle stuff */
final static int RADIUS = 100;
final static int XSTART = 500;
final static int YSTART = 500;
final static int DELAYMS = 1;
final static int ROUNDS = 5;
public static void main(String args[]) {
long startT = System.currentTimeMillis();
Robot bot = null;
try {
bot = new Robot();
} catch (Exception failed) {
System.err.println("Failed instantiating Robot: " + failed);
}
int mask = InputEvent.BUTTON1_DOWN_MASK;
int howMany = 360 * ROUNDS;
while (howMany > 0) {
int x = getX(howMany);
int y = getY(howMany);
bot.mouseMove(x, y);
bot.delay(DELAYMS);
System.out.println("x:" + x + " y:" + y);
howMany--;
}
long endT = System.currentTimeMillis();
System.out.println("Duration: " + (endT - startT));
}
/**
*
* #param angle
* in degree
* #return
*/
private static int getX(int angle) {
double radians = Math.toRadians(angle);
Double x = RADIUS * Math.cos(radians) + XSTART;
int result = x.intValue();
return result;
}
/**
*
* #param angle
* in degree
* #return
*/
private static int getY(int angle) {
double radians = Math.toRadians(angle);
Double y = RADIUS * Math.sin(radians) + YSTART;
int result = y.intValue();
return result;
}
}
Based on the answer above from Daniel, here's my take using Python3.
import numpy
def circlepoints(points,radius,center):
shape = []
slice = 2 * 3.14 / points
for i in range(points):
angle = slice * i
new_x = center[0] + radius*numpy.cos(angle)
new_y = center[1] + radius*numpy.sin(angle)
p = (new_x,new_y)
shape.append(p)
return shape
print(circlepoints(100,20,[0,0]))

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