How to interpolate hue values in HSV colour space? - javascript

I'm trying to interpolate between two colours in HSV colour space to produce a smooth colour gradient.
I'm using a linear interpolation, eg:
h = (1 - p) * h1 + p * h2
s = (1 - p) * s1 + p * s2
v = (1 - p) * v1 + p * v2
(where p is the percentage, and h1, h2, s1, s2, v1, v2 are the hue, saturation and value components of the two colours)
This produces a good result for s and v but not for h. As the hue component is an angle, the calculation needs to work out the shortest distance between h1 and h2 and then do the interpolation in the right direction (either clockwise or anti-clockwise).
What formula or algorithm should I use?
EDIT: By following Jack's suggestions I modified my JavaScript gradient function and it works well. For anyone interested, here's what I ended up with:
// create gradient from yellow to red to black with 100 steps
var gradient = hsbGradient(100, [{h:0.14, s:0.5, b:1}, {h:0, s:1, b:1}, {h:0, s:1, b:0}]);
function hsbGradient(steps, colours) {
var parts = colours.length - 1;
var gradient = new Array(steps);
var gradientIndex = 0;
var partSteps = Math.floor(steps / parts);
var remainder = steps - (partSteps * parts);
for (var col = 0; col < parts; col++) {
// get colours
var c1 = colours[col],
c2 = colours[col + 1];
// determine clockwise and counter-clockwise distance between hues
var distCCW = (c1.h >= c2.h) ? c1.h - c2.h : 1 + c1.h - c2.h;
distCW = (c1.h >= c2.h) ? 1 + c2.h - c1.h : c2.h - c1.h;
// ensure we get the right number of steps by adding remainder to final part
if (col == parts - 1) partSteps += remainder;
// make gradient for this part
for (var step = 0; step < partSteps; step ++) {
var p = step / partSteps;
// interpolate h, s, b
var h = (distCW <= distCCW) ? c1.h + (distCW * p) : c1.h - (distCCW * p);
if (h < 0) h = 1 + h;
if (h > 1) h = h - 1;
var s = (1 - p) * c1.s + p * c2.s;
var b = (1 - p) * c1.b + p * c2.b;
// add to gradient array
gradient[gradientIndex] = {h:h, s:s, b:b};
gradientIndex ++;
}
}
return gradient;
}

You should just need to find out which is the shortest path from starting hue to ending hue. This can be done easily since hue values range from 0 to 255.
You can first subtract the lower hue from the higher one, then add 256 to the lower one to check again the difference with swapped operands.
int maxCCW = higherHue - lowerHue;
int maxCW = (lowerHue+256) - higherHue;
So you'll obtain two values, the greater one decides if you should go clockwise or counterclockwise. Then you'll have to find a way to make the interpolation operate on modulo 256 of the hue, so if you are interpolating from 246 to 20 if the coefficient is >= 0.5f you should reset hue to 0 (since it reaches 256 and hue = hue%256 in any case).
Actually if you don't care about hue while interpolating over the 0 but just apply modulo operator after calculating the new hue it should work anyway.

Although this answer is late, the accepted one is incorrect in stating that hue should be within [0, 255]; also more justice can be done with clearer explanation and code.
Hue is an angular value in the interval [0, 360); a full circle where 0 = 360. The HSV colour space is easier to visualize and is more intuitive to humans then RGB. HSV forms a cylinder from which a slice is shown in many colour pickers, while RGB is really a cube and isn't really a good choice for a colour picker; most ones which do use it would have to employ more sliders than required for a HSV picker.
The requirement when interpolating hue is that the smaller arc is chosen to reach from one hue to another. So given two hue values, there are four possibilities, given with example angles below:
Δ | ≤ 180 | > 180
--|---------|---------
+ | 40, 60 | 310, 10
− | 60, 40 | 10, 310
if Δ = 180 then both +/− rotation are valid options
Lets take + as counter-clockwise and − as clockwise rotation. If the difference in absolute value exceeds 180 then normalize it by ± 360 to make sure the magnitude is within 180; this also reverses the direction, rightly.
var d = h2 - h1;
var delta = d + ((Math.abs(d) > 180) ? ((d < 0) ? 360 : -360) : 0);
Now just divide delta by the required number of steps to get the weight of each loop iteration to add to the start angle during interpolating.
var new_angle = start + (i * delta);
Relevant function excerpted from the complete code that follows:
function interpolate(h1, h2, steps) {
var d = h2 - h1;
var delta = (d + ((Math.abs(d) > 180) ? ((d < 0) ? 360 : -360) : 0)) / (steps + 1.0);
var turns = [];
for (var i = 1; d && i <= steps; ++i)
turns.push(((h1 + (delta * i)) + 360) % 360);
return turns;
}
"use strict";
function interpolate(h1, h2, steps) {
var d = h2 - h1;
var delta = (d + ((Math.abs(d) > 180) ? ((d < 0) ? 360 : -360) : 0)) / (steps + 1.0);
var turns = [];
for (var i = 1; d && i <= steps; ++i)
turns.push(((h1 + (delta * i)) + 360) % 360);
return turns;
}
function get_results(h1, h2, steps) {
h1 = norm_angle(h1);
h2 = norm_angle(h2);
var r = "Start: " + h1 + "<br />";
var turns = interpolate(h1, h2, steps);
r += turns.length ? "Turn: " : "";
r += turns.join("<br />Turn: ");
r += (turns.length ? "<br />" : "") + "Stop: " + h2;
return r;
}
function run() {
var h1 = get_angle(document.getElementById('h1').value);
var h2 = get_angle(document.getElementById('h2').value);
var steps = get_num(document.getElementById('steps').value);
var result = get_results(h1, h2, steps);
document.getElementById('res').innerHTML = result;
}
function get_num(s) {
var n = parseFloat(s);
return (isNaN(n) || !isFinite(n)) ? 0 : n;
}
function get_angle(s) {
return get_num(s) % 360;
}
function norm_angle(a) {
a %= 360;
a += (a < 0) ? 360 : 0;
return a;
}
<h1 id="title">Hue Interpolation</h1>
Angle 1
<input type="text" id="h1" />
<br />Angle 2
<input type="text" id="h2" />
<br />
<br />Intermediate steps
<input type="text" id="steps" value="5" />
<br />
<br/>
<input type="submit" value="Run" onclick="run()" />
<p id="res"></p>

Related

What am I misunderstanding about this fluid dynamics calculation?

The last question was deleted for being too short and vague, so this time I will try to be as verbose as possible.
I am using Javascript to implement a basic physically accurate SPH following this paper. All I want in the beginning is to get the basics working, just to get through the first algorithm.
I start with a test case of just two particles at (0.05, 0.05) and (0.06999, 0.05). Their sample radii just touch, and I am already getting weird numbers.
This is how I calculate the densities and pressures:
for (var i = 0; i < particles.length; i++)
{
var pi = particles[i];
pi.density = 0;
for (var j = 0; j < pi.neighbors.length; j++)
{
var n = pi.neighbors[j];
var pj = n.p;
var h = (pi.sampleDist + pj.sampleDist) / 2;
n.r = subv(pi.pos, pj.pos); // subv() subtracts first vector from second
n.dist = length(n.r);
n.q = n.dist/h;
pi.density += pj.mass * W(n.q, h, 2);
}
pi.pressure = pi.stiffness * (Math.pow(pi.density / pi.restDensity, 7) - 1);
}
using the smoothing kernel
function W(q, h, d)
{
var f = 0;
if (0 <= q && q < 1) f = 2/3 - sqr(q) + 1/2*cube(q); // 2/3 - q² + 1/2q³
else if (1 <= q && q < 2) f = 1/6 * cube(2 - q); // 1/6(2-q)³
return 3/Tau / Math.pow(h, d) * f;
}
These are the settings I'm using:
mass: 0.001 kg
h: 0.01 m
rest density: 1000 kg/m³
stiffness: 0.2
In the case of just two particles with these settings, each particle's density becomes
pi.density = pj.mass * W(n.q, h, 2); = 0.001 * W(1.99899, 0.01, 2) = 0.001 * 0.00000079577 = 0.00000000079...
This number makes sense to me — there is almost nothing around, so the density is almost zero.
But now I go to calculate the pressure force, which I do like this:
Here is my implementation:
var Fp = {x:0, y:0};
if (pi.density !== 0)
{
var Vp = valueGradient(pi, 'pressure');
Fp.x = -pi.mass/pi.density * Vp.x;
Fp.y = -pi.mass/pi.density * Vp.y;
}
function valueGradient(pi, A) // A is a scalar
{
var vA = {x: 0, y: 0};
for (var j = 0; j < pi.neighbors.length; j++)
{
var n = pi.neighbors[j];
var pj = n.p;
var h = (pi.sampleDist + pj.sampleDist) / 2;
var f = pj.mass / pj.density * pj[A];
var vw = VW(n, h);
vA.x += f * vw.x;
vA.y += f * vw.y;
}
return vA;
}
function VW(n, h)
{
var dw =
dW(n.q, h, 2)
/ (n.dist * h);
return {
x: dw * n.r.x,
y: dw * n.r.y };
}
function dW(q, h, d) // derivative of kernel
{
var f = 0;
if (0 <= q && q < 1) f = -2*q + 3/2*sqr(q); // -2q + 3/2q²
else if (1 <= q && q < 2) f = -1/2 * sqr(2 - q); // -1/2(2-q)²
return 3/Tau / Math.pow(h, d) * f;
}
But something is going wrong, because when the particles sample radii just touch, the repulsive force is enormous, and when they're close it goes to zero. I get why this would be happening, because given my settings there are lots of 0.000... everywhere, and with the density so tiny, dividing by all these 0.000... will give huge numbers in the end. So it makes sense why those numbers are there, and so I suspect that I'm misunderstanding something fundamental.
This feels... backwards. What mistake am I making? I keep looking at the kernel derivative...
Another thing is that the forces are not strictly symmetrical as they should be, they tend to stay diagonal until the difference in one of the coordinates is around zero, where they suddenly flip.
I feel like I'm messing up the differential equations somewhere...
A live version of this code that can be stepped through is at www.bourt.com/particles. The code in question is in simulation.js, physics.js and kernel.js.

javascript algorithm: Check x/y-points to get if component is a circle or not

I have an matrix like this, so I hope you can get the circle outline:
EDIT 1: What about the outline? The outline doesn't includes spaces (so every y-value gets minimum 2 x-values)
EDIT 2: What is a circle? Searching for an more ore less "exact circles" like the example below! (nearly same radius at every point)
00000000000000000000000000000000
00000000000001111111100000000000
00000000000100000000010000000000
00000000010000000000000100000000
00000000100000000000000010000000
00000000100000000000000010000000
00000001000000000000000001000000
00000010000000000000000000100000
00000010000000000000000000100000
00000010000000000000000000100000
00000100000000000000000000100000
00000100000000000000000000100000
00000100000000000000000000100000
00000100000000000000000000100000
00000100000000000000000000100000
00000010000000000000000000100000
00000010000000000000000000100000
00000010000000000000000001000000
00000001000000000000000010000000
00000000100000000000000010000000
00000000100000000000000100000000
00000000010000000000001000000000
00000000000111111111100000000000
00000000000000000000000000000000
00000000000000000000000000000000
and I also have an array like this including all positions of the outline:
var coordinates = [
[13,1],[14,1],[15,1],[16,1],[17,1],[18,1],[19,1],[20,1],
[11,2],[21,2],
[9,3],[23,3],
[8,4],[24,4],
[8,5],[24,5],
[7,6],[25,6],
[6,7],[26,7],
[6,8],[26,8],
[6,9],[26,9],
[5,10],[26,10],
[5,11],[26,11],
[5,12],[26,12],
[5,13],[26,13],
[5,14],[26,14],
[6,15],[26,15],
[6,16],[26,16],
[6,17],[25,17],
[7,18],[24,18],
[8,19],[24,19],
[8,20],[23,20],
[9,21],[22,21],
[11,22],[12,22],[13,22],[14,22],[15,22],[16,22],[17,22],[18,22],[19,22],[20,22]]
What is a good way to check if the coordinates are matching to be a circle?
My first idea was using some code like this but tbh I'm sure there is a way more efficient & working way:
var circle = [[13,1],[14,1],[15,1],[16,1],[17,1],[18,1],[19,1],[20,1],[11,2],[21,2],[9,3],[23,3],[8,4],[24,4],[8,5],[24,5],[7,6],[25,6],[6,7],[26,7],[6,8],[26,8],[6,9],[26,9],[5,10],[26,10],[5,11],[26,11],[5,12],[26,12],[5,13],[26,13],[5,14],[26,14],[6,15],[26,15],[6,16],[26,16],[6,17],[25,17],[7,18],[24,18],[8,19],[24,19],[8,20],[23,20],[9,21],[22,21],[11,22],[12,22],[13,22],[14,22],[15,22],[16,22],[17,22],[18,22],[19,22],[20,22]]
var no_circle= [[13,1],[14,1],[25,4]]
Array.prototype.is_circle = function() {
var min = {
'x': Infinity,
'y': Infinity
};
var max = {
'x': 0,
'y': 0
};
var center = {
'x': 0,
'y': 0
};
var radius;
this.forEach(function(a) {
a[0] = a[0]
a[1] = a[1]
if (a[0] > max.x) max.x = a[0];
if (a[0] < min.x) min.x = a[0];
if (a[1] > max.y) max.y = a[1];
if (a[1] < min.y) min.y = a[1];
});
center.x = (max.x + min.x) / 2;
center.y = (max.y + min.y) / 2;
radius = []
radius[0] = max.x - center.x
radius[1] = center.x - min.x
radius[2] = center.y - min.y
radius[3] = max.y - center.y
r = (radius[0] + radius[1] + radius[2] + radius[3]) / 4;
if ((radius[0] > r-1 && radius[0] < r+1) && (radius[1] > r-1 && radius[1] < r+1) && (radius[2] > r-1 && radius[2] < r+1) && (radius[3] > r-1 && radius[3] < r+1)) return true;
return false
}
var result1 = circle.is_circle();
console.log(result1)
var result2 = no_circle.is_circle();
console.log(result2)
Your algorithm seems to check only four most distant points on X and Y axis. I guess that if you provide points that represent square shape it will also pass is_circle test.
I propose that you do two-stage test with some extra roundness margin called e. Walk over whole set of points and remember x_min, y_min, x_max and y_max. Then make check if difference between delta of X and delta of Y is in the error margin i.e. abs((x_max-x_min) - (y_max-y_min)) <= e. That checks squarness of shape so that it can be circle, not oval. If that test passes then calculate center point c at (x_c, y_c) = (x_min+(x_max-x_min)/2, y_min+(y_max-y_min)/2) and for each point calculate if radius (distance from any point to center c) is within error margin e. To save on expensive calculations check if squared radius of each point is within error margin i.e. if abs((x-x_c)^2 + (y-y_c)^2 - r^2) <= e, where r^2 is computed for center c and first point on the list.

How to detect if a user has drawn a circle on a touch device using canvas and javascript?

I am creating a Tangram puzzle game using Javascript. And I need to detect when a user has drawn a circle (or circle like shape) with their finger. I have been able to gather hundreds (if not thousands) of x and y points with:
var touchX = event.targetTouches[0].pageX - canvas.offsetLeft;
var touchY = event.targetTouches[0].pageY - canvas.offsetTop;
I then push each x and y coordinate into an array:
touchMoveX.push(touchX);
touchMoveY.push(touchY);
I then loop through each array and create two points:
for(var i = 0; i < touchMoveX.length; i++)
{
for(var l=0; l < touchMoveY.length; l++)
{
var xPosition = touchMoveX[i];
var yPosition = touchMoveY[l];
var v1x = touchMoveX[i];
var v2x = touchMoveX[i + 1];
var v1y = touchMoveY[l];
var v2y = touchMoveY[l + 1];
Then using those two points, I use the following formula to figure out the angle between these two points in degrees:
var v1 = {x: v1x, y: v1y}, v2 = {x: v2x, y: v2y},
angleRad = Math.acos( (v1.x * v2.x + v1.y * v2.y) /
(Math.sqrt(v1.x*v1.x + v1.y*v1.y) * Math.sqrt(v2.x*v2.x + v2.y*v2.y) ) ),
angleDeg = angleRad * 180 / Math.PI;
I then sum up all of the angles and see if they are around 360 degrees.
But the above code I have described isn't working very well. Does someone out there have a better way to do this? Thank you very much.
yeah compute the average of all points (giving you a cheaply approximated center) then check if more than a certain percent of points are within a certain threshold. You can tune those values to adjust the precision until it feels right.
edit: Didn't consider that the circle could have multiple sizes, but you could just add another step computing the average of all distances. Adjusted the example for that.
var totalAmount = touchMoveX.length;
// sum up all coordinates and divide them by total length
// the average is a cheap approximation of the center.
var averageX = touchMoveX.reduce( function ( previous, current) {
return previous + current;
} ) / totalAmount ;
var averageY = touchMoveY.reduce( function ( previous, current) {
return previous + current;
} ) / totalAmount ;
// compute distance to approximated center from each point
var distances = touchMoveX.map ( function ( x, index ) {
var y = touchMoveY[index];
return Math.sqrt( Math.pow(x - averageX, 2) + Math.pow(y - averageY, 2) );
} );
// average of those distance is
var averageDistance = distances.reduce ( function ( previous, current ) {
return previous + current;
} ) / distances.length;
var min = averageDistance * 0.8;
var max = averageDistance * 1.2;
// filter out the ones not inside the min and max boundaries
var inRange = distances.filter ( function ( d ) {
return d > min && d < max;
} ).length;
var minPercentInRange = 80;
var percentInRange = inRange.length / totalAmount * 100;
// by the % of points within those boundaries we can guess if it's circle
if( percentInRange > minPercentInRange ) {
//it's probably a circle
}

Weighted Random Number Generator in Javascript

I am using the following code to generate a random number:
function getRandomInt (min, max) {
return Math.floor((Math.random() * (max - min + 1)) + min;
}
What I want to do is add a weighting that favours the numbers at the lower end of the range.
I thought about maybe trying to multiply the numbers by 1/cosine.
Would this work and does anyone know how I might go about it?
Many thanks!
First Solution
You need a function which contains the points (0, 0) and (1, 1). For instance: x^n when n > 0
Math.pow(1, n) === 1
And
Math.pow(0, n) === 0
Therefore, you would just change n depending on how you want the weighting to work.
When n = 1 : y === x
When n > 1 : y <= x
When 0 < n < 1 : y >= x
So, if you want lower values to be favored over higher values, simply use n > 1.
var weighted = Math.pow(Math.random(), 2);
Then you can scale the result as usual.
var scaled = Math.floor(weighted * (max - min + 1)) + min;
Other Functions
Likewise, you could use any continuous function which contains the points (0, 0), (1, 1), and has range and domain of [0, 1].
Sine
y = sin(xπ/2)
Cosine
y = 1 - cos(xπ/2)
EDIT: there was a type in the final formula, log(2+log(x)) is incorrect it should have been log(1+log(x))+1, its fixed now.
If you are using logarithmic weighting, using something like
var x = Math.random();
var weighted = x * Math.log(1+x);
would make 0.5 weigh in at around 0.2, but 1 would only weigh in at around 0.69.
Using this
var x = Math.random();
var weighted = x * Math.log(2 + Math.log(x));
would allow 1 to weigh in at 1. So combine them, and this
var x = Math.random();
var weighted = (x <= 0.5) ? x * Math.log(1 + x) : x * Math.log(1 + Math.log(x))+1;
should do the trick

How can I calculate the area of a bezier curve?

Given the following path (for example) which describes a SVG cubic bezier curve: "M300,140C300,40,500,40,500,140",
and assuming a straight line connecting the end points 300,140 to 500,140 (closing the area under the curve), is it possible to calculate the area so enclosed?
Can anyone suggest a formula (or JavaScript) to accomplish this?
Convert the path to a polygon of arbitrary precision, and then calculate the area of the polygon.
Interactive Demo: Area of Path via Subdivision (broken)
                     
At its core the above demo uses functions for adaptively subdividing path into a polygon and computing the area of a polygon:
// path: an SVG <path> element
// threshold: a 'close-enough' limit (ignore subdivisions with area less than this)
// segments: (optional) how many segments to subdivisions to create at each level
// returns: a new SVG <polygon> element
function pathToPolygonViaSubdivision(path,threshold,segments){
if (!threshold) threshold = 0.0001; // Get really, really close
if (!segments) segments = 3; // 2 segments creates 0-area triangles
var points = subdivide( ptWithLength(0), ptWithLength( path.getTotalLength() ) );
for (var i=points.length;i--;) points[i] = [points[i].x,points[i].y];
var doc = path.ownerDocument;
var poly = doc.createElementNS('http://www.w3.org/2000/svg','polygon');
poly.setAttribute('points',points.join(' '));
return poly;
// Record the distance along the path with the point for later reference
function ptWithLength(d) {
var pt = path.getPointAtLength(d); pt.d = d; return pt;
}
// Create segments evenly spaced between two points on the path.
// If the area of the result is less than the threshold return the endpoints.
// Otherwise, keep the intermediary points and subdivide each consecutive pair.
function subdivide(p1,p2){
var pts=[p1];
for (var i=1,step=(p2.d-p1.d)/segments;i<segments;i++){
pts[i] = ptWithLength(p1.d + step*i);
}
pts.push(p2);
if (polyArea(pts)<=threshold) return [p1,p2];
else {
var result = [];
for (var i=1;i<pts.length;++i){
var mids = subdivide(pts[i-1], pts[i]);
mids.pop(); // We'll get the last point as the start of the next pair
result = result.concat(mids)
}
result.push(p2);
return result;
}
}
// Calculate the area of an polygon represented by an array of points
function polyArea(points){
var p1,p2;
for(var area=0,len=points.length,i=0;i<len;++i){
p1 = points[i];
p2 = points[(i-1+len)%len]; // Previous point, with wraparound
area += (p2.x+p1.x) * (p2.y-p1.y);
}
return Math.abs(area/2);
}
}
// Return the area for an SVG <polygon> or <polyline>
// Self-crossing polys reduce the effective 'area'
function polyArea(poly){
var area=0,pts=poly.points,len=pts.numberOfItems;
for(var i=0;i<len;++i){
var p1 = pts.getItem(i), p2=pts.getItem((i+-1+len)%len);
area += (p2.x+p1.x) * (p2.y-p1.y);
}
return Math.abs(area/2);
}
Following is the original answer, which uses a different (non-adaptive) technique for converting the <path> to a <polygon>.
Interactive Demo: http://phrogz.net/svg/area_of_path.xhtml (broken)
                 
At its core the above demo uses functions for approximating a path with a polygon and computing the area of a polygon.
// Calculate the area of an SVG polygon/polyline
function polyArea(poly){
var area=0,pts=poly.points,len=pts.numberOfItems;
for(var i=0;i<len;++i){
var p1 = pts.getItem(i), p2=pts.getItem((i+len-1)%len);
area += (p2.x+p1.x) * (p2.y-p1.y);
}
return Math.abs(area/2);
}
// Create a <polygon> approximation for an SVG <path>
function pathToPolygon(path,samples){
if (!samples) samples = 0;
var doc = path.ownerDocument;
var poly = doc.createElementNS('http://www.w3.org/2000/svg','polygon');
// Put all path segments in a queue
for (var segs=[],s=path.pathSegList,i=s.numberOfItems-1;i>=0;--i)
segs[i] = s.getItem(i);
var segments = segs.concat();
var seg,lastSeg,points=[],x,y;
var addSegmentPoint = function(s){
if (s.pathSegType == SVGPathSeg.PATHSEG_CLOSEPATH){
}else{
if (s.pathSegType%2==1 && s.pathSegType>1){
x+=s.x; y+=s.y;
}else{
x=s.x; y=s.y;
}
var last = points[points.length-1];
if (!last || x!=last[0] || y!=last[1]) points.push([x,y]);
}
};
for (var d=0,len=path.getTotalLength(),step=len/samples;d<=len;d+=step){
var seg = segments[path.getPathSegAtLength(d)];
var pt = path.getPointAtLength(d);
if (seg != lastSeg){
lastSeg = seg;
while (segs.length && segs[0]!=seg) addSegmentPoint( segs.shift() );
}
var last = points[points.length-1];
if (!last || pt.x!=last[0] || pt.y!=last[1]) points.push([pt.x,pt.y]);
}
for (var i=0,len=segs.length;i<len;++i) addSegmentPoint(segs[i]);
for (var i=0,len=points.length;i<len;++i) points[i] = points[i].join(',');
poly.setAttribute('points',points.join(' '));
return poly;
}
I hesitated to just make a comment or a full reply. But a simple Google search of "area bezier curve" results in the first three links (the first one being this same post), in :
http://objectmix.com/graphics/133553-area-closed-bezier-curve.html (archived)
that provides the closed form solution, using the divergence theorem. I am surprised that this link has not been found by the OP.
Copying the text in case the website goes down, and crediting the author of the reply Kalle Rutanen:
An interesting problem. For any piecewise differentiable curve in 2D,
the following general procedure gives you the area inside the curve /
series of curves. For polynomial curves (Bezier curves), you will get
closed form solutions.
Let g(t) be a piecewise differentiable curve, with 0 <= t <= 1. g(t)
is oriented clockwise and g(1) = g(0).
Let F(x, y) = [x, y] / 2
Then div(F(x, y)) = 1 where div is for divergence.
Now the divergence theorem gives you the area inside the closed curve
g (t) as a line integral along the curve:
int(dot(F(g(t)), perp(g'(t))) dt, t = 0..1)
= (1 / 2) * int(dot(g(t), perp(g'(t))) dt, t = 0..1)
perp(x, y) = (-y, x)
where int is for integration, ' for differentiation and dot for dot
product. The integration has to be pieced to the parts corresponding
to the smooth curve segments.
Now for examples. Take the Bezier degree 3 and one such curve with
control points (x0, y0), (x1, y1), (x2, y2), (x3, y3). The integral
over this curve is:
I := 3 / 10 * y1 * x0 - 3 / 20 * y1 * x2 - 3 / 20 * y1 * x3 - 3 / 10 *
y0 * x1 - 3 / 20 * y0 * x2 - 1 / 20 * y0 * x3 + 3 / 20 * y2 * x0 + 3 /
20 * y2 * x1 - 3 / 10 * y2 * x3 + 1 / 20 * y3 * x0 + 3 / 20 * y3 * x1
+ 3 / 10 * y3 * x2
Calculate this for each curve in the sequence and add them up. The sum
is the area enclosed by the curves (assuming the curves form a loop).
If the curve consists of just one Bezier curve, then it must be x3 =
x0 and y3 = y0, and the area is:
Area := 3 / 20 * y1 * x0 - 3 / 20 * y1 * x2 - 3 / 20 * y0 * x1 + 3 /
20 * y0 * x2 - 3 / 20 * y2 * x0 + 3 / 20 * y2 * x1
Hope I did not do mistakes.
--
Kalle Rutanen
http://kaba.hilvi.org
I had the same problem but I am not using javascript so I cannot use the accepted answer of #Phrogz. In addition the SVGPathElement.getPointAtLength() which is used in the accepted answer is deprecated according to Mozilla.
When describing a Bézier curve with the points (x0/y0), (x1/y1), (x2/y2) and (x3/y3) (where (x0/y0) is the start point and (x3/y3) the end point) you can use the parametrized form:
(source: Wikipedia)
with B(t) being the point on the Bézier curve and Pi the Bézier curve defining point (see above, P0 is the starting point, ...). t is the running variable with 0 ≤ t ≤ 1.
This form makes it very easy to approximate a Bézier curve: You can generate as much points as you want by using t = i / npoints. (Note that you have to add the start and the end point). The result is a polygon. You can then use the shoelace formular (like #Phrogz did in his solution) to calculate the area. Note that for the shoelace formular the order of the points is important. By using t as the parameter the order will always be correct.
To match the question here is an interactive example in the code snippet, also written in javascript. This can be adopted to other languages. It does not use any javascript (or svg) specific commands (except for the drawings). Note that this requires a browser which supports HTML5 to work.
/**
* Approximate the bezier curve points.
*
* #param bezier_points: object, the points that define the
* bezier curve
* #param point_number: int, the number of points to use to
* approximate the bezier curve
*
* #return Array, an array which contains arrays where the
* index 0 contains the x and the index 1 contains the
* y value as floats
*/
function getBezierApproxPoints(bezier_points, point_number){
if(typeof bezier_points == "undefined" || bezier_points === null){
return [];
}
var approx_points = [];
// add the starting point
approx_points.push([bezier_points["x0"], bezier_points["y0"]]);
// implementation of the bezier curve as B(t), for futher
// information visit
// https://wikipedia.org/wiki/B%C3%A9zier_curve#Cubic_B%C3%A9zier_curves
var bezier = function(t, p0, p1, p2, p3){
return Math.pow(1 - t, 3) * p0 +
3 * Math.pow(1 - t, 2) * t * p1 +
3 * (1 - t) * Math.pow(t, 2) * p2 +
Math.pow(t, 3) * p3;
};
// Go through the number of points, divide the total t (which is
// between 0 and 1) by the number of points. (Note that this is
// point_number - 1 and starting at i = 1 because of adding the
// start and the end points.)
// Also note that using the t parameter this will make sure that
// the order of the points is correct.
for(var i = 1; i < point_number - 1; i++){
let t = i / (point_number - 1);
approx_points.push([
// calculate the value for x for the current t
bezier(
t,
bezier_points["x0"],
bezier_points["x1"],
bezier_points["x2"],
bezier_points["x3"]
),
// calculate the y value
bezier(
t,
bezier_points["y0"],
bezier_points["y1"],
bezier_points["y2"],
bezier_points["y3"]
)
]);
}
// Add the end point. Note that it is important to do this
// **after** the other points. Otherwise the polygon will
// have a weird form and the shoelace formular for calculating
// the area will get a weird result.
approx_points.push([bezier_points["x3"], bezier_points["y3"]]);
return approx_points;
}
/**
* Get the bezier curve values of the given path.
*
* The returned array contains objects where each object
* describes one cubic bezier curve. The x0/y0 is the start
* point and the x4/y4 is the end point. x1/y1 and x2/y2 are
* the control points.
*
* Note that a path can also contain other objects than
* bezier curves. Arcs, quadratic bezier curves and lines
* are ignored.
*
* #param svg: SVGElement, the svg
* #param path_id: String, the id of the path element in the
* svg
*
* #return array, an array of plain objects where each
* object represents one cubic bezier curve with the values
* x0 to x4 and y0 to y4 representing the x and y
* coordinates of the points
*/
function getBezierPathPoints(svg, path_id){
var path = svg.getElementById(path_id);
if(path === null || !(path instanceof SVGPathElement)){
return [];
}
var path_segments = splitPath(path);
var points = [];
var x = 0;
var y = 0;
for(index in path_segments){
if(path_segments[index]["type"] == "C"){
let bezier = {};
// start is the end point of the last element
bezier["x0"] = x;
bezier["y0"] = y;
bezier["x1"] = path_segments[index]["x1"];
bezier["y1"] = path_segments[index]["y1"];
bezier["x2"] = path_segments[index]["x2"];
bezier["y2"] = path_segments[index]["y2"];
bezier["x3"] = path_segments[index]["x"];
bezier["y3"] = path_segments[index]["y"];
points.push(bezier);
}
x = path_segments[index]["x"];
y = path_segments[index]["y"];
}
return points;
}
/**
* Split the given path to the segments.
*
* #param path: SVGPathElement, the path
*
* #return object, the split path `d`
*/
function splitPath(path){
let d = path.getAttribute("d");
d = d.split(/\s*,|\s+/);
let segments = [];
let segment_names = {
"M": ["x", "y"],
"m": ["dx", "dy"],
"H": ["x"],
"h": ["dx"],
"V": ["y"],
"v": ["dy"],
"L": ["x", "y"],
"l": ["dx", "dy"],
"Z": [],
"C": ["x1", "y1", "x2", "y2", "x", "y"],
"c": ["dx1", "dy1", "dx2", "dy2", "dx", "dy"],
"S": ["x2", "y2", "x", "y"],
"s": ["dx2", "dy2", "dx", "dy"],
"Q": ["x1", "y1", "x", "y"],
"q": ["dx1", "dy1", "dx", "dy"],
"T": ["x", "y"],
"t": ["dx", "dy"],
"A": ["rx", "ry", "rotation", "large-arc", "sweep", "x", "y"],
"a": ["rx", "ry", "rotation", "large-arc", "sweep", "dx", "dy"]
};
let current_segment_type;
let current_segment_value;
let current_segment_index;
for(let i = 0; i < d.length; i++){
if(typeof current_segment_value == "number" && current_segment_value < segment_names[current_segment_type].length){
let segment_values = segment_names[current_segment_type];
segments[current_segment_index][segment_values[current_segment_value]] = d[i];
current_segment_value++;
}
else if(typeof segment_names[d[i]] !== "undefined"){
current_segment_index = segments.length;
current_segment_type = d[i];
current_segment_value = 0;
segments.push({"type": current_segment_type});
}
else{
delete current_segment_type;
delete current_segment_value;
delete current_segment_index;
}
}
return segments;
}
/**
* Calculate the area of a polygon. The pts are the
* points which define the polygon. This is
* implementing the shoelace formular.
*
* #param pts: Array, the points
*
* #return float, the area
*/
function polyArea(pts){
var area = 0;
var n = pts.length;
for(var i = 0; i < n; i++){
area += (pts[i][1] + pts[(i + 1) % n][1]) * (pts[i][0] - pts[(i + 1) % n][0]);
}
return Math.abs(area / 2);
}
// only for the demo
(function(){
document.getElementById('number_of_points').addEventListener('change', function(){
var svg = document.getElementById("svg");
var bezier_points = getBezierPathPoints(svg, "path");
// in this example there is only one bezier curve
bezier_points = bezier_points[0];
// number of approximation points
var approx_points_num = parseInt(this.value);
var approx_points = getBezierApproxPoints(bezier_points, approx_points_num);
var doc = svg.ownerDocument;
// remove polygon
var polygons;
while((polygons = doc.getElementsByTagName("polygon")).length > 0){
polygons[0].parentNode.removeChild(polygons[0]);
}
// remove old circles
var circles;
while((circles = doc.getElementsByTagName("circle")).length > 0){
circles[0].parentNode.removeChild(circles[0]);
}
// add new circles and create polygon
var polygon_points = [];
for(var i = 0; i < approx_points.length; i++){
let circle = doc.createElementNS('http://www.w3.org/2000/svg', 'circle');
circle.setAttribute('cx', approx_points[i][0]);
circle.setAttribute('cy', approx_points[i][1]);
circle.setAttribute('r', 1);
circle.setAttribute('fill', '#449944');
svg.appendChild(circle);
polygon_points.push(approx_points[i][0], approx_points[i][1]);
}
var polygon = doc.createElementNS('http://www.w3.org/2000/svg', 'polygon');
polygon.setAttribute("points", polygon_points.join(" "));
polygon.setAttribute("stroke", "transparent");
polygon.setAttribute("fill", "#cccc00");
polygon.setAttribute("opacity", "0.7");
svg.appendChild(polygon);
doc.querySelector("output[name='points']").innerHTML = approx_points_num;
doc.querySelector("output[name='area']").innerHTML = polyArea(approx_points);
});
var event = new Event("change");
document.getElementById("number_of_points").dispatchEvent(event);
})();
<html>
<body>
<div style="width: 100%; text-align: center;">
<svg width="250px" height="120px" viewBox="-5 -5 45 30" id="svg">
<path d="M 0 0 C 10 15 50 40 30 0 Z" fill="transparent" stroke="black" id="path" />
</svg>
<br />
<input type="range" min="3" max="100" value="5" class="slider" id="number_of_points">
<br />
Approximating with
<output name="points" for="number_of_points"></output>
points, area is
<output name="area"></output>
</div>
</body>
</html>
I like the solution in the accepted answer by Phrogz, but I also looked a little further and found a way to do the same with Paper.js using the CompoundPath class and area property. See my Paper.js demo.
The result (surface area = 11856) is the exact same as with Phrogz's demo when using threshold 0, but the processing seems a lot quicker! I know it's overkill to load Paper.js just to calculate the surface area, but if you are considering implementing a framework or feel like investigating how Paper.js does it...
Firstly, I am not so familiar with Bézier curves, but I know that they are continuous functions. If you ensure that your cubic curve does not intersect itself, you may integrate it in closed form (I mean by using analytic integrals) on the given enclosing domain ([a-b]) and subtract the area of triangle that is formed by the the end joining straight line and the X axis. In case of intersection with the Bézier curve and end joining straight line, you may divide into sections and try to calculate each area separately in a consistent manner..
For me suitable search terms are "continuous function integration" "integrals" "area under a function" "calculus"
Of course you may generate discrete data from your Bézier curve fn and obtain discrete X-Y data and calculate the integral approximately.
Couldn't you use an application of Gauss's magic shoelace theorem by getting a set of data points by changing T, then simply inputting that into the equation?
Here's a simple video demo https://www.youtube.com/watch?v=0KjG8Pg6LGk&ab_channel=Mathologer
And then here's the wiki https://en.wikipedia.org/wiki/Shoelace_formula
I can suggest a formula to do this numerically.
Starting with the general.
Cubic Bezier Equation
You can expand it out and you will end up with
this.
You can sub in your coordinates and simplify, then integrate with this formula.
This should give you the area between the curve and the x-axis. You can then subtract the area under the line,using standard integration, and this should give you the area enclosed.
Credit for the integration formula (image 3) and further info:https://math.libretexts.org/Courses/University_of_California_Davis/UCD_Mat_21C%3A_Multivariate_Calculus/10%3A_Parametric_Equations_and_Polar_Coordinates/10.2%3A_Calculus_with_Parametric_Curves#:~:text=The%20area%20between%20a%20parametric,%E2%80%B2(t)dt.
Inspired by James Godfrey-Kittle's suggestion in this bézierInfo thread: add section: area under a bézier curve I've wrapped this concept in a js helper function, that will get svg <path> and other elements' areas.
It's based on the same formula as suggested in #nbonneel's answer.
The main steps:
Parse and normalize a path's d attribute to an array of absolute and cubic commands. For this task, I'm using Jarek Foksa's path-data polyfill. The polyfill allows us to retrieve absolute coordinates from any path by its getPathData({normalize:true}) option. This way we don't have to bother about relative, cubic or shorthand commands.
Calculate the area for each curve segment (b0 and b1).
/**
* James Godfrey-Kittle#jamesgk
* https://github.com/Pomax/BezierInfo-2/issues/238
*/
function getBezierArea(coords) {
let x0 = coords[0];
let y0 = coords[1];
//if is cubic command
if (coords.length == 8) {
let x1 = coords[2];
let y1 = coords[3];
let x2 = coords[4];
let y2 = coords[5];
let x3 = coords[6];
let y3 = coords[7];
let area = (
x0 * (-2 * y1 - y2 + 3 * y3) +
x1 * (2 * y0 - y2 - y3) +
x2 * (y0 + y1 - 2 * y3) +
x3 * (-3 * y0 + y1 + 2 * y2)
) * 3 / 20;
return area;
} else {
return 0;
}
}
x0, y0 are the last coordinates of the command preceding the current C command. x1, y1, x2, y2, x3, y3 are the current pathdata values.
Since we don't need a polygon approximation based on the rather expensive getPointAtLength() method – the calculation is comparatively fast.
Add the remaining polygon's area to the bézier areas (p0). This step will also use the shoelace formula.
Example 1: semi circle with a radius of 50 (svg user units)
We can easily check, if the calculation works, since the expected result should be:
π·50²/2 = 3926.99
//example 1:
let svg = document.querySelector("svg");
let path = svg.querySelector("path");
let pathArea = getshapeAreaSimple(path);
let result = document.getElementById("result");
result.textContent = 'area: ' + pathArea;
function getshapeAreaSimple(el) {
let totalArea = 0;
let polyPoints = [];
let type = el.nodeName.toLowerCase();
let log = [];
let bezierArea = 0;
let pathData = el.getPathData({
normalize: true
});
pathData.forEach(function(com, i) {
let [type, values] = [com.type, com.values];
if (values.length) {
let prevC = i > 0 ? pathData[i - 1] : pathData[0];
let prevCVals = prevC.values;
let prevCValsL = prevCVals.length;
let [x0, y0] = [prevCVals[prevCValsL - 2], prevCVals[prevCValsL - 1]];
// C commands
if (values.length == 6) {
let area = getBezierArea([
x0,
y0,
values[0],
values[1],
values[2],
values[3],
values[4],
values[5]
]);
//push points to calculate inner/remaining polygon area
polyPoints.push([x0, y0], [values[4], values[5]]);
bezierArea += area;
}
// L commands
else {
polyPoints.push([x0, y0], [values[0], values[1]]);
}
}
});
let areaPoly = polygonArea(polyPoints, false);
//values have the same sign - subtract polygon area
if ((areaPoly < 0 && bezierArea < 0) || (areaPoly > 0 && bezierArea > 0)) {
totalArea = Math.abs(bezierArea) - Math.abs(areaPoly);
} else {
totalArea = Math.abs(bezierArea) + Math.abs(areaPoly);
}
return totalArea;
}
function getPathArea(pathData) {
let totalArea = 0;
let polyPoints = [];
pathData.forEach(function(com, i) {
let [type, values] = [com.type, com.values];
if (values.length) {
let prevC = i > 0 ? pathData[i - 1] : pathData[0];
let prevCVals = prevC.values;
let prevCValsL = prevCVals.length;
let [x0, y0] = [prevCVals[prevCValsL - 2], prevCVals[prevCValsL - 1]];
// C commands
if (values.length == 6) {
let area = getBezierArea([
x0,
y0,
values[0],
values[1],
values[2],
values[3],
values[4],
values[5]
]);
//push points to calculate inner/remaining polygon area
polyPoints.push([x0, y0], [values[4], values[5]]);
totalArea += area;
}
// L commands
else {
polyPoints.push([x0, y0], [values[0], values[1]]);
}
}
});
let areaPoly = polygonArea(polyPoints);
totalArea = Math.abs(areaPoly) + Math.abs(totalArea);
return totalArea;
}
/**
* James Godfrey-Kittle#jamesgk
* https://github.com/Pomax/BezierInfo-2/issues/238
*/
function getBezierArea(coords) {
let x0 = coords[0];
let y0 = coords[1];
//if is cubic command
if (coords.length == 8) {
let x1 = coords[2];
let y1 = coords[3];
let x2 = coords[4];
let y2 = coords[5];
let x3 = coords[6];
let y3 = coords[7];
let area =
((x0 * (-2 * y1 - y2 + 3 * y3) +
x1 * (2 * y0 - y2 - y3) +
x2 * (y0 + y1 - 2 * y3) +
x3 * (-3 * y0 + y1 + 2 * y2)) *
3) /
20;
return area;
} else {
return 0;
}
}
function polygonArea(points, absolute = true) {
let area = 0;
for (let i = 0; i < points.length; i++) {
const addX = points[i][0];
const addY = points[i === points.length - 1 ? 0 : i + 1][1];
const subX = points[i === points.length - 1 ? 0 : i + 1][0];
const subY = points[i][1];
area += addX * addY * 0.5 - subX * subY * 0.5;
}
if (absolute) {
area = Math.abs(area);
}
return area;
}
svg {
max-height: 20em;
max-width: 100%;
border: 1px solid #ccc;
fill: #ccc;
}
<p> Expected area: <br /> π·50²/2 = 3926.99</p>
<p id="result"></p>
<svg viewBox="0 0 100 50">
<path d="M50,0C22.383,0,0,22.385,0,49.998h100C100,22.385,77.613,0,50,0z" />
</svg>
<script src="https://cdn.jsdelivr.net/npm/path-data-polyfill#1.0.3/path-data-polyfill.min.js"></script>
Example 2: get areas of primitives and compound paths
For a more versatile helper function, we can include primitives like <circle>, <ellipse>, <polygon> etc. and skip the bézier calculation for these element types.
Compound paths – so shapes like the letters O or i will require to calculate the areas for each sub path. If a sub path is within the boundaries of another shape like the letter O, we also need to subtract inner shapes from the total area.
function getshapeArea(el, decimals = 0) {
let totalArea = 0;
let polyPoints = [];
let type = el.nodeName.toLowerCase();
switch (type) {
// 1. paths
case "path":
let pathData = el.getPathData({
normalize: true
});
//check subpaths
let subPathsData = splitSubpaths(pathData);
let isCompoundPath = subPathsData.length > 1 ? true : false;
let counterShapes = [];
// check intersections for compund paths
if (isCompoundPath) {
let bboxArr = getSubPathBBoxes(subPathsData);
bboxArr.forEach(function(bb, b) {
//let path1 = path;
for (let i = 0; i < bboxArr.length; i++) {
let bb2 = bboxArr[i];
if (bb != bb2) {
let intersects = checkBBoxIntersections(bb, bb2);
if (intersects) {
counterShapes.push(i);
}
}
}
});
}
subPathsData.forEach(function(pathData, d) {
//reset polygon points for each segment
polyPoints = [];
let bezierArea = 0;
let pathArea = 0;
let multiplier = 1;
pathData.forEach(function(com, i) {
let [type, values] = [com.type, com.values];
if (values.length) {
let prevC = i > 0 ? pathData[i - 1] : pathData[0];
let prevCVals = prevC.values;
let prevCValsL = prevCVals.length;
let [x0, y0] = [
prevCVals[prevCValsL - 2],
prevCVals[prevCValsL - 1]
];
// C commands
if (values.length == 6) {
let area = getBezierArea([
x0,
y0,
values[0],
values[1],
values[2],
values[3],
values[4],
values[5]
]);
//push points to calculate inner/remaining polygon area
polyPoints.push([x0, y0], [values[4], values[5]]);
bezierArea += area;
}
// L commands
else {
polyPoints.push([x0, y0], [values[0], values[1]]);
}
}
});
//get area of remaining polygon
let areaPoly = polygonArea(polyPoints, false);
//subtract area by negative multiplier
if (counterShapes.indexOf(d) !== -1) {
multiplier = -1;
}
//values have the same sign - subtract polygon area
if (
(areaPoly < 0 && bezierArea < 0) ||
(areaPoly > 0 && bezierArea > 0)
) {
pathArea = (Math.abs(bezierArea) - Math.abs(areaPoly)) * multiplier;
} else {
pathArea = (Math.abs(bezierArea) + Math.abs(areaPoly)) * multiplier;
}
totalArea += pathArea;
});
break;
// 2. primitives:
// 2.1 circle an ellipse primitives
case "circle":
case "ellipse":
totalArea = getEllipseArea(el);
break;
// 2.2 polygons
case "polygon":
case "polyline":
totalArea = getPolygonArea(el);
break;
// 2.3 rectancle primitives
case "rect":
totalArea = getRectArea(el);
break;
}
if (decimals > 0) {
totalArea = +totalArea.toFixed(decimals);
}
return totalArea;
}
function getPathArea(pathData) {
let totalArea = 0;
let polyPoints = [];
pathData.forEach(function(com, i) {
let [type, values] = [com.type, com.values];
if (values.length) {
let prevC = i > 0 ? pathData[i - 1] : pathData[0];
let prevCVals = prevC.values;
let prevCValsL = prevCVals.length;
let [x0, y0] = [prevCVals[prevCValsL - 2], prevCVals[prevCValsL - 1]];
// C commands
if (values.length == 6) {
let area = getBezierArea([
x0,
y0,
values[0],
values[1],
values[2],
values[3],
values[4],
values[5]
]);
//push points to calculate inner/remaining polygon area
polyPoints.push([x0, y0], [values[4], values[5]]);
totalArea += area;
}
// L commands
else {
polyPoints.push([x0, y0], [values[0], values[1]]);
}
}
});
let areaPoly = polygonArea(polyPoints);
totalArea = Math.abs(areaPoly) + Math.abs(totalArea);
return totalArea;
}
/**
* James Godfrey-Kittle/#jamesgk : https://github.com/Pomax/BezierInfo-2/issues/238
*/
function getBezierArea(coords) {
let x0 = coords[0];
let y0 = coords[1];
//if is cubic command
if (coords.length == 8) {
let x1 = coords[2];
let y1 = coords[3];
let x2 = coords[4];
let y2 = coords[5];
let x3 = coords[6];
let y3 = coords[7];
let area =
((x0 * (-2 * y1 - y2 + 3 * y3) +
x1 * (2 * y0 - y2 - y3) +
x2 * (y0 + y1 - 2 * y3) +
x3 * (-3 * y0 + y1 + 2 * y2)) *
3) /
20;
return area;
} else {
return 0;
}
}
function polygonArea(points, absolute = true) {
let area = 0;
for (let i = 0; i < points.length; i++) {
const addX = points[i][0];
const addY = points[i === points.length - 1 ? 0 : i + 1][1];
const subX = points[i === points.length - 1 ? 0 : i + 1][0];
const subY = points[i][1];
area += addX * addY * 0.5 - subX * subY * 0.5;
}
if (absolute) {
area = Math.abs(area);
}
return area;
}
function getPolygonArea(el) {
// convert point string to arra of numbers
let points = el
.getAttribute("points")
.split(/,| /)
.filter(Boolean)
.map((val) => {
return parseFloat(val);
});
let polyPoints = [];
for (let i = 0; i < points.length; i += 2) {
polyPoints.push([points[i], points[i + 1]]);
}
let area = polygonArea(polyPoints);
return area;
}
function getRectArea(el) {
let width = el.getAttribute("width");
let height = el.getAttribute("height");
let area = width * height;
return area;
}
function getEllipseArea(el) {
let r = el.getAttribute("r");
let rx = el.getAttribute("rx");
let ry = el.getAttribute("ry");
//if circle – take radius
rx = rx ? rx : r;
ry = ry ? ry : r;
let area = Math.PI * rx * ry;
return area;
}
//path data helpers
function splitSubpaths(pathData) {
let pathDataL = pathData.length;
let subPathArr = [];
let subPathMindex = [];
pathData.forEach(function(com, i) {
let [type, values] = [com["type"], com["values"]];
if (type == "M") {
subPathMindex.push(i);
}
});
//split subPaths
subPathMindex.forEach(function(index, i) {
let end = subPathMindex[i + 1];
let thisSeg = pathData.slice(index, end);
subPathArr.push(thisSeg);
});
return subPathArr;
}
function getSubPathBBoxes(subPaths) {
let ns = "http://www.w3.org/2000/svg";
let svgTmp = document.createElementNS(ns, "svg");
svgTmp.setAttribute("style", "position:absolute; width:0; height:0;");
document.body.appendChild(svgTmp);
let bboxArr = [];
subPaths.forEach(function(pathData) {
let pathTmp = document.createElementNS(ns, "path");
svgTmp.appendChild(pathTmp);
pathTmp.setPathData(pathData);
let bb = pathTmp.getBBox();
bboxArr.push(bb);
});
svgTmp.remove();
return bboxArr;
}
function checkBBoxIntersections(bb, bb1) {
let [x, y, width, height, right, bottom] = [
bb.x,
bb.y,
bb.width,
bb.height,
bb.x + bb.width,
bb.y + bb.height
];
let [x1, y1, width1, height1, right1, bottom1] = [
bb1.x,
bb1.y,
bb1.width,
bb1.height,
bb1.x + bb1.width,
bb1.y + bb1.height
];
let intersects = false;
if (width * height != width1 * height1) {
if (width * height > width1 * height1) {
if (x < x1 && right > right1 && y < y1 && bottom > bottom1) {
intersects = true;
}
}
}
return intersects;
}
svg {
max-height: 20em;
max-width: 100%;
border: 1px solid #ccc;
fill: #ccc;
}
<p><button type="button" onclick="getSingleArea(path0)">Get this area</button></p>
<svg class="svg0" viewBox="300 51.399147033691406 215.8272705078125 98.6994857788086">
<path id="curve" d="M 300 140 C 300 40 505 16 480 113 C544 47 523 235 411 100Z" />
</svg>
<p class="result0"></p>
<svg class="svg1" xmlns="http://www.w3.org/2000/svg" viewBox="0 0 280 25">
<path id="singleCurve" d="M0,12.667h25C25-4.222,0-4.222,0,12.667z" />
<path id="circle-two-quarter" d="M37.5,12.667c0,6.904,5.596,12.5,12.5,12.5c0-6.511,0-12.5,0-12.5l12.5,0c0-6.903-5.597-12.5-12.5-12.5
v12.5L37.5,12.667z" />
<path id="circle-three-quarters" d="M75,12.667c0,6.904,5.596,12.5,12.5,12.5c6.903,0,12.5-5.597,12.5-12.5
c0-6.903-5.597-12.5-12.5-12.5v12.5L75,12.667z" />
<circle id="circle" cx="125" cy="12.667" r="12.5" />
<ellipse id="ellipse" cx="162.5" cy="13.325" rx="12.5" ry="6.25" />
<rect id="rect" x="187.5" y="0.167" width="25" height="25" />
<polygon id="hexagon" points="231.25,23.493 225,12.667 231.25,1.842 243.75,1.842 250,12.667 243.75,23.493 " />
<path id="compound" d="M268.951,10.432c-3.452,0-6.25,2.798-6.25,6.25s2.798,6.25,6.25,6.25s6.25-2.798,6.25-6.25
S272.403,10.432,268.951,10.432z M268.951,19.807c-1.726,0-3.125-1.399-3.125-3.125s1.399-3.125,3.125-3.125
s3.125,1.399,3.125,3.125S270.677,19.807,268.951,19.807z M272.076,4.968c0,1.726-1.399,3.125-3.125,3.125s-3.125-1.399-3.125-3.125
c0-1.726,1.399-3.125,3.125-3.125S272.076,3.242,272.076,4.968z" />
</svg>
<p class="result1"></p>
<p><button type="button" onclick="getAllAreas(areaEls)">Get all areas</button></p>
<!--Dependency: path data polyfill -->
<script src="https://cdn.jsdelivr.net/npm/path-data-polyfill#1.0.3/path-data-polyfill.min.js"></script>
<script>
// 1st example: single path area
let svg0 = document.querySelector('.svg0');
let path0 = svg0.querySelector('path');
let result0 = document.querySelector('.result0');
function getSingleArea(shape) {
let shapeArea = getshapeArea(shape, 3);
result0.textContent = 'area: ' + shapeArea;
}
// 2nd example: multiple shape areas
let svg1 = document.querySelector('.svg1');
let areaEls = svg1.querySelectorAll('path, polygon, circle, ellipse, rect');
let result1 = document.querySelector('.result1');
//benchmark
let [t0, t1] = [0, 0];
function getAllAreas(areaEls) {
let results = []
perfStart();
areaEls.forEach(function(shape, i) {
let type = shape.nodeName.toLowerCase();
let id = shape.id ? '#' + shape.id : '<' + type + '/> [' + i + ']';
let shapeArea = getshapeArea(shape, 3);
let resultString = `<strong>${id}:</strong> ${shapeArea}`;
results.push(resultString);
let title = document.createElementNS('http://www.w3.org/2000/svg', 'title');
title.textContent = `${id}: ${shapeArea}`;
shape.appendChild(title);
});
let totalTime = perfEnd();
result1.innerHTML = results.join('<br />') + '<br /><br /><strong>time: </strong>' + totalTime + 'ms';
}
/**
* helpers for performance testing
*/
function adjustViewBox(svg) {
let bb = svg.getBBox();
let [x, y, width, height] = [bb.x, bb.y, bb.width, bb.height];
svg.setAttribute('viewBox', [x, y, width, height].join(' '));
}
function perfStart() {
t0 = performance.now();
}
function perfEnd(text = '') {
t1 = performance.now();
total = t1 - t0;
return total;
}
</script>
Codepen example
Square area covered by radius vector of a point moving in 2D plane is 1/2*integral[(x-xc)*dy/dt - (y-yc)*dx/dt]dt. Here xc and yc are coordinates of the origin point (center). Derivation for the case of Bezier curves is rather cumbersome but possible. See functions squareAreaQuadr and squareAreaCubic below. I have tested and retested these formulae, rather sure, that there are no mistakes. This signature gives positive square area for clockwise rotation in SVG coordinates plane.
var xc=0.1, yc=0.2, x0=0.9, y0=0.1, x1=0.9, y1=0.9, x2=0.5, y2=0.5, x3=0.1, y3=0.9
var cubic = document.getElementById("cubic");
cubic.setAttribute("d", "M "+xc*500+" "+yc*500+" L "+x0*500+" "+y0*500+" C "+x1*500+" "+y1*500+" "+x2*500+" "+y2*500+" "+x3*500+" "+y3*500+" L "+xc*500+" "+yc*500);
var center1 = document.getElementById("center1");
center1.setAttribute("cx", xc*500);
center1.setAttribute("cy", yc*500);
function squareAreaCubic(xc, yc, x0, y0, x1, y1, x2, y2, x3, y3)
{
var s;
s = 3/4*( (x0-xc)*(y1-y0) + (x3-xc)*(y3-y2) ) +
1/4*(x3-x0)*(y1+y2-y0-y3) +
1/8*( (x0+x3-2*xc)*(3*y2-3*y1+y0-y3) + (x1+x2-x0-x3)*(y1-y0+y3-y2) ) +
3/40*( (2*x1-x0-x2)*(y1-y0) + (2*x2-x1-x3)*(y3-y2) ) +
1/20*( (2*x1-x0-x2)*(y3-y2) + (2*x2-x1-x3)*(y1-y0) + (x1+x2-x0-x3)*(3*y2-3*y1+y0-y3) ) +
1/40*(x1+x2-x0-x3)*(3*y2-3*y1+y0-y3) -
3/4*( (y0-yc)*(x1-x0) + (y3-yc)*(x3-x2) ) -
1/4*(y3-y0)*(x1+x2-x0-x3) -
1/8*( (y0+y3-2*yc)*(3*x2-3*x1+x0-x3) + (y1+y2-y0-y3)*(x1-x0+x3-x2) ) -
3/40*( (2*y1-y0-y2)*(x1-x0) + (2*y2-y1-y3)*(x3-x2) ) -
1/20*( (2*y1-y0-y2)*(x3-x2) + (2*y2-y1-y3)*(x1-x0) + (y1+y2-y0-y3)*(3*x2-3*x1+x0-x3) ) -
1/40*(y1+y2-y0-y3)*(3*x2-3*x1+x0-x3) ;
return s;
}
var s = squareAreaCubic(xc, yc, x0, y0, x1, y1, x2, y2, x3, y3);
document.getElementById("c").innerHTML = document.getElementById("c").innerHTML + s.toString();
<html>
<body>
<h1>Bezier square area</h1>
<p id="q">Quadratic: S = </p>
<svg height="500" width="500">
<rect width="500" height="500" style="fill:none; stroke-width:2; stroke:black" />
<path id="quadr" fill="lightgray" stroke="red" stroke-width="1" />
<circle id="q_center" r="5" fill="black" />
</svg>
<script>
var xc=0.1, yc=0.2, x0=0.9, y0=0.1, x1=0.9, y1=0.9, x2=0.1, y2=0.9;
var quadr = document.getElementById("quadr");
quadr.setAttribute("d", "M "+xc*500+" "+yc*500+" L "+x0*500+" "+y0*500+" Q "+x1*500+" "+y1*500+" "+x2*500+" "+y2*500+" L "+xc*500+" "+yc*500);
var center = document.getElementById("q_center");
q_center.setAttribute("cx", xc*500);
q_center.setAttribute("cy", yc*500);
function squareAreaQuadr(xc, yc, x0, y0, x1, y1, x2, y2)
{
var s = 1/2*( (x0-xc)*(y1-y0) + (x2-xc)*(y2-y1) - (y0-yc)*(x1-x0) - (y2-yc)*(x2-x1) ) +
1/12*( (x2-x0)*(2*y1-y0-y2) - (y2-y0)*(2*x1-x0-x2) );
return s;
}
var s = squareAreaQuadr(xc, yc, x0, y0, x1, y1, x2, y2);
document.getElementById("q").innerHTML = document.getElementById("q").innerHTML + s.toString();
</script>
<p id="c">Cubic: S = </p>
<svg height="500" width="500">
<rect width="500" height="500" style="fill:none; stroke-width:2; stroke:black" />
<path id="cubic" fill="lightgray" stroke="red" stroke-width="1" />
<circle id="center1" r="5" fill="black" />
</svg>
</body>
</html>

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