I have searched and haven't found anything really on how to draw spirals in canvas using JavaScript.
I thought it might be possible to do it with the bezier curve and if that didn't work use lineTo()
, but that seemed a lot harder.
Also, to do that I'm guessing I would have to use trigonometry and graphing with polar coordinates and its been a while since I did that. If that is the case could you point开发者_开发技巧 me in the right direction on the math.
The Archimedean spiral is expressed as r=a+b(angle)
. Convert that into x, y coordinate, it will be expressed as x=(a+b*angle)*cos(angle)
, y=(a+b*angle)*sin(angle)
. Then you can put angle in a for loop and do something like this:
for (i=0; i< 720; i++) {
angle = 0.1 * i;
x=(1+angle)*Math.cos(angle);
y=(1+angle)*Math.sin(angle);
context.lineTo(x, y);
}
Note the above assumes a = 1 and b = 1.
Here is a jsfiddle link: http://jsfiddle.net/jingshaochen/xJc7M/
This is a slightly-changed, javascript-ified version of a Java spiral I once borrowed from here
It uses lineTo()
and its not all that hard.
<!DOCTYPE HTML>
<html><body>
<canvas id="myCanvas" width="300" height="300" style="border:1px solid #c3c3c3;"></canvas>
<script type="text/javascript">
var c=document.getElementById("myCanvas");
var cxt=c.getContext("2d");
var centerX = 150;
var centerY = 150;
cxt.moveTo(centerX, centerY);
var STEPS_PER_ROTATION = 60;
var increment = 2*Math.PI/STEPS_PER_ROTATION;
var theta = increment;
while( theta < 40*Math.PI) {
var newX = centerX + theta * Math.cos(theta);
var newY = centerY + theta * Math.sin(theta);
cxt.lineTo(newX, newY);
theta = theta + increment;
}
cxt.stroke();
</script></body></html>
Here's a function I wrote for drawing Archimedean spirals:
CanvasRenderingContext2D.prototype.drawArchimedeanSpiral =
CanvasRenderingContext2D.prototype.drawArchimedeanSpiral ||
function(centerX, centerY, stepCount, loopCount,
innerDistance, loopSpacing, rotation)
{
this.beginPath();
var stepSize = 2 * Math.PI / stepCount,
endAngle = 2 * Math.PI * loopCount,
finished = false;
for (var angle = 0; !finished; angle += stepSize) {
// Ensure that the spiral finishes at the correct place,
// avoiding any drift introduced by cumulative errors from
// repeatedly adding floating point numbers.
if (angle > endAngle) {
angle = endAngle;
finished = true;
}
var scalar = innerDistance + loopSpacing * angle,
rotatedAngle = angle + rotation,
x = centerX + scalar * Math.cos(rotatedAngle),
y = centerY + scalar * Math.sin(rotatedAngle);
this.lineTo(x, y);
}
this.stroke();
}
there is a fine free tool that will help if you have illustrator ai2canvas
it will create all the curves to javascript in html canvas tag for you!
(if you are looking for archmedes spiral than you will first have to get it from coreldraw and copy that to illustrator, because the default spiral tool enlarges the angle with each point)
this is example of drawing spiral using function below:
spiral(ctx, {
start: {//starting point of spiral
x: 200,
y: 200
},
angle: 30 * (Math.PI / 180), //angle from starting point
direction: false,
radius: 100, //radius from starting point in direction of angle
number: 3 // number of circles
});
spiral drawing code:
spiral = function(ctx,obj) {
var center, eAngle, increment, newX, newY, progress, sAngle, tempTheta, theta;
sAngle = Math.PI + obj.angle;
eAngle = sAngle + Math.PI * 2 * obj.number;
center = {
x: obj.start.x + Math.cos(obj.angle) * obj.radius,
y: obj.start.y + Math.sin(obj.angle) * obj.radius
};
increment = 2 * Math.PI / 60/*steps per rotation*/;
theta = sAngle;
ctx.beginPath();
ctx.moveTo(center.x, center.y);
while (theta <= eAngle + increment) {
progress = (theta - sAngle) / (eAngle - sAngle);
tempTheta = obj.direction ? theta : -1 * (theta - 2 * obj.angle);
newX = obj.radius * Math.cos(tempTheta) * progress;
newY = obj.radius * Math.sin(tempTheta) * progress;
theta += increment;
ctx.lineTo(center.x + newX, center.y + newY);
}
ctx.stroke();
};
The following code approximates a spiral as a collection of quarters of a circle each with a slightly larger radius. It might look worse than an Archimedes spiral for small turning numbers but it should run faster.
function drawSpiral(ctx, centerx, centery, innerRadius, outerRadius, turns=2, startAngle=0){
ctx.save();
ctx.translate(centerx, centery);
ctx.rotate(startAngle);
let r = innerRadius;
let turns_ = Math.floor(turns*4)/4;
let dr = (outerRadius - innerRadius)/turns_/4;
let cx = 0, cy = 0;
let directionx = 0, directiony = -1;
ctx.beginPath();
let angle=0;
for(; angle < turns_*2*Math.PI; angle += Math.PI/2){
//draw a quarter arc around the center point (x, cy)
ctx.arc( cx, cy, r, angle, angle + Math.PI/2);
//move the center point and increase the radius so we can draw a bigger arc
cx += directionx*dr;
cy += directiony*dr;
r+= dr;
//rotate direction vector by 90 degrees
[directionx, directiony] = [ - directiony, directionx ];
}
//draw the remainder of the last quarter turn
ctx.arc( cx, cy, r, angle, angle + 2*Math.PI*( turns - turns_ ))
ctx.stroke();
ctx.restore();
}
Result:
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