使用JavaScript在HTML画布上绘制螺旋

时间:2011-07-26 01:49:16

标签: javascript html html5 math canvas

我已经搜索过并且没有找到任何关于如何使用JavaScript在画布中绘制螺旋的真正信息。

我认为有可能用bezier曲线来做,如果不起作用,请使用lineTo(),但这似乎要困难得多。

另外,为了做到这一点,我猜我将不得不使用三角法和极坐标图表,并且我已经有一段时间了。如果是这种情况,你可以指出我在数学上的正确方向。

6 个答案:

答案 0 :(得分:39)

阿基米德螺旋线表示为r=a+b(angle)。将其转换为x,y坐标,它将表示为x=(a+b*angle)*cos(angle)y=(a+b*angle)*sin(angle)。然后你可以把角度放在for循环中并做这样的事情:

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);
}

注意上面假设a = 1且b = 1。

这是一个jsfiddle链接:http://jsfiddle.net/jingshaochen/xJc7M/

答案 1 :(得分:3)

这是我曾经从here

借用的Java螺旋的略微改变的javascript-ified版本

它使用lineTo()并不是那么难。

<!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>

答案 2 :(得分:2)

这是我为绘制Archimedean spirals编写的函数:

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();
        }

答案 3 :(得分:1)

有一个很好的免费工具,如果你有插图画家会有所帮助 ai2canvas

它将为你创建html canvas标签中javascript的所有曲线!

(如果您正在寻找拱形螺旋线,而不是首先从coreldraw获取它并将其复制到插图画面,因为默认的螺旋工具会扩大每个点的角度)

答案 4 :(得分:1)

这是使用以下函数绘制螺旋的示例:

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
});

螺旋绘图代码:

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();
};

答案 5 :(得分:0)

以下代码将螺旋近似为半径稍大的四分之一圆的集合。对于小转数,它可能看起来比阿基米德螺旋更糟糕,但它应该运行得更快。

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();
}

结果:

enter image description here