Question

我想创建一个交互式数字画布，它将在网格中生成一系列近似方形的多边形。下面的jsFiddle显示了所述多边形的2x2网格系统。如果您检查代码或刷新页面，您将看到中心顶点是一个半随机生成的点，4个相邻的多边形共享。

我想将此网格缩放到16x16附近的某个位置，每个内部顶点都是半随机生成的，但在当前状态下，完成的代码效率低，不灵活且不可伸缩。我知道有一个相对简单的解决方案，但考虑到我对for循环和数组的经验不足，它目前超出了我的范围。非常感谢任何帮助。

HTML：

<body>  
    <canvas id="myCanvas" width="200" height="200"></canvas>
</body>

JS：

var canvas = document.getElementById('myCanvas');
var context = canvas.getContext('2d');

var randomX = Math.floor((Math.random()-0.5)*30);
var randomY = Math.floor((Math.random()-0.5)*30);

var x1 = canvas.width/2 + randomX;
var x2 = canvas.width;
var y1 = canvas.height/2 + randomY;
var y2 = canvas.height;


//background
context.beginPath();
context.rect(0,0,canvas.height,canvas.width);
context.fillStyle = '#A3DCEE';
context.fill();


//top left polygon
context.beginPath();
context.lineTo(0,0); //top left quadrant
context.lineTo(canvas.width/2,0); //top right quadrant
context.lineTo(x1, y1); //bottom right quadrant
context.lineTo(0, canvas.width/2); //bottom left quadrant

context.closePath();
context.fillStyle = '#ABE2EF';
context.fill();


//top right polygon
context.beginPath();
context.lineTo(x2/2,0); //top left quadrant
context.lineTo(x2,0); //top right quadrant
context.lineTo(x2, y2/2); //bottom right quadrant
context.lineTo(x1, y1); //bottom left quadrant

context.closePath();
context.fillStyle = '#A3DCEE';
context.fill();


//bottom left polygon
context.beginPath();
context.lineTo(0,y2/2); //top left quadrant
context.lineTo(x1,y1); //top right quadrant
context.lineTo(x2/2, y2); //bottom right quadrant
context.lineTo(0, y2); //bottom left quadrant

context.closePath();
context.fillStyle = '#8CD6F6';
context.fill();


//bottom right polygon
context.beginPath();
context.lineTo(x1,y1); //top left quadrant
context.lineTo(x2,y2/2); //top right quadrant
context.lineTo(x2, y2); //bottom right quadrant
context.lineTo(x2/2, y2); //bottom left quadrant

context.closePath();
context.fillStyle = '#85D2ED';
context.fill();

CSS：

html, body {
    background-color: #fff;
     width:  100%;
    height: 100%;
    margin: 0px;
    padding: 0px;
    position: relative;
    }

canvas {
    margin: auto;
    position: absolute;
    top: 0; left: 0; bottom: 0; right: 0;
    }

Answer 1

如何绘制具有轻微随机偏移交叉点的网格

enter image description here

将网格上的每个交叉点视为圆的中心点。

然后，您可以使用随机半径和随机角度计算该中心点的偏移量。

这将随机偏移每个交叉点。

这是一个如何计算0-2.5像素“摆动”的网格点的例子，它是“通常”的位置：

  // make each grid cell's side 20px 
  // (you can alter this to fit your needs)

  var sideLength=20;


  // allow the point to wobble up to 2.5 pixels from its "usual" position

  var radius=2.5*Math.random();


  // allow that point to wobble at a random angle around its "usual" position

  var radianAngle=2*Math.PI*Math.random();


  // do the calculation of the new wobbled intersection point
  // the x & y are the cell's horizontal(x) and vertical(y) position on the grid

  var cx=x*sideLength+radius*Math.cos(radianAngle);
  var cy=y*sideLength+radius*Math.sin(radianAngle);

以下是示例代码和小提琴：http://jsfiddle.net/m1erickson/Wqhb9/

<!doctype html>
<html>
<head>
<link rel="stylesheet" type="text/css" media="all" href="css/reset.css" /> <!-- reset css -->
<script type="text/javascript" src="http://code.jquery.com/jquery.min.js"></script>

<style>
    body{ background-color: ivory; }
    #canvas{border:1px solid red;}
</style>

<script>
$(function(){

    var canvas=document.getElementById("canvas");
    var ctx=canvas.getContext("2d");

    // declare an array
    var points=new Array(17);
    for(var i=0;i<points.length;i++){
        points[i]=new Array(17);
    }

    var sideLength=20;

    // fill the array with 
    for(var y=0;y<17;y++){
        for(var x=0;x<17;x++){
            // create a random semi-offset grid point
            var radius=2.5*Math.random();
            var radianAngle=2*Math.PI*Math.random();
            var cx=x*sideLength+radius*Math.cos(radianAngle);
            var cy=y*sideLength+radius*Math.sin(radianAngle);
            // if this is a sidepoint, don't offset (sides are straight)
            if(x==0){cx=0;}
            if(y==0){cy=0;}
            if(x==16){cx=x*sideLength;}
            if(y==16){cy=y*sideLength;}
            // add this gridpoint to the points array
            points[x][y]={x:cx+10,y:cy+10};
        }
    }

    // stroke the 4 sides of each cell
    for(var y=0;y<16;y++){
        for(var x=0;x<16;x++){
            strokeCell(x,y);
        }
    }

    // draw the 4 sides of the cell
    function strokeCell(x,y){
        var pt0=points[x][y];
        var pt1=points[x+1][y];
        var pt2=points[x+1][y+1];
        var pt3=points[x][y+1];

        ctx.beginPath();
        ctx.moveTo(pt0.x,pt0.y);
        ctx.lineTo(pt1.x,pt1.y);
        ctx.lineTo(pt2.x,pt2.y);
        ctx.lineTo(pt3.x,pt3.y);
        ctx.closePath();
        ctx.stroke();
    }


}); // end $(function(){});
</script>

</head>

<body>
    <canvas id="canvas" width=400 height=400></canvas>
</body>
</html>

使用for循环的可伸缩多边形网格

1 个答案: