javascript - Canvas ArcTo 困惑

Question

所以我再次处理环形扇区，这不是我的强项。我可以在 Canvas 上很好地使用 .arc 方法，问题来自需要我的弧线成为路径的一部分。

例如:

 ctx.save();
 ctx.arc(centerX, centerY, radius, startAngle, endAngle, true);
 ctx.stroke();
 ctx.restore();

工作正常。但现在我需要它作为路径的一部分，所以我有这样的东西:

 var pointArray = [...]; //this contains all four corner points of the annular sector
 ctx.save();
 ctx.moveTo(pointArray[0].x, pointArray[0].y);
 ctx.lineTo(pointArray[1].x, pointArray[1].y); //so that draws one of the flat ends
 ctx.arcTo(?, ?, pointArray[2].x pointArray[2].y, radius);

切线坐标的切线快把我逼疯了。另外，我有一个严重的担忧: http://www.dbp-consulting.com/tutorials/canvas/CanvasArcTo.html听起来好像用 arcTo 绘制的弧永远无法覆盖 180 度或更多的圆，有时我的环形扇区会大于 180 度。

感谢stackoverflow高手几何思维的帮助!

更新好的，所以我必须在这里做 svg canvas inter-polarity，并使用 coffee-script，实际的生产代码如下!

 annularSector : (startAngle,endAngle,innerRadius,outerRadius) ->

    startAngle  = degreesToRadians startAngle+180
    endAngle    = degreesToRadians endAngle+180
    p           = [ 
        [ @centerX+innerRadius*Math.cos(startAngle),    @centerY+innerRadius*Math.sin(startAngle) ]
        [ @centerX+outerRadius*Math.cos(startAngle),    @centerY+outerRadius*Math.sin(startAngle) ]
        [ @centerX+outerRadius*Math.cos(endAngle),      @centerY+outerRadius*Math.sin(endAngle) ]
        [ @centerX+innerRadius*Math.cos(endAngle),      @centerY+innerRadius*Math.sin(endAngle) ] 
    ]
    angleDiff   = endAngle - startAngle
    largeArc    = (if (angleDiff % (Math.PI * 2)) > Math.PI then 1 else 0)

    if @isSVG

        commands    = []

        commands.push "M" + p[0].join()
        commands.push "L" + p[1].join()
        commands.push "A" + [ outerRadius, outerRadius ].join() + " 0 " + largeArc + " 1 " + p[2].join()
        commands.push "L" + p[3].join()
        commands.push "A" + [ innerRadius, innerRadius ].join() + " 0 " + largeArc + " 0 " + p[0].join()
        commands.push "z"

        return commands.join(" ")

    else

        @gaugeCTX.moveTo p[0][0], p[0][1]
        @gaugeCTX.lineTo p[1][0], p[1][1]
        #@gaugeCTX.arcTo 
        @gaugeCTX.arc @centerX, @centerY, outerRadius, startAngle, endAngle, false
        #@gaugeCTX.moveTo p[2][0], p[2][1]
        @gaugeCTX.lineTo p[3][0], p[3][1]
        @gaugeCTX.arc @centerX, @centerY, innerRadius, startAngle, endAngle, false          

解决方案

        @gaugeCTX.moveTo p[0][0], p[0][1]
        @gaugeCTX.lineTo p[1][0], p[1][1]
        @gaugeCTX.arc @centerX, @centerY, outerRadius, startAngle, endAngle, false
        @gaugeCTX.lineTo p[3][0], p[3][1]
        @gaugeCTX.arc @centerX, @centerY, innerRadius, endAngle, startAngle, true #note that this arc is set to true and endAngle and startAngle are reversed!

Answer 1

我最近发现自己对 arcTo() 方法感到失望(它实际上应该被称为 roundedCorner() )。我决定为那些也想使用 cx,cy,r,theta1,theta2 表达式的人想出一个通用的解决方法:

http://www.purplefrog.com/~thoth/art/paisley/arcTo.html

将重要的代码复制到这里:

/**
   if code is "move" then we will do a moveTo x0,y0
   if code is "line" then we will do a lineTo x0,y0
   if code is anything else, we'll assume the cursor is already at x0,y0
*/
function otherArcTo(ctx, cx, cy, r, theta1, theta2, code)
{
    console.log([cx,cy,r,theta1, theta2, code])
    var x0 = cx + r*Math.cos(theta1)
    var y0 = cy + r*Math.sin(theta1)
    if (code=="move") {
        ctx.moveTo(x0,y0)
    } else if (code=="line") {
        ctx.lineTo(x0,y0)
    }

    var dTheta = theta2-theta1
    var nChunks = Math.ceil( Math.abs(dTheta) / (0.67*Math.PI) )
    if (nChunks <=1) {
        var theta3 = theta1 + dTheta/2
        var r3 = r/Math.cos(dTheta/2)
        var x1 = cx + r3*Math.cos(theta3)
        var y1 = cy + r3*Math.sin(theta3)
        var x2 = cx + r*Math.cos(theta2)
        var y2 = cy + r*Math.sin(theta2)
        ctx.arcTo(x1,y1,x2,y2, r)
    } else {
        for (var i=0; i<nChunks; i++) {
            var code2 = null
            if (i==0)
                code2 = code
            otherArcTo(ctx, cx, cy, r,
                       theta1 + dTheta*i/nChunks,
                       theta1 + dTheta*(i+1)/nChunks, code2)
        }
    }
}