Question

我正在进行一项代码/数学任务，我需要将一个形状旋转到0度，然后再回到它所处的任何角度。

以下方法：

    function rotateAroundOrigin(xPos:Number, yPos:Number, angle:Number, origin:Point):Point {
        var point:Point = new Point();
        angle *= Math.PI / 180;
        point.x = Math.cos(angle) * (xPos - origin.x) + Math.sin(angle) * (yPos - origin.y) + origin.x;
        point.y = Math.sin(angle) * (xPos - origin.x) - Math.cos(angle) * (yPos - origin.y) + origin.y;
        return point;
    }

这样调用两次：

var rotatePoint:Point = rotateAroundOrigin(30, 30, -90, midPoint);
var rotatePointBack:Point = rotateAroundOrigin(30, 30, -90, midPoint);

它工作正常，但它没有意义，因为度数值是“-90”它旋转的两倍。单向旋转应为“-90”，然后旋转“90”旋转回原位。

任何人都有任何想法如何修改“rotateAroundOrigin”方法，以便您可以调用“-90”然后使用“90”调用？

var rotatePoint:Point = rotateAroundOrigin(30, 30, -90, midPoint);
var rotatePointBack:Point = rotateAroundOrigin(30, 30, 90, midPoint);

Answer 1

var rotatePoint:Point = rotateAroundOrigin(30, 30, -90, midPoint);
var rotatePointBack:Point = rotateAroundOrigin(30, 30, -90, midPoint);

看起来你只做了两次相同的事情：围绕midPoint转动点（30,30）。考虑一下：

var rotatePoint:Point = rotateAroundOrigin(30, 30, -90, midPoint);
var rotatePointBack:Point = rotateAroundOrigin(rotatePoint.x, rotatePoint.y, 90, midPoint);

编辑：我还注意到你在旋转方法中混淆了一些迹象。试试这个：

function rotateAroundOrigin(xPos:Number, yPos:Number, angle:Number, origin:Point):Point {
    var point:Point = new Point();
    angle *= Math.PI / 180;
    point.x = Math.cos(angle) * (xPos - origin.x) - Math.sin(angle) * (yPos - origin.y) + origin.x;
    point.y = Math.sin(angle) * (xPos - origin.x) + Math.cos(angle) * (yPos - origin.y) + origin.y;
    return point;
}

当我旋转对象时，如何向后旋转？

1 个答案: