C# 围绕另一点旋转一点-方向错误

我现在有一段代码，可以围绕另一个点旋转一个点。问题是，当用户输入'90'时，我希望它是水平轴的旋转，当前90度指向下（垂直轴）。下面是一张示意图来说明：

这是我目前掌握的代码：

    public static Point RotatePoint(Point pointToRotate, Point centerPoint, int angleInDegrees) {
        double angleInRadians = angleInDegrees * (Math.PI / 180);
        double cosTheta = Math.Cos(angleInRadians);
        double sinTheta = Math.Sin(angleInRadians);
        int x = (int)
                (cosTheta * (pointToRotate.X - centerPoint.X) -
                sinTheta * (pointToRotate.Y - centerPoint.Y) + centerPoint.X);
        int y = (int)
                (sinTheta * (pointToRotate.X - centerPoint.X) +
                cosTheta * (pointToRotate.Y - centerPoint.Y) + centerPoint.Y);
    }

---

我尝试过使y=x，反之亦然，也尝试过在输入的过程中伪造数据，但没有成功。任何帮助都将不胜感激

如通过注释所述，包含函数中的计算很好：旋转点按预期计算（负角度表示顺时针旋转）。您所追求的是基于极坐标系的旋转点的定义，这必须在稍后阶段完成。“所需”中描述的参考系相对于“默认参考系”有+90度的间隙，因此您只需将+90添加到此类系统的计算中即可

您所追求的角度可以通过以下函数计算：

public static double angleFromPoint(Point inputPoint, Point centerPoint)
{
    double varX1 = Math.Abs(inputPoint.X - centerPoint.X);
    double varY1 = Math.Abs(inputPoint.Y - centerPoint.Y);

    double outAngle = 180 * Math.Atan(varY1 / varX1) / Math.PI; //Angle from 0 to 90 which has to be updated on account of the quadrant it is in and the chosen syst

    int curQuadrant = determineQuadrant(inputPoint, centerPoint);

    //Modifications to account for the default system of reference
    if (curQuadrant == 1)
    {
        outAngle = 180 - outAngle;
    }
    else if (curQuadrant == 3)
    {
        outAngle = 360 - outAngle;
    }
    else if (curQuadrant == 4)
    {
        outAngle = 180 + outAngle;
    }

    //Over-modification to account for the system of reference "Desired", +90 the default system of reference
    outAngle = outAngle + 90;


    if (outAngle > 360)
    {
        outAngle = outAngle - 360;
    }

    return outAngle;
}

//Moving clockwisely, the first quadrant is located between 180 and 90 degrees in the default system of reference 
public static int determineQuadrant(Point inputPoint, Point centerPoint)
{
    int curQuadrant = 0;

    if (inputPoint.X < centerPoint.X && inputPoint.Y >= centerPoint.Y)
    {
        //Default system of reference -> 180 to 90
        curQuadrant = 1;
    }
    else if (inputPoint.X >= centerPoint.X && inputPoint.Y >= centerPoint.Y)
    {
        //Default system of reference -> 90 to 0/360
        curQuadrant = 2;
    }
    else if (inputPoint.X >= centerPoint.X && inputPoint.Y < centerPoint.Y)
    {
        //Default system of reference -> 0/360 to 270
        curQuadrant = 3;
    }
    else if (inputPoint.X < centerPoint.X && inputPoint.Y < centerPoint.Y)
    {
        //Default system of reference -> 270 to 180
        curQuadrant = 4;
    }

    return curQuadrant;
}

---

你不能从用户输入中减去90吗？有两个问题：首先，我已经测试了代码，它没有显示“Current”上描述的行为，即顺时针方向为正，但正好相反，因此@ChrisSinclair的建议应该可以接受。第二个问题是：参考角的确切点是什么？（0、90等）其唯一实用程序用于极坐标系（例如，点位于距离X和Y度处），但函数的输出为笛卡尔坐标。因此，唯一真正重要的是方向（+表示顺时针或逆时针-）@ChrisSinclair-Nope，不太管用；仅适用于希望当前水平线指向下方的情况。从垂直线减去90度，使其指向左侧。@varocarbas我使用该函数围绕“起点”旋转和“终点”，然后从“起点->终点”绘制一条线。目前的实施是按常规轮流进行的。虽然当渲染时，它显示为CW，因为它被绘制到画布上：（0,0）是左上角，（最大X，0）是右上角，（0，最大Y）是左下角，我认为你需要澄清一下你的想法；请看一下我的答案。

Point rotatedPoint = RotatePoint(curPointnew, centerPoint, rotationAngle);
double angleRotatedPoint = angleFromPoint(rotatedPoint, centerPoint);