不同平面直角坐标系之间的转换公式的推导及C#代码实现

    本篇讨论的主题是在平面坐标系中如何将一个坐标系(目标坐标系,以下简称目标系)中的所有点投射到另一个坐标系(基坐标系)中。平面坐标系之间的转化一般有三步操作:1、平移;2、旋转;3、拉伸。

    在转化的过程中需要的几个已知条件分别是:1、目标系的一个已知点(特征点A)对应于基坐标系中的点(特征点A’)。2、目标系的原点(O)对应于基坐标系中的原点(O')。3、基座标系的原点(O‘’)。

一、坐标系拉伸

①、计算两坐标系X和Y轴分别对应的拉伸比例:


②、将A点按缩放比例映射到基坐标系中(A''):


二、坐标系旋转和平移

①:坐标系旋转


②坐标系平移加旋转


③得出公式

根据坐标系旋转和平移的规则,可以得出公式

我们的目标是求出θ。

设m = cos(θ),n=sin(θ),根据已知量可得出公式:


得出方程组的增广矩阵,根据高斯消除元法即可求得m,n:



再根据反三角函数求得θ,此时θ、a、b、T(坐标系倍数)的值都已经得到。

故任意目标系坐标(x、y)可通过公式映射到基坐标系(f(x)、f(y)):

C#代码:

public class CoorPoint
    {
        public double x { get; set; }
        public double y { get; set; }

        public CoorPoint()
        { }

        public CoorPoint(double x, double y)
        {
            this.x = x;
            this.y = y;
        }

        public CoorPoint(CoorPoint pt)
        {
            this.x = pt.x;
            this.y = pt.y;
        }

        public static CoorPoint operator +(CoorPoint src, CoorPoint tar)
        {
            CoorPoint result = new CoorPoint(src);
            result.x += tar.x;
            result.y += tar.y;
            return result;
        }

        public static CoorPoint operator -(CoorPoint src, CoorPoint tar)
        {
            CoorPoint result = new CoorPoint(src);
            result.x -= tar.x;
            result.y -= tar.y;
            return result;
        }

        public static CoorPoint operator /(CoorPoint src, CoorPoint tar)
        {
            CoorPoint result = new CoorPoint(src);
            result.x /= tar.x;
            result.y /= tar.y;
            return result;
        }

        public static CoorPoint operator *(CoorPoint src, CoorPoint tar)
        {
            CoorPoint result = new CoorPoint(src);
            result.x *= tar.x;
            result.y *= tar.y;
            return result;
        }
    }

    class CoordinateSystemChg
    {
        /// <summary>
        /// 基坐标系:特征点(A')
        /// </summary>
        public CoorPoint baseCoor_basePoint { get; set; }
        /// <summary>
        /// 基坐标系:原点(O)
        /// </summary>
        public CoorPoint baseCoor_originPoint { get; set; }
        /// <summary>
        /// 现实坐标系:特征点(A)
        /// </summary>
        public CoorPoint realistic_basePoint { get; set; }
        /// <summary>
        /// 现实坐标系对应基坐标系拉伸后的点(A'')
        /// </summary>
        public CoorPoint realistic_basePointAfterZoom { get; set; }
        /// <summary>
        /// 现实坐标系:原点(O)
        /// </summary>
        public CoorPoint realistic_originPoint { get; set; }
        /// <summary>
        /// 现实坐标系对应基坐标系:原点(O')
        /// </summary>
        public CoorPoint realistic_originAsBaseCoorPoint { get; set; }   //定义原点时 已经统一缩放比例

        //第一步的结果
        //位移量:a(x轴),b(y轴)
        double a = 0;
        double b = 0;
        //坐标系倍数
        CoorPoint coorTimes;
        //坐标系夹角
        double offsetAngle;

        //第一步:得到坐标系之间的角度θ
        /// <summary>
        /// 计算出两坐标系的X,Y比例
        /// </summary>
        public void GetSale()
        {
            coorTimes = (realistic_basePoint - realistic_originPoint) / (baseCoor_basePoint - realistic_originAsBaseCoorPoint);
            if (coorTimes.x < 0)
                coorTimes.x = -coorTimes.x;
            if (coorTimes.y < 0)
                coorTimes.y = -coorTimes.y;
        }
        /// <summary>
        /// ①按比例缩放(实际坐标系的基点),使之与基坐标系统一
        /// </summary>
        public void DoZoom()
        {
            realistic_basePointAfterZoom = realistic_basePoint / coorTimes;
        }

        /// <summary>
        /// ②得到偏移角,入参实际坐标系原点坐标在基坐标系上的坐标
        /// </summary>
        public void GetOffsetAngle()
        {
            double[] x = new double[2];
            a = (realistic_originAsBaseCoorPoint.x - baseCoor_originPoint.x);
            b = (realistic_originAsBaseCoorPoint.y - baseCoor_originPoint.y);

            double[,] g = {
                { realistic_basePointAfterZoom.x, realistic_basePointAfterZoom.y, (baseCoor_basePoint.x - a) },
                { realistic_basePointAfterZoom.y, -realistic_basePointAfterZoom.x, (baseCoor_basePoint.y - b) }
            };

            GaussianElimination(g, x);

            // x[0]:θ角的余弦值;x[1]:θ角的正弦值
            offsetAngle = Math.Acos(x[0]);
            Console.WriteLine(x[0] + ":" + Math.Acos(x[0]));
            Console.WriteLine(x[1] + ":" + Math.Asin(x[1]));
        }

        //第二步:根据θ角求出在基坐标系中点的映射
        /// <summary>
        /// 将实际坐标系中的点转换到基坐标系中的点
        /// </summary>
        /// <param name="src"></param>
        /// <param name="rlt"></param>
        public void GetCoordinateBaseSystem(CoorPoint src, out CoorPoint rlt)
        {
            double db1 = Math.Cos(offsetAngle);
            double db2 = Math.Sin(offsetAngle);
            rlt = new CoorPoint();
            rlt.x = ((src.x / coorTimes.x) * Math.Cos(offsetAngle) + (src.y / coorTimes.y) * Math.Sin(offsetAngle) + a);
            rlt.y = ((src.y / coorTimes.y) * Math.Cos(offsetAngle) - (src.x / coorTimes.x) * Math.Sin(offsetAngle) + b);
        }
        /// <summary>
        /// 将基坐标系中的点转换到实际坐标系中的点
        /// </summary>
        /// <param name="src"></param>
        /// <param name="rlt"></param>
        public void GetCoordinateRealisticSystem(CoorPoint src, out CoorPoint rlt)
        {
            double[] x = new double[2];

            double[,] g = {
                {  Math.Cos(offsetAngle), Math.Sin(offsetAngle),(src.x - a)*coorTimes.x },
                { -Math.Sin(offsetAngle), Math.Cos(offsetAngle),(src.y - b)*coorTimes.y }
            };

            GaussianElimination(g, x);

            rlt = new CoorPoint();
            rlt.x = x[0];
            rlt.y = x[1];
        }
        #region 数学方法
        //简单的高斯消元法;
        //输入要求解的扩展矩阵g[m,n];和存放结果的数组x[n];
        //返回值为计算结果数组x[n];
        public static double[] GaussianElimination(double[,] g, double[] x)
        {
            int m = g.GetLength(0);//获得扩展矩阵的行(方程个数);
            int n = g.GetLength(1);//获得扩展矩阵的列(未知数个数+1);
                                   //========================================================
                                   //消元过程;
            for (int i = 1; i < m; i++)
            {
                for (int j = i; j < m; j++)
                {
                    for (int k = n - 1; k > i - 2; k--)
                    {
                        g[j, k] = g[j, k] - (g[j, i - 1] / g[i - 1, i - 1]) * (g[i - 1, k]);
                    }
                }
            }
            //回代过程;
            //第一步:翻转;
            //换行;
            double tem;
            for (int i = 0; i < m / 2; i++)
                for (int j = 0; j < n; j++)
                {
                    tem = g[i, j];
                    g[i, j] = g[m - i - 1, j];
                    g[m - i - 1, j] = tem;
                }
            //倒序
            for (int i = 0; i < m; i++)
                for (int j = 0; j < n / 2; j++)
                {
                    tem = g[i, j];
                    g[i, j] = g[i, n - 2 - j];
                    g[i, n - 2 - j] = tem;
                }
            //第二步:消元;
            for (int i = 1; i < m; i++)
            {
                for (int j = i; j < m; j++)
                {
                    for (int k = n - 1; k > i - 2; k--)
                    {
                        g[j, k] = g[j, k] - (g[j, i - 1] / g[i - 1, i - 1]) * (g[i - 1, k]);
                    }
                }
            }
            //第三步:翻回;
            //重新换行;
            for (int i = 0; i < m / 2; i++)
                for (int j = 0; j < n; j++)
                {
                    tem = g[i, j];
                    g[i, j] = g[m - i - 1, j];
                    g[m - i - 1, j] = tem;
                }
            //重新倒序;
            for (int i = 0; i < m; i++)
                for (int j = 0; j < n / 2; j++)
                {
                    tem = g[i, j];
                    g[i, j] = g[i, n - 2 - j];
                    g[i, n - 2 - j] = tem;
                }
            //取结果(这里是正序结果哦);
            for (int i = 0; i < m; i++)
                x[i] = g[i, n - 1] / g[i, i];
            return x;//返回计算结果;
        }
        #endregion
    }




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转载自blog.csdn.net/dai815904261/article/details/78612149