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  • 三角形相关算法--求解三角形顶点坐标

    问题描述:

      已知三角形两个顶点p1,p2的坐标,p1与p0的距离len,p0-p1边与p1-p2边的夹角angle,求解p0的坐标。

    求解思路:

      1、根据p1与p0的距离可以列出一个等式:

    (p1.x-p0.x)2+(p1.y-p0.y)2=len2

      2、p1-p0向量为v1=(p0.x-p1.x, p0.y-p1.y),p1-p2向量为v2=(p2.x-p1.x, p2.y-p1.y),两向量的夹角为angle。根据向量夹角公式可得到一个等式:

    cos(angle)=(v1•v2)/(|v1|*|v2|)

    v1•v2=(p0.x-p1.x)*(p2.x-p1.x)+(p0.y-p1.y)*(p2.y-p1.y)

    |v1|=sqrt((p1.x-p0.x)2+(p1.y-p0.y)2)=len

    |v2|=sqrt((p2.x-p1.x)2+(p2.y-p1.y)2)

      根据这两个等式,可推导p0.x、p0.y的求解公式,一般应有两组解。

      3、在推导过程中,(p2.x-p1.x)被用作分母,因此,在实际计算时,需要将p2.x=p1.x的情况(即p1-p2边与纵轴平行)单独考虑。

    C#算法:

    /// <summary>
    /// 三角形的顶点为p0、p1、p2,
    /// 已知三角形两个顶点p1,p2的坐标,p1与p0的距离len,p0-p1边与p1-p2边的夹角angle,
    /// 求解p0的坐标(结果应有2组坐标)
    /// </summary>
    /// <param name="p1">顶点p1的坐标</param>
    /// <param name="p2">顶点p2的坐标</param>
    /// <param name="len">顶点p1与p0的距离</param>
    /// <param name="angle">p0-p1边与p1-p2边的夹角</param>
    /// <returns>p0的坐标(若len不等于0,则结果应有2组;否则结果为p1)</returns>
    public static TwoDimPoint[] GetVertex(TwoDimPoint p1, TwoDimPoint p2, double len, double angle)
    {            
        TwoDimPoint[] result;
    
        //若p1与p0的距离为0,则p0直接取p1的坐标
        if (Utility.IsEqualZero(len))
        {
            result = new TwoDimPoint[1];
            result[0] = new TwoDimPoint(p1);
        }
        else
        {
            if (Utility.IsEqualZero(TwoDimPoint.GetDistance(p1, p2)))
            {
                throw new NotImplementedException("已知两顶点的坐标一致,无法计算未知顶点的坐标!");
            }
            angle %= 2 * Math.PI;
            if (Utility.IsEqualZero(p2.X - p1.X))//p1-p2边与纵轴平行
            {
                if (angle % Math.PI == 0)//p0与p1、p2在一条直线上
                {
                    result = new TwoDimPoint[1];
                    result[0] = new TwoDimPoint();
                    result[0].X = p1.X;
    
                    if (angle == 0)
                    {
                        if (p1.Y > p2.Y)
                            result[0].Y = p1.Y - len;
                        else
                            result[0].Y = p1.Y + len;
                    }
                    else
                    {
                        if (p1.Y > p2.Y)
                            result[0].Y = p1.Y + len;
                        else
                            result[0].Y = p1.Y - len; 
                    }
                }
                else
                {
                    result = new TwoDimPoint[2];
                    result[0] = new TwoDimPoint();
                    result[0].X = p1.X + len * Math.Sin(angle);
                    if (p1.Y > p2.Y)
                        result[0].Y = p1.Y - len * Math.Cos(angle);
                    else
                        result[0].Y = p1.Y + len * Math.Cos(angle);
    
                    result[1] = new TwoDimPoint();
                    result[1].X = p1.X - len * Math.Sin(angle);
                    result[1].Y = result[0].Y;
                }
            }
            else
            {
                double A = TwoDimPoint.GetDistance(p1, p2);//p1,p2的距离
                double B = (p2.Y - p1.Y) / (p2.X - p1.X);
                double C = (len * A * Math.Cos(angle) + p1.Y * p2.Y - Utility.Square(p1.Y)) / (p2.X - p1.X);
                double D = Utility.Square(B) + 1;
                double E = (B * C + p1.Y) * 2;
                double F = Utility.Square(C) + Utility.Square(p1.Y) - Utility.Square(len);
                double G = Utility.Square(E) - 4 * D * F;
    
                if (Utility.IsEqualZero(G))
                {
                    result = new TwoDimPoint[1];
    
                    result[0] = new TwoDimPoint();
                    result[0].Y = E / (2 * D);
                    result[0].X = C + p1.X - B * result[0].Y;
                }
                else
                {
                    result = new TwoDimPoint[2];
    
                    result[0] = new TwoDimPoint();
                    result[0].Y = (E + Math.Sqrt(G)) / (2 * D);
                    result[0].X = C + p1.X - B * result[0].Y;
    
                    result[1] = new TwoDimPoint();
                    result[1].Y = (E - Math.Sqrt(G)) / (2 * D);
                    result[1].X = C + p1.X - B * result[1].Y;
                }
            }
        }            
        return result;
    }
    /// <summary>
    /// 二维点
    /// </summary>
    public class TwoDimPoint
    {
        public double X;
        public double Y;
    
        public TwoDimPoint()
        {
            this.X = 0;
            this.Y = 0;
        }
    
        public TwoDimPoint(double x, double y)
        {
            this.X = x;
            this.Y = y;
        }
    
        public TwoDimPoint(TwoDimPoint p)
        {
            this.X = p.X;
            this.Y = p.Y;
        }
    
        /// <summary>
        /// 计算两点的距离
        /// </summary>
        /// <param name="p1"></param>
        /// <param name="p2"></param>
        /// <returns></returns>
        public static double GetDistance(TwoDimPoint p1, TwoDimPoint p2)
        {
            return Math.Sqrt(Utility.Square(p1.X - p2.X) + Utility.Square(p1.Y - p2.Y));
        }
    }
    TwoDimPoint
    public class Utility
    {
        /// <summary>
        /// 判断数值是否等于0
        /// </summary>
        /// <param name="value"></param>
        /// <returns></returns>
        public static bool IsEqualZero(double value)
        {
            return Math.Abs(value) < Constant.MinPositiveNum;
        }
    
        /// <summary>
        /// 计算数值的平方
        /// </summary>
        /// <param name="value"></param>
        /// <returns></returns>
        public static double Square(double value)
        {
            return Math.Pow(value, 2);
        }
    }
    Utility
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  • 原文地址:https://www.cnblogs.com/hibernation/p/3289014.html
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