使用四点相交的线段

发布于 2024-12-01 19:01:00 字数 189 浏览 0 评论 0 原文

我有四个点 p0 、 p1 、p1' 、 p2' ，每个点由 x,y,z 分量定义，并且全部位于一条线上，如图

在此处输入图像描述

我想从四个点之间的交点获得线段（虚线部分）结果

任何建议，或使用 C# 的示例代码

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没有伤那来痛 2024-12-08 19:01:00

这或多或少与 Howard 给出的答案相同，但被压入 C# 中......我希望这对您的代码库有所帮助。

这段代码片段应该可以解决这个问题（从 4 个点中找到中点，但前提是所有点都是共线的） - 另请注意，我不检查真正的交点，您可以通过检查答案和您的点来轻松地自己完成此操作。
我没有花时间以合理的方式实现 Vector3D 结构（运算符，...） - 您也可以轻松做到这一点。
另请注意，这不仅适用于 4 点，而且请牢记您的图表。

private struct Vector3D
{
    public double X { get; set; }
    public double Y { get; set; }
    public double Z { get; set; }
}
static class Vectors
{
    static public double ScalProd(Vector3D v1, Vector3D v2)
    {
        return v1.X*v2.X + v1.Y*v2.Y + v1.Z*v2.Z;
    }

    static public Vector3D Minus(Vector3D v1, Vector3D v2)
    {
        return new Vector3D {X = v1.X - v2.X, Y = v1.Y - v2.Y, Z = v1.Z - v2.Z};
    }

    static public Vector3D Normalize(Vector3D v)
    {
        var len = Math.Sqrt(ScalProd(v, v));
        return new Vector3D {X = v.X/len, Y = v.Y/len, Z = v.Z/len};
    }
}

private Vector3D[]  FindIntersectionOnCoLinearVectors(params Vector3D[] input)
{
    if (input.Length < 2) throw new Exception("you need a minimum of two vectors");
    var v0 = input[0];
    var direction = Vectors.Normalize(Vectors.Minus(input[1], v0));
    Func<Vector3D, double> projectOntoLineStartingAtv0 =
        v => Vectors.ScalProd(direction, Vectors.Minus(v, v0));
    var mapped = input.OrderBy(projectOntoLineStartingAtv0).ToArray();
    return new Vector3D[] {mapped[1], mapped[2] };
}

This is more or less the same answer as Howard gave but pressed into C# ... I hope this helps with your code-base.

This code snippet should do the trick (finding the mid-points from your 4, but only if all are colinear) - also note I don't check for real intersection, you can do this easily youself by inspecting the answer and your points.
I did not take the time and implement the Vector3D struct in a sensible manner (operators, ...) - you can do this easily too.
Also note that this will work for not only 4 points but keep your diagram in mind.

private struct Vector3D
{
    public double X { get; set; }
    public double Y { get; set; }
    public double Z { get; set; }
}
static class Vectors
{
    static public double ScalProd(Vector3D v1, Vector3D v2)
    {
        return v1.X*v2.X + v1.Y*v2.Y + v1.Z*v2.Z;
    }

    static public Vector3D Minus(Vector3D v1, Vector3D v2)
    {
        return new Vector3D {X = v1.X - v2.X, Y = v1.Y - v2.Y, Z = v1.Z - v2.Z};
    }

    static public Vector3D Normalize(Vector3D v)
    {
        var len = Math.Sqrt(ScalProd(v, v));
        return new Vector3D {X = v.X/len, Y = v.Y/len, Z = v.Z/len};
    }
}

private Vector3D[]  FindIntersectionOnCoLinearVectors(params Vector3D[] input)
{
    if (input.Length < 2) throw new Exception("you need a minimum of two vectors");
    var v0 = input[0];
    var direction = Vectors.Normalize(Vectors.Minus(input[1], v0));
    Func<Vector3D, double> projectOntoLineStartingAtv0 =
        v => Vectors.ScalProd(direction, Vectors.Minus(v, v0));
    var mapped = input.OrderBy(projectOntoLineStartingAtv0).ToArray();
    return new Vector3D[] {mapped[1], mapped[2] };
}

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清晨说晚安 2024-12-08 19:01:00

您可以按如下方式进行：

步骤 1：转换为一维问题

定义t(P) = (P-P0).(P1-P0) / (P1-P0).(P1-P0) 其中点表示标量积
t 是通过 P1 和 P0 的直线上的线性度量，
因此我们有t(P0)=0, t(P1)=1

步骤 2：解决一维问题

我们假设 t(P0') <= t( P1') （否则交换以下行中的 P0' 和 P1'）
现在可能出现以下情况
- <代码>t(P1') < 0 =>没有交集
- <代码>1 < t(P0') =>没有交集
- t(P0') <= 0 <= t(P1') <= 1 =>交集是线段(P0,P1')
- t(P0') <= 0 < 1 < t(P1') =>交集是线段(P0,P1)
- 0 <= t(P0') <= t(P1') <= 1 =>交集是线段(P0',P1')
- <代码>0 <= t(P0') <= 1 <= t(P1') =>交集是线段(P0',P1)
或者，如果您只对 t 值感兴趣，则交集由 t0 = max 之间的线段给出(0, t(P0')) 和 t1 = min(1, t(P1')) iff t0 <= t1