3D矩形面减运算：如何计算生成的8个面的顶点？

给定两个3D矩形面ABCD（主面）与WXYZ（子面），执行面减运算后需生成8个新面，目标是计算这些新面的顶点坐标。

已尝试的思路 #

• 向量求交法：逻辑可行，但计算量过大，效率不足
• 三角形/矩形相似性几何思路：难以找到通用落地方式，推进受阻

C#代码尝试（非最终方案） #

代码说明：

1. Face实体包含6个索引，其中4个为唯一顶点索引，通过6个索引构成的双三角形网格表示四边形面
2. 代码可能存在计算错误，欢迎指正

using System.Collections.Generic;
using UnityEngine;

public class Face
{
    public int[] indices; // 6 indices for two triangles (quad face)
    public Vector3[] vertices; // 4 unique vertices

    public Face(int[] indices, Vector3[] vertices)
    {
        this.indices = indices;
        this.vertices = vertices;
    }
}

public class FaceSubtraction
{
    public List<Face> SubtractFaces(Face mainFace, Face subFace)
    {
        List<Face> resultFaces = new List<Face>();

        // TODO: Implement the face subtraction logic here
        // Current attempt uses vector intersection but has performance issues
        // and possible calculation errors

        // Example placeholder logic (not functional)
        foreach (Vector3 v in mainFace.vertices)
        {
            if (!IsPointInsideSubFace(v, subFace))
            {
                // Create new faces based on intersection points and remaining vertices
                // This is a simplified placeholder
            }
        }

        return resultFaces;
    }

    private bool IsPointInsideSubFace(Vector3 point, Face subFace)
    {
        // TODO: Implement point-in-polygon check for 3D plane
        // Current implementation may have errors
        Vector3 planeNormal = Vector3.Cross(subFace.vertices[1] - subFace.vertices[0], subFace.vertices[2] - subFace.vertices[0]).normalized;
        float dot = Vector3.Dot(point - subFace.vertices[0], planeNormal);
        if (Mathf.Abs(dot) > 0.001f)
            return false; // Point not on the plane

        // Project to 2D and check point in polygon
        Vector3 axis1 = subFace.vertices[1] - subFace.vertices[0];
        Vector3 axis2 = subFace.vertices[3] - subFace.vertices[0];
        Vector3 localPoint = point - subFace.vertices[0];
        float u = Vector3.Dot(localPoint, axis1) / axis1.sqrMagnitude;
        float v = Vector3.Dot(localPoint, axis2) / axis2.sqrMagnitude;

        return u >= 0 && u <= 1 && v >= 0 && v <= 1;
    }
}

内容的提问来源于stack exchange，提问作者Ragnarok Fate