我如何获得4D网格的3D截面？ [英] How do I get a 3D cross section of a 4D mesh?
问题描述
我有一个以四维网格表示的 polychoron ，顶点法。所有的面都是三角形。我怎样才能得到这个图的三维横截面？
我发现的最接近的东西是这个问题，但它是一维短小的。
使用4个维度有点困难。
解决这个问题的唯一方法是找到尺寸类比。
我们从2D开始
凸二维多边形具有凸一维边：线段。
填充的凸多边形的横截面是线段。
计算交点你的poligons边与相交线的点，你会得到一个凸多边形的两个交点，而横截面将是一个线段。
为了做到这一点很容易转换坐标，以便在坐标系的Y轴上执行横截面。边的两点是A和B.它们的坐标是a1，a2和b1，b2。
如果a1和b1的符号相同（又名a1 * b1> 0），则边缘不会与Y轴相交。
$ b
否则计算k = a1 /（a1-b1）。然后交点坐标为：（0;（1-k）* a2 + k * b2）
$ b
正如我所说的凸多边形，您将获得两个交点，将两个点连接以获得一维横截面。
现在让我们推广到3D
一个凸3D网格具有凸二维边：三角形。
转换坐标以使操作更容易。让我们在Y-Z平面上得到网格的横截面，所以X坐标将再次为零。
获取三角形的横截面。使用上面的算法为它们的每个边缘。
由于我们有3个维度，所以边缘的端点将具有坐标a1，a2，a3和b1，b2，b3。如果a1 * b1 <0，我们有一个交点。所以
让k = a1 /（a1 - b1）
交点的坐标是（0 ;（1-k）* a2 + k * b2;（1-k）* a3 + k * b3）。存储该坐标，同时存储网格的A和B点的索引（边缘索引）。稍后会有用。
对于每个三角形，这将产生一条线段。
现在您需要将这些横截面线段连接到一个多边形。这就是为什么我们将边缘索引与交叉点一起存储的原因。你有一组线，你可以通过存储的边索引来匹配它们的端点，所以你可以将它们连接成一个多边形。
现在让我们推广到4D
凸4D网格具有凸3D边：四面体。 （注意你的脸部顶点表示是不正确的）
再次，转换坐标以使操作更容易。我们来看Y-Z-W超平面上网格的横截面，所以X坐标将再次为零。
获取四面体的横截面。使用上面的算法为它们的每个面。
由于我们有4个维度，边缘的端点将具有坐标a1，a2，a3，a4和b1，b2，b3，b4。如果a1 * b1 <0，我们有一个交点。所以
让k = a1 /（a1 - b1）
交点的坐标是（0 ;（1-k）* a2 + k * b2;（1-k）* a3 + k * b3;（1-k）* a4 + k * b4）。
对于四面体的每个三角形，这将产生一条线段。对于每个四面体，这将产生一个三角形。对于这些三角形的每个边缘还存储线段起源于的3D网格三角形（面部索引）的A，B和C点的索引。稍后会有用。
现在您需要将这些横截面三角形连接到3D网格。这就是为什么我们将脸部索引与相交边缘一起存储的原因。你有一组三角形，你可以通过存储的脸部索引匹配它们的边缘，所以你可以将它们连接成一个三角形网格。
对于凹形4D网格，您可能有获得多个3D网格。
希望你能看到类比。
具体的实现会有点困难，你需要关心所有的角落情况（除以零情况，浮点错误等）。
I have a polychoron represented as a four-dimensional mesh, stored with the face-vertex method. All the faces are triangles. How can I get a three-dimensional cross-section of the figure?
The closest thing I've found is this question, but it's one dimension short.
Working with 4 dimensions are a bit difficult.
The only way to solve the problem is finding dimensional analogies.
Let's start from 2D
A convex 2D polygon has convex 1D sides: line segments.
The cross-section of a filled convex polygon is a line segment.
Calculate the intersection points of your poligons edges with the intersecting line, you'll get two intersections for a convex polygon, and the cross section will be a line segment.
To do this easily transform the coordinates so you do the cross section on the Y axis of the coordinate system. The two points of the edge is A and B. Their coordinates are a1, a2 and b1, b2.
If the signs of a1 and b1 is the same, (aka a1*b1 > 0) the edge won't intersect the Y axis.
Otherwise calculate k = a1/(a1-b1).
Then the coordinate of the intersection point will be: (0; (1-k)*a2+k*b2)
As I said for convex polygons you'll get two intersection points, connect the two point to get the 1D cross section.
Now let's generalize to 3D
A convex 3D mesh has convex 2D sides: triangles.
Again, transform the coordinates to make the operation easier. Let's get the cross section of the mesh on the Y-Z plane, so the X coordinates will be zero again.
Get the cross sections of the triangles. Using the algorithm above for each edge of them. Since we have 3 dimensions the endpoints of the edge will have the coordinates a1, a2, a3 and b1, b2, b3. If a1*b1<0 we have an intersection point. So
Let k = a1 / (a1 - b1)
The coordinate of the intersection point is (0; (1-k)*a2+k*b2; (1-k)*a3+k*b3). Store this coordinate, but also store the index of A and B points of the mesh (the edge index). It'll be useful later.
For each triangle this will yield a line segment.
Now you'll need to connect these cross section line segments to a polygon. That's why we stored the edge indexes along with the intersection points. You have a set of lines and you can match their endpoints by the stored edge index, so you can connect them into a polygon.
Now let's generalize to 4D
A convex 4D mesh has convex 3D "sides": tetrahedrons. (note probably your face-vertex representation is incorrect)
Again, transform the coordinates to make the operation easier. Let's get the cross section of the mesh on the Y-Z-W hyperplane, so the X coordinates will be zero again.
Get the cross sections of the tetrahedrons. Using the algorithm above for each face of them. Since we have 4 dimensions the endpoints of the edge will have the coordinates a1, a2, a3, a4 and b1, b2, b3, b4. If a1*b1<0 we have an intersection point. So
Let k = a1 / (a1 - b1)
The coordinate of the intersection point is (0; (1-k)*a2+k*b2; (1-k)*a3+k*b3; (1-k)*a4+k*b4).
For each triangle of a tetrahedron this will yield a line segment. For each tetrahedron this will yield a triangle. For each edge of these triangles also store the index of A, B and C points of the triangle of the 3D mesh (the face index) the line segment originates from. It'll be useful later.
Now you'll need to connect these cross section triangles to a 3D mesh. That's why we stored the face indexes along with the intersection edges. You have a set of triangles and you can match their edges by the stored face index, so you can connect them into a triangle mesh.
For concave 4D meshes you may have get multiple 3D meshes.
Hope you see the analogy.
The concrete implementation will be a bit diffcult, you'll need to take care of all corner cases (division by zero cases, floating point errors, etc).
这篇关于我如何获得4D网格的3D截面？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
