移除闭合网格上被遮挡的面
我需要删除封闭三角形网格内的一些隐藏面。我找到了一种使用 MeshLab 来做到这一点的方法，但有两个问题：1）文章说它在某些情况下可能会失败（你能想象在哪些情况下吗？）和2）我想以编程方式执行此操作（你知道任何现有的吗？ 执行？）。我什至不知道这种简化过滤器的名称。
我真的很感激任何提示。
I need to remove some hidden faces that are inside a closed triangular mesh. I've found a way to do it using MeshLab, but there are two problems: 1) the article says that it could fail in some cases (can you image in which cases?) and 2) I would like to do it programatically (do you know of any existing implementation?). I don't even know the name of this kind of simplification filters.
I would really appreciate any hint.
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最明显的（但几乎肯定不是最有效的）算法如下：
The most obvious (but almost certainly not the most efficient one) algorithm to do that is as follows:
您想要做的是一劳永逸地删除几何图形内的面。
如果您只想从一个方向查看对象，罗马解决方案可能是可以接受的。
由于不可见的表面可能会变得可见，因此如果旋转相机，这不是可接受的解决方案。
使用洪水填充适用于多维数组，但它也适用于 3D 坐标吗？由于表面并不总是 (int) 并且具有除 (1,0,0)、(0,1,0) (0,0,1) 之外的其他倾角，因此在我看来，这种方法也不起作用。
算法可能失败的情况：
计算出的遮挡值可以告诉您表面的顶点是否在对象内部。然而：虽然所有顶点都在对象内部，但表面的一部分可能会显示出来。
考虑一下：您有三个提升的对象，每个对象内部都包含一个顶点。虽然未显示顶点，但它们之间跨越的表面部分可见。
我自己无法帮助你，因为我也在寻找一种可以做到这一点的算法。
问候
What you want to do is deleting faces wich are inside a geometric figure once and for all.
Romans solution may be acceptable if you only want to look at the object from 1 direction.
Since non visible surfaces may become visible, if you rotate the camera, it is not an acceptable solution.
Using floodfill works with multi dimensional arrays, but does it also work with 3D Koordinates? since surfaces are not always (int) and have other inclination than (1,0,0), (0,1,0) (0,0,1), this approach, in my oppinion, does not work either.
the cases in wich the algorithm may fail:
The occlusion value computed tells you wheter or not the vertices of your surface are inside your object. However: although all vertices are inside the object, part of the surface may show anyway.
Consider this: You have three elevated Objects, each containing a Vertex inside. While the vertices are not shown, the surface spanned between them is partially visible.
I myselve cannot help you, since I am also looking for an algorithm wich can do this.
greets