我需要与所有的icosphere边相同的长度。目前，我生成一个二十面体一个从地址，然后线性插入每一个三角形。所有教程我能找到使用相同的线性插值。后来有人建议只使用从搅拌机，它的工作原理，在icosphere虽然也采用了线性插值...

I need an icosphere with all edges the same length. Currently I generate one starting from an icosahedron and then linearly interpolate each triangle. All tutorials I could find used the same linear interpolation. Someone later suggested just using the icosphere from blender, which works, though also uses the linear interpolation...

因此​​，任何人对如何使一个icosphere，并确保所有各方都是平等的一个很好的教程？ （这也将是确定的，如果你知道一个程序，它可以使这个并将其导出为一个.OBJ什么的。）

So anyone has a good tutorial on how to make an icosphere and make sure all the sides are equal? (It would also be ok if you know a program which can make this and export it as a .obj or something.)

推荐答案

这是不可能有从等边三角形任意precise领域。如果您需要等边三角形，最好你能得到的是二十面体（你可能已经拥有）。之所以叫角度缺陷：

This is not possible to have an arbitrarily precise sphere from equilateral triangles. If you require equilateral triangles, the best you can get is the icosahedron (which you probably already have). The reason is called "angle defect":

角度缺陷为360度围绕一个顶点顶点角度之和。任何多面体必须具有这样的总顶点缺陷等于720度：

The angle defect is 360 degrees the sum of vertex angles around a vertex. Any polyhedron must have a total vertex defect equal to 720 degrees:

如果有五个等边三角形的顶点相遇，他们积累36度的缺陷。如果六个等边三角形的顶点相遇，它们有助于没什么缺陷。

If five equilateral triangles meet at a vertex, they accumulate 36 degrees of defect. If six equilateral triangles meet at a vertex, they contribute nothing to the defect.

如果六个等边三角形的顶点满足，他们必须形成无论是

If six equilateral triangles meet at a vertex, they must form either

有一个多面体叫pentakis二十面体有32个顶点，但如果你做出来的等边三角形，你会得到一个非凸多面体。

There is a polyhedron called pentakis icosahedron with 32 vertices but if you make it out of equilateral triangles, you'll get a non-convex polyhedron.

所以，如果你想要一个任意precise领域，你需要允许不同长度的边缘，使角缺陷可能均匀地分布在表面上。任意precision等边是不可能的。

So, if you want an arbitrarily precise sphere, you need to allow different length edges so that the angle defect may distribute evenly across the surface. Equal sides with arbitrary precision are impossible.