在c++中计算立方体的每个顶点的平均值。

I'm struggling calculating the normals for each of the 8 vertices for my cube in c++. I thought it should be going like this: - calculate the normals for each of the 6 faces of the cube - each vertex touches 3 faces, so calculating the normalized vector for all the 3 faces

我正在努力计算我的立方体在c++中每8个顶点的法线。我认为它应该是这样的:-计算立方体的6个面的法线——每个顶点接触3个面，所以计算所有3个面的归一化向量。

m_iNoVerts=8;
Vertex verts[8];
verts[0].position = XMFLOAT3(-1.0f, -1.0f,  1.0f);
verts[1].position = XMFLOAT3(-1.0f,  1.0f,  1.0f);
verts[2].position = XMFLOAT3( 1.0f, -1.0f,  1.0f);
verts[3].position = XMFLOAT3( 1.0f,  1.0f,  1.0f);
verts[4].position = XMFLOAT3(-1.0f, -1.0f, -1.0f);
verts[5].position = XMFLOAT3(-1.0f,  1.0f, -1.0f);
verts[6].position = XMFLOAT3( 1.0f, -1.0f, -1.0f);
verts[7].position = XMFLOAT3( 1.0f,  1.0f, -1.0f);

m_iNoIndices = 36;
int indices[36] = {0, 1, 2, // front face   0
                  1, 2, 3,

                  4, 5, 6,  // back face    1
                  5, 6, 7,

                  4, 5, 0,  // left face    2
                  5, 0, 1,

                  2, 3, 6,  // right face   3
                  3, 6, 7,

                  1, 5, 3,  // top face     4
                  5, 3, 7,

                  0, 4, 2,  // bottom face  5
                  4, 2, 6};

// Calculate the normals for each face
XMVECTOR result[6]; // cross product of vec1 and vec2 represents the normal of the face
XMVECTOR vec1;      // vec1 = B - A example first calc: B = verts[1] and A = verts[0]
XMVECTOR vec2;      // vec2 = C - B
int ii = 0;
for(int i = 0; i < 6; ++i)
{
    vec1.x = verts[indices[ii+1]].position.x - verts[indices[ii]].position.x;
    vec1.y = verts[indices[ii+1]].position.y - verts[indices[ii]].position.y;
    vec1.z = verts[indices[ii+1]].position.z - verts[indices[ii]].position.z;

    vec2.x = verts[indices[ii+2]].position.x - verts[indices[ii+1]].position.x;
    vec2.y = verts[indices[ii+2]].position.y - verts[indices[ii+1]].position.y;
    vec2.z = verts[indices[ii+2]].position.z - verts[indices[ii+1]].position.z;

    // calculate the cross product
    //result[i].x = (vec1.y * vec2.z) - (vec1.z * vec2.y);
    //result[i].y = (vec1.z * vec2.x) - (vec1.x * vec2.z);
    //result[i].z = (vec1.x * vec2.y) - (vec1.y * vec2.x);
    //result[i].w = 0;
    result[i] = XMVector3Cross(vec1, vec2);

    ii += 6;    // increasing the counter for the indices to jump to the next face 
}

// calculating the normals of each vertex
XMVECTOR normal[8];
// building the resulting vector of the 3 sites on each vertex
normal[0] = result[0] + result[2] + result[5];
normal[1] = result[0] + result[2] + result[4];
normal[2] = result[0] + result[3] + result[5];
normal[3] = result[0] + result[3] + result[4];
normal[4] = result[1] + result[2] + result[5];
normal[5] = result[1] + result[2] + result[4];
normal[6] = result[1] + result[3] + result[5];
normal[7] = result[1] + result[3] + result[4];

for(int i = 0; i < m_iNoVerts; ++i)
{
    normal[i] = XMVector3Normalize(normal[i]);  // normalization of the vector
    verts[i].normal.x = normal[i].x;
    verts[i].normal.y = normal[i].y;
    verts[i].normal.z = normal[i].z;
}

when I check the normals in the debugger, the values are +-0.577.. The given values from my teacher are

当我检查调试器中的法线时，值是+-0.577。我的老师给出的价值观是。

0.0  0.5  0.5
0.0  0.5  0.5
0.0 -0.5  0.5
0.0 -0.5  0.5
0.0  0.5 -0.5
0.0  0.5 -0.5
0.0 -0.5 -0.5
0.0 -0.5 -0.5

Whats the thing I'm doing wrong? Thanks!

我做错什么了?谢谢!

2 个解决方案

#1

The normals of faces of a cube are simply

立方体面的法线是简单的。

[+-1, 0, 0],   [0,+-1,0],    [0,0,+-1]
(Left/Right), (Top/Bottom), (Front/Back)

The 8 normals for the vertices, if such a concept is meaningful, are even more simply vertex - center or vertex it self in this case, since center is at origin.

对于顶点的8个法线，如果这个概念是有意义的，那么在这个例子中更简单的是顶点中心或顶点，因为中心是原点。

That can be of course calculated by adding the 8 linear combinations of the face normals and normalizing, giving

当然，这可以通过增加8个线性组合的面法线和正态，给予来计算。

[+-1, +-1, +-1] / sqrt(3)

But in essence, there's not that much to calculate...

但从本质上说，没有那么多可以计算的……

#2

Each face can be defined by just 3 vertices. You have a list of 6, why?

每个面只能由3个顶点定义。你有6个清单，为什么?
If you do this: vec1 = B - A; vec2 = C - A then in the case of a cube they will be orthogonal :)

如果你这样做:vec1 = B - A;向量= C - A在立方体中它们是正交的:)
The normals can be scaled by any constant, they will still be normal. It is probably convention to scale them to be unit vectors. To do that, just divide each by its norm (vector length). 0.577 is sqrt(1/3), probably. (+-0.577, +-0.577, +-0.577) are unit vectors.

法线可以按任何常数缩放，它们仍然是正常的。很可能是将它们缩放成单位向量的惯例。要做到这一点，只需按它的模(向量长度)除以它们。0.577是sqrt(1/3),可能。(+-0.577，+-0.577)是单位向量。

#1