原文地址：http://www.terathon.com/code/tangent.html
為一個(gè)任意網(wǎng)格模型計(jì)算其切線空間的基本向量（即切線空間的T B N三個(gè)向量）
Modern bump mapping (also known as normal mapping) requires that tangent plane basis vectors be calculated for each vertex in a mesh. This article presents the theory behind the computation of per-vertex tangent spaces for an arbitrary triangle mesh and provides source code that implements the proper mathematics.
現(xiàn)在的bump mapping(或者normal mapping)需要每個(gè)頂點(diǎn)的切面的基本向量。這篇文章描述了逐頂點(diǎn)計(jì)算任意三角模型的切線空間原理，并且提供了實(shí)現(xiàn)這個(gè)數(shù)學(xué)理論的源代碼。
Mathematical Derivation 數(shù)學(xué)來(lái)源
[This derivation also appears in Mathematics for 3D Game Programming and Computer Graphics, 2nd ed., Section 6.8.]
We want our tangent space to be aligned such that the x axis corresponds to the u direction in the bump map and the y axis corresponds to the v direction in the bump map. That is, if Q represents a point inside the triangle, we would like to be able to write
我們想我們的切線空間像這樣對(duì)齊，比如； X軸相當(dāng)于bump圖中的U方向，而Y軸相當(dāng)于bump圖中的V方向。于是，如果Q表示三角形上的一點(diǎn)，則我們可以得出如下等式
Q − P0 = (u − u0)T + (v − v0)B,
where T and B are tangent vectors aligned to the texture map, P0 is the position of one of the vertices of the triangle, and (u0, v0) are the texture coordinates at that vertex. The letter B stands for bitangent, but in many places it is stilled called binormal because of a mix-up in terms when tangent-space bump mapping first became widespread. (See “Bitangent versus Binormal” below.)
等式中的T是與紋理對(duì)應(yīng)的切線向量，P0 是三角形的其中一個(gè)頂點(diǎn)。(u0, v0) 是對(duì)應(yīng)頂點(diǎn)的紋理坐標(biāo)。 B表示副（雙）切線（bitangent），但是由于在切線空間的bump mapping第一次被傳播開來(lái)的時(shí)候，有人混淆這個(gè)定義，所以在許多地方它始終被稱為副法線（binormal ）。（參見下面的Bitangent VS Binormal）
Suppose that we have a triangle whose vertex positions are given by the points P0, P1, and P2, and whose corresponding texture coordinates are given by (u0, v0), (u1, v1), and (u2, v2). Our calculations can be made much simpler by working relative to the vertex P0, so we let
假設(shè)三角形的三個(gè)頂點(diǎn)分別為P0, P1, and P2, 其分別對(duì)應(yīng)的紋理坐標(biāo)為(u0, v0), (u1, v1), and (u2, v2). 。那我們的計(jì)算就可以簡(jiǎn)化下為下面的方程式組
Q1 = P1 − P0
Q2 = P2 − P0
設(shè)
(s1, t1) = (u1 − u0, v1 − v0)
(s2, t2) = (u2 − u0, v2 − v0).
We need to solve the following equations for T and B.
于是，我們最終要解決的就是下面這個(gè)方程式
Q1 = s1T + t1B
Q2 = s2T + t2B
This is a linear system with six unknowns (three for each T and B) and six equations (the x, y, and z components of the two vector equations). We can write this in matrix form as follows.
這個(gè)有著6個(gè)未知數(shù)的線性方程組（T,B是向量，每個(gè)有三個(gè)量）。我們可以把它寫成如下的矩陣方式
Multiplying both sides by the inverse of the (s, t) matrix, we have
兩邊都乘以(s,t)的逆矩陣
.
This gives us the (unnormalized) T and B tangent vectors for the triangle whose vertices are P0, P1, and P2. To find the tangent vectors for a single vertex, we average the tangents for all triangles sharing that vertex in a manner similar to the way in which vertex normals are commonly calculated. In the case that neighboring triangles have discontinuous texture mapping, vertices along the border are generally already duplicated since they have different mapping coordinates anyway. We do not average tangents from such triangles because the result would not accurately represent the orientation of the bump map for either triangle.
由此，我們解出等式后，就得到了未單位化的T和B。 為了找到單個(gè)頂點(diǎn)的切線向量，我們平均共享這個(gè)頂點(diǎn)的所有切三角形的的切線。法線也可以通過類似的方式計(jì)算出來(lái)。 而在這種情況下，相鄰的兩個(gè)三角形則會(huì)形成不連續(xù)的紋理映射，處于邊緣的頂點(diǎn)由于有不同的紋理坐標(biāo)而經(jīng)常被復(fù)制。我們沒有平均這些三角形的切線，因?yàn)槲覀兊挠?jì)算結(jié)果就可以不精確地描述每個(gè)三角形的切線朝向。
Once we have the normal vector N and the tangent vectors T and B for a vertex, we can transform from tangent space into object space using the matrix
一旦我們得到了一個(gè)頂點(diǎn)的法向量N和切線T和B。我們就可以用它們來(lái)構(gòu)造一個(gè)由正切空間到對(duì)象空間的矩陣。
To transform in the opposite direction (from object space to tangent space—what we want to do to the light direction), we can simply use the inverse of this matrix. It is not necessarily true that the tangent vectors are perpendicular to each other or to the normal vector, so the inverse of this matrix is not generally equal to its transpose. It is safe to assume, however, that the three vectors will at least be close to orthogonal, so using the Gram-Schmidt algorithm to orthogonalize them should not cause any unacceptable distortions. Using this process, new (still unnormalized) tangent vectors T′ and B′ are given by
為了向反方向變換（從對(duì)象空間變換到正切空間，[法線貼圖時(shí)]我們需這樣處理光照方向），我們可以簡(jiǎn)單地使用這個(gè)矩陣的逆矩陣。 但我們要注意切線向量并非總是與其它兩個(gè)向量，或法向量垂直的。于是，我們的逆矩陣并非總是等于其轉(zhuǎn)置矩陣。但是，可以安全地認(rèn)為，這三個(gè)向量是非常接近正交關(guān)系的。 所以使用我們可以使用Gram-Schmidt算法來(lái)對(duì)其進(jìn)行正交化，就不會(huì)出現(xiàn)不可接受的誤差了。
T′ = T − (N · T)N
B′ = B − (N · B)N − (T′ · B)T′/T′2
Normalizing these vectors and storing them as the tangent and bitangent for a vertex lets us use the matrixto transform the direction to light from object space into tangent space. Taking the dot product of the transformed light direction with a sample from the bump map then produces the correct Lambertian diffuse lighting value.
單位化這些向量并將其存為一個(gè)頂點(diǎn)的tangent 和bitangent 。
于是，我們可以用下面這個(gè)矩陣將光方向從對(duì)象空間轉(zhuǎn)換到切線空間。然后將轉(zhuǎn)換后的光照方向和bump圖中的采樣值點(diǎn)乘，然后再處理某些矯正因子就可以完成光照的計(jì)算。
It is not necessary to store an extra array containing the per-vertex bitangent since the cross product N × T′ can be used to obtain mB′, where m = ±1 represents the handedness of the tangent space. The handedness value must be stored per-vertex since the bitangent B′ obtained from N × T′ may point in the wrong direction. The value of m is equal to the determinant of the matrix in Equation (*). One may find it convenient to store the per-vertex tangent vector T′ as a four-dimensional entity whose w coordinate holds the value of m. Then the bitangent B′ can be computed using the formula
我們也不必要存儲(chǔ)bitangent這個(gè)值。因?yàn)?strong>N × T′可以得出我們的mB′，這里的m = ±1，表示切空間的左右手坐標(biāo)系習(xí)慣。表示左右手坐標(biāo)系習(xí)慣的這個(gè)值必須逐頂點(diǎn)存儲(chǔ)。因?yàn)閎itangent B′ 是由N × T′計(jì)算得來(lái)，就可能指向錯(cuò)誤的方向。m的值必須要和(*)中的決定保持一致。 所以我們可以用T′的w分量來(lái)存儲(chǔ)m值。最后B′的值就可以向下面一樣計(jì)算
B′ = T’w·(N × T′),
where the cross product ignores the w coordinate. This works nicely for vertex programs by avoiding the need to specify an additional array containing the per-vertex m values.
向量的叉乘忽略了w分量，因此不會(huì)造成任何影響。這樣就可以很好地寫頂點(diǎn)程序，而不用另外用空間來(lái)存儲(chǔ)m值。
Bitangent versus Binormal
The term binormal is commonly used as the name of the second tangent direction (that is perpendicular to the surface normal and u-aligned tangent direction). This is a misnomer. The term binormal pops up in the study of curves and completes what is known as a Frenet frame about a particular point on a curve. Curves have a single tangent direction and two orthogonal normal directions, hence the terms normal and binormal. When discussing a coordinate frame at a point on a surface, there is one normal direction and two tangent directions, which should be called the tangent and bitangent.
（上面講的就是binormal 和bitangent的區(qū)別）。
Source Code
The code below generates a four-component tangent T in which the handedness of the local coordinate system is stored as ±1 in the w-coordinate. The bitangent vector B is then given by B = (N × T) · Tw.
下面的代碼產(chǎn)生了一個(gè)4維的T，其第４個(gè)分量用于存儲(chǔ)左右手坐標(biāo)系習(xí)慣。B 是由 B = (N × T) · Tw.計(jì)算而來(lái)。
#include "Vector4D.h"
struct Triangle
{
unsigned short index[3];
};
void CalculateTangentArray(long vertexCount, const Point3D *vertex, const Vector3D *normal,
const Point2D *texcoord, long triangleCount, const Triangle *triangle, Vector4D *tangent)
{
Vector3D *tan1 = new Vector3D[vertexCount * 2];
Vector3D *tan2 = tan1 + vertexCount;
ZeroMemory(tan1, vertexCount * sizeof(Vector3D) * 2);
for (long a = 0; a < triangleCount; a++)
{
long i1 = triangle->index[0];
long i2 = triangle->index[1];
long i3 = triangle->index[2];
const Point3D& v1 = vertex[i1];
const Point3D& v2 = vertex[i2];
const Point3D& v3 = vertex[i3];
const Point2D& w1 = texcoord[i1];
const Point2D& w2 = texcoord[i2];
const Point2D& w3 = texcoord[i3];
float x1 = v2.x - v1.x;
float x2 = v3.x - v1.x;
float y1 = v2.y - v1.y;
float y2 = v3.y - v1.y;
float z1 = v2.z - v1.z;
float z2 = v3.z - v1.z;
float s1 = w2.x - w1.x;
float s2 = w3.x - w1.x;
float t1 = w2.y - w1.y;
float t2 = w3.y - w1.y;
float r = 1.0F / (s1 * t2 - s2 * t1);
Vector3D sdir((t2 * x1 - t1 * x2) * r, (t2 * y1 - t1 * y2) * r,
(t2 * z1 - t1 * z2) * r);
Vector3D tdir((s1 * x2 - s2 * x1) * r, (s1 * y2 - s2 * y1) * r,
(s1 * z2 - s2 * z1) * r);
tan1[i1] += sdir;
tan1[i2] += sdir;
tan1[i3] += sdir;
tan2[i1] += tdir;
tan2[i2] += tdir;
tan2[i3] += tdir;
triangle++;
}
for (long a = 0; a < vertexCount; a++)
{
const Vector3D& n = normal[a];
const Vector3D& t = tan1[a];
// Gram-Schmidt orthogonalize
tangent[a] = (t - n * Dot(n, t)).Normalize();
// Calculate handedness
tangent[a].w = (Dot(Cross(n, t), tan2[a]) < 0.0F) ? -1.0F : 1.0F;
}
delete[] tan1;
}
How to cite this article
Lengyel, Eric. “Computing Tangent Space Basis Vectors for an Arbitrary Mesh”. Terathon Software 3D Graphics Library, 2001. http://www.terathon.com/code/tangent.html