如何从头开始用人类可读的角度组成旋转矩阵？

Question

总是阻碍我做3D编程的一件事是无法理解数学是如何工作的.我可以使用方法和函数在编程流程中使用数学,然后它对我来说都是清晰和合乎逻辑的,但在数学符号中,我无法从它做出正面或反面.

我一直在阅读网站,观看研究所试图解释这个问题的视频,但他们都使用数学符号,我只是迷失在其中,我的思想不会将其转化为可理解的东西.我可能有缺陷.

另外,只是使用某人的代码不是我的兴趣,我想了解它背后的机制,逻辑.我很乐意使用其他人的代码,但我真的想了解它是如何工作的.

这个问题

你能用简单的术语向我解释没有数学符号,只是编程符号/函数/伪代码,如何沿所有3轴实现矩阵变换？

理想情况下我想要的是编写方法/对象的材料/理解,我可以在其中定义3个轴的角度,类似于glRotate,以旋转我所拥有的四边形/三角形的集合.(我正在尝试编写立方体形状的3D旋转而无需访问OpenGL函数来为我执行此操作,因为每次在显示列表中发生更改时,都会在一次绘制调用中完成.)

我做了什么？

我试图制作一个90度的变换函数来获得数学的悬念,但是在制作一个理论上最简单的矩阵时却完全失败了.你可以在http://jsfiddle.net/bLfg0tj8/5/看到我失败的尝试.

Vec3 = function(x,y,z) {
    this.x = x;
    this.y = y;
    this.z = z;
}
Matrix = function Matrix() {
    this.matrixPoints = new Array();    
    this.rotationPoint = new Vec3(0,0,0);
    this.rotationAngle = 90;
}
Matrix.prototype.addVector = function(vector) {
    this.matrixPoints.push(vector);
}
Matrix.prototype.setRotationPoint = function(vector) {
    this.rotationPoint = vector; 
}
Matrix.prototype.setRotationAngle = function(angle) {
    this.rotationAngle = angle;
}
Matrix.prototype.populate = function() {
    translateToOrigin =     [[1,0,0-this.rotationPoint.x],
                                  [0,1,0-this.rotationPoint.y],
                                  [0,0,0-this.rotationPoint.z]];
    rotationMatrix =         [[0,-1,0],
                                  [0,1,0],
                                  [0,0,1]];
    translateEnd =         [[1,0,this.rotationPoint.x],
                                  [0,1,this.rotationPoint.y],
                                  [0,0,this.rotationPoint.z]];
    currentColumn = 0;
    currentRow = 0;
    this.combomatrix = this.mergeMatrices(this.mergeMatrices(translateEnd,rotationMatrix),
                                          translateToOrigin);
}
Matrix.prototype.transform = function() {
    newmatrix = new Array();
    for(c = 0;c<this.matrixPoints.length;c++) {
        newmatrix.push(this.applyToVertex(this.matrixPoints[c]));
    }
    return newmatrix;
}
Matrix.prototype.applyToVertex = function(vertex) {
    ret = new Vec3(vertex.x,vertex.y,vertex.z);
    ret.x = ret.x + this.combomatrix[0][0] * vertex.x +
            this.combomatrix[0][1] * vertex.y +
            this.combomatrix[0][2] * vertex.z;
    
    ret.y = ret.y + this.combomatrix[1][0] * vertex.x +
            this.combomatrix[1][1] * vertex.y +
            this.combomatrix[1][2] * vertex.z;
    
    ret.z = ret.z + this.combomatrix[2][0] * vertex.x +
            this.combomatrix[2][1] * vertex.y +
            this.combomatrix[2][2] * vertex.z;
    return ret;
}
Matrix.prototype.mergeMatrices = function(lastStep, oneInFront) {
    step1 = [[0,0,0],[0,0,0],[0,0,0]];
    step1[0][0] =   lastStep[0][0] * oneInFront[0][0] + 
                    lastStep[0][1] * oneInFront[1][0] + 
                    lastStep[0][2] * oneInFront[2][0];
    
    step1[0][1] =   lastStep[0][0] * oneInFront[0][1] + 
                    lastStep[0][1] * oneInFront[1][1] + 
                    lastStep[0][2] * oneInFront[2][1];
    
    step1[0][2] =   lastStep[0][0] * oneInFront[0][2] + 
                    lastStep[0][1] * oneInFront[1][2] + 
                    lastStep[0][2] * oneInFront[2][2];
    //============================================================
    step1[1][0] =   lastStep[1][0] * oneInFront[0][0] + 
                    lastStep[1][1] * oneInFront[1][0] + 
                    lastStep[1][2] * oneInFront[2][0];
    
    step1[1][1] =   lastStep[1][0] * oneInFront[0][1] + 
                    lastStep[1][1] * oneInFront[1][1] + 
                    lastStep[1][2] * oneInFront[2][1];
    
    step1[1][2] =   lastStep[1][0] * oneInFront[0][2] + 
                    lastStep[1][1] * oneInFront[1][2] + 
                    lastStep[1][2] * oneInFront[2][2];
    //============================================================
    step1[2][0] =   lastStep[2][0] * oneInFront[0][0] + 
                    lastStep[2][1] * oneInFront[1][0] + 
                    lastStep[2][2] * oneInFront[2][0];
    
    step1[2][1] =   lastStep[2][0] * oneInFront[0][1] + 
                    lastStep[2][1] * oneInFront[1][1] + 
                    lastStep[2][2] * oneInFront[2][1];
    
    step1[2][2] =   lastStep[2][0] * oneInFront[0][2] + 
                    lastStep[2][1] * oneInFront[1][2] + 
                    lastStep[2][2] * oneInFront[2][2];
    return step1;
}
Matrix.prototype.getCurrentMatrix = function() {
    return this.matrixPoints;
}
myvectors = [new Vec3(50,50,0), new Vec3(20,80,0), new Vec3(80, 80, 0)];

function drawVectors(vectors,color) {
    for(c=0;c<vectors.length;c++) {
        document.getElementById("whoa").innerHTML += '<div style="color:'+color+';position:absolute;left:'+vectors[c].x+'px; top:'+vectors[c].y+'px;z-index:'+vectors[c].z+';">('+c+').</div>';
    }
}
matrix = new Matrix();
for(c=0;c<myvectors.length;c++) {
    matrix.addVector(myvectors[c]);
}
matrix.setRotationPoint(new Vec3(50,70,0));
matrix.populate();
somematrix = matrix.transform();
drawVectors(matrix.getCurrentMatrix(),"lime"); // draw current matrix that was hand coded
drawVectors([matrix.rotationPoint],'white'); // draw rotation point
drawVectors(somematrix,"red"); // transformed matrix... somehow two points merge

<div id="whoa" style="position:relative;top:50px;left:150px;background-color:green;color:red;width:400px;height:300px;">
    &nbsp;
</div>

绿色文本是原始三角形,白色点是中心点,红色指向失败的变换(我认为,因为它没有围绕中心点对齐).我在教程中想到如何将矩阵组合成一个组合矩阵,但我想我搞砸了.

正如我所说,我真的很难理解数学符号并说话.没有帮助的是,大多数教师都会跳过部分解释.单独花了我2个小时来了解何时需要将每个步骤相加而不是仅仅继续相乘.是的解释.

我与之合作/想要合作的一个实际例子

例如,我有一个立方体,从位于世界的wavefront obj文件加载

x = 50
y = 100
z = 200

使用四边形和一些uv映射绘制立方体.这里没问题.它渲染得很漂亮,所有纹理都能正确显示.

这些是使用四边形绘制的立方体的每个"面"的位置坐标.

// Front face
-1.0, -1.0,  1.0,
 1.0, -1.0,  1.0,
 1.0,  1.0,  1.0,
-1.0,  1.0,  1.0,

// Back face
-1.0, -1.0, -1.0,
-1.0,  1.0, -1.0,
 1.0,  1.0, -1.0,
 1.0, -1.0, -1.0,

// Top face
-1.0,  1.0, -1.0,
-1.0,  1.0,  1.0,
 1.0,  1.0,  1.0,
 1.0,  1.0, -1.0,

// Bottom face
-1.0, -1.0, -1.0,
 1.0, -1.0, -1.0,
 1.0, -1.0,  1.0,
-1.0, -1.0,  1.0,

// Right face
 1.0, -1.0, -1.0,
 1.0,  1.0, -1.0,
 1.0,  1.0,  1.0,
 1.0, -1.0,  1.0,

// Left face
-1.0, -1.0, -1.0,
-1.0, -1.0,  1.0,
-1.0,  1.0,  1.0,
-1.0,  1.0, -1.0

所以这一切都很棒.但是如果我希望这个立方体沿x轴旋转90度并绕z轴旋转45度呢？我不能使用glRotate,因为目前我将数据传递给tesselator对象,我无法通过opengl函数进行任何矩阵变换,因为它只是接收数据,而不是实际渲染它本身.

数据的存储方式如下:

WaveFrontObject()
   |
   |-> Groups(String groupname)
        |
        |-> Faces()
              |
              |-> Vertex(float x, float y, float z)[] 
              |-> Float UVmap[] corresponding to each vertex
              |-> drawFace() // Draws the face as a quad or triangle

因此,我给出的上述每个坐标都存储为"立方体"组中波前对象的一个​​面.

当立方体被添加到tesselator时,它被转换为世界上的右坐标并且它呈现正常.

然而它总是呈现相同的.如果我希望它以一个角度渲染,我必须在此刻制作一个单独的波前对象才能够做到这一点.在我的意见中,当它可以通过一些数学解决时,它是疯狂的.

需要答案

1. 逐步解释如何构建翻译矩阵并尝试向我解释数学.

2. 解释如何将平移矩阵应用于面部中的四边形/三角形,它们保持围绕其位置中心的方向

   x = 50.5 y = 100.5 z = 200.5

3. 一些示例/伪代码与解释一起.

用于解释的已使用的编程语言只要在C系列中具有实际意义

请尽量远离数学符号/说话.我不知道alpha beta,thetha是什么,我知道x轴,y轴和z轴是什么.我知道角度是什么,但我不知道数学家找到它的名字.

如果您想使用数学名称,请向我解释它们在3D世界/代码中的含义以及它们的形成/计算方式.

我只是想制作一个方法/对象

Matrix.transformVertices(vertices[], 90deg x, 45 deg y, 0 deg z);

Answer 1

所以问题确实是理解4x4同质变换矩阵

没有数学背后的唯一剩下的就是几何表示/意义,这对于人类的抽象/理解来说要好得多.

那么4x4矩阵是什么？

它是一些笛卡尔坐标系的表示,它由以下部分组成:

1. 3 基础向量(每个轴一个)红色,绿色,蓝色

   因此,如果红色,绿色,蓝色矢量彼此垂直,则坐标系是正交的.如果它们也是单位向量,那么它是正交的(例如单位矩阵).

2. 原点灰色

3. 投影和同质侧(矩阵的未标记的底部剩余部分)

   这部分只用于一次启用旋转和平移,因此使用的点必须是同质的,即形式(x,y,z,w=1).如果它只是(x,y,z,w=0)那时矩阵将是(x,y,z)并且这不足以进行翻译.我不会使用他们不易解释几何的任何投影.

这个布局来自OpenGL表示法,那里也有转置表示(向量是行而不是列)

现在如何将任何点转换为此坐标系:

g=M*l;
l=Inverse(M)*g;

哪里:

• 3x3 是变换矩阵
• M是l局部坐标系点(LCS)
• M 是全球坐标系点(GCS)

对于转置版本(DirectX),它是:

l=M*g;
g=Inverse(M)*l;

这是因为转置正交旋转矩阵本身也是反转的

• 有关更多信息,请参阅变换矩阵解剖和3D图形管道

2.如何形象化

是的,您可以绘制矩阵编号但是它们在第一眼看起来没有意义,特别是如果数字正在改变,那么绘制轴矢量如上图所示.其中每个轴线是一条线从g至origin

3.如何构建它

只需计算轴向量和原点并将它们放在矩阵中.确保正交性利用跨产品(但要注意乘法器的顺序以使用正确的方向)

效果

• 通过旋转轴完成旋转,这样您就可以通过参数圆方程计算每个轴...
• 通过将比例因子乘以轴来完成缩放
• 倾斜只是使用非垂直轴

5.旋转

For most cases the incremental rotation is used. There are two types

• local rotation origin + line_size*axis_vector it rotates around local coordinate axises like you will control plane or car or player ... Most engines/games do not use these and fake it with euler angles instead which is a cheap solution (have many quirks and problems) because most people who using OpenGL do not even know this is possible and rather stack list of M'=M*rotation_matrix calls...

• global rotation glRotate/glTranslate it rotates around global coordinate system axises.

where M'=Inverse(Inverse(M)*rotation_matrix) is any standard rotation transform matrix.

If you have different matrix layout (transposed) then the rotations local and global are computed the other way around ...

You can also compute your rotation_matrix from rotation_matrix angles like:

rotation_matrix=rotation_around_x(ax)*rotation_around_y(ay)*rotation_around_z(az);

see Wiki rotation matrices the 3D 3 from Rx,Ry,Rz are what you need. As you can see they are just unit circle parametric equation really. The order of multiplication change how the angles converge to target position. This is called Euler angles and I do not use it (I integrate step changes instead which has no restrictions if done properly not to mention it is simpler).

6. glRotate

If you want Basic rotations then you should use quaternions instead because that is rotation around axis not by 3 angles! There is workaround:

1. create transform matrix glRotate for that axis
2. then transform your matrix N to it
3. rotate M by angle
4. then transform N back from M to global coordinates

Or you can use Rodrigues_rotation_formula instead

To transform Matrix to/from Matrix in this case just transform axises as points and leave the origin as is but the origin of N must be (0,0,0)!!!

7. usage

Transformations are cumulative that means:

• N is the same as w=0

So if you have many points to transform then you precompute all transformations to single matrix and use just it. Do not need to multiply points by all subsequent matrices. OK now the concept:

you should have p'=M1*M2*M3*M4*p; coordinate systems:

• camera M=M1*M2*M3*M4; p'=M*p
• world (usually unit matrix)
• object 3 (each object have its own matrix)

so if you have cube with C vertexes O then you have to perform transformation on each point from object local coordinates to camera local coordinates. Some gfx api do some of it so you apply only what you have to so you really need:

• 8

the transforms are cumulative and unit matrix does not change anything so:

• p0,...,p7

so before drawing compute p(i)'=inverse(C)*unit*M*p(i); for drawed object then take each point Q=inverse(C)*M; p(i)'=Q*p(i); of the object and compute the transformed Q and draw/use the transformed one ... The p(i) is in local camera coordinate system (x,y of the screen) but there is no perspective there so before drawing you can also add any of the projection matrices and divide by p(i)' cordinate at the end ... The projection is also cumulative so it can be also inside p(i)'

[edit1] C++ example

//$$---- Form CPP ----
//---------------------------------------------------------------------------
// apart from math.h include you can ignore this machine generated VCL related code
#include <vcl.h>
#pragma hdrstop
#include "win_main.h"
#include <math.h>
//---------------------------------------------------------------------------
#pragma package(smart_init)
#pragma resource "*.dfm"
TMain *Main; // pointer to main window ...
//---------------------------------------------------------------------------
// Here is the important stuff some math first
//---------------------------------------------------------------------------
const double deg=M_PI/180.0;
double divide(double x,double y);
void  matrix_mul       (double *c,double *a,double *b); // c[16] = a[16] * b[16]
void  matrix_mul_vector(double *c,double *a,double *b); // c[ 4] = a[16] * b[ 4]
void  matrix_subdet    (double *c,double *a);           // c[16] = all subdets of a[16]
double matrix_subdet   (          double *a,int r,int s);//      = subdet(r,s) of a[16]
double matrix_det      (          double *a);           //       = det of a[16]
double matrix_det      (          double *a,double *b); //       = det of a[16] and subdets b[16]
void  matrix_inv       (double *c,double *a);           // c[16] = a[16] ^ -1
//---------------------------------------------------------------------------
double divide(double x,double y)
        {
        if (!y) return 0.0;
        return x/y;
        }
void  matrix_mul       (double *c,double *a,double *b)
        {
        double q[16];
        q[ 0]=(a[ 0]*b[ 0])+(a[ 1]*b[ 4])+(a[ 2]*b[ 8])+(a[ 3]*b[12]);
        q[ 1]=(a[ 0]*b[ 1])+(a[ 1]*b[ 5])+(a[ 2]*b[ 9])+(a[ 3]*b[13]);
        q[ 2]=(a[ 0]*b[ 2])+(a[ 1]*b[ 6])+(a[ 2]*b[10])+(a[ 3]*b[14]);
        q[ 3]=(a[ 0]*b[ 3])+(a[ 1]*b[ 7])+(a[ 2]*b[11])+(a[ 3]*b[15]);
        q[ 4]=(a[ 4]*b[ 0])+(a[ 5]*b[ 4])+(a[ 6]*b[ 8])+(a[ 7]*b[12]);
        q[ 5]=(a[ 4]*b[ 1])+(a[ 5]*b[ 5])+(a[ 6]*b[ 9])+(a[ 7]*b[13]);
        q[ 6]=(a[ 4]*b[ 2])+(a[ 5]*b[ 6])+(a[ 6]*b[10])+(a[ 7]*b[14]);
        q[ 7]=(a[ 4]*b[ 3])+(a[ 5]*b[ 7])+(a[ 6]*b[11])+(a[ 7]*b[15]);
        q[ 8]=(a[ 8]*b[ 0])+(a[ 9]*b[ 4])+(a[10]*b[ 8])+(a[11]*b[12]);
        q[ 9]=(a[ 8]*b[ 1])+(a[ 9]*b[ 5])+(a[10]*b[ 9])+(a[11]*b[13]);
        q[10]=(a[ 8]*b[ 2])+(a[ 9]*b[ 6])+(a[10]*b[10])+(a[11]*b[14]);
        q[11]=(a[ 8]*b[ 3])+(a[ 9]*b[ 7])+(a[10]*b[11])+(a[11]*b[15]);
        q[12]=(a[12]*b[ 0])+(a[13]*b[ 4])+(a[14]*b[ 8])+(a[15]*b[12]);
        q[13]=(a[12]*b[ 1])+(a[13]*b[ 5])+(a[14]*b[ 9])+(a[15]*b[13]);
        q[14]=(a[12]*b[ 2])+(a[13]*b[ 6])+(a[14]*b[10])+(a[15]*b[14]);
        q[15]=(a[12]*b[ 3])+(a[13]*b[ 7])+(a[14]*b[11])+(a[15]*b[15]);
        for(int i=0;i<16;i++) c[i]=q[i];
        }
void  matrix_mul_vector(double *c,double *a,double *b)
        {
        double q[3];
        q[0]=(a[ 0]*b[0])+(a[ 1]*b[1])+(a[ 2]*b[2])+(a[ 3]);
        q[1]=(a[ 4]*b[0])+(a[ 5]*b[1])+(a[ 6]*b[2])+(a[ 7]);
        q[2]=(a[ 8]*b[0])+(a[ 9]*b[1])+(a[10]*b[2])+(a[11]);
        for(int i=0;i<3;i++) c[i]=q[i];
        }
void  matrix_subdet    (double *c,double *a)
        {
        double   q[16];
        int     i,j;
        for (i=0;i<4;i++)
         for (j=0;j<4;j++)
          q[j+(i<<2)]=matrix_subdet(a,i,j);
        for (i=0;i<16;i++) c[i]=q[i];
        }
double matrix_subdet    (         double *a,int r,int s)
        {
        double   c,q[9];
        int     i,j,k;
        k=0;                            // q = sub matrix
        for (j=0;j<4;j++)
         if (j!=s)
          for (i=0;i<4;i++)
           if (i!=r)
                {
                q[k]=a[i+(j<<2)];
                k++;
                }
        c=0;
        c+=q[0]*q[4]*q[8];
        c+=q[1]*q[5]*q[6];
        c+=q[2]*q[3]*q[7];
        c-=q[0]*q[5]*q[7];
        c-=q[1]*q[3]*q[8];
        c-=q[2]*q[4]*q[6];
        if (int((r+s)&1)) c=-c;       // add signum
        return c;
        }
double matrix_det       (         double *a)
        {
        double c=0;
        c+=a[ 0]*matrix_subdet(a,0,0);
        c+=a[ 4]*matrix_subdet(a,0,1);
        c+=a[ 8]*matrix_subdet(a,0,2);
        c+=a[12]*matrix_subdet(a,0,3);
        return c;
        }
double matrix_det       (         double *a,double *b)
        {
        double c=0;
        c+=a[ 0]*b[ 0];
        c+=a[ 4]*b[ 1];
        c+=a[ 8]*b[ 2];
        c+=a[12]*b[ 3];
        return c;
        }
void  matrix_inv       (double *c,double *a)
        {
        double   d[16],D;
        matrix_subdet(d,a);
        D=matrix_det(a,d);
        if (D) D=1.0/D;
        for (int i=0;i<16;i++) c[i]=d[i]*D;
        }
//---------------------------------------------------------------------------
// now the object representation
//---------------------------------------------------------------------------
const int pnts=8;
double pnt[pnts*3]=     // Vertexes for 100x100x100 cube centered at (0,0,0)
    {
    -100.0,-100.0,-100.0,
    -100.0,+100.0,-100.0,
    +100.0,+100.0,-100.0,
    +100.0,-100.0,-100.0,
    -100.0,-100.0,+100.0,
    -100.0,+100.0,+100.0,
    +100.0,+100.0,+100.0,
    +100.0,-100.0,+100.0,
    };
const int facs=6;
int fac[facs*4]=        // faces (index of point used) no winding rule
    {
    0,1,2,3,
    4,5,6,7,
    0,1,5,4,
    1,2,6,5,
    2,3,7,6,
    3,0,4,7,
    };
double rep[16]=        // 4x4 transform matrix of object (unit from start) at (0,0,+100)
    {
    1.0,0.0,0.0,  0.0,
    0.0,1.0,0.0,  0.0,
    0.0,0.0,1.0,100.0,
    0.0,0.0,0.0,1.0,
    };
double eye[16]=        // 4x4 transform matrix of camera at (0,0,-150)
    {
    1.0,0.0,0.0,   0.0,
    0.0,1.0,0.0,   0.0,
    0.0,0.0,1.0,-150.0,
    0.0,0.0,0.0,1.0,
    };
//---------------------------------------------------------------------------
// this is how to draw it
//---------------------------------------------------------------------------
void obj(double *pnt,int pnts,int *fac,int facs,double *rep,double *ieye)
    {
    // variables for drawing
    int i;
    double p0[3],p1[3],p2[3],p3[3],m[16],d;
    // gfx api variables (change to your stuff) Main is the main form of this application
    TCanvas *scr=Main->bmp->Canvas;
    double xs2=Main->ClientWidth/2,ys2=Main->ClientHeight/2;
    double v=xs2*tan(30.0*deg); // 60 degree viewing angle perspective projection

    matrix_mul(m,ieye,rep);             // cumulate all needed transforms

    for (i=0;i<facs*4;)                 // go through all faces
        {
        // convert all points of face
        matrix_mul_vector(p0,m,&pnt[fac[i]*3]); i++;
        matrix_mul_vector(p1,m,&pnt[fac[i]*3]); i++;
        matrix_mul_vector(p2,m,&pnt[fac[i]*3]); i++;
        matrix_mul_vector(p3,m,&pnt[fac[i]*3]); i++;
        // here goes perspective divide by z coordinate if needed
        d=divide(v,p0[2]); p0[0]*=d; p0[1]*=d;
        d=divide(v,p1[2]); p1[0]*=d; p1[1]*=d;
        d=divide(v,p2[2]); p2[0]*=d; p2[1]*=d;
        d=divide(v,p3[2]); p3[0]*=d; p3[1]*=d;
        // here is viewport transform (just translate (0,0) to middle of screen in this case
        p0[0]+=xs2; p0[1]+=ys2;
        p1[0]+=xs2; p1[1]+=ys2;
        p2[0]+=xs2; p2[1]+=ys2;
        p3[0]+=xs2; p3[1]+=ys2;
        // draw quad
        // I use VCL GDI TCanvas you use what you have ...
        // and wireframe only to keep this simple (no Z buffer,winding culling,...)
        scr->Pen->Color=clAqua;     // perimeter wireframe
        scr->MoveTo(p0[0],p0[1]);
        scr->LineTo(p1[0],p1[1]);
        scr->LineTo(p2[0],p2[1]);
        scr->LineTo(p3[0],p3[1]);
        scr->LineTo(p0[0],p0[1]);
//      scr->Pen->Color=clBlue;     // face cross to visualy check if I correctly generate the fac[]
//      scr->MoveTo(p0[0],p0[1]);
//      scr->LineTo(p2[0],p2[1]);
//      scr->MoveTo(p1[0],p1[1]);
//      scr->LineTo(p3[0],p3[1]);
        }
    }
//---------------------------------------------------------------------------
//---------------------------------------------------------------------------
void TMain::draw()
    {
    if (!_redraw) return;
    bmp->Canvas->Brush->Color=clBlack;
    bmp->Canvas->FillRect(TRect(0,0,xs,ys));

    // compute inverse of camera need to compute just once for all objects
    double ieye[16];
    matrix_inv(ieye,eye);
    // draw all objects
    obj(pnt,pnts,fac,facs,rep,ieye);

    Main->Canvas->Draw(0,0,bmp);
    _redraw=false;
    }
//---------------------------------------------------------------------------
__fastcall TMain::TMain(TComponent* Owner) : TForm(Owner)
    {
    // window constructor you can ignore this ... (just create a backbuffer bitmap here)
    bmp=new Graphics::TBitmap;
    bmp->HandleType=bmDIB;
    bmp->PixelFormat=pf32bit;
    pyx=NULL;
    }
//---------------------------------------------------------------------------
void __fastcall TMain::FormDestroy(TObject *Sender)
    {
    // window destructor release memory ... also ignoe this
    if (pyx) delete pyx;
    delete bmp;
    }
//---------------------------------------------------------------------------
void __fastcall TMain::FormResize(TObject *Sender)
    {
    // on resize event ... just resize/redraw backbuffer also can ignore this
    xs=ClientWidth;  xs2=xs>>1;
    ys=ClientHeight; ys2=ys>>1;
    bmp->Width=xs;
    bmp->Height=ys;
    if (pyx) delete pyx;
    pyx=new int*[ys];
    for (int y=0;y<ys;y++) pyx[y]=(int*) bmp->ScanLine[y];
    _redraw=true;
    }
//---------------------------------------------------------------------------
void __fastcall TMain::FormPaint(TObject *Sender)
    {
    // repaint event can ignore
    _redraw=true;
    }
//---------------------------------------------------------------------------
void __fastcall TMain::tim_redrawTimer(TObject *Sender)
    {
    // timer event to animate the cube ...
    _redraw=true;

    // rotate the object to see it in motion
    double ang,c,s;

    ang=5.0*deg; c=cos(ang); s=sin(ang);    // rotate baround z by 5 degrees per timer step
    double rz[16]= { c, s, 0, 0,
                    -s, c, 0, 0,
                     0, 0, 1, 0,
                     0, 0, 0, 1 };

    ang=1.0*deg; c=cos(ang); s=sin(ang);    // rotate baround x by 1 degrees per timer step
    double rx[16]= { 1, 0, 0, 0,
                     0, c, s, 0,
                     0,-s, c, 0,
                     0, 0, 0, 1 };
    matrix_mul(rep,rep,rz);
    matrix_mul(rep,rep,rx);

    draw();
    }
//---------------------------------------------------------------------------

这是它的样子:

和GIF动画背面剔除:

[笔记]

如果您有更多问题,请评论我...

[Edit2]经常需要基本的3D矢量操作

如果您不知道如何计算矢量/点积或绝对值等矢量运算,请参阅:

// cross product: W = U x V
W.x=(U.y*V.z)-(U.z*V.y)
W.y=(U.z*V.x)-(U.x*V.z)
W.z=(U.x*V.y)-(U.y*V.x)
// dot product: a = (U.V)
a=U.x*V.x+U.y*V.y+U.z*V.z
// abs of vector a = |U|
a=sqrt((U.x*U.x)+(U.y*U.y)+(U.z*U.z))

这里我的C++矢量数学:

static double vector_tmp[3];
double divide(double x,double y) { if ((y>=-1e-30)&&(y<=+1e-30)) return 0.0; return x/y; }
double* vector_ld(double x,double y,double z)          { double *p=vector_tmp; p[0]=x; p[1]=y; p[2]=z; return p;}
double* vector_ld(double *p,double x,double y,double z) {                      p[0]=x; p[1]=y; p[2]=z; return p;}
void  vector_copy(double *c,double *a)         { for(int i=0;i<3;i++) c[i]=a[i];       }
void  vector_abs(double *c,double *a)          { for(int i=0;i<3;i++) c[i]=fabs(a[i]); }
void  vector_one(double *c,double *a)
        {
        double l=divide(1.0,sqrt((a[0]*a[0])+(a[1]*a[1])+(a[2]*a[2])));
        c[0]=a[0]*l;
        c[1]=a[1]*l;
        c[2]=a[2]*l;
        }
void  vector_len(double *c,double *a,double l)
        {
        l=divide(l,sqrt((a[0]*a[0])+(a[1]*a[1])+(a[2]*a[2])));
        c[0]=a[0]*l;
        c[1]=a[1]*l;
        c[2]=a[2]*l;
        }
void  vector_neg(double *c,double *a)          { for(int i=0;i<3;i++) c[i]=-a[i];      }
void  vector_add(double *c,double *a,double *b) { for(int i=0;i<3;i++) c[i]=a[i]+b[i]; }
void  vector_sub(double *c,double *a,double *b) { for(int i=0;i<3;i++) c[i]=a[i]-b[i]; }
void  vector_mul(double *c,double *a,double *b) // cross
        {
        double   q[3];
        q[0]=(a[1]*b[2])-(a[2]*b[1]);
        q[1]=(a[2]*b[0])-(a[0]*b[2]);
        q[2]=(a[0]*b[1])-(a[1]*b[0]);
        for(int i=0;i<3;i++) c[i]=q[i];
        }
void  vector_mul(double *c,double *a,double  b) { for(int i=0;i<3;i++) c[i]=a[i]*b; }
void  vector_mul(double *c,double  a,double *b) { for(int i=0;i<3;i++) c[i]=a*b[i]; }
double vector_mul(         double *a,double *b) { double c=0; for(int i=0;i<3;i++) c+=a[i]*b[i]; return c; } // dot
double vector_len(double *a) { return sqrt((a[0]*a[0])+(a[1]*a[1])+(a[2]*a[2])); }
double vector_len2(double *a) { return (a[0]*a[0])+(a[1]*a[1])+(a[2]*a[2]); }

[Edit3]通过键盘进行摄像机和对象控制的局部旋转

最近这里有很多问题,我的一些示例答案是我的演示:

• 固定摄像机视图控制(部分伪逆矩阵)
• 相机和播放器控制(逆矩阵)
• 如何通过累积变换随时间保持精度(全伪逆矩阵)
• 圆形风格简单的OpenGL/C++/VCL播放器控件实例