三维变换之斜投影透视投影

Posted 2020-09-20

编译器：VS2013

原理：依旧是矩阵相乘，只要求得矩阵就可以很方便的求出结果

注意！注意！注意！

楼主数学学得不好，齐次坐标这里没好好听，齐次坐标的第四个坐标值一直为1，楼主懵逼的也按物体坐标等比例放大导致错误调了两天

错误：

1 int a[8][4] = { 0, 0, 0, 200, 200, 0, 0,200, 200, 200, 0,200, 0, 200, 0,200, 0, 0, 200,200, 200, 0, 200,200, 200, 200, 200,200, 0, 200, 200,200};

 

正确：

1 int a[8][4] = { 0, 0, 0, 1, 200, 0, 0, 1, 200, 200, 0, 1, 0, 200, 0, 1, 0, 0, 200, 1, 200, 0, 200, 1, 200, 200, 200, 1, 0, 200, 200, 1 };

 

主函数块

 1 #include "stdafx.h"
 2 #include<stdio.h>
 3 #include"graphics.h"
 4 #include<stdlib.h>
 5 #include<math.h>
 6 
 7 #define PI 3.14159
 8 
 9 //函数声明
10 void apoint(int a[][4]);//一点透视
11 void oblique2_30(int a[][4]);//斜二测45°
12 void oblique2_45(int a[][4]);//斜二测30°
13 void twopoint(int a[][4]);//两点透视
14 void threepoint(int a[][4]);//三点透视
15 void putline(double b[][4]);//画线函数
16 
17 int main()
18 {
19     int gdriver = DETECT, gmove;
20     int a[8][4] = { 0, 0, 0, 1, 200, 0, 0, 1, 200, 200, 0, 1, 0, 200, 0, 1, 0, 0, 200, 1, 200, 0, 200, 1, 200, 200, 200, 1, 0, 200, 200, 1 };
21     
22     initgraph(&gdriver, &gmove, "");
23     setcolor(YELLOW);
24 
25     apoint(a);//一点透视
26     //oblique2_30(a);//斜二测30°
27     //oblique2_45(a);//斜二测45°
28     //twopoint(a);//两点透视
29     //threepoint(a);//三点透视
30 
31     system("pause");
32 
33     closegraph();
34 
35     return 0;
36 }

 

 

一点透视：

 1 //一点透视
 2 void apoint(int a[][4])
 3 {
 4     double k = 160, m = 220, n = 400, d = 500,b[8][4];
 5     int i;
 6 
 7     for (i = 0; i < 8; i++)
 8     {
 9         b[i][3] = (d + n + a[i][2]) / d*1.0;//其次变换
10         b[i][0] = (a[i][0] + k) / b[i][3]*1.0;//x变换
11         b[i][1] = (a[i][1] + m) / b[i][3]*1.0;//y变换
12         b[i][2] = 0;//z变换
13         b[i][3] = 1;//齐次坐标赋1
14     }
15 
16     putline(b);//画线函数

画线函数

 1 void putline(double b[][4])
 2 {
 3     int i;
 4 
 5     //小图层
 6     for (i = 0; i < 3; i++)
 7         line(b[i][0], b[i][1], b[i + 1][0], b[i + 1][1]);
 8 
 9     //小图层尾连接
10     line(b[3][0], b[3][1], b[0][0], b[0][1]);
11 
12     //大图层
13     for (i = 4; i < 7; i++)
14         line(b[i][0], b[i][1], b[i + 1][0], b[i + 1][1]);
15 
16     //大图层首尾连接
17     line(b[4][0], b[4][1], b[7][0], b[7][1]);
18 
19     //大小图层连接
20     for (i = 0; i < 4; i++)
21         line(b[i][0], b[i][1], b[i + 4][0], b[i + 4][1]);
22 }

 

 

二点透视

 1 //两点透视
 2 void twopoint(int a[][4])
 3 {
 4     double p = 0.002, r = 0.002, A = PI / 6.0, k = 150, n = 30, m = 50;
 5     //二维数组c为转换矩阵
 6     double b[8][4] = { 0 };
 7     //转换矩阵
 8     double c[4][4] = { { cos(A), 0, 0, p*cos(A) - r*sin(A) }, { 0, 1, 0, 0 }, { sin(A), 0, 0, p*sin(A) + r*cos(A) }, { k*cos(A) + n*sin(A), m, 0, p*(k*cos(A) + n*sin(A)) + r*(n*cos(A) - k*sin(A)) + 1 } };
 9     int i, j, x;
10 
11     //矩阵相乘
12     for (i = 0; i < 8; i++)
13         for (j = 0; j < 4; j++)
14             for (x = 0; x < 4; x++)
15                 b[i][j] += a[i][x] * c[x][j];
16             
17     for (i = 0; i < 8; i++)
18         for (j = 0; j < 4; j++)
19         {
20             b[i][j] /= b[i][3];//使齐次坐标变为1
21             b[i][j] = (int)b[i][j];//取整
22         }
23 
24     putline(b);//画线函数
25 }

 

 

三点透视

 1 //三点透视
 2 void threepoint(int a[][4])
 3 {
 4     double p = 0.0015,q=0.0015, r = 0.0015, A = PI / 6.0,B=PI/45, k = 150, n = 30, m = 30;
 5     //二维数组c为转换矩阵
 6     double b[8][4] = { 0 };
 7     //转换矩阵
 8     double c[4][4] = { { cos(A),sin(A)*sin(B),0,p }, { 0,cos(B),0,q }, { sin(A),-cos(A)*sin(B),0,r }, { k*cos(A)+n*sin(A), m*cos(B)+sin(B)*(k*sin(B)-n*cos(A)),0,k*p+m*q+n*r} };
 9     int i, j, x;
10 
11     //矩阵相乘
12     for (i = 0; i < 8; i++)
13         for (j = 0; j < 4; j++)
14             for (x = 0; x < 4; x++)
15                 b[i][j] += a[i][x] * c[x][j];
16 
17     for (i = 0; i < 8; i++)
18         for (j = 0; j < 4; j++)
19         {
20             b[i][j] /= b[i][3];//使齐次坐标变为1
21             b[i][j] = (int)b[i][j];//取整
22         }
23 
24     putline(b);//画线函数，括号内强制转换成int
25 }

 

 

斜二测30°

 1 //斜二测30°
 2 void oblique2_45(int a[][4])
 3 {
 4     double b[8][4];
 5     int i;
 6 
 7     for (i = 0; i < 8; i++)
 8     {
 9         b[i][0] = a[i][0] + a[i][2] * cos(PI / 4)*sqrt(5.0) / 5 * 1.0;//x变换
10         b[i][1] = a[i][1] + a[i][2] * sin(PI / 4)*sqrt(5.0) / 5 * 2.0;//y变换
11         b[i][2] = 0;//z变换
12         b[i][3] = 1;//齐次坐标变换
13     }
14 
15     putline(b);//画线函数
16 }

 

 

斜二测45°

 1 //斜二测45°
 2 void oblique2_30(int a[][4])
 3 {
 4     double b[8][4];
 5     int i;
 6 
 7     for (i = 0; i < 8; i++)
 8     {
 9         b[i][0] = a[i][0] + a[i][2] * cos(PI / 6)*sqrt(5.0) / 5 * 1.0;//x变换
10         b[i][1] = a[i][1] + a[i][2] * sin(PI / 6)*sqrt(5.0) / 5 * 2.0;//y变换
11         b[i][2] = 0;//z变换
12         b[i][3] = 1;//齐次坐标变换
13     }
14 
15     putline(b);//画线函数
16 }