2015-02-04 46 views
1

我正在尝试3D投影几个点。从我读过的内容来看,在将投影矩阵应用到顶点之后,顶点在剪辑空间中结束。此时如果-w < x,y,z < w顶点可见,否则它位于可见区域之外,需要剪切。我的问题是,我不能得到一个顶点是-w < x,y,z < w。我一定在做错事,但我不知道它是什么。我尽量坚持使用openGL约定(坐标系等)。请看看这个SSCCE。剪辑空间中的3D透视投影坐标

#include <stdio.h> 
#include <stdlib.h> 
#include <math.h> 

#define DEG2RAD 0.01745329f 

// multiplies two matrices 
void m3d_matrixMul(const float matrix1[], const float matrix2[], float destination[]) 
{ 
    int i, row, col; 

    for (i = 15; i >= 0; i--) 
    { 
     row = i % 4; 
     col = i/4 * 4; 
     destination[i] = matrix1[row] * matrix2[col] + matrix1[row + 4] * matrix2[col + 1] + matrix1[row + 8] * matrix2[col + 2] + matrix1[row + 12] * matrix2[col + 3]; 
    } 
} 

// multiplies matrix with vector/vertex 
void m3d_matVecMul(const float matrix[], const float vector[], float destination[]) 
{ 
    float x = vector[0], y = vector[1], z = vector[2], w = vector[3]; 

    destination[0] = x * matrix[0] + y * matrix[4] + z * matrix[8] + w * matrix[12]; 
    destination[1] = x * matrix[1] + y * matrix[5] + z * matrix[9] + w * matrix[13]; 
    destination[2] = x * matrix[2] + y * matrix[6] + z * matrix[10] + w * matrix[14]; 
    destination[3] = x * matrix[3] + y * matrix[7] + z * matrix[11] + w * matrix[15]; 
} 

// creates the projection matrix (column major) 
void m3d_getFrustum(float left, float right, float bottom, float top, float near, float far, float matrix[]) 
{ 
    matrix[0] = 2.0f * near/(right - left); 
    matrix[1] = 0.0f; 
    matrix[2] = 0.0f; 
    matrix[3] = 0.0f; 
    matrix[4] = 0.0f; 
    matrix[5] = 2.0f * near/(top - bottom); 
    matrix[6] = 0.0f; 
    matrix[7] = 0.0f; 
    matrix[8] = (right + left)/(right - left); 
    matrix[9] = (top + bottom)/(top - bottom); 
    matrix[10] = -(far + near)/(far - near); 
    matrix[11] = -1.0f; 
    matrix[12] = 0.0f; 
    matrix[13] = 0.0f; 
    matrix[14] = -2.0f * far * near/(far - near); 
    matrix[15] = 0.0f; 
} 

void m3d_getPerspective(float vfov, float aspect, float near, float far, float matrix[]) 
{ 
    float height = tanf(vfov * 0.5f * DEG2RAD) * near; 
    float width = aspect * height; 
    m3d_getFrustum(-width, width, -height, height, near, far, matrix); 
} 

void m3d_getIdentity(float matrix[]) 
{ 
    matrix[0] = 1.0f; 
    matrix[1] = 0.0f; 
    matrix[2] = 0.0f; 
    matrix[3] = 0.0f; 
    matrix[4] = 0.0f; 
    matrix[5] = 1.0f; 
    matrix[6] = 0.0f; 
    matrix[7] = 0.0f; 
    matrix[8] = 0.0f; 
    matrix[9] = 0.0f; 
    matrix[10] = 1.0f; 
    matrix[11] = 0.0f; 
    matrix[12] = 0.0f; 
    matrix[13] = 0.0f; 
    matrix[14] = 0.0f; 
    matrix[15] = 1.0f; 
} 

void m3d_translate(float x, float y, float z, float matrix[]) 
{ 
    matrix[12] += x; 
    matrix[13] += y; 
    matrix[14] += z; 
} 

int main(void) 
{ 
    float vertex[4] = {0.5f, 0.5f, 0.5f, 1.0f}; 
    float modelView[16]; 
    float projection[16]; 
    float mvp[16]; 
    char buffer[4]; 

    // initialize model-view matrix with identity matrix 
    m3d_getIdentity(modelView); 
    // add z-translation to model-view matrix 
    m3d_translate(0.0f, 0.0f, -2.0f, modelView); 
    // get perspective projection matrix 
    m3d_getPerspective(45.0f, 800.0f/600.0f, -0.5f, -1000.0f, projection); 
    // fuse projection and model-view matrix into one matrix 
    m3d_matrixMul(projection, modelView, mvp); 
    // apply model-view-projection matrix to vertex 
    m3d_matVecMul(mvp, vertex, vertex); 

    printf("Projected vertex: %f, %f, %f, %f\n", vertex[0], vertex[1], vertex[2], vertex[3]); 
    printf("Press ENTER to quit."); 
    fgets(buffer, 4, stdin); 
    return EXIT_SUCCESS; 

    // Projected vertex: 0.905330, 1.207107, 2.502001, 1.500000 
} 

thx提前。

+0

至少'm3d_matVecMul(MVP,顶点,顶点)'覆盖与结果的各部分输入顶点的乘法期间,如果我没有弄错。数组总是通过指针传递,所以你不能通过const向量修改顶点,但是当目的地指向同一个地址时,你可以通过目的地。我建议你检查其他方法,并用笔和纸来验证至少一个非平凡例子(或matlab)的矩阵计算。 – Thomas

+0

嗨! thx为您的评论。实际上这是打算。这些值将被覆盖,因为旧值不再需要。最后我只想要投影坐标。据我所知,不应该是一个问题,对吧?生病尝试手动验证计算,但我不确定是否会导致解决方案。 – user3563584

回答

0

好的,我再次查看了您的代码,我认为您的投影矩阵计算有问题。我用gluPerspective的定义替换了m3d_getPerspective方法,并添加了printMatrix(pm)方法。对于一些示例点,结果现在看起来是合理的(下面的代码)。

我没有具体找到你的错误,因为我不确定你为什么正好做这个平截头体 - 解决方法。一个提示可能是你似乎使用tanf在哪里我现在使用cotangens,但这并不是我想的完全问题。此外,设置接近和远离负值不直观,因为它们应该表示距离。你的乘法似乎没问题,我没有意识到你首先将顶点拷贝到x,y,z,w。

#include <stdio.h> 
#include <stdlib.h> 
#include <math.h> 

#define DEG2RAD 0.01745329f 

void m3d_matrixMul(const float matrix1[], const float matrix2[], float destination[]) 
{ 
    int i, row, col; 

    for (i = 15; i >= 0; i--) 
    { 
     row = i % 4; 
     col = i/4 * 4; 
     destination[i] = matrix1[row] * matrix2[col] + matrix1[row + 4] * matrix2[col + 1] + matrix1[row + 8] * matrix2[col + 2] + matrix1[row + 12] * matrix2[col + 3]; 
    } 
} 

void m3d_matVecMul(const float matrix[], const float vector[], float destination[]) 
{ 
    float x = vector[0], y = vector[1], z = vector[2], w = vector[3]; 

    destination[0] = x * matrix[0] + y * matrix[4] + z * matrix[8] + w * matrix[12]; 
    destination[1] = x * matrix[1] + y * matrix[5] + z * matrix[9] + w * matrix[13]; 
    destination[2] = x * matrix[2] + y * matrix[6] + z * matrix[10] + w * matrix[14]; 
    destination[3] = x * matrix[3] + y * matrix[7] + z * matrix[11] + w * matrix[15]; 
} 

//not used anymore 
void m3d_getFrustum(float left, float right, float bottom, float top, float near, float far, float matrix[]) 
{ 
    matrix[0] = 2.0f * near/(right - left); 
    matrix[1] = 0.0f; 
    matrix[2] = 0.0f; 
    matrix[3] = 0.0f; 
    matrix[4] = 0.0f; 
    matrix[5] = 2.0f * near/(top - bottom); 
    matrix[6] = 0.0f; 
    matrix[7] = 0.0f; 
    matrix[8] = (right + left)/(right - left); 
    matrix[9] = (top + bottom)/(top - bottom); 
    matrix[10] = -(far + near)/(far - near); 
    matrix[11] = -1.0f; 
    matrix[12] = 0.0f; 
    matrix[13] = 0.0f; 
    matrix[14] = -2.0f * far * near/(far - near); 
    matrix[15] = 0.0f; 
} 

void m3d_getPerspective(float vfov, float aspect, float near, float far, float matrix[]) 
{ 
    float f = 1.0f/tanf(vfov * 0.5f * DEG2RAD); 

    matrix[0] = f/aspect; 
    matrix[1] = 0.0f; 
    matrix[2] = 0.0f; 
    matrix[3] = 0.0f; 

    matrix[4] = 0.0f; 
    matrix[5] = f; 
    matrix[6] = 0.0f; 
    matrix[7] = 0.0f; 

    matrix[8] = 0; 
    matrix[9] = 0; 
    matrix[10] = (near+far)/(near-far); 
    matrix[11] = -1; 

    matrix[12] = 0.0f; 
    matrix[13] = 0.0f; 
    matrix[14] = 2.0f * far * near/(near-far); 
    matrix[15] = 0.0f; 



} 

void m3d_getIdentity(float matrix[]) 
{ 
    matrix[0] = 1.0f; 
    matrix[1] = 0.0f; 
    matrix[2] = 0.0f; 
    matrix[3] = 0.0f; 
    matrix[4] = 0.0f; 
    matrix[5] = 1.0f; 
    matrix[6] = 0.0f; 
    matrix[7] = 0.0f; 
    matrix[8] = 0.0f; 
    matrix[9] = 0.0f; 
    matrix[10] = 1.0f; 
    matrix[11] = 0.0f; 
    matrix[12] = 0.0f; 
    matrix[13] = 0.0f; 
    matrix[14] = 0.0f; 
    matrix[15] = 1.0f; 
} 

void m3d_translate(float x, float y, float z, float matrix[]) 
{ 
    matrix[12] += x; 
    matrix[13] += y; 
    matrix[14] += z; 
} 

void pm(float* m) { 
    printf("%f, %f, %f, %f\n", m[0], m[4], m[8], m[12]); 
    printf("%f, %f, %f, %f\n", m[1], m[5], m[9], m[13]); 
    printf("%f, %f, %f, %f\n", m[2], m[6], m[10], m[14]); 
    printf("%f, %f, %f, %f\n", m[3], m[7], m[11], m[15]); 
    printf("\n"); 
} 

int main(void) 
{ 
    //float vertex[4] = {0.0f, 0.0f, -1000.0f, 1.0f}; //point at center & far 
    //float vertex[4] = {0.0f, 0.0f, -0.5f, 1.0f}; //point at center & near 
    float vertex[4] = {(800.0f/600.0f)*1.0f, 1.0f, -2.0f, 1.0f}; //point at one quater of the x and y range 
    float modelView[16]; 
    float projection[16]; 
    float mvp[16]; 
    char buffer[4]; 

    m3d_getIdentity(modelView); 
    pm(modelView); 
    m3d_translate(0.0f, 0.0f, 0.0f, modelView); 
    pm(modelView); 
    m3d_getPerspective(90.0f, 800.0f/600.0f, 0.5f, 1000.0f, projection); 
    pm(projection); 
    m3d_matrixMul(projection, modelView, mvp); 
    pm(mvp); 
    m3d_matVecMul(mvp, vertex, vertex); 

    printf("Projected vertex: %f, %f, %f, %f\n", vertex[0], vertex[1], vertex[2], vertex[3]); 
    printf("Press ENTER to quit."); 
    fgets(buffer, 4, stdin); 
    return EXIT_SUCCESS; 
} 
+0

感谢您的耐心等待!我目前很忙,所以我现在无法测试,但我很快就会。你已经帮了我很多。至少现在我知道在哪里看。 “平截头体 - 解决方法”只是一种方便的方法。我从[scratchapixel](http://www.scratchapixel.com/)那里得到了。我使用负z值,因为据我所知,在openGL中,摄像机处于0,0,0处,并且沿着z轴向下看,所以如果近平面和远平面应位于摄像机前方,则需要是负面的。可能是错误的... – user3563584

+0

“截锥体解决方法”的代码也可以在[OpenGL wiki](https://www.opengl.org/wiki/GluPerspective_code)中找到 – user3563584

+0

zNear和zFar的负z值似乎是问题。它以完美的价值完美地工作。 Thx再次。 – user3563584