在OpenGL中计算摄像机到立方体平面的距离

kninwzqo  于 11个月前  发布在  其他
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我试图让透明度混合在我的立方体上正常工作,但遇到了一点障碍。我已经通过简单地按正确的顺序绘制多个立方体来实现混合,但让它在立方体本身的表面上工作要困难得多。
我想解决这个问题的方法是重新排序EBO中的索引,以便它以正确的顺序绘制面。我已经让它重新排序索引,问题是它们没有被正确排序。
我这样做的方法是这样的:

void GECCube::SortIndicies(glm::mat4 ModelMatrix)
{

glm::vec4 View = glm::vec4{0.0f, 0.0f, -3.0f, 0.f};
unsigned int NewIndices[36];
std::multimap<float, std::vector<unsigned int>> SortedMap;

for (int i = 0; i < 12; i += 2)
    {
    
    glm::vec4 Tri1 = glm::vec4{
        (Vertices[Indices[i * 3] * 3] + Vertices[Indices[(i * 3) + 1] * 3] + Vertices[Indices[(i * 3) + 2] * 3]) / 3, //X
        (Vertices[Indices[i * 3] * 4] + Vertices[Indices[(i * 3) + 1] * 4] + Vertices[Indices[(i * 3) + 2] * 4]) / 3, //Y
        (Vertices[Indices[i * 3] * 5] + Vertices[Indices[(i * 3) + 1] * 5] + Vertices[Indices[(i * 3) + 2] * 5]) / 3, //Z
        0.f
        };

    glm::vec4 Tri2 = glm::vec4{
        (Vertices[Indices[(i + 1) * 3] * 3] + Vertices[Indices[((i + 1) * 3) + 1] * 3] + Vertices[Indices[((i + 1) * 3) + 2] * 3]) / 3, //X
        (Vertices[Indices[(i + 1) * 3] * 4] + Vertices[Indices[((i + 1) * 3) + 1] * 4] + Vertices[Indices[((i + 1) * 3) + 2] * 4]) / 3, //Y
        (Vertices[Indices[(i + 1) * 3] * 5] + Vertices[Indices[((i + 1) * 3) + 1] * 5] + Vertices[Indices[((i + 1) * 3) + 2] * 5]) / 3, //Z
        0.f
        };

    glm::vec4 PlaneMed = glm::vec4{(Tri1.x + Tri2.x) / 2, (Tri1.y + Tri2.y) / 2 , (Tri1.z + Tri2.z) / 2, 0 };

    float Distance = glm::length(View - (ModelMatrix * PlaneMed));

    std::vector<unsigned int> TransportVector = std::vector<unsigned int>{ Indices[i * 3] , Indices[(i * 3) + 1] , Indices[(i * 3) + 2],
        Indices[(i + 1) * 3] , Indices[((i + 1) * 3) + 1] , Indices[((i + 1) * 3) + 2] };

    SortedMap.insert(std::pair<float, std::vector<unsigned int>>(Distance, TransportVector));
    }

auto MapIterator = SortedMap.begin();
for (int i = 0; i < 12; i += 2)
    {

    NewIndices[i * 3] = MapIterator->second[0];
    NewIndices[(i * 3) + 1] = MapIterator->second[1];
    NewIndices[(i * 3) + 2] = MapIterator->second[2];
    NewIndices[(i + 1) * 3] = MapIterator->second[3];
    NewIndices[((i + 1) * 3) + 1] = MapIterator->second[4];
    NewIndices[((i + 1) * 3) + 2] = MapIterator->second[5];

    if (i != 10) { MapIterator++; }
    
    }

std::copy(std::begin(NewIndices), std::end(NewIndices), std::begin(Indices));

glBindBuffer(GL_ELEMENT_ARRAY_BUFFER, EBO);
glBufferData(GL_ELEMENT_ARRAY_BUFFER, sizeof(Indices), Indices, GL_STATIC_DRAW);
}

字符串
我还没有做一些更复杂的相机,所以现在它只是在位置0,0,-3.0
我认为我做错了的地方是当涉及到计算相机和飞机之间的距离。即:

float Distance = glm::length(View - (ModelMatrix * Plane));


最初我只打算使用View - Plane,但我意识到这将为每个索引给予相同的值,因此我考虑如何将Model集成到距离中。我的理由是,既然着色器使用模型矩阵在正确的位置绘制,那么我应该能够使用它来找到模型空间中平面的正确位置。但我认为我错了。
所有这些都在draw函数中集合在一起(它绘制多个立方体):

...    
glm::mat4 model = glm::mat4(1.0f);
model = glm::translate(model, it->second * 0.9f * std::clamp((float) sinf(( float )SDL_GetTicks() / 1000.f) + 1.5f, 0.f, 3.f));
float angle = 20.0f * ((i + 1)/20) * ( float )SDL_GetTicks() / 200 * glm::radians(50.0f);
model = glm::rotate(model, glm::radians(angle), glm::vec3(1.0f, 0.3f, 0.5f));
    
OGLCMain->MainCube.SortIndicies(model);
            
int modelLoc = glGetUniformLocation(OGLCMain->ShaderProgram, "model");
glUniformMatrix4fv(modelLoc, 1, GL_FALSE, glm::value_ptr(model));
glDrawElements(GL_TRIANGLES, 36, GL_UNSIGNED_INT, 0);
...

8fsztsew

8fsztsew1#

所以我通过改变距离代码来使系统工作,该距离代码将使用相机前向量上的向量的点积来计算

Plane = ModelMatrix * Plane
float Distance = glm::dot((CameraPos - Plane), CameraFront);

字符串
并且通过将平面med的计算方式改变为:

glm::vec4 Plane = glm::vec4{
        (Vertices[(Indices[(i) * 3] * 5)] + Vertices[(Indices[((i) * 3) + 1] * 5)] + Vertices[(Indices[((i) * 3) + 2] * 5)] + Vertices[(Indices[((i + 1) * 3) + 1] * 5)]) / 4.f,
        (Vertices[(Indices[(i) * 3] * 5) + 1] + Vertices[(Indices[((i) * 3) + 1] * 5) + 1] + Vertices[(Indices[((i) * 3) + 2] * 5) + 1] + Vertices[(Indices[((i + 1) * 3) + 1] * 5) + 1]) / 4.f,
        (Vertices[(Indices[(i) * 3] * 5) + 2] + Vertices[(Indices[((i) * 3) + 1] * 5) + 2] + Vertices[(Indices[((i) * 3) + 2] * 5) + 2] + Vertices[(Indices[((i + 1) * 3) + 1] * 5) + 2]) / 4.f,
        1.f };


我还颠倒了它放回去的顺序:

auto MapIterator = SortedMap.begin();
for (int i = 11; i > 0; i -= 2)
    {

    NewIndices[i * 3] = MapIterator->second[0];
    NewIndices[(i * 3) + 1] = MapIterator->second[1];
    NewIndices[(i * 3) + 2] = MapIterator->second[2];
    NewIndices[(i - 1) * 3] = MapIterator->second[3];
    NewIndices[((i - 1) * 3) + 1] = MapIterator->second[4];
    NewIndices[((i - 1) * 3) + 2] = MapIterator->second[5];

    MapIterator++; 
    
    }


并将距离计算改为:

float Distance = -glm::dot((CameraPos - Plane), CameraFront);

**但是!**这应该没关系,因为这两个都颠倒了顺序,因此净变化为0。

为了清楚起见,我在这个项目中使用的顶点和索引数组看起来像这样:

GLfloat TempVertices[] = {
        -0.5f, -0.5f, -0.5f,  0.0f, 0.0f,  // A 0
        0.5f, -0.5f, -0.5f,  1.0f, 0.0f,  // B 1
        0.5f,  0.5f, -0.5f,  1.0f, 1.0f,  // C 2
        -0.5f,  0.5f, -0.5f,  0.0f, 1.0f,  // D 3
        -0.5f, -0.5f,  0.5f,  0.0f, 0.0f,  // E 4
        0.5f, -0.5f,  0.5f,  1.0f, 0.0f,   // F 5
        0.5f,  0.5f,  0.5f,  1.0f, 1.0f,   // G 6
        -0.5f,  0.5f,  0.5f,  0.0f, 1.0f,   // H 7

        -0.5f,  0.5f, -0.5f,  0.0f, 0.0f,  // D 8
        -0.5f, -0.5f, -0.5f,  1.0f, 0.0f,  // A 9
        -0.5f, -0.5f,  0.5f,  1.0f, 1.0f,  // E 10
        -0.5f,  0.5f,  0.5f,  0.0f, 1.0f,  // H 11
        0.5f, -0.5f, -0.5f,  0.0f, 0.0f,   // B 12
        0.5f,  0.5f, -0.5f,  1.0f, 0.0f,   // C 13
        0.5f,  0.5f,  0.5f,  1.0f, 1.0f,   // G 14
        0.5f, -0.5f,  0.5f,  0.0f, 1.0f,   // F 15

        -0.5f, -0.5f, -0.5f,  0.0f, 0.0f,  // A 16
        0.5f, -0.5f, -0.5f,  1.0f, 0.0f,   // B 17
        0.5f, -0.5f,  0.5f,  1.0f, 1.0f,   // F 18
        -0.5f, -0.5f,  0.5f,  0.0f, 1.0f,  // E 19
        0.5f,  0.5f, -0.5f,   0.0f, 0.0f,  // C 20
        -0.5f,  0.5f, -0.5f,  1.0f, 0.0f,  // D 21
        -0.5f,  0.5f,  0.5f,  1.0f, 1.0f,  // H 22
        0.5f,  0.5f,  0.5f,   0.0f, 1.0f,  // G 23
        };
    
    unsigned int TempIndices[] = {
        // front and back
        0, 3, 2,
        2, 1, 0,
        4, 5, 6,
        6, 7 ,4,
        // left and right
        11, 8, 9,
        9, 10, 11,
        12, 13, 14,
        14, 15, 12,
        // bottom and top
        16, 17, 18,
        18, 19, 16,
        20, 21, 22,
        22, 23, 20
        };

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