如何解決 OpenGL/GLSL 上的觀察矩陣問題

Question

我有以下代碼用於我自己的觀察矩陣（矩陣的乘法和向量的叉積工作完美，我檢查了它）：

template<typename Type>
void setLookAt(Matrix4x4<Type>& matrix, const Vector3<Type> eye, const Vector3<Type> center, const Vector3<Type> up) noexcept
{
    Math::Vector3f right = Math::cross(center, up).normalize();
    Matrix4x4f lookAt({
        right.getX(), right.getY(), right.getZ(), 0.0,
        up.getX(), up.getY(), up.getZ(), 0.0,
        center.getX(), center.getY(), center.getZ(), 0.0,
        0.0, 0.0, 0.0, 1.0
    });
    Matrix4x4f additionalMatrix({
        0.0, 0.0, 0.0, -(eye.getX()),
        0.0, 0.0, 0.0, -(eye.getY()),
        0.0, 0.0, 0.0, -(eye.getZ()),
        0.0, 0.0, 0.0, 1.0
    });
    lookAt.mul(additionalMatrix);
    matrix = lookAt;
}

template<typename Type>
void setPerspectiveMatrix(Matrix4x4<Type>& matrix, Type fov, Type aspect, Type znear, Type zfar) noexcept
{
    const Type yScale = static_cast<Type>(1.0 / tan(RADIANS_PER_DEGREE * fov / 2));
    const Type xScale = yScale / aspect;
    const Type difference = znear - zfar;
    matrix  = {
        xScale, 0, 0, 0,
        0, yScale, 0, 0,
        0, 0, (zfar + znear) / difference, 2 * zfar * znear / difference,
        0, 0, -1, 0
    };
}

矩陣乘法實現：

// static const std::uint8_t ROW_SIZE = 4;
// static const std::uint8_t MATRIX_SIZE = ROW_SIZE * ROW_SIZE;
// static const std::uint8_t FIRST_ROW = 0;
// static const std::uint8_t SECOND_ROW = ROW_SIZE;
// static const std::uint8_t THIRD_ROW = ROW_SIZE + ROW_SIZE;
// static const std::uint8_t FOURTH_ROW = ROW_SIZE + ROW_SIZE + ROW_SIZE;

template<class Type>
void Matrix4x4<Type>::mul(const Matrix4x4& anotherMatrix) noexcept
{
    Type currentElements[MATRIX_SIZE];
    std::copy(std::begin(mElements), std::end(mElements), currentElements);
    const Type* otherElements = anotherMatrix.mElements;
    for (std::uint8_t i = 0; i < MATRIX_SIZE; i += ROW_SIZE) 
    {
        mElements[i] = currentElements[i] * otherElements[FIRST_ROW] +
            currentElements[i + 1] * otherElements[SECOND_ROW] +
            currentElements[i + 2] * otherElements[THIRD_ROW] +
            currentElements[i + 3] * otherElements[FOURTH_ROW];
        mElements[i + 1] = currentElements[i] * otherElements[FIRST_ROW + 1] +
            currentElements[i + 1] * otherElements[SECOND_ROW + 1] +
            currentElements[i + 2] * otherElements[THIRD_ROW + 1] +
            currentElements[i + 3] * otherElements[FOURTH_ROW + 1];
        mElements[i + 2] = currentElements[i] * otherElements[FIRST_ROW + 2] +
            currentElements[i + 1] * otherElements[SECOND_ROW + 2] +
            currentElements[i + 2] * otherElements[THIRD_ROW + 2] +
            currentElements[i + 3] * otherElements[FOURTH_ROW + 2];
        mElements[i + 3] = currentElements[i] * otherElements[FIRST_ROW + 3] +
            currentElements[i + 1] * otherElements[SECOND_ROW + 3] +
            currentElements[i + 2] * otherElements[THIRD_ROW + 3] +
            currentElements[i + 3] * otherElements[FOURTH_ROW + 3];
    }
}

跨產品實施：

template<typename Type>
Math::Vector3<Type> cross(Vector3<Type> vector, Vector3<Type> anotherVector) noexcept
{
    const Type x = vector.getY()*anotherVector.getZ() - vector.getZ()*anotherVector.getY();
    const Type y = -(vector.getX()*anotherVector.getZ() - vector.getZ()*anotherVector.getX());
    const Type z = vector.getX()*anotherVector.getY() - vector.getY()*anotherVector.getX();
    return { x, y, z };
}

使用它：

// OpenGL

glUseProgram(mProgramID);
Matrix4x4f lookAt;
setLookAt(lookAt, { 0.0f, 0.0f, 3.0f }, { 0.0f, 0.0f, -1.0f }, { 0.0f, 1.0f, 0.0f });
glUniformMatrix4fv(glGetAttribLocation(mProgramID, "viewMatrix"), 1, GL_TRUE, lookAt);
Matrix4x4f projection;
setPerspectiveMatrix(projection, 45.0f, width / height, -0.1, 100.0f);
glUniformMatrix4fv(glGetAttribLocation(mProgramID, "projectionMatrix "), 1, GL_TRUE, projection);

// GLSL

layout (location = 0) in vec3 position;

uniform mat4 viewMatrix;
uniform mat4 projectionMatrix;

void main()
{
    gl_Position = projectionMatrix * viewMatrix * vec4(position, 1.0f);
}

使用此代碼后，我得到一個空白屏幕，盡管我必須繪制一個立方體。 問題出在矩陣本身，所以其他矩陣可以正常工作（偏移、旋轉等），但我可以確切地理解在哪里。 你能告訴我可能是什么問題嗎？

Answer 1

"projectionMatrix"和"viewMatrix"是統一變量。 統一位置可以通過glGetUniformLocation而不是glGetAttribLocation ，這將返回活動屬性的屬性索引：

GLint projLoc = glGetUniformLocation( mProgramID, "projectionMatrix" );
GLint viewLoc = glGetUniformLocation( mProgramID, "viewMatrix" );

在透視投影中，投影矩陣描述了從針孔相機看到的世界中的 3D 點到視口的 2D 點的映射。
相機視錐體（截棱錐）中的眼睛空間坐標被映射到立方體（標准化設備坐標）。
在透視投影中，視圖空間（體積）由一個截錐體（截棱錐）定義，其中棱錐的頂部是觀察者的位置。 視線方向（視線）以及遠近距離定義了將金字塔截斷為截頭錐體的平面（視線方向是該平面的法向矢量）。
這意味着到近平面的距離和到遠平面的距離這兩個值都必須是正值：

Matrix4x4f lookAt;
setLookAt(lookAt, { 0.0f, 0.0f, 3.0f }, { 0.0f, 0.0f, -1.0f }, { 0.0f, 1.0f, 0.0f });
glUniformMatrix4fv(viewLoc, 1, GL_TRUE, lookAt);

Matrix4x4f projection;
setPerspectiveMatrix(projection, 45.0f, width / height, 0.1f, 100.0f); // 0.1f instead of -0.1f
glUniformMatrix4fv(projLoc, 1, GL_TRUE, projection);

視圖空間是局部系統，由場景中的視點定義。 視圖的位置、視線和視圖的向上方向，定義了一個相對於世界坐標系的坐標系。
視圖矩陣必須從世界空間轉換到視圖空間，因此視圖矩陣是視圖坐標系的逆矩陣。
如果視圖空間的坐標系是右手坐標系，其中 X 軸指向左側，Y 軸指向上方，則 Z 軸指向視圖外（注意在右手坐標系中Z 軸是 X 軸和 Y 軸的叉積）。

z 軸視線是從視點eye到目標center的向量：

template<typename Type>
void setLookAt(Matrix4x4<Type>& matrix, const Vector3<Type> eye, const Vector3<Type> center, const Vector3<Type> up) noexcept
{
    Vector3f mz( { eye.getX()-center.getX(), eye.getY()-center.getY(), eye.getZ()-center.getZ() } );
    mz = mz.normalize();
    Vector3f my = up.normalize();
    Vector3f mx = cross(my, mz).normalize();

    Type tx = dot( mx, eye );
    Type ty = dot( my, eye );
    Type tz = -dot( mz, eye );

    matrix = {
        mx.getX(), mx.getY(), mx.getZ(), tx,
        my.getX(), my.getY(), my.getZ(), ty,
        mz.getX(), mz.getY(), mz.getZ(), tz,
        0.0,       0.0,       0.0,       1.0
    };
}

template<typename Type>
Vector3<Type> cross(Vector3<Type> vector, Vector3<Type> anotherVector) noexcept
{
    const Type x = vector.getY()*anotherVector.getZ() - vector.getZ()*anotherVector.getY();
    const Type y = -(vector.getX()*anotherVector.getZ() - vector.getZ()*anotherVector.getX());
    const Type z = vector.getX()*anotherVector.getY() - vector.getY()*anotherVector.getX();
    return { x, y, z };
}

template<typename Type>
Vector3<Type> Vector3<Type>::normalize(void) const
{ 
  Type len = std::sqrt(mV[0]*mV[0] + mV[1]*mV[1] + mV[2]*mV[2]);
  return { mV[0] / len, mV[1] / len, mV[2] / len }; 
}

template<typename Type>
Type dot(Vector3<Type> vector, Vector3<Type> anotherVector) noexcept
{
    Type ax = vector.getX(), ay = vector.getY(), az = vector.getZ();
    Type bx = anotherVector.getX(), by = anotherVector.getY(), bz = anotherVector.getZ();
    return ax*bx + ay*by + az*bz;
}

透視投影矩陣可以由一個截錐體定義。
距離left 、 right 、 bottom和top是在近平面上從視圖中心到截錐體側面的距離。 near和far指定到截錐體上近平面和遠平面的距離。

r = right, l = left, b = bottom, t = top, n = near, f = far

x              y              z                t
2*n/(r-l)      0              (r+l)/(r-l)      0
0              2*n/(t-b)      (t+b)/(t-b)      0
0              0             -(f+n)/(f-n)     -2*f*n/(f-n)
0              0             -1                0

如果投影是對稱的，其中視線是視錐體的對稱軸，則矩陣可以簡化：

x              y              z                t
1/(ta*a)       0              0                0
0              1/ta           0                0
0              0             -(f+n)/(f-n)     -2*f*n/(f-n)
0              0             -1                0

在哪里：

a = w / h
ta = tan( fov_y / 2 );

2 * n / (r-l) = 1 / (ta * a)
2 * n / (t-b) = 1 / ta

此外，投影矩陣從右手系切換到左手系，因為 z 軸旋轉。

template<typename Type>
void setPerspectiveMatrix(Matrix4x4<Type>& matrix, Type fov, Type aspect, Type znear, Type zfar) noexcept
{
    const Type yScale = static_cast<Type>(1.0 / tan(RADIANS_PER_DEGREE * fov / 2));
    const Type xScale = yScale / aspect;
    const Type difference = zfar - znear;
    matrix  = {
        xScale, 0,       0,                            0,
        0,      yScale,  0,                            0,
        0,      0,      -(zfar + znear) / difference, -2 * zfar * znear / difference,
        0,      0,      -1,                            0
    };
}