如何在C++中将球面几何转换为椭球几何并生成顶点

我有一段生成球面顶点的C++代码，但需要修改为生成满足椭球方程 ((x²)/(a²)) + ((y²)/(b²)) + ((z²)/(c²)) = 1 的顶点，其中a、b、c是椭球三个轴的长度。现有球面生成代码如下：

Sphere::Sphere(const LengthT radius, const LengthT a, const LengthT b, const LengthT c, std::uint16_t sectorCount, std::uint16_t stackCount) :
    m_radius(radius),
    m_sectorCount(sectorCount),
    m_stackCount(stackCount) {
    const double sectorStep{2 * std::numbers::pi / m_sectorCount};
    const double stackStep{std::numbers::pi / m_stackCount};
    constexpr double half_pi{std::numbers::pi / 2};
    double xy, sectorAngle, stackAngle;
    const auto lengthInv{double(1) / m_radius};
    for(std::uint16_t i{0}; i <= stackCount; i++) {
        Vertex vertex;
        stackAngle = (half_pi) - (i * stackStep);
        xy = m_radius * cos(stackAngle);
        vertex.pos.z = m_radius * sin(stackAngle);
        for(std::uint16_t j{0}; j <= sectorCount; j++) {
            sectorAngle = j * sectorStep;
            vertex.pos.x = xy * cos(sectorAngle);
            vertex.pos.y = xy * sin(sectorAngle);
            vertex.normals.x = vertex.pos.x * lengthInv;
            vertex.normals.y = vertex.pos.y * lengthInv;
            vertex.normals.z = vertex.pos.z * lengthInv;
            vertex.texCords.x = double(j) / sectorCount;
            vertex.texCords.y = double(i) / stackCount;
            verticies.push_back(vertex);
        }
    }
}

核心思路是将球面坐标按椭球三个轴的长度进行缩放，同时修正法线计算（椭球法线与球面法线逻辑不同，需基于椭球方程推导）。修改后的代码如下：

Sphere::Sphere(const LengthT radius, const LengthT a, const LengthT b, const LengthT c, std::uint16_t sectorCount, std::uint16_t stackCount) :
    m_radius(radius),
    m_sectorCount(sectorCount),
    m_stackCount(stackCount) {
    const double sectorStep{2 * std::numbers::pi / m_sectorCount};
    const double stackStep{std::numbers::pi / m_stackCount};
    constexpr double half_pi{std::numbers::pi / 2};
    double sectorAngle, stackAngle;
    // 预计算轴长的倒数平方，用于法线计算
    const double invA2 = 1.0 / (static_cast<double>(a) * static_cast<double>(a));
    const double invB2 = 1.0 / (static_cast<double>(b) * static_cast<double>(b));
    const double invC2 = 1.0 / (static_cast<double>(c) * static_cast<double>(c));

    for(std::uint16_t i{0}; i <= stackCount; i++) {
        Vertex vertex;
        stackAngle = half_pi - (i * stackStep);
        // 先计算单位球面的坐标分量
        const double sphereXy = cos(stackAngle);
        const double sphereZ = sin(stackAngle);

        for(std::uint16_t j{0}; j <= sectorCount; j++) {
            sectorAngle = j * sectorStep;
            // 缩放球面坐标到椭球尺寸（若需保留radius整体缩放，可乘以m_radius）
            vertex.pos.x = static_cast<double>(a) * sphereXy * cos(sectorAngle);
            vertex.pos.y = static_cast<double>(b) * sphereXy * sin(sectorAngle);
            vertex.pos.z = static_cast<double>(c) * sphereZ;

            // 计算椭球法线：基于椭球方程的梯度(x/a², y/b², z/c²)，归一化后得到单位法线
            glm::vec3 normal{
                vertex.pos.x * invA2,
                vertex.pos.y * invB2,
                vertex.pos.z * invC2
            };
            normal = glm::normalize(normal);
            vertex.normals.x = normal.x;
            vertex.normals.y = normal.y;
            vertex.normals.z = normal.z;

            vertex.texCords.x = static_cast<double>(j) / sectorCount;
            vertex.texCords.y = static_cast<double>(i) / stackCount;
            verticies.push_back(vertex);
        }
    }
}

关键改动说明 #

• 顶点坐标生成：不再用原radius统一缩放，而是将单位球面坐标分别乘以a、b、c三个轴长，直接生成满足椭球方程的顶点。若需要保留radius作为整体缩放因子，可在每个坐标计算时额外乘以m_radius。
• 法线修正：椭球的法线不能复用球面的归一化坐标，正确逻辑是取椭球方程的梯度方向(x/a², y/b², z/c²)，归一化后得到单位法线向量。
• 简化冗余逻辑：移除原代码中lengthInv、xy等冗余变量，改用更直观的单位球面分量，提升代码可读性。

更新：此方案已开发并测试通过，已被选为有效答案。

内容的提问来源于stack exchange，提问作者Peter F