Question

基于MATLAB link的相机校准的外在矩阵应该是4x3矩阵（包括方向和平移）意味着我们需要12个元素，但是，根据Tango documentation中的解释，我们只得到3个数字用于翻译和4个用于轮换的数字。如何用这7个数字创建4x3矩阵？

谢谢，瓦希德。

Answer 1

View Matrix用于保存这些值。它是一个4x4矩阵，允许操纵3D姿势（位置+方向）。

您可以在此处找到有关此矩阵的更多详细信息： http://www.3dgep.com/understanding-the-view-matrix/

请注意，Tango java库基于Rajawali 3D库。您可以在此处查看MatrixX44的结构：

https://github.com/Rajawali/Rajawali/blob/master/rajawali/src/main/java/org/rajawali3d/math/Matrix4.java

特别是，以下方法显示了如何存储7个值。为了便于阅读，您可以假设比例矢量为（1,1,1）

public Matrix4 setAll(@NonNull Vector3 position, @NonNull Vector3 scale, @NonNull Quaternion rotation) {
    // Precompute these factors for speed
    final double x2 = rotation.x * rotation.x;
    final double y2 = rotation.y * rotation.y;
    final double z2 = rotation.z * rotation.z;
    final double xy = rotation.x * rotation.y;
    final double xz = rotation.x * rotation.z;
    final double yz = rotation.y * rotation.z;
    final double wx = rotation.w * rotation.x;
    final double wy = rotation.w * rotation.y;
    final double wz = rotation.w * rotation.z;

    // Column 0
    m[M00] = scale.x * (1.0 - 2.0 * (y2 + z2));
    m[M10] = 2.0 * scale.y * (xy - wz);
    m[M20] = 2.0 * scale.z * (xz + wy);
    m[M30] = 0;

    // Column 1
    m[M01] = 2.0 * scale.x * (xy + wz);
    m[M11] = scale.y * (1.0 - 2.0 * (x2 + z2));
    m[M21] = 2.0 * scale.z * (yz - wx);
    m[M31] = 0;

    // Column 2
    m[M02] = 2.0 * scale.x * (xz - wy);
    m[M12] = 2.0 * scale.y * (yz + wx);
    m[M22] = scale.z * (1.0 - 2.0 * (x2 + y2));
    m[M32] = 0;

    // Column 3
    m[M03] = position.x;
    m[M13] = position.y;
    m[M23] = position.z;
    m[M33] = 1.0;
    return this;
}

探戈相机外在

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