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bonus gimbal lock中,旋转顺序定为x-y-z,按照坐标系旋转的图示,这里似乎是所谓的「内旋」,即第n次旋转时的旋转轴 是 第n-1次旋转后的坐标系中的某个轴。与此相对的是「外旋」,即按外部固定旋转轴旋转。就我个人对线性代数的理解,内旋的变换矩阵复合应该是右乘,即:如果按照先x后y最后z的旋转顺序,复合的变换矩阵应该是RxRyRz,与此相对的外旋变换矩阵则是RzRyRx。
不知这部分是否有些许问题,期待讨论,多谢!
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我也这样认为,内旋按描述的顺序应是 Rx(a)Ry(pi/2)Rz(b)=Rx(a+b)Ry(pi/2)。如图示一样,z轴的自由度没了。
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bonus gimbal lock中,旋转顺序定为x-y-z,按照坐标系旋转的图示,这里似乎是所谓的「内旋」,即第n次旋转时的旋转轴 是 第n-1次旋转后的坐标系中的某个轴。与此相对的是「外旋」,即按外部固定旋转轴旋转。就我个人对线性代数的理解,内旋的变换矩阵复合应该是右乘,即:如果按照先x后y最后z的旋转顺序,复合的变换矩阵应该是RxRyRz,与此相对的外旋变换矩阵则是RzRyRx。
不知这部分是否有些许问题,期待讨论,多谢!
The text was updated successfully, but these errors were encountered: