使用四元数消除向量 [英] De-Rotating vectors using Quaternions

问题描述

我在x，y，z方向(传感器主体框架)上具有一组加速度值.在操作过程中，传感器多次倾斜和改变方向.使用陀螺仪，磁力计和加速度计，我能够以四元数格式融合并提取侧倾角，俯仰角和偏航角(校准为Q).

I have a set of acceleration values along the x, y, z directions (sensor body frame). The sensor tilted and changed orientation many times during operation. Using the gyroscope, magnetometer and accelerometer, I was able to fuse and extract the Roll, Pitch and Yaw angles all in quaternion format ( cal it Q).

现在，将加速度值旋转到全局X，Y，Z位置.我只是简单地1.通过将标量部分设置为0(称为Acc)将加速度值转换为四元数2.将加速四元数乘以以下公式:Qtranspose * Acc * Q ??

Now, to derotate the acceleration values into global X, Y, Z positions. Do I just simply 1. convert the acceleration values into quaternion by setting the scalar part to 0 (call it Acc) 2. Multiply the acceleration quaternion as follows : Qtranspose * Acc * Q??

还是我的结论不对?我们如何旋转值?

Or is my conclusion wrong? how do we derotate the values?

谢谢

推荐答案

如果您已经具有正确的四元数方向 q ，那么加速度的未旋转"矢量就是您所说的:

If you already have an accurate quaternion orientation q, then the 'unrotated' vector of the accelerations is as you said:

Ag = q* A q

或

Ag = q A q*

取决于四元数乘法所使用的约定.(只需尝试两者，看看哪一个能给您正确的重力矢量).

depending on which convention you are using for quaternion multiplication. (just try both and see which one give you the right gravity vector).

请注意，A和Ag是纯四元数"-这是一个四元数，其标量分量为0，轴分量为加速度矢量.换句话说，如果您的四元数为q = [w x y z]，则A = [0 ax ay az].然后，您可以舍弃Ag的标量分量，以获得未旋转的矢量.确保加速度矢量的顺序与四元数矢量的顺序一致.

NOTE that A and Ag are 'pure quaternions' - this is a quaternion with the scalar component as 0, and the axis component as your acceleration vector. In other words if your quaternion is q = [w x y z], then A = [0 ax ay az]. You can then discard the scalar component of Ag to get the unrotated vector. Make sure the order of the acceleration vector is consistent with your quaternion vector order.

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