加速度计加速度的双重积分来查找位置？
我试图使用在单轴上来回移动的加速度计的输出来计算其当前位置。
我尝试过使用欧拉积分，但速度和位置误差变得太大太快。读完之后，我想知道是否可以应用 RK4 来解决这个问题以最小化错误？
I am attempting to use output from an accelerometer moving back and forth on a single axis to calculate its current position.
I have tried using Euler integrations, but the velocity and position errors become too large too quickly. After some reading, I am wondering whether RK4 could be applied to this problem to minimise the error?
如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱，如果其他用户或者作者回复了您的评论，将不能在第一时间通知您！
发布评论
评论（2）
加速度计通常存在一些漂移。您需要测量实际位置以消除静态误差。
There is usually some drift in accelerometers. You need a measurement of the actual position to remove static errors.
RK4 可能会有所帮助。
您是否也在测量位置，这就是您对累积误差的评估的来源？
也可能是由于时间增量太大。也许您已经超出了系统的稳定性限制。更精细的时间步长和隐式积分方法可能有助于解决这些错误。
RK4 might help.
Are you also measuring position, and that's where your assessment of accumulated error comes from?
It might also be due to time increments being too large. Perhaps you've exceeded a stability limit for your system. A finer time step and an implicit integration method might help with those errors.