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Removal of 1 g from accelerometer

  1. Jun 28, 2010 #1

    I was wondering if it is possible to remove the 1g (sqrt(x^2+y^2+z^2)) from an accelerometer when in static positions.

    When the device is tilted to different positions the total g-force does not equal 1 and differs quite a bit from position to position.

    I would like to be able to have the device at a particular angle which gives a reading of 0,0,0 as the device is static (angle not important) and when linear acceleration is experienced thay it just shows on the relevant axis.

    Is there a formula that can achieve this?

  2. jcsd
  3. Jun 28, 2010 #2
    It can likely be "zeroed", or calibrated in a way that you redefine the force of gravity to be zero g's. But that's not really what you're asking about, and I hate to disappoint, but, as Einstein would tell you, rest in the presence of a gravitational field and linear acceleration are indistinguishable. There is no way to teach an accelerometer to learn the difference between sitting on a table and accelerating through space in a rocket ship at 9.8 m/s^2. However, robotics engineers have done some clever things by adding gyroscopes and other components to their setup.
  4. Jun 28, 2010 #3
    I'm not really to worried about rest and linear acceleration as i have a method that determines the position. What I really want is to have a reading of 0,0,0 at rest and then if it accelerates lets say for example in the z-plane that the reading might look something like 0,0,0.2 or 0,0,-0.7. Can this be done.

  5. Jun 28, 2010 #4


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    If you have orientation of the accelerometer, yes. What you need to know is which way the true vertical points relative to the X, Y, and Z of accelerometer.
  6. Jun 28, 2010 #5
    with the values like:

    0.6, 0.6, 0.53.

    The magnitude is 1 so therefore could just subtract each element from itself to get:

    0, 0, 0.

    but if there acceleration in z direction which gives something like:

    0.6, 0.6, 0.6, or 0.6, 0.6, 0.8. and to get a normalized value:


    How can I solve for z?
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