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Visitor II
May 20, 2021
Question

What is the vibration range in velocity mm/s?

  • May 20, 2021
  • 1 reply
  • 2008 views

If the vibration range in acceleration of iis3dwb is 39m/s^2 (4g), what is its vibration range in velocity mm/s? Please advise. Thanks!

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    1 reply

    ST Employee
    May 21, 2021

    Hi @JTam.1​ ,

    You can use the standard harmonic oscillator equations to derive the speed value from the acceleration.

    Stated that, at the peak, the followings hold:

    • acc = w^2 * A_peak
    • speed = w * A_peak

    you get:

    • speed = acc / w

    and w is the , equal to 2 * pi * freq, where freq is the frequency of interest, for example 1kHz.

    The calculation leads to:

    • speed = 39m/s^2 / (2*pi*1000Hz) = 0.006 m/s = 6 mm/s

    These value are close to the ones declared in the UM2566 user manual, that you also reported in this post.

    If my reply answered your question, please click on Select as Best at the bottom of this post. This will help other users with the same issue to find the answer faster.

    -Eleon

    JTam.1Author
    Visitor II
    July 20, 2021

    Hi Elleon,

    You said the speed range can attain 6 mm/s. However, I found that the IIS3DWB sensor can measure speed of 16.45mm/s RMS or higher. Why? Please advise. Thanks!

    ST Employee
    July 20, 2021

    Hi @JTam.1​ ,

    my example was related specifically to a 4g (rms) - 1kHz stimulus.

    In order to achieve higher speeds, considering the formula speed = acc / (2*pi*freq), you can act on two parameters:

    • Increasing the acceleration, and specifically the Full scale (that is selectable up to +-16g = 157m/s^2 on IIS3DWB sensor) --> 157m/s^2 / (2*pi*1000Hz) = 0.025 m/s = 25 mm/s RMS
    • Decreasing the frequency, keeping constant the Full scale --> 39m/s^2 / (2*pi*200Hz) = 0.031 m/s = 31 mm/s RMS
    • A mix of the two above.

    Hope this can clarify a bit more.

    -Eleon