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Accelerometer force or the square please

  1. Jan 28, 2008 #1

    IMK

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    Hello,
    I am using a tri axial accelerometer for a project that I have just about completed but have just realized that some of math may be flawed with respect to the device output and actual acceleration forces. So the question; as I am using vector subtraction and vector dot products to manipulate my results should I be using the output directly (which I believe is the sine) from the accelerometer or the square of the output which is the actual g force on that vector.... Or have I missed something?

    Many thanks IMK
     
  2. jcsd
  3. Jan 29, 2008 #2
    I don't quite follow your post and I might be able to help if I understand more. Normally the output of an accelerometer will be so many mV/g of acceleration, not force. And, that output will be linear, not sinusoidal.

    Are you doing some processing in the instrument amp that gives the signal you describe? Or, do you have a special type of accelerometer? Or, more likely, am I just not following your explanation?
     
  4. Jan 29, 2008 #3

    IMK

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    TVP45 Many thanks for your post as I am more than just a little confused on this issue.
    Yes the output of the device is mv/g thus 1v = 1g and 0v = 0g.

    http://www.analog.com/Analog_Root/static/library/techArticles/mems/Sensor971/sld27.html

    But it is the measurements in-between that I am a confused about, for example why is 45 degrees not 0.5 g, the link below almost makes it clear but still leaves me wondering if I should be using the actual accelerometer output or the square of the output to get the correct values thus: 45 degrees = 0.707 and 0.707^2 = 0.5.

    The math (vector dot and vector cross products) in my current implementation just uses the accelerometer output values, but I am just wondering if I should square them to get the linear value before I do the vector math.

    The link below makes seems to make some sense but I still just can’t see it:

    https://www.physicsforums.com/archive/index.php/t-177804.html

    Many thanks in advance IMK
    PS really understanding this would be a major breakthrough for me as I seem to have really missed something in my understanding of force vectors.
     
  5. Jan 29, 2008 #4
    Now I see (or so I think). Accelerometers are made to give you vector values (at least the good ones). So, if you imagine looking down on it and forward is +y and to the right is +x, and your acceleration a is at 30 degrees from the x axis between the two, you will read 0.50 a on the y channel and 0.866 a on the x channel. In other words, x will be a cos (theta) - I gotta learn LaTeX - and y will be a sin (theta) where theta is the standard angle. So, to get the actual acceleration, take the sqr of the sum of the squares
    ((0.5a)^2 + (0.866a)^2)^0.5 = a
    And if you have only the values of x and y, theta = arctan (y/x)

    And, of course, you can extend all this to 3 components (x,y,z). Depending on the software you're using, you may want to just leave them in component form (i.e., the direct reading), but that's up to you.

    For all this to work, you have to check that there is no significant cross-coupling between the axes. The accelerometer manufacturer should provide that information. if not, you have to do sensitivity tests with a known acceleration.
     
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