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Homework Help: Understanding use of accelerometers

  1. Mar 19, 2015 #1
    1. The problem statement, all variables and given/known data
    Imagine this: A rod is attached at one end to a rotating shaft. So, as the shaft rotates from 0 to 90 degrees back-and-forth (in the xy plane), so does the rod. My goal is to measure the instantaneous acceleration of the rod while its moving. I have a few 3-axis accelerometers, but I feel like I'm really confused. I guess I'm not really understanding how accelerometers can work best in this scenario because it's not really a "linear" acceleration. I think its a rotational or angular acceleration (are those the same?). Also, even though the rod is only moving in the xy plane, since it rotating, doesn't that mean that it's rotating about the z-axis? Although I'm assuming I only need data in the x and y dimensions.

    I guess here are my more specific questions: Should I use more than one accelerometer here? If so, where would be the best locations to mount them? How should I mathematically combine these data to give me an overall acceleration (perhaps just sum the squares, and then take the square root)? And lastly, let's say the rod ramps up from 0 degrees, stays at a constant velocity, and then ramps down. How should I expect the data to look like in this simple case?

    I'm hoping someone can shed some light on the use of accelerometers in this particular scenario. It got to the point where doing more research online has only confused me more, so I'd really appreciate any help I can get here.

    2. Relevant equations
    If I need two dimensions, then to find the total "rotational" acceleration, I could do sqrt(ax^2 +ay^2) maybe?

    3. The attempt at a solution
    I've used one accelerometer, but cant make sense of the data.
  2. jcsd
  3. Mar 19, 2015 #2


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    A rotation around the z axis happens in the xy plane, and without acceleration in z-direction.

    If you have a purely rotational motion, one accelerometer at the end of the rod can be sufficient - either measuring centripetal force (to be sensitive to velocity) or the force along the direction of motion (to be sensitive to acceleration). With two, you can monitor both at the same time and perform cross-checks.

    What exactly do you want to measure?
  4. Mar 19, 2015 #3


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    Hello MG,

    I can find only one dimension in your description: the angle. Am I correct in assuming the shaft doesn't go up and down and only rotates around its axis ? Any point on the rod then only experiences acceleration in the x-y plane. In polar coordinates that can be expressed as (equation (4) in this link ) (*) : $$
    \vec a = (\ddot r− r\dot \theta^2) {\bf \hat r} + (r\ddot\theta + 2\dot r \dot \theta) {\bf \hat θ}
    $$If the accelerometer is fixed to the rod, r is a constant and hence ##\dot r## and ##\ddot r## are both zero, and the device reads $$
    \vec a = − r\dot \theta^2 \; {\bf \hat r}$$ in the radial direction (which we generally call the centripetal acceleration), and $$
    r\ddot\theta \;{\bf \hat θ}$$in the tangential direction. So after dividing by the fixed r, one of the two gives you ##\;\dot\theta^2 =\vec \omega^2\;## and the other ##\;\ddot\theta = \alpha\;##.

    I am quite curious to see if you can deduct these form your preliminary results !

    Yes. The usual symbol is ##\vec \alpha##.


    As the above should make clear, one (2-axis) accelerometer should be enough. Summing squares isn't such a good idea, I think/ And you can answer the other questions yourself now, I hope (thus being able to show your 'attempt at solution' :smile: )

    (*) Nicely presented derivation here
    Last edited: Mar 19, 2015
  5. Mar 19, 2015 #4
    Ultimately, I'm interested in the measuring the instantaneous torque of the rod (the shaft is that of a motor, so the motor is moving everything). So basically, I want to measure the torque the motor is producing using acceleration.
  6. Mar 19, 2015 #5
    1) you're correct about the bolded. The shaft does not move up and down. It indeed only rotates about its axis.
    2) With your derivation, are you trying to show me that the acceleration should be the second derivative of the position. I think that makes sense...
    3) Thanks for clearing up the notation and z-axis thing. That eases my confusion.
    4) Can you expand on why taking the square root of the sum of squares is a bad thing? You say it's one dimension problem (which I think makes sense now), but what if I pick up data on another axis due to drift etc... ?
    5) If I get my system working and acquire data, can I post the data here for interpretation?

    Thanks for your help either way.
  7. Mar 19, 2015 #6


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    I said I don't think it's a good idea, because it gives you ##r\sqrt{\omega^4+\alpha^2}## and what good is that ? It probably is the acceleration in the x-y plane but not particularly useful.

    The tangential component ##r\ddot\theta=r\alpha\;##, however, is a lot more interesting, since the torque you mention is ##\tau = I\alpha## (with I the moment of inertia).

    Could you now try to answer your own question
    to reassure us you understand what to expect ? In exchange I'll gladly look at the results :smile:
  8. Mar 20, 2015 #7
    OK thanks for your comment (also sorry for the late reply). If it's ramping up, then then I think at that point the acceleration should be linearly increasing, and then when it's ramping down, it should linearly decrease. If its moving constantly, that means that velocity is a constant number, and the acceleration is the derivative of velocity, so here the acceleration will be zero. Is this right?

    I'm gonna go ahead and setup my system to collect some data now. I'll get back to take you up on your offer :-)
  9. Mar 20, 2015 #8


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    That's a possibility. The simplest case would be applying a constant torque - so the angular velocity ##\omega## would increase linearly. Then no torque -- hence constant ##\omega## and finally putting on a constant braking torque -- hence linearly decreasing ##\omega##. In theory.

    In practice you probably want to drive something and there will be some friction to overcome. And of course your motor has some torque vs rotational speed curve - which is what you want to determine.
  10. Mar 20, 2015 #9
    OK. So I just finished gathering data. I put the accelerometer at the end of the rod, but I got some weird things. First, note that I just used a basic smoothing filter to get an idea about the shapes, but I'll look more closely about how I should filter it later on. I have three plots attached, showing the data gathered from all three dimensions. Note that what we normally consider the z-axis (the up and down direction in real life), is actually the x-axis in these plots. The most confusing thing is why those bell shaped curves for the y-axis are positive, then negative, and then positive again and so on. I have no idea where this is coming from. However, I think it makes sense that the z axis shows near zero acceleration-as you said it seems like I'm dealing with a one dimensional problem. However, I'm also concerned about that offset. I wondered where it might be coming from, and I thought that it may have something to do with the orientation of the sensor-I figured that I might not have had it perfectly straight. So I decided to test the same sequence I did two more times, but this time I purposefully tilted the sensor at different angles (they were in opposite directions). The results are the next two figures. This time, the offset for the z-axis is huge, and now I'm seeing bell shaped curves for the x-axis as well. Why is this the case? My rod is never moving up and down-ever. Also, how can I mathematically quantify that offset? I want to do it each time so I can have the absolute most precise measurement of acceleration. I really appreciate your help and interpretation by the way.

    I suppose my most important question is how I should go about quantifying everything so that I get the most true representation for the (one-dimensional) acceleration in what you see as the y-dimension, EVEN if my sensor is slightly tilted (unless I'm wrong about that).

    Attached Files:

  11. Mar 20, 2015 #10
    Oh yes. That makes sense too.
  12. Mar 20, 2015 #11


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    Positive is acceleration in one direction, then the velocity is constant for a while, then you brake which is an acceleration in the opposite direction.

    Tilting gives an offset, right, as gravity gets some more complicated direction in the coordinate system of the accelerometer. This also means the acceleration from the rotation (previously in y-direction only) gets some components in the other axes. x is not "up/down" any more, but some new direction in the old x/y plane.

    You can use rotation matrices to transform the measurements to a different coordinate system - one where one axis feels an acceleration of g and the rotation of the rod influences another axis only.
  13. Mar 20, 2015 #12


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    Well, I hope I didn't raise false expectations, I'm a physicist but have never dealt with these devices (that's why I find it an interesting thread :smile: !).

    Anyway, to begin with the straight one. You don't provide much context, so I have to ask or guess.
    My guess is that y is perpendicular to the rod and z along the rod, so green is angular acceleration and blue is angular speed (with appropriate scale factors).

    And I guess you let the motor turn from 0 to 90 degrees in 2 seconds, come to a stop and immediately return to 0 degrees in another 2 seconds, etc.

    It's nice to see how your description of what's to be seen (in post #7) is much better than my simplest case in post #8!

    I've tried to idealize a little:


    Where the bottom one is crude and stretched vertically. (It's also wishful thinking wrt. the actual plot :wink:)

    To go from 0 to 90 in 2 seconds (ignoring startup/stop) is approx. 0.8 rad/s.
    From 0 to 0.8 rad/s happens in approx 0.2 s, so 4 rad/s2.

    From integrating the green one I guess the average acceleration ##\alpha r ## is 0.2 m/s2 (linear from 0 to 0.2g in 0.2 s so average 0.1g x 0.2 s = 0.2 m/s2) and 0.2 m/s2 / 4 rad/s2 = 0.05 m.

    That would place your accelerometer about 5 cm from the axis. ##\omega^2r## would then be around 0.04 m/s2 or 0.004 g, which may or may not be considered in agreement with the plot -- it looks as if the resolution at such low acceleration is somewhat cumbersome.

    Can you follow this crude reasoning a little ? With better info and more detailed calculation you can set up a calibration that way.


    When you tilt, the x value goes to cosine of the tilting angle (23 degrees ?, opposite 25 degrees ? ), whilst the y value picks up the sine (some 0.45 g) . This is beyond my understanding. It looks as if this is pretty much a constant. In other words: the orientation of the rod wrt horizontal is almost unchanged. So you don't turn over 90 degrees but only over a very small angle. That would mean my whole guessing story is based on quicksand ! Egg on face smiley, or is there a way to save me from that ?

    Perhaps these guys can help with the tilting -- but I find it hard to read
  14. Mar 23, 2015 #13
    Hmm thanks for your interpretation. I think you probably have the right idea but I'm not sure about your actual numerical answers. First, I'm not sure why you suggest the blue curve is the velocity. The blue and the green are on the same plane as the rod, the only part perpendicular to the rod (the up and down dimension) is the red. I think this makes sense because it's reading 1 G. Am I wrong?
    Also you were very very close to the angle, its closer to 100 degrees but it's pretty awesome that you made a guess that was so close. Also, the acceleromter is actually pretty far from the center of rotation-like over 30 cm if I remember correctly.
    Lastly, are you implying at the end there that I need to measure the angle the accelerometer is titled with respect to the vertical/horizontal axis in order to quantify its effect on the data? If so that sucks. I dont have a means of measuring that yet.
    I'll take a look at the link you posted-it might help. Thanks.
  15. Mar 23, 2015 #14


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    Well, shows my calculations can be discarded utterly.

    But it could work as follows:

    100 degrees in less than 2 sec means about 1 rad/s, so ##\omega^2 r## would be 0.3 m/s2 or 0.03 g, so the blue line could well fit that.

    Then: changing direction in 2/3 sec makes ##{\Delta\omega\over\Delta t}\approx 3 ## rad/s2 so that average ##\alpha ## is around 3 rad/s2 and average ##\alpha r## arounf 1 m/s2 or 0.1 g.

    So my
    was no good at all (not even dimensionally o:) )

    Blue curve is not the velocity, but corresponds to centripetal acceleration, so it should be ##\omega^2 r##. See earlier posts.
  16. Mar 23, 2015 #15


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    You can extract it from measurement values if you know the precise distance to the center of rotation well enough. Some calibration data would help, like constant spinning at a known (fast) rate.
  17. Mar 25, 2015 #16
    Good news: I understand the bolded! :D But I'm assuming that you want me to ignore the fact that the blue curve is negative.

    About the red: Not sure how you ended up with that answer. What did you plug in for the change in speed and time? Lastly, does multiplying the acceleration by the radius essentially give you the "full" acceleration? I'm trying to picture the actual physics of what's going on in my system.

  18. Mar 25, 2015 #17
    I think I accidentally ignored this post?! But it seems like you really know what you're talking about. I was recently doing some research online about my issue and rotation matrices showed up. The (potential?) problem: I'm not sure how the rotation matrices idea works (I'm hoping you can help me with that :-) ), but I'm not tracking the angle that the acclerometer is tilted at throughout the movement of the rod. If I have to measure it only once before movement onset, then maybe I could use a level or something before every time it runs and use that. But if I'm supposed to be tracking the instantaneous angle it is titled at (although that shouldn't really change...?) then I can't do that. If rotation matrices don't need the angle the accelerometer is tilted at then ignore all this :/ Either way, can you help me get set up with them? Don't really know where to start, and I haven't been able to find a good resource online so my understanding is still low.

    OK, well as of today I'm actually tracking the instantaneous position and time of the rod! So I can calculate velocity by dividing the change in velocity by the change in time. And I do in fact know the exact distance from the center of rotation. Where should I go from there?

    Thanks for your interesting input either way.
  19. Mar 25, 2015 #18


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    1. Good :smile:

    2. No. From post #3: ##\
    \vec a = − r\dot \theta^2 \; {\bf \hat r}\quad
    ##and in post #12 I drew it "upside down" (zero line is indicated!) :smile: So yes, it's negative (towards the rotation axis).

    3. ##\omega## from + 1 rad/s to -1 rad/s in approx 2/3 of a second. 2 rad/s / (2/3) s = 3 rad/s2.

    4. multipying the angular acceleration by the radius gives you the radial component of "the" acceleration (case r fixed). Posts #3, #6 (among others).

    Re rotation matrices: study the derivation in the link in post #2 . Once you can follow that check out the Wiki lemma (which you must have found already :rolleyes: ). As you can see there basic rotations keep one coordinate unchanged. Ideally, yours is ##R_x##, x stays the same.

    6a. When the axis isn't perfectly vertical, life becomes more complicated. But if the axis is fixed, you only have to determine it once.
    6b. Gives me the impression the axis isn't the same from run to run. Why wouldn't it be ?

    7. See 6a. Once is enough if it's fixed.

    8. Unfortunately, they do (as you've seen in your data). By the way, in post #13 you say you have no means to measure. Not true. You can make a good guess by estimating; that's also a measurement, albeit a less accurate one. Were the tilt angles around + and - 25 degrees ?

    9. Some studying required.

    Does that mean you now have 4 signals (x,y,z acceleration and ##\theta##) ?

    You can calculate velocity by dividing the change in velocity angle by the change in time.
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