0
$\begingroup$

I am having great difficult understand my readout from an accelerometer that I placed on a rotation wheel. The wheel spins at various speeds and i measured the radius from the center. I think I mentioned before that I had done some calculations and I was expecting $z = 1$g (which I know many disputed). And then I was thinking that the displacement in the x and y direction would be

$$ x(t) = r \cos(ω t) = r \cos(2 π f t) $$

$$ y(t) = r \sin(ω t) = r \sin(2 π f t) $$

and so acceleration would be this differentiated twice.

When I looked at the recording on my accelerometer it was strange as $z$ was indeed 1g, $x$ remained at 0 and $y$ does vary but not in the way I was expecting. ie like a sin or cos wave over time in fact the signal oscillated about a particular value in the same was $x$ and $z$ does. however, this value is not quite as much as I would expect. Any help or advice to offer insight into this would be greatly appreciated you guys

Improve this question
$\endgroup$
7
  • $\begingroup$ The accelerometer rotates with the wheel, right? $\endgroup$
    – Colin K
    Commented Jun 24, 2014 at 16:46
  • $\begingroup$ Is the accelerometer aligned with the z axis in the radial direction? If so you'd expect constant $z$ and zero $x$ and $y$. If the wheel was slightly out of true you could get variations in $x$ and/or $y$ at the frequency of revolution. $\endgroup$
    – John Rennie
    Commented Jun 24, 2014 at 16:49
  • $\begingroup$ yes it is rotating as it is placed on the wheel $\endgroup$
    – branny12000
    Commented Jun 24, 2014 at 16:50
  • 2
    $\begingroup$ Could you include a picture or diagram of the set up? That would help a lot. $\endgroup$
    – NeutronStar
    Commented Jun 24, 2014 at 16:50
  • $\begingroup$ the z axis points downwards. The experiment has been a variation on one where the sensor was placed on a bar between two rotating wheels. I thought originally that i wont see accelerations in the x and y as it is moving at a constant velocity however the paper I took it from suggests that Acceleration can be generalised as 8*rpifreq^2. I then tried to solve this in x and y $\endgroup$
    – branny12000
    Commented Jun 24, 2014 at 16:52

1 Answer 1

Reset to default
1
$\begingroup$

Let me replace the accelerometer by a whole smartphone because it's easy to visualize it.

Your comment that the $a_z$ acceleration was always equal to $1g$ implies that the wheel was rotating in the horizontal plane. The vertical $z$ direction contributes the gravitational acceleration $1g$ and this direction always goes along the same direction of the smartphone that you have called $z$.

To see what the other components $a_x,a_y$ of the accelerometer will show, it is convenient to work in the rotating frame associated with the wheel. In that frame, there is still the gravity $1g$ that goes in the direction that is still called $z$. But the $x,y$ axes are rotating into each other in such a way that the outward radial direction is called $x$ by the accelerometer.

There is a centrifugal acceleration – otherwise indistinguishable from gravity – which will be shown as $a_x$ while $a_y$ will be zero as it goes in the angular direction. The centrifugal acceleration is $$a_x = r\omega^2 = 4\pi^2 r f^2$$ It is irrelevant that the speed (absolute value of the velocity) is constant. There is acceleration because the velocity is changing the direction. There is also an acceleration from the changing orientation of the accelerometer. Using the fictitious forces in the rotating frame is the simplest way to see the things.

So the accelerometer will show components $$ \vec a = (r\omega^2, 0 , -1g) $$ If they are permuted, it means that you must adjust your idea which components of the accelerometer carry which names.

If signs are wrong, you must adapt to the conventions that the accelerometer is using about the sign (I wrote the expectation that the $a_z$ component is negative because "down" should mean a negative $z$, but the accelerometer produces whatever signs it produces: you must find the conventions experimentally). If the $y$ component is nonzero, it means that the accelerometer is rotated relatively to the simple position above. Note that if the smartphone is seemingly oriented exactly radially, it doesn't mean that the accelerometer is oriented radially. If the accelerometer resides away from the center of the smartphone, then its $x$ axis fails to be parallel to the radial direction of the wheel, and one gets the usual mixture/spilling in between the components $a_x$ and $a_y$.

As long as the accelerometer is tightly attached to the wheel, there is no Coriolis' force. The latter is proportional to the velocity of the object relatively to the rotating frame. But if the accelerometer is attached to the wheel, its velocity to our chosen wheel's frame of reference is zero.

Incidentally, it's common that some tablets have switched conventions what their accelerometers call $a_y$ and $a_z$ relatively to smartphones. So water level apps sometimes show the gravity that is rotated by 90 degrees on the display relatively to the reality etc. One must be careful about all these conventions and their possible dependence on the particular device. If the screen is allowed to switch from the portrait mode to the landscape mode, there is one more issue to worry about. Apps like that should be separately tested on smartphones, tablets, and for both, both in the landscape and portrait mode.

Improve this answer
$\endgroup$
6
  • 1
    $\begingroup$ Lubos!!! I cant tell you how good your explanantion was! Thank you so much for being so clear for me! I was also wondering if you could help with something else. I want to work out the error from my signal as you can imagine over the three minute segment the values go up and down a bit and do not stay at whatever value the signal oscillates a bit. What would be the best way to estimate error? $\endgroup$
    – branny12000
    Commented Jun 25, 2014 at 10:59
  • $\begingroup$ Nice to hear that/if my comment at least looked helpful to you, @branny12000! Concerning the error, I think that I may need - and you may need to think about it as well - to know which quantity's error you want to quantify. There are various types of systematic/statistical errors, some of them stay the same at all times, some of them are noise that tends to average out over many readings, and so on. Some of these errors are more relevant for some purposes, others for others etc. I would need to know more what you want to do... $\endgroup$
    – Luboš Motl
    Commented Jun 25, 2014 at 16:08
  • $\begingroup$ Well I would like to work out the error in the readings so perhaps I should take the sum of the errors (ie difference between readings and expected value) and divide by n (number of values)? When would coefficient of variance be appropriate? $\endgroup$
    – branny12000
    Commented Jun 25, 2014 at 23:37
  • $\begingroup$ Dear Branny, it's always more natural to compute the variance etc. - to sum not the absolute values of the errors but the squared errors (EE), and then divide by N to find the average EE, and then take square root to find average (more precisely "root mean square") error. Absolute values are unnatural, squares are natural. It's probably OK to assume that the accelerometer gives you $A(t)+E1+E2(t)$ where $A$ is the right value, $E1$ is an error that is independent of time, a systematic one, and E2 is a random error. Both E1,E2 may be estimated from many measurements. $\endgroup$
    – Luboš Motl
    Commented Jun 26, 2014 at 4:09
  • $\begingroup$ Are e1 and e2 worked out differently? $\endgroup$
    – branny12000
    Commented Jun 26, 2014 at 9:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.