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Calculate amplitude(displacement) from accelerometer

by Pivskid
Tags: accelerometer
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Pivskid
#1
May6-13, 05:12 AM
P: 8
Hi there
Given a result from an accelerometer, mounted on a vibrating machinery.
I would like to be able to calculate the physical amplitude. That is the actual movement of the vibration.
I'm sure it should be possible, but my attempt to manipulate the formula of acceleration has not resulted in valid results.

Given is a sinusodial measurement of 20hz, 4g "peak to peak".

The acceleration fomula is fairly simple: a=v/t
But how do I convert it into using hz and g as input values and output the displacement (amplitude) ?
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LastOneStanding
#2
May6-13, 05:28 AM
P: 718
The formula ##v = at## is only true for constant acceleration. In vibrational motion, you have a changing force (e.g. changing linearly in simple harmonic motion) and so acceleration is not constant. In the general case, you need to integrate: ##v(t) = \int _{t_0} ^t a(t) dt + v(t_0)##. Since you will have a set of numerical data points, you will need to do use some technique of numerical integration. With ##t_0## chosen so that ##v(t_0)=0## (the turn-around point), you can ignore the integration constant. Since integrating over a single cycle of periodic motion is equivalent wherever in the cycle you start, you can just thrown this term out. Now, having ##v## as a function of time, you can numerically integrate it again to get displacement as a function of time.
Pivskid
#3
May6-13, 06:37 AM
P: 8
Can you outline for me more specifically how it would look like ?
I assume it is the "length" (time) from turn-around-point(=t0) to peak acceleration, that is necessary.
That time should be =(1/hz)/4
How to get from there ? ...integrate from t0 to (1/hz)/4 ?

AlephZero
#4
May6-13, 07:08 AM
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Calculate amplitude(displacement) from accelerometer

The usual practical way to do this is to assume the vibration is simple harmonic motion, not to integrate the acceleration numerically.

If the displacement is ##A\sin \omega t##, the acceleration is ##-\omega^2 A \sin \omega t##.

You need to convert "g" into units of meters/sec^2 or inches/sec^2, and ##\omega = 2 \pi f## .
Pivskid
#5
May6-13, 08:54 AM
P: 8
If ω=2π*frequency
Then what is t ? 1/f ?
AlephZero
#6
May6-13, 09:44 AM
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t is time.

If the maximum displacement is ##A##, and the maximum acceleration is ##\omega^2 A##.
LastOneStanding
#7
May6-13, 10:43 AM
P: 718
Quote Quote by AlephZero View Post
The usual practical way to do this is to assume the vibration is simple harmonic motion, not to integrate the acceleration numerically.
Given this a physical experiment that is actually measuring the acceleration, that would rather seem to defeat the purpose. Why bother even using an accelerometer if you're just going to assume SHM?
AlephZero
#8
May6-13, 11:50 AM
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Quote Quote by LastOneStanding View Post
Given this a physical experiment that is actually measuring the acceleration, that would rather seem to defeat the purpose. Why bother even using an accelerometer if you're just going to assume SHM?
First, most "vibrating machinery" does move in SHM, to a good approximation. It is responding to a steady-state sinusoidal applied force.

Second, mesuring relatively large accelerations is cheap and simple, compared with measuring small displacements directly.

For example at 50 Hz, or 3000 RPM for a rotating machine, a peak accleration of 1g corresponds to a peak displacement of about 0.01mm.
LastOneStanding
#9
May6-13, 12:05 PM
P: 718
Quote Quote by AlephZero View Post
First, most "vibrating machinery" does move in SHM, to a good approximation. It is responding to a steady-state sinusoidal applied force.
Yes, I am perfectly aware of that. However, the OP asked how to calculate displacement given a series of acceleration measurements. Making assumptions—no matter how well founded—about the form the displacement will take defeats the entire purpose of doing an empirical measurement. This would be like if one asked, "How can I accurately measure the variation in the earth's gravitational field near its surface?" and you responded, "Well, we usually assume the gravitational field is constant near the earth's surface, which it is to good approximation. Just use ##F=mg##." It's completely circular.

Quote Quote by AlephZero View Post
Second, mesuring relatively large accelerations is cheap and simple, compared with measuring small displacements directly.

For example at 50 Hz, or 3000 RPM for a rotating machine, a peak accleration of 1g corresponds to a peak displacement of about 0.01mm.
I don't understand what you're saying. I am not proposing measuring the peak displacement directly, I am explaining how to calculate it from the acceleration measurements, exactly as the OP requested. As you have said, measuring accelerations of the order OP will be looking at is easy and cheap, and hence provides a nice way of directly determining the (otherwise hard to measure) displacements.

Numerical integration is easy and, if the accelerometer has sufficient time resolution, very accurate. It will give OP his answer directly without having to make any assumptions about the linearity of material. In fact, it will provide a way of checking to what degree the motion deviates from SHM, if that's something OP is interested to know.

I can't for the life of me understand why someone would advocate relying on a theoretical (and imperfect) model for determining something you could actually directly determine using the available tools. It makes no sense at all.
LastOneStanding
#10
May6-13, 12:20 PM
P: 718
Quote Quote by Pivskid View Post
Can you outline for me more specifically how it would look like ?
I assume it is the "length" (time) from turn-around-point(=t0) to peak acceleration, that is necessary.
That time should be =(1/hz)/4
How to get from there ? ...integrate from t0 to (1/hz)/4 ?
Your acceleration measurements should be fairly cyclic (I say "fairly" because friction means they will decay over time). So, the easiest thing to do will be to pick out the peaks in the acceleration profile. These will also correspond to the peak displacement since the restorative force in most oscillatory motion is generally at a peak at the maximum displacement. Hence, if you numerically integrate (twice) from one crest to the next you will get the displacement between the two points of greatest displacement from equilibrium. That is, twice the amplitude of oscillation.

This looks to be a nice overview of the techniques of numerical integration for definite integrals. Software packages like MATLAB and Python have built in modules for doing this as well.
AlephZero
#11
May6-13, 01:50 PM
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Quote Quote by LastOneStanding View Post
Yes, I am perfectly aware of that. However, the OP asked how to calculate displacement given a series of acceleration measurements.
The OP also said
Given is a sinusodial measurement of 20hz, 4g "peak to peak".
Which part of "sinusoidal" don't you understand?

The OP never said anything about "a series of acceleration measurements" either. You don't need digitallly sampled data to do this sort of vibration monitoring, in the real world.
cjl
#12
May6-13, 02:24 PM
P: 1,017
Quote Quote by LastOneStanding View Post
I can't for the life of me understand why someone would advocate relying on a theoretical (and imperfect) model for determining something you could actually directly determine using the available tools. It makes no sense at all.
Because it is entirely possible (and I would guess even fairly likely) that the error introduced in the numerical integration (depending on the sample rate and resolution of the equipment) may in fact be much larger than the error introduced by assuming SHM and calculating based on the peak (or RMS) acceleration value.
Pivskid
#13
May7-13, 07:21 AM
P: 8
Hi again
I found the solution.
http://www.spaceagecontrol.com/calcsinm.htm

Very simple, and it conforms to my actual "real-life" measurements.

Now I just wonder how the sin() and integration appeared on the scene.


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