Physical significance of RMS of acceleration

[EngForm78]

What is the physical significance of the RMS of acceleration data. I have read that it is a measure of energy. What would the units be? Is there an integral function to describe this? Can anyone expand on this a little?

Thanks in advance

 

[IRstuff]

Just go by the mathematical definition of rms, the integral of the square of the function gives you power, which divided by the period, gives you average power.

TTFN

 

[CESSNA1]

ENGFORM78: RMS is the root mean square of the acceleration. To get the peak acceleration multiuply by the square root of 2 (1.414). RMS acceleration is used because it is a vector and the data can be added and subtracted. The peak values cannot be added or subtracted. To add two or more RMS data signals use the equation:

Result^2 = (A1^2 + A2^2)

OR Result = (A1^2 + A2^2)^.5

Typically this is used in the Power Spectral Density (PSD) plots or amplitude (Grms^2)verese frequency (Hertz).

Regards
Dave

 

[IRstuff]

Multiplying by square of 2 ONLY applies to a pure sine wave.

TTFN

 

[EngForm78]

Thanks for the posts.

So, I have 2 sets of vibration data collected over a set time and there are a number of frequency components that stand out. I want to compare the 2 sets to determine in which case the vibration is worse. Will the RMS of acceleration give me this?

Is it correct to say that the RMS of acceleration over a time interval is the energy transmitted in that mount of time, or power.

Would I solve for a RMS value at each frequency and add them all together or can I sum the squares of all data points in the spectrum to get an overall power transmitted.

Thanks

 

[IRstuff]

It is proportional to the energy. The RMS will NOT tell you which one is worse, other than that one has a lower energy content. However, "worse" is related to the specific weakness of the UUT, and only YOU know anything about that.

TTFN

 

[CESSNA1]

ENGFORM78: Assuming what you have are two PSD curves. The energy recorded by the accelerometer is the area under the PSD curve. Be careful though the PSD curve is a log-log curve and special equations a needed to integrate it. A straight integration will get you in the ballpark, but will be in error. If you are in the USA and reply with your fax number, I will fax you the equations. I suggest you look into classes at "

http://www.tti4edu.com".

These classes are the best kept secret in engineering.

Regards
Dave

 

[electricpete]

So, I have 2 sets of vibration data collected over a set time and there are a number of frequency components that stand out. I want to compare the 2 sets to determine in which case the vibration is worse. Will the RMS of acceleration give me this?

Is it correct to say that the RMS of acceleration over a time interval is the energy transmitted in that mount of time, or power.

Would I solve for a RMS value at each frequency and add them all together or can I sum the squares of all data points in the spectrum to get an overall power transmitted.

You can compute RMS as square-root of sum of squares of rms values of individual frequency components.

for example
a(t) = sqrt(2)*A1*cos(w1*t) + sqrt(2)*A2*cos(w2*t)
Then
rms(a(t)) = sqrt(<a(t)^2>) = sqrt(A1^2 + A2^2)

Also if you have a reasonably large representative time sample set {a[n]} n=1..N sampled at a frequency which is not harmonically related to the data you can compute rms:
rms({a[n]) = sqrt< sum(a[n]^2,n=1..N)/N>

I agree with IRStuff it is important to distinguish an “overall” acceleration expressed on a peak basis from a true peak acceleration level. (“overall” is a term used in condition monitoring).

Overall Acceleration expressed on a peak basis is sqrt(2) * rms acceleration.
True peak acceleration = Crest Factor * rms acceleration.
True peak is most directly determined from the time waveform itself (the highest value of instantaneous acceleration).

I would be very careful in relating the terms power and energy to physical power = work/time and energy = work. I don’t think there is a straightforward way to do that unless you postulate a system model which could relate instantaneous power to instantaneous acceleration. At any rate “average” power transfer of periodic vibrations in a mass spring system over time is zero since energy oscillates back and forth.


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[IRstuff]

Just to expand a bit on EP's post. It's also highly dependent on the specific frequency content and susceptibility of the UUT.

Does your PSD excite resonances that will cause fasteners to loosen or materials to fatigue or solder joints to fatigue?

TTFN

 

[GregLocock]

If you are talking about energy you should look at RMS^2, or 20log(RMS)

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[electricpete]

While that may be a perfectly valid signal processing useage of the word energy, I don't find it adequate in at a thread asking about the physical significance of acceleration measurements, followed up by a question about "power transmitted".

Maybe someone can explain to me what it has to do with physical power = work / time and energy = work.


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[GregLocock]

The kinetic energy in the system is proportional to velocity squared, which at a given frequency is proportional to acceleration squared.

The acosutic power radiated will be proportional to velocity squared

the potential energy in teh springs is proportional to displacement squared, which at a given frequency is proportional to acceleration squared.

So, it seems to be a square law.



Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[electricpete]

Well, for one thing, there is not only one frequency present here (see 9 Aug 06 8:58). Two systems with multiple frequenc content having the same rms acceleration will have differing rms velocities and different rms displacements.

For another thing... I'm not sure but I still don't like it.

I would like to know what kind of system the original poster is studying and what is he trying to accomplish and why he is interested in energy and power transmitted?

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[electricpete]

The question was the physical significance of rms acceleraiton.

In my personal opinion, power and energy take us off track.

I think IRStuff had it right. If you know what rms is and you know what acceleration is, you know what rms acceleration is.

I'm pretty sure you know acceleration, so on to rms.

The rms of a waveform a(t) is the root of the mean of the square of that function. i.e. sqrt(<a(t)^2>).

You can compute it from the time waveform. But it is also a very handy way for "adding" together multiple frequency components. The reason can be seen by looking at mean-square <a(t)^2>

If a(t) has multiple frequency components. Forget the phase angle for simplicity
a(t) = sum(Ai*cos(wi*t))
Then a(t)^2 = sum(Ai*cos(wi*t)) * sum(Ai*cos(wi*t))
a(t)^2 = sum(Ai*Aj*cos(wi*t)*cos(wj*t) i=1..N and j=1.N
<a(t)^2> = <sum(Ai*Aj*cos(wi*t)*cos(wj*t)>

But the cross terms i <> j are zero due to orthagonality of sinusoids of different frequencies. (The average value of the product of two sinusoids of different frequencies is 0).

So
<a(t)^2> = <sum(Ai^2*cos(wi*t)^2> = sum<Ai^2*cos(wi*t)^2> = sum of the mean square of the individual components.

Thus the result: the mean square of a waveform is the sum of the mean square of the components. This provides a logical means to roll together the contributios of the components.

I can draw one analogy in electrical engineering. I can create a wide variety of current waveforms that have an rms value of 1A. All of them will have the exact same heating effect when injected into a 1 ohm resistor. For applications where the square of the variable is important (such as power = I^2 * R), preserving the rms preserves the average value of the square of the variable. In this case the square happens to be proportional to power.. ouch there's that word again.

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[GregLocock]

Hang on, are we talking about a spectrum with a y axis scale in RMS(m/s^2), or a single total RMS value?

Ok, what do I think the physical significance of a single rms acceleration figure is for a broadband signal, in general?

It means just about nothing.

Rather like an overall sound measurement of 45.6 dB Lin, it tells you almost nothing useful until you add some description of what is being measured.

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[IRstuff]

Based on the OP's second post, we're talking both. He has a PSD and an RMS value of that PSD.

TTFN

 

[electricpete]

Who said he has a PSD?

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[electricpete]

To clarify some terminology as commonly used in rotating equipment condition monitoring (which are not necessarily the way terms are used elsewhere).

The output of the accelerometer is sampled to give us a time waveform. The units would be g's (or m/sec^2 or equivalent).

An FFT (plus windowing) is performed to give us a spectrum.

Now one tricky part, mathematically the units of the spectrum should be different than the units of the time waveform. But we gloss over this and display the spectrum in units of g's. I am happy to accept that simplification.

The "analog" overall (a strange term) can be computed from the time waveform samples. When expressed in rms units it is sqrt(1/N * sum (x[n]^2). To express on a peak basis we multiply by sqrt(2) even though that is not a true peak value.

The "digital" overall can be computed from the spectrum (FFT) magnitudes. When expressed in rms units it is sqrt(sum (x[k]^2) where x[k] are the rms spectral components. To express on a peak basis we multiply by sqrt(2).

Since the signal was anti-alias filtered, the digital overall is typically slightly less than analog overall. If we increase Fmax above the highest frequency content in the original signal, digital overall will become the same as analog overall.

The term rms has mathematical definitions as described above. You can see in a sense that any of the methods for forming an overall ("digital" or "analog") rely first upon determining an rms value. Then sometimes the overall is converted to a peak or peak-to-peak value which again is not a true peak or peak/peak but would represent the peak (or peak-to-peak) value of a single-sinusoid that had the same rms content.

PSD is not used in rotating equipment monitoring. But mathematically PSD represents a square of a spectrum. I didn't see anything in the original post to indicate the square of a spectrum.

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[spciesla]

Pete,

"...Now one tricky part, mathematically the units of the spectrum should be different than the units of the time waveform. But we gloss over this and display the spectrum in units of g's. I am happy to accept that simplification..."

I'm unfamiliar with this simplification. How does the transform change the units?

Steve

 

[IRstuff]

"2 sets of vibration data collected over a set time and there are a number of frequency components that stand out"

That implies an acceleration vs frequency data, hence the square of that is proportional to the power spectral density

TTFN