Physical significance of RMS of acceleration
Physical significance of RMS of acceleration
(OP)
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
Thanks in advance





RE: Physical significance of RMS of acceleration
TTFN
RE: Physical significance of RMS of acceleration
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
RE: Physical significance of RMS of acceleration
TTFN
RE: Physical significance of RMS of acceleration
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
RE: Physical significance of RMS of acceleration
TTFN
RE: Physical significance of RMS of acceleration
Regards
Dave
RE: Physical significance of RMS of acceleration
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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RE: Physical significance of RMS of acceleration
Does your PSD excite resonances that will cause fasteners to loosen or materials to fatigue or solder joints to fatigue?
TTFN
RE: Physical significance of RMS of acceleration
Cheers
Greg Locock
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
RE: Physical significance of RMS of acceleration
Maybe someone can explain to me what it has to do with physical power = work / time and energy = work.
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RE: Physical significance of RMS of acceleration
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.
RE: Physical significance of RMS of acceleration
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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RE: Physical significance of RMS of acceleration
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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RE: Physical significance of RMS of acceleration
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.
RE: Physical significance of RMS of acceleration
TTFN
RE: Physical significance of RMS of acceleration
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RE: Physical significance of RMS of acceleration
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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RE: Physical significance of RMS of acceleration
"...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
RE: Physical significance of RMS of acceleration
That implies an acceleration vs frequency data, hence the square of that is proportional to the power spectral density
TTFN
RE: Physical significance of RMS of acceleration
The reason I was confused was by by comparison with the continuos Fourier transform.
X(w) = Integral{x(t)*exp(-j*w*t)dt
x(t) = 1/(2Pi)*Integral{X(w)*exp(j*w*t)dw
Those are tricky units. X(w) has an extra unit of time compared to x(t). I don't think that same complication applies to discrete fourier transforms that we work with (FFT's).
(I'm interested to hear if anyone has other comments to clear that up).
Thanks for the correction.
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RE: Physical significance of RMS of acceleration
IRStuff - I was objecting to your statement "He has a PSD and an RMS value of that PSD."
I think he has a spectrum, not a PSD. I agree he could calculate a PSD if he wanted. (but that's only another complication not needed here in my opinion).
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RE: Physical significance of RMS of acceleration
Time series has units of seconds for the independent variable. FT has units of radians/second for the independent variable. That's a direct consequence of the j*w*t in the exponent of the transform kernel.
TTFN
RE: Physical significance of RMS of acceleration
I'm not talking about the independent variable but about the units we apply to waveform x(t) and continuous spectrum X(w).
If x(t) were g's, then X(w) would be g*sec as can be seen from the definition above.
We can apply Parseval's theorem:
Int(x(t)^2 dt = Int(X(w)^2 dt (integral -inf to +inf both cases)
g^2 * sec = (g*sec)^2 * (1/sec)
Parseval is happy.
I personally view FFT as an approximation to the fourier transform of the underlying continuous signal. I think many others do as well. That's why the funky units of continuous fourier transform seem a little bit tricky to me.
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RE: Physical significance of RMS of acceleration
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