help needed. vibration, calculation
help needed. vibration, calculation
(OP)
how to calculation the vibration amplitude of a random vibration?
one step further, if there are several superimposed sinusoidal vibrations?
i have one test requirement. the test lab told me that it exceeded their table DA.
thanks a lot.
one step further, if there are several superimposed sinusoidal vibrations?
i have one test requirement. the test lab told me that it exceeded their table DA.
thanks a lot.





RE: help needed. vibration, calculation
Do you have a Power Spectral Density (PSD) plot? There is a method to take a value from the PSD and figure out the average amplitude of each frequency.
RE: help needed. vibration, calculation
lab told me, this one exceeds 2" DA.
random
10-100 Hz, 3.01 dB/Oct
100-300 Hz, 0.01 g*g/Hz
300-500 Hz, -13.569 dB/Oct
seven sinu
4.3Hz/0.43g; 8.6Hz/0.86g; 17.2Hz/1.72g; 3.380Hz/0.338g, 6.767 Hz/0.677g, 10.150 Hz/1.015g, and 13.533 Hz/1.353g
RE: help needed. vibration, calculation
Assume a sample rate of 1024 points/sec and an FFT size of 2048 points. This gives a frame length of 2 sec or a delta f (resolution) of 1/2 or .5 hertz. For a 100 hz sine wave, a PSD value of .01 g*g/Hz would give an energy of .01 g*g/Hz * .5 hertz = .005 g*g. A square root produces a peak amplitude of .071 g's. The displacement needed would be:
x = accel/(omega**2) = .071*386/(6.28*100)**2 = .000069 inches - very small!!
Your numbers will depend on the sample rate and FFT size used to generate the numbers you gave. Doesn't appear though that this particular frequency band requires much stroke.
For your sinusoidal tests, again use x = accel/(omega**2). For the 4.3 hertz,
x = .43*386/((6.28*4.3)**2) = .227 inches - again, not much displacement.
Caution, I make a lot of math errors! Find out what test is the one they are having problems with.
RE: help needed. vibration, calculation
2) people were talking about 3 sigma. the g*g/Hz will be three times bigger at some time.
3) the problem is the random with 7 sinusoidal SUPERIMPOSED on it. not each one individually causes excessive DA.
thanks.
RE: help needed. vibration, calculation
2) Three Sigma usually refers to Guassian distributions of random data. I think you need more info from the supplier of the accel data.
3) Add the 7 sine waves together (with the appropriate phases between them). Figure out how much stroke is needed for that acceleration sum.
I take it you are getting accel data from your customer and need to test it using your's or someone else's test equipment?
RE: help needed. vibration, calculation
to educate myself, i just want to know how to calculate it.
add 7 sin is not difficult. they are two groups. one is 3.38 HZ and its multiples. the other 4.3 Hz. at one point if the test time is long enough, they will just all add up with no phase diff.
is it just a arithmatic DA sum to add sin and random together?
RE: help needed. vibration, calculation
I don't know what the 3.01 dB/Octive means for the 10-100 hz data. Seems to me they need a reference acceleration as part of part of the spec. My guess is that, because it contains the lower frequencies, that is where the stroke limitations come from. The one stroke estimate I made earlier for the 100-300 hertz band (for the stroke at just 100 hertz) was so low that I'd guess you could probably add all the strokes in that band (one for each frequency line - depends on sample rate and FFT size), assume they were all in phase (they aren't for random noise), and still have a small stroke.
See if your customer can clarify what the db/Octive means and also find out how your test lab interpreted it.
RE: help needed. vibration, calculation
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RE: help needed. vibration, calculation
say freq_1, 10 Hz; freq_2, 100 Hz
dB = 10
Oct=LOG(ASD_2/ASD_1)/LOG(2)
ASD_1=.001 g*g/Hz; ASD_2=.01 g*g/Hz
Oct=3.32
so the "slope" of the first leg from 10 to 100 Hz is
10/3.32=3.01 dB/Oct
(see attached.)
thanks.
RE: help needed. vibration, calculation
x = ((.001*.5)**.5 x 386)/(6.28*10)**2 = .0022 inches single amplitude.
@20 hertz,
x = ((.002*.5)**.5 x 386)/(6.28*20)**2 = .00077 inches.
If you had 180 frequency lines between 10 and 100 hertz and they each needed .002" of stroke and they were all in phase then this portion of the random data would need 180 * .002" = .36".
These numbers assume a sampling rate of 1024 pts/sec and an FFT of 2048 points (delta f = .5 hz).
Add this to the .704" needed for the discrete sine waves and you get about plus/minus 1.1". This doesn't include the high frequency stuff (300-500 hz - probably don't need any stroke for that) but is very conservative in that it assumes all frequencies are in phase and their peaks occur at the same time.
Sure seem like small stroke values. I'm surprised your test house is having trouble. Can they send you an ASD of the drive they came up with?