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phanikumar (Mechanical)
15 Jun 05 18:46
I am trying to figure out how to calculate octaves/minute for one of the shaker tests I am running. The original test request said that I needed to run a shaker test with a logarithmic Sine sweep from 10 to 2000 Hz and back to 10 Hz in a period of 20Minutes (10minutes/sweep). Now I want to change the log sine sweep from 10 to 200 Hz and back to 10 Hz (instead of 2000 Hz), now I want to find out what time it would have taken to go from 10 to 200Hz (had I swept it to 2000Hz in 10minutes). The software asks me to enter Octaves per Minute (instead of time) for new test requirement. Having a Master's in vibration I am a bit ashamed to say that I never seen this units of convention before in vibrations. Can someone help me how to figure out the octave per minute calculation(I somehow came to know it is about .764 for this particular case).
Helpful Member!  GregLocock (Automotive)
15 Jun 05 20:36
10 hz to 2000 Hz is n octaves, where 2^n=2000/10

so

n log 2 =log 2000-log 10

so

n=(log(2000)-log(10))/log(2)

and similarly for the new sweep.

Cheers

Greg Locock

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

phanikumar (Mechanical)
15 Jun 05 21:34
Thanks for the reply Greg,but still it doesnot explain how I would end up with a .764 number (I got this number from a report on the same test done previously at a different test lab   -test was outsourced). Using your formula
n=(log(200)-log(10))/log(2)=4.321
considering it is 4.321 octave/sec
n would be .0720321 octave/minute
so where in the world .764 come from? Am I missing any more data?
GregLocock (Automotive)
15 Jun 05 22:58
That formula is telling you the number of octaves, not the sweep rate. Your original test was run at a sweep rate of .764 octaves per minute.



Cheers

Greg Locock

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

phanikumar (Mechanical)
15 Jun 05 23:24
Never mind I got it, n=(log(2000)-log(10))/log(2)=7.64385
So if they are 7.64 octaves between 10 to 2000 and I need it to sweep from 10 to 2000Hz in 10minutes, it needs to sweep at the rate of 7.64/10 octaves/minute=0.764octaves/minute and thats how I got the sweep rate
Got it. Thanks a lot for your help. I needed to know this though I already started the test I was trying to figure out where the number came from?
Thanks to your formula now its clear.
Can you tell me in which book can I find this formula and the derivation behind it. I would really appreciate it
GregLocock (Automotive)
15 Jun 05 23:41
It doesn't come from  a book, I worked it out from the definition of an octave.

Cheers

Greg Locock

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

phanikumar (Mechanical)
15 Jun 05 23:56
So I guess octave is an acoustic equivalent of an harmonic in vibrations so they are even multiples of 2 so the octaves between 10 to 2000 hz would be
20,40,80,160,320,640,1280 so they are 7+x Octaves right?
x should be less than 1 since octave of 1280 would be 2560 which is greater than 2000 so x should be a fraction leass than 1
but how do u get 'x' by this method
it can't be (1280)^x=(2000-1280)????
what is it then?
GregLocock (Automotive)
15 Jun 05 23:59
well, if you plug 2000 and 1280 into that formula it tells you.

There's no 'easier' way, that's just how it works.

Cheers

Greg Locock

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

TPL (Mechanical)
17 Jun 05 7:43
Well done Greg...for demonstrating the 'patience of a saint' principle


Masters in Vibration and not understanding how to figure out octaves????   Ask for your money back!!!
phanikumar (Mechanical)
17 Jun 05 7:49
well sir, with all due respect to you, octaves is not a term extensively used in vibrations, its very rarely used and no wonder I didnt pay much attention to it. but still i said i am ashamed bcos its no reason not to know for my background.
Rob45 (Automotive)
17 Jun 05 9:14
The most basic book I could find, "Industrial Noise and Vibration Control," by Irwin and Graf, has a workable definition of octaves much like the one expressed above by Mr. Locock.
Perhaps you could review your undergraduate texts.
tomirvine (Mechanical)
17 Jun 05 10:52
I have written a program that performs various sine sweep calculations.  Please contact me via my website if you would like this program, which I will provide as a free sample.

Tom Irvine
www.vibrationdata.com
phanikumar (Mechanical)
17 Jun 05 11:04
In my defense
"Octave
An octave is a frequency interval having a ratio of two. It is called an octave from the music tradition where an octave spans eight notes of the scale. The second harmonic of a spectral component is one octave above the fundamental. In acoustical measurements, sound pressure level is often measured in octave bands, and the center frequencies of these bands are defined by the ISO. Vibration measurements are seldom expressed as octave band levels, but the US Navy has used 1/3 octave band analysis for vibration measurements on submarines for a long time. "
Source: http://www.vibinst.org/vglos_f-o.htm#o
I hope you know what a vibration institue is

I am not ashamed to say "I dont know when I dont know" b'cos being an engineer I know its always being on a learning curve. Octave is seldom used in vibrations its associated with acoustics, since a lot of times acoustics and vibrations go together, definitions are interchanged easily.

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