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Unit Conversion (Vibration - micro in./s TO mm/s RMS)

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Jason SBM

Civil/Environmental
May 15, 2019
2
Hello. I was wondering if Eng-Tips could help me. I'll confess--I seem to be having a "herp-a-derp" moment. I'm reading a report and I just don't understand how the people converted from VdB re 1 micro inch per second to millimetre per second RMS. I am sure this is a laughable failure on my part.

Per the FTA "Noise and Vibration Manual," Fig 10-1 one can find a generalized ground surface vibration and apply it to their current situation.
(p. 171)

Per the MOE/TTC "Draft Protocol for Noise and Vibration Assessment for the Proposed Scarborough Rapid Transit Extension" May 11, 1993
(a summarized copy is included near the end of the document, p. 58 onward)

In the MOE draft p. 6 (p. 62 of the PDF), they state that "If the predicted rms vertical vibration velocity from the Line exceeds 0.1mm/sec, mitigation methods shall be applied during the detailed design to meet this criterion to the extent technologically, economically, and administratively feasible."

I'm reading a report and I'm not understanding how they arrived at their number (please see attached calcs).
Conversion_hyeyw2.png

When I convert micro inch to mm, I go by 1 micro inch = 0.0000254mm

So (77 micro in./sec) * (0.0000254 mm/micro in.) = .0019558 mm/sec

The report calls it out as 0.19mm/sec. So what am I missing? Is it not just a simple distance conversion for VdB re 1 micro in./s to mm/s? Or is there a formula I'm missing?


They also have this statement: "These standards indicate that the appropriate criteria for office buildings are 84 dBV (referred to 1 micro inch per second) or 0.4 mm/s RMS for vibrations."

I'm even more lost on this one. I presume the "83 dBV re 1 micro in./sec" from the previously attached pic also is calculated to 0.379 is a similar situation.

I'm guess I'm lost because, by Fig 10-1, converting the way I am there is no situation where the rail vibration exceeds TTC standards. I mean, 10 ft away when the train is doing 95mph passes the 0.1mm/s benchmark. So, I think I'm just missing something. Any help greatly appreciated.
 
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I'm not fully confident in this response because my quick calculator conversions result in: 77 dBV -> ~0.180 mm/sec and 83 dBV -> ~0.360 mm/sec vice the results they provide as equivalents. This may be due to rounding or decimal places though.

The only 77 I see in the image above is the total vibration level. That value is a unitless quantity since its dB, and therefore if you desire to perform the conversion (it appears the results have already done this though) you have to consider the dB reference (re), which they state is 1 micro-inch/ sec (shorthand u-in/s)..

dBV = 20*log (V [u-in/s] / V re [u-in/s])

Substituting 77 (or 83) for the dBV, raising both sides by 10^ to cancel the log, and then performing unit conversion from u-in/s to mm/sec results in the numbers I provided in the opening sentence so long as I didn't make a mistake.
 
Possibly rounding: I got 0.1798 mm/s and 0.3588 mm/s

see: for a reference on db conversion
also
Note that 20 log (v1/v0) is used because you are supposedly interested in the energy contained in the vibration, e.g., the power spectral density, not just the amplitude.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! faq731-376 forum1529 Entire Forum list
 
raar and IRstuff,

Thanks so much for the replies, the additional reading, and the definitions. They were of immense help.

Math below to help any future thread searches on how the numbers were achieved:

Log info:
Code:
But Sometimes There Is Confusion ... !
Mathematicians use "log" (instead of "ln") to mean the natural logarithm. This can lead to confusion:

[u]Example[/u]	     [u]Engineer Thinks[/u]    [u]Mathematician Thinks[/u]  [u]Result[/u]	 
log(50)	     log[i][sub]10[/sub][/i](50)	        log[i][sub]e[/sub][/i](50)	      confusion
ln(50)	     log[i][sub]e[/sub][/i](50)	        log[i][sub]e[/sub][/i](50)	      no confusion
log[i][sub]10[/sub][/i](50)    log[i][sub]10[/sub][/i](50)	        log[i][sub]10[/sub][/i](50)	      no confusion


1 micro inch = 0.0000254mm

77 dBV = 20*log[sub]10[/sub] (x/0.0000254mm/s)
(77/20) = log[sub]10[/sub] (x/0.0000254)
x/0.0000254 = 10^(77/20)
x = ~0.18mm/s

Thanks again! I guess it's refreshing that, yes, the problem was PEBKAC. Cheers.
 
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