A question asked about to fans at 60 dB of sound.
If one fan is removed, what is the dB
The answer said half the sound would be a change of 3 dB, but I always read the change in half the sound is 10 dB.
I put 50 dB but the blackspecs answer is 57 dB. Can you please explain this to me?
@gloriaw92 sure thing!
When it comes to acoustics and having multiple fans with similar sound levels the equation that we would be looking at is Lt = Ls + 10log(n) where n = the number of fans, Lt = the total sound observed or 60dB and Ls is the decibel level of an individual fan. In our case the equation would look like 60dB = Ls + 10log(2) or Ls + 3 since 10log(2) = 3. Which translates to 60dB = Ls+3 or Ls = 57dB.
This works for 3 fans, 5 fans, 20 fans etc. as long as the fans are equal in sound intensity.
Hope this helps!
would this equation be in the references tab provided while taking the exam?
@wildlandguy - Unfortunately this formula is likely not going to be in the references on the exam. NCARB’s ARE guidelines does provide you with a list of the various formulas that are available in the exam and it is broken out by discipline.
I have copied the link here below for reference.
To clarify what you heard previously, a decrease of 10 dB is perceived by listeners as being half as loud. This relationship is subjective. -10 dB actually corresponds to a 10-fold decrease in acoustical energy.
For simplicity, just remember this.
1/2 = -3 dB and 2x = +3 dB
1/10 = -10 dB and 10x = +10 dB
To get other values, multiple values on the left side together and add the corresponding values on the right together. For example:
1/2 * 1/2 = 1/4, therefore -3 dB + -3 dB = -6 dB (a quarter the energy is -6 dB)
2 * 10 = 20, therefore 3 dB + 10 dB = 13 dB (20x the energy is +13 dB)