0

So basically I had an early question and found out that if I have 15 machines running at 80 dB in a room it should end up being around 10*Log(15(10^7.5)) dB approx. 91.76 dB.

But how would the room size and the dispersion of the machines affect this number? As well as distance from the machines.

Edit: You can assume the machines run at a constant 75 dB and have the sound/pitch of a computer fan. For the sake of trying to find a general formula you can assume any other variables in your answer as long as you explain them.

Edit: I understand that there may be many more variables that come into play when determining the sound intensity in a room. But I am trying to find a rough estimate.

12
  • This feels like it could turn into the audio version of the Coastline Paradox (en.wikipedia.org/wiki/Coastline_paradox). - Could you tell us what problem you're trying to solve here? The solution might be less complex than you think. May 17 at 19:54
  • Yes, so basically I have x machines running in a room, and they are running at 75 decibels. I do not have specific dimensions, but I plan to insert some assumptions that include 1000-2000 square-foot rooms. Let me know if this helps and if certain parts should be added to the question itself.
    – Joe
    May 17 at 20:00
  • 1
    I got that. But what problem are you trying to solve by answering this question? Why do you need to know how loud this room will be? May 17 at 20:10
  • 1
    But that only works in open space. It doesn't take into account the reflections from walls multiplying or nullifying amplitude at different frequencies and different wall distances. Then there's absorption and diffusion... Please just tell us why you want to make these calculations - what problem will they solve for you? Then we might be able to give you a sensible answer. May 18 at 19:36
  • 1
    In that case, you might be better off in an IT/Computing forum where you can ask someone with experience how to write those estimates - You're unlikely to be the first to do so and you'd probably get a much better answer than you will here May 19 at 18:40

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.