I think that is wrong, when source of sound is multiplied by two, loudness will increase only by +3 dB.
One cat = 65-75 dB; two cats = 68-78; four cats = 71-81....

And there is a limit to how loud something can get based on the density of the air. For a sustained sound, the limit is 194Db. However even a billion cats meowing at the same time couldn't add up to that.

Correct me if I'm wrong, but...
If there are 500 million cats in the world and one meow's intensity level L(1)=65dB, the correct answer should be around 150dB. Why? Because intensity level is logarithmic and here's the solution: L=10xlog(I/I0) in which I0=10^-12W/m^2. We get one meow's intensity I(1)=3,16...x10^-6W/m^2. If there are 500 million cats, the total intensity I(tot)=500x10^6x3,16...x10^-6=1581,1...W/m^2.
The total intensity level L(tot) is then 152dB.

yea but the dude forgot the recursive formula, which states that sound is exponentially inverse to the amount of things making the noise. 1 thing makes 1dB, 2 things make 1.5dB. ish

It doesn't work that way. In order to increase the loudness all cats would have to meow the exact same way. Same with humans. My psychology class went to a museum about senses and there was this small glass chamber where one could go in and shout as loud as possible. I went in alone and shouted "F*****CK". After me 5 others got in and screamed as loud as they could. The actual loudness was the same. Think of it like throwing stones in a lake. If you throw 1000 stones in they would not create a giant wave but many small ones. Some would amplify each other, some would cancel each other out. So 2bn cats would be 65/75 dB loud but at 2 billion different frequencies, driving every person with (for example) Asperger autism completely made but not creating a sound loud enough to put (presumably) the FRIGGING ZAR BOMB (the biggest ever detonated nuclear bomb) at shame

tinbucket69· 8 years ago · FIRSTsomeone· 8 years agoboredatwork· 8 years agonsa· 8 years agoyouwillhateme· 8 years agonsa· 8 years agoMillion

youwillhateme· 8 years agoguest· 8 years agofizzlesticks23· 8 years agocanyounawt· 8 years agoguest· 8 years agoOne cat = 65-75 dB; two cats = 68-78; four cats = 71-81....

guest· 8 years agobiscuit0883· 8 years agoguest· 8 years agoguest· 8 years agoIf there are 500 million cats in the world and one meow's intensity level L(1)=65dB, the correct answer should be around 150dB. Why? Because intensity level is logarithmic and here's the solution: L=10xlog(I/I0) in which I0=10^-12W/m^2. We get one meow's intensity I(1)=3,16...x10^-6W/m^2. If there are 500 million cats, the total intensity I(tot)=500x10^6x3,16...x10^-6=1581,1...W/m^2.

The total intensity level L(tot) is then 152dB.

undead_red· 6 years agoguest· 5 years agoguest· 4 years agoguest· 4 years ago