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kick@ss
12-02-2000, 08:03 PM
Does anyone know of a website that tells you how to calculate the total decibals of numerous things at numerous decibals?

coleslaw
12-02-2000, 08:34 PM
what exactly are you referring to? home speakers? auto speakers? oh, and another thing, it is decibels, not decibals (actually, it started out as deciBell (after Alexander G. Bell and got simplified to decibel) :D

coleslaw
12-02-2000, 08:37 PM
... Oh yeah, some useful websites are here (http://personal.cityu.edu.hk/~bsapplec/manipula.htm) and here (http://www.loudspeakers.net/technote/dbcalc.html)

http://personal.cityu.edu.hk/~bsapplec/Fire/Image261.gif

So basically, it's 10 times the log of the sum of the inverse logs of each decibel value over 10.

[Edited by coleslaw on 12-02-2000 at 05:47 PM]

kick@ss
12-02-2000, 10:03 PM
Fan X @ 30 dBa
Fan Y @ 27 dBa
Fan z @ 21 dBa
Lets say i have fans X, Y, and Z in my computer case and I want to calculate the total amount of dBa. I think some of these sites may help me. This is for overclocking...

coleslaw
12-03-2000, 01:29 AM
A fan of a certain decibel reading simply means that that is how loud it is, correct? Or does decibels for fans mean the measure of the air pressure supplied?

coleslaw
12-03-2000, 01:31 AM
Actually, I don't think it really matters what the measurement represents. If something is based on a logarithmic scale such as decibels, you should use the above equation to sum them.

kick@ss
12-03-2000, 01:52 AM
It represents how loud it is...
Now, if some VB Coder would help me make a program that would do all this... I was making attempts at it but couldnt get it to work.

i6s1
12-03-2000, 02:39 AM
DBs are the log of the ratio of the amount of power recieved per area from sound waves to the threshold of human hearing.

I'll use metric units.

db = 10log[( X watts/metre^2)/( y watts/metre^2)]

Where X is the amount of energy recived from changes in pressure per second per square meter, and Y is the threshold of human hearing. Its a constant, but I don't remember what it is.

But coleslaw's right, it doesn't matter, we don't need to know that.

Total db = 10log( 10^3 + 10^2.7 + 10^2.1 ) = 32.11dbs, not accounting for noise cancelation.

By the way, all this is from memory, so if it's wrong, I'm sorry. Feel free to correct it.