ONE fan vs. FOUR fans . . . .

[Ct. Strangelove]

Will four fans rated 27.39dBA be louder than one 120mm fan rated at 42dBA?

Will four fans rated @ 32.5 CFM push more air than a single fan (120MM) @ 108CFM.

sure the numbers are greater but a 80mm fan can not push as well as a 120mm.

does any one have that dBA addition program?


*edit*

thank you!

[Cjwinnit]

1/ no, the 120mm will be louder. (four fans rated 27.39dBA is about 33.59 dbA)

2/ Four of the smaller fans will push more..

[Ct. Strangelove]

how did you come up with 33.59dBA?

[Cjwinnit]

Ok.....

1/ Get your noise sources and divide them all by 10, so 2.739 Bels.

2/ Now, where x is your previous answer, get a new answer y where y = 10^x. (so new value is 10^2.739).

3/ You have 4 sources, so times the new answer by 4.
now 4*(10^2.739).

4/ Now take the common log of the answer, so log (4*10^2.739)

That's the new value in bels, so now times it by 10 to get Decibels.

My original post was a guess, the actual answer I get with a calculator is 33.41 dBA.

[Blueacid]

Another victim of the maths ninja :D

[arnoldma]

hah just download a program here is my answer:

33.4106dbA

no brain beats the power of a ms-dos program :)

[snyper1982]

depends on who wrote the prog :)

[Since87]

Or easier than that, is remember that two noise identical noise sources increase the noise about 3 dB over a single noise source.

So 1 fan = 27.4 dB

1 fan + 1 fan = 27.4 +3 = 30.4 dB

2 fans + 2 fans = 30.4 + 3 = 33.4 dB

Cjwinnit's method is better, but this shortcut can be handy.

[Since87]

Originally posted by arnoldma
hah just download a program here is my answer:

33.4106dbA

no brain beats the power of a ms-dos program :)

hah, I was within 0.04% of the actual value and I bet I spent less time coming up with it, than you did downloading the software.

[Cjwinnit]

Originally posted by arnoldma
hah just download a program here is my answer:

33.4106dbA

no brain beats the power of a ms-dos program :)

Apart from the guy who wrote the program.

Oh, and I'm right :)

[Cjwinnit]

That took a few minutes in Paint and is still weird..

A is the total dBA.

http://iupload.net/052003/poo.jpg

In this case:

n = 4.

F = {2.739, 2.739, 2.739, 2.739}

E = {548.27..., 548.27..., 548.27..., 548.27...}

(10 ^ 2.739 = 548.27...)

Then A = 10 (log (548.27... + 548.27... + 548.27... + 548.27...))

= 10 (log (2193.107...))

= 10 * 3.3410....

= 33.41 dBA. (2 decimal places)

:)

[Toysrme]

LoL

[arnoldma]

Originally posted by Since87


hah, I was within 0.04% of the actual value and I bet I spent less time coming up with it, than you did downloading the software.

truly doubt that, the program is about 170k, im on 1mbit cable :P do the maths i download at 120k/s :D

[Since87]

Originally posted by arnoldma


truly doubt that, the program is about 170k, im on 1mbit cable :P do the maths i download at 120k/s :D

Ok, but throw in the mouse clicks involved and I beat ya. :D

[anvil82]

Originally posted by Cjwinnit
That took a few minutes in Paint and is still weird..

A is the total dBA.

..................


The bible says don't steal bandwidth, might want to fix that ;)

[Cjwinnit]

Originally posted by anvil82


The bible says don't steal bandwidth, might want to fix that ;)

I used it on another foru[M] as well ;)

[Ct. Strangelove]

wow thx for the math....

got the program linky?

lol


thx