ONE fan vs. FOUR fans . . . .

Ct. Strangelove · Jun 2, 2003

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 · Jun 2, 2003

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 · Jun 2, 2003

how did you come up with 33.59dBA?

Cjwinnit · Jun 2, 2003

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 · Jun 2, 2003

Another victim of the maths ninja

arnoldma · Jun 2, 2003

hah just download a program here is my answer:

33.4106dbA

no brain beats the power of a ms-dos program

snyper1982 · Jun 2, 2003

depends on who wrote the prog

Since87 · Jun 2, 2003

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 · Jun 2, 2003

arnoldma said:
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 · Jun 2, 2003

arnoldma said:
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 · Jun 2, 2003

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

A is the total dBA.

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 · Jun 2, 2003

arnoldma · Jun 2, 2003

Since87 said:
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

do the maths i download at 120k/s

Since87 · Jun 2, 2003

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

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

anvil82 · Jun 2, 2003

Cjwinnit said:
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 · Jun 2, 2003

anvil82 said:
The bible says don't steal bandwidth, might want to fix that

I used it on another foru[M] as well

Ct. Strangelove · Jun 3, 2003

wow thx for the math....

got the program linky?

lol

thx

ONE fan vs. FOUR fans . . . .

Ct. Strangelove

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Cjwinnit

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Ct. Strangelove

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Blueacid

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arnoldma

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