Questions about dB Options

[Sebastian Mares]

Hello!

I have two questions:

1. Are two fans at 12 dB louder than one fan at 12 dB?
2. Is 24 dB twice as loud as 12 dB?

Regards,
Sebastian Mares

[Otto42]

1. Are two fans at 12 dB louder than one fan at 12 dB?
2. Is 24 dB twice as loud as 12 dB?

1. I don't think it's additive that way, although you could get it in stereo.

2. No. 22 dB is twice as loud as 12 dB. Decibels are logarithmic in nature. Adding 10 dB = double the volume.

[Omion]

1. Are two fans at 12 dB louder than one fan at 12 dB?
2. Is 24 dB twice as loud as 12 dB?

1. I don't think it's additive that way, although you could get it in stereo.

2. No. 22 dB is twice as loud as 12 dB. Decibels are logarithmic in nature. Adding 10 dB = double the volume.

2. You're right that decibels are logarithmic, but you got the scale wrong. Adding 10dB will give you 10 times the volume. Double the volume is around 3dB.

(The root unit is actually the bel, which is 10 decibels. +1 bel = 10 times the volume. Nobody uses bels, though, because they're too coarse)

[breez]

10dB, yes, 10 times the Sound Pressure Level, but human ear perceives it as "double as loud as".

[Gabriel]

I'd say 12dB + 12dB = 15dB

[Jebus]

Just because you got 3 conflicting responses, I should assert that Omion is correct. And because +3db is doubling, so is Gabriel.

This post has been edited by Jebus: Jul 26 2004, 17:09

[sPeziFisH]

I found this interesting stuff:

Dezibel @ wikipedia.org (german)

Look at the paragraph before 'Siehe Auch'.

If you are interested in and follow the given links you will also find sth. with regard to question 1 under weblinks at the 'Schalldruckpegel'-page:

noncoherent source

coherent source

so it should be sth. between 15 and 18 dB

(Similar informations are also in the english wikipedia:
dezibel @ english wikipedia)

Die Seite is echt schon ganz brauchbar, schaue da in letzter Zeit öfter rein

[Otto42]

Sorry, but you're all wrong. 3dB doubles the needed POWER to produce that SPL level in a speaker, but it does not double the perceived volume.

An 85dB sensitivity speaker will have power/dB ratios as follows:
Power SPL
1 watt 85 db
2 watts 88 db
4 watts 91 db
8 watts 94 db
16 watts 97 db
32 watts 100 db
64 watts 103 db
...

And so on. But power != volume, and perceived volume will double at a roughly 10dB increase, not 3 dB.

Edit: if you still think I'm wrong, I suggest that a few minutes with an SPL meter and an amplifier will enlighten you. Really.

This post has been edited by Otto42: Jul 26 2004, 17:23

[Gabriel]

Summary:

Two 12dB fans will produce 15dB.

You will not perceive your two fans to be twice as loud as a single one.

[Omion]

Sorry, but you're all wrong. 3dB doubles the needed POWER to produce that SPL level in a speaker, but it does not double the perceived volume.

An 85dB sensitivity speaker will have power/dB ratios as follows:
Power SPL
1 watt 85 db
2 watts 88 db
4 watts 91 db
8 watts 94 db
16 watts 97 db
32 watts 100 db
64 watts 103 db
...

And so on. But power != volume, and perceived volume will double at a roughly 10dB increase, not 3 dB.

Hmm... I think you may be right. I was under the impression that volume was power, but looking around the internet, I see that you may be right. (although some web sites think that 6dB is double the volume, since it is double the amplitude. That's just completely wrong.) There seems to be many conflicting opinions of this around.

[cabbagerat]

As people said above, two 12dB fans will make 15.0103dB. This corresponds to double the sound energy being emitted. You can calculate this yourself by saying:
Energy=10^(12/10)
new dB = 10*log10(2*Energy)

For the percieved volume to double (in general, it's frequency and audience specific, but it's a good rule nonetheless) the sound energy needs to increase by 1 Bel or 10 deciBels. This means you would need ten fans to sound twice as loud.
Because:
fan dB = 10*log10(fan Energy)
to double loudness:
fan dB+10 = 10*log10(fan Energy)+10
fan dB+10 = 10*[log10(fan Energy)+1]
fan dB+10 = 10*[log10(10*fan Energy)]

To directly answer the question:

1. Are two fans at 12 dB louder than one fan at 12 dB?

Yes. Two fans produce 15dB which is perceptibly louder than 12dB

2. Is 24 dB twice as loud as 12 dB?

No, 22dB is twice as loud as 12dB.

This post has been edited by cabbagerat: Jul 26 2004, 17:44

[rfarris]

Some random comments about this thread:

Doubling power is an increase of 3 dB. Doubling amplitude is an increase of 6 dB.

In electrical engineering terms, where power is measured in watts and amplitude is measured in volts:

dB = 10 log p1/p2

and

dB = 20 log v1/v2

The log (base 10) of 2 is about .3, so when the ratio is 2::1, the formulas degenerate into 10 * 0.3 and 20 * 0.3, leading to the aforementioned 3 dB and 6 dB. When you see conflicts where one source claims 3 dB and the other claims 6 dB, it's because one is referring to power and the other to amplitude.

However, as Otto42 mentioned, a doubling in power doesn't lead to a doubling in audible perception.

[WmAx]

As rfarris pionted out, observation needs to be taken of power vs. amplitude. SInce SPL is represented as amplitude....

Two fans would likely = 3dB louder then a single unit. The acoustic output is not correlated(in phase). If they acoustic output was correlated, they would = 6dB louder output, which is double actual SPL/amplitude.

It was determined many years ago in controlled audibility testing, that the following rules were generally accurate among the population:

6dB SPL increase is percieved as an approx. 50% increase in volume by a sample group.

10dB SPL increase is percieved as an approx. 100% increase in volume by a sample group.

-Chris

Some random comments about this thread:

Doubling power is an increase of 3 dB.  Doubling amplitude is an increase of 6 dB. 

In electrical engineering terms, where power is measured in watts and amplitude is measured in volts:

dB = 10 log p1/p2

and

dB = 20 log v1/v2

The log (base 10) of 2 is about .3, so when the ratio is 2::1, the formulas degenerate into 10 * 0.3 and 20 * 0.3, leading to the aforementioned 3 dB and 6 dB.  When you see conflicts where one source claims 3 dB and the other claims 6 dB, it's because one is referring to power and the other to amplitude.

However, as Otto42 mentioned, a doubling in power doesn't lead to a doubling in audible perception.

[Pio2001]

If they acoustic output was correlated, they would = 6dB louder output, which is double actual SPL/amplitude.

The power is proportional to the square of the amplitude (for example P = U^2/R).
If the amplitude doubles, the power becomes 4 times the initial one.
If you manage to double the amplitude, and get 4 watts from two 1 watt fans, where do the 2 extra watts come from ?
Just asking.

[WmAx]

For doubilng of amplitude, I have always used:

6dB: 20 * log10(2/1) = 20 * 0.3 = 6

INdded, in application, this has always held true as long as the vector of both correlated(in phase) SPL sources were localized to within a small proportion relative to the wavelengths examined). Sound pressure change is only an amplitude change.

Here is an easilly illustrated electrical example of double amplitude = double power:

V * I = W(power)

Assume a 10 volt peak to peak AC signal @ 10 amperes/current.

10 * 10 = 100(power/watts)

Double voltage amplitude(voltage=20), same current/amperes:

20 * 10 = 200(power/watts)


-Chris

If they acoustic output was correlated, they would = 6dB louder output, which is double actual SPL/amplitude.

The power is proportional to the square of the amplitude (for example P = U^2/R).
If the amplitude doubles, the power becomes 4 times the initial one.
If you manage to double the amplitude, and get 4 watts from two 1 watt fans, where do the 2 extra watts come from ?
Just asking.

This post has been edited by WmAx: Jul 26 2004, 19:46

[Omion]

If they acoustic output was correlated, they would = 6dB louder output, which is double actual SPL/amplitude.

The power is proportional to the square of the amplitude (for example P = U^2/R).
If the amplitude doubles, the power becomes 4 times the initial one.
If you manage to double the amplitude, and get 4 watts from two 1 watt fans, where do the 2 extra watts come from ?
Just asking.

Say you have two sound sources, each at the same volume and frequency. There are positions where you can measure a 6dB increase in power (=in-phase), but there are just as many where you get a complete cancelling of power (=out-of-phase). Baseically, there are points that "get" 4 watts, but some that don't get any, so it all averages out in the end to 2. You're just borrowing power from certain points and moving them to others.


Would I be correct in summarizing the thread so far:
3dB = twice the power
6dB = twice the amplitude
~10dB = twice the perceived volume
Adding up two 12dB noise sources will get you, on average, 15dB (which will not sound twice as loud)

This post has been edited by Omion: Jul 26 2004, 20:18

[rfarris]

The power is proportional to the square of the amplitude (for example P = U^2/R).
If the amplitude doubles, the power becomes 4 times the initial one.
If you manage to double the amplitude, and get 4 watts from two 1 watt fans, where do the 2 extra watts come from ?

You're a sly one, Pio, but in your example, when you add the extra fan, you add the extra R, also, and being on the bottom, that halves the power, so instead of the 4 watts you were hoping for, you ended up with the 2 watts from the two 1-watt fans, not 2-watts a piece.