Doubling speakers vs. Doubling Watts: What Happens?

[Koop]

I won't argue the math, but please remember we're dealing with the human ear and perception of sound. I know all of my amps sound louder to me through a 2x12 versus a 1x12. This may be an illusion, but you know what they say - my perception is my reality.

[Stu Blue]

Originally Posted by pdf64

Why do you think that the situation you described results in half the SPL?
Pete

See post 43 (again) Comparing a 1x12 to 2x12 cab each speaker in the latter gets half the amps energy (and therefore produces half the SPL which the internet calculators all say twice/half SPL =+/- 6db (6.02 to be exact)

Watts x speaker efficiency =SPL

Double cone area changes speaker efficiency... I'm disputing the +3db across all frequencies figure commonly given.. I "know" that's not realistic except for low (PA subs) frequencies... I don't expect any gain in the high mids for instance.

[Onioner]

Originally Posted by Koop

I won't argue the math, but please remember we're dealing with the human ear and perception of sound. I know all of my amps sound louder to me through a 2x12 versus a 1x12. This may be an illusion, but you know what they say - my perception is my reality.

Hah. I totally agree, except I put a "don't" after the word "amps," though I'd allow for a "significantly" before the "louder."

[pdf64]

Double the speakers and each gets half the watts which is minus 3db(edit sound power) to each speaker ie -6db (edit half SPL =-6db) has gone before you start adding up

Halving the power doesn't result in half the pressure, see
http://www.sengpielaudio.com/calculator-levelchange.htm

Doubling the sound pressure (voltage) corresponds to a measured level change of 6 dB.
Doubling of acoustic power (sound intensity) corresponds to a calculated level change of 3 dB.

To follow on from my previous post, power is a directionless scalar quantity, pressure is a vector quantity, it has direction.
See http://www.sengpielaudio.com/SoundPr...SoundPower.pdf sengpiel makes a useful analogy, power is the 'cause', pressure is the 'effect'.

So when we add a second speaker, the power to the first is halved, and so the pressure from it drops 3dB. Note that the pressure from the first hasn't halved.
Also, the pressure from the second speaker has now been added.
The overall power stays the same, but the overall pressure increases 3dB, due to mutual coupling between the two speakers.

Consider that if instead of a second speaker, we add a resistor (value equal to the nominal speaker impedance), the overall electrical power will be the same, the overall sound power will be half, the sound pressure will be -3dB.
Similarly if we add the second speaker in a different (sound proofed) room, the overall electrical and sound power will be the same, but the pressure (in the first room) will be -3dB (the pressure in the second room will have increased, eg from a background of say 40dB to 100dB, or whatever).
Pete

[GT100]

Ok, folks
Sent this question to the Dr Decibel over at Celestion -they got to know what's going on.

Here is his response:

"What a question! It gets very involved when you actually start to think it through. As is typical with loudspeakers you can't ever 'prove' it because you can't double the number of speakers without changing many other parameters. If you assume a theoretically perfect world though the output would be the same because -

1x V30 at 1W@1m = 100dB
In a 2x12 each speaker gets 0.5W
1x V30 at 0.5W@1m = 97dB and two similar sources increases by 3dB so total = 100dB

But this can NEVER happen due to coupling, compression, amplifer design, cabinet size & vibration, cable loss, distance to ear, dispersion, room conditions, effect of series v parallel wiring, speaker distortion and how the ear perceives it, and probably many many more.

Regards
Doc"

Basically he is saying that you can't create energy doing this.
I am surprised that he says two identical sound sources increase the sound by 3db and not 6db according to the math I shared a link to.
But probably he is using sound power level as opposed to sound pressure level. Which probably means the V30s would only produce 94db at 1/2 a watt each (sound pressure level) and when you add them you are back to 100db. The same as a single V30.

So I think the real world answer is that multiple speakers will produce slightly less volume than a single speaker (phase cancellation) with a more pronounced bass response due to inclosure size and low frequency coupling.
The real reason 4x12 cabs were invented had to do with speakers frying...

Lloyd

[pdf64]

Lloyd, in Feb 2010 I asked -
'I've long been of the understanding that doubling the number of speakers increases the sensitivity by 3dB, ie 1 greenback 97dB/watt, 2 greenbacks 100dB/watt, 4 103dB/watt. However, I can't remember where I read it, and can't google anything useful to confirm it. Can you advise whether this is correct, and point me at a textbook where I could try to understand the theory?
With the response -
'This is correct. I looked on line for a decent reference but couldn't find one. Something like Martin Colloms 'High Performance Loudspeakers' would give you some background and your local library might have a copy.'
It seems we may have a clash of opinion, or perhaps the new DrdB isn't up to speed regarding mutual coupling?

Basically he is saying that you can't create energy doing this

Creating energy - no, as explained above. Rather mutual coupling of 2 speakers is more effective at turning that scalar quantity into a vector quantity, directing it in front of the speaker. The total energy (power) is the same, just less of it goes off to the sides.

It gets very involved when you actually start to think it through

I guess that's been amply demonstrated here in the hundreds of posts over the various threads; it has been a beneficial exercise though, for me at least.
Pete

[GT100]

Originally Posted by pdf64

Lloyd, in Feb 2010 I asked -
'I've long been of the understanding that doubling the number of speakers increases the sensitivity by 3dB, ie 1 greenback 97dB/watt, 2 greenbacks 100dB/watt, 4 103dB/watt. However, I can't remember where I read it, and can't google anything useful to confirm it. Can you advise whether this is correct, and point me at a textbook where I could try to understand the theory?
With the response -
'This is correct. I looked on line for a decent reference but couldn't find one. Something like Martin Colloms 'High Performance Loudspeakers' would give you some background and your local library might have a copy.'
It seems we may have a clash of opinion, or perhaps the new DrdB isn't up to speed regarding mutual coupling?


Creating energy - no, as explained above. Rather mutual coupling of 2 speakers is more effective at turning that scalar quantity into a vector quantity, directing it in front of the speaker. The total energy (power) is the same, just less of it goes off to the sides.


I guess that's been amply demonstrated here in the hundreds of posts over the various threads; it has been a beneficial exercise though, for me at least.
Pete

I also discussed this with my brother who is a Civil Engineer (Timber frames) -he also has a Physics degree.
He was going from the top of his head but he thinks with multiple speakers you end up at the same volume -citing that you can't create energy without putting additional energy into the system.
I think the mutual coupling is minimal - certainly not 3db. No way. That is too much of a gain in energy!
Even if it was true you would in the very least have to see a law of diminishing returns. Otherwise a one watt amp could fill a stadium with enough speakers...

Lloyd

[The Kid]

I didn't read the whole thread, but here's my experience.

I had a head and a 2X12. I bought an additional identical 2X12 to be louder. It wasn't louder. It was a tad fuller, but not perceivably louder. It was a pretty expensive realization.

Although you have more speakers moving more air, the wattage is divided evenly between speakers and the excursion of the speaker doesn't go as far with the added cabinet. This results in a more "relaxed" sound, fuller lows, and a little high end roll off.

A 100W head into a 2X12 results in 50W per speaker. In a 4X12 of identical speakers, each speaker would see 25W. The speakers in the 4X12 wouldn't be pushed as hard.

A Blues Jr through two 4X12's (or fifty 4X12's if it were possible) isn't going to be nearly as loud as a JCM 800 1X12 combo. Adding more wattage (through identical speakers) will be louder.

This is all an oversimplification obviously, and hey,... what the hell do I know anyway? I'm just a guy on an internet forum. Hope everyone is well regardless what side of the debate you're on.

[The Kid]

Wow, did I solve the debate???

Yes!

(just kidding)

[James Freeman]

Die thread DDDIIIIIEEEEEE !!!!!!!!!!!!

[The Kid]

Originally Posted by ilya-v

Die thread DDDIIIIIEEEEEE !!!!!!!!!!!!

I think I killed it!

[Stu Blue]

...still alive and with new evidence here http://www.thegearpage.net/board/sho...2#post13867942

[dspellman]

Originally Posted by SatelliteAmps

Didn't bother reading the old thread, but here's a fast answer:

If the same total volume was achieved with the same amount of wattage into either one or two speakers, then there would be no reason to run a 4x12, a 2x12 or anything other than a single 1x12.

Well, wait just a minute. We move more than a sheet of air -- we move a volume of air, so there's another dimension beyond just cone area to produce "volume." So there's cone excursion ( and volume of displacement) to consider, or Vd. A speaker that has a larger Vd can move more air than the same size speaker with a smaller one. But wait, there's more. A *properly* designed closed cabinet with a tuned port (not just a random hole) can almost double the Vd at the lower frequencies.

At some limit of displacement/power handling, you'll have reached the limits to the total volume of air that can be moved by the speaker in its cabinet at a given frequency. Let's say that limit is 500W. Let's also say that the amplifier doing the work is incapable of putting out more than 500W. And let's also assume that the speaker hasn't lost any efficiency at its limits as opposed to its cruise mode.

In that case, simply adding another speaker will reduce the work that each speaker is doing (because there's no more wattage available), but it can't increase the total amount of work done, or the volume of air moved.

Coupling (acoustic, mechanical, etc.) is a whole 'nother story and far more complicated than the basic premise. But suffice it to say that you can do some interesting things with it. ISP Technologies, for example, has a subwoofer that uses a pair of very high displacement 12" speakers and a folded horn cabinet to produce far more output with 1000W than you can get with a pair of 18" speakers in a reflex cabinet with 2000W. That's a LOT more cone area and twice the wattage, but high volume displacement (and some coupling) can out-shout it at subwoofer frequencies. Coupling a pair of those 2x12s will enable you to go noticeably lower and louder, but simply adding more of these beyond, say, four total, runs into diminishing returns very quickly.

[Jef Bardsley]

At the frequency to which a port is tuned, the driver's excursion is greatly reduced. Guitar speakers do lose efficiency approaching their excursion limits. Horns are much more efficient than direct radiators but their size and directivity makes them unsuitable for stage monitors (which is the main purpose of guitar cabs now we have PAs). The speakers on the edge of an array don't get the full benefit of mutual coupling, so you need at least nine to have the center one double its output. Gains increase beyond that point.