Mutual coupling question

[dooper]

I found this in some ElectroVoice product literature:

"When two speaker systems are placed side by side, the woofer cones “mutually couple,” causing the two systems to act as one system with twice the effective cone area at very-low frequencies, giving an additional 3-dB increase in maximum acoustic output. Mutual coupling will occur when the frequency is such that the center-to-center distance between the two woofer manifolds is less than about one-half wavelength."

I have two identical subwoofer cabinets, placed side by side in a corner of my listening room. They are crossed over actively at 90 Hz. I've been trying to get my head around the mutual coupling thing as it applies to these boxes. Here are the numbers:

Center-to-center distance = 44.7675 cm

Estimated Speed of Sound = 34500 cm per second

Frequency = Speed of Sound / Wavelength

For Wavelength = 89.535 cm, Frequency = 34500 / 89.535 = 385.3 Hz

What I get from this is that mutual coupling only occurs above 385.3 Hz. I wouldn't call these "very-low frequencies"! What am I missing here?

EDIT: I did a search here and now I'm even more confised....

I found this formula:

Fo = c/(d*sqrt(n))

Where

"c" is the speed of sound in the air in metres per second (normally 343 is used, but can vary slightly according to air temperature and humidity).

"d" is the distance in metres between the two speakers

"n" is the number of sources.

Fo = 343 / ( .447675 * 1.4142 ) = 541.77 Hz

Help?

Thanks,

dooper

[Bill Fitzpatrick]

The mutual coupling occurs below 385Hz, not above.

P.S. The corner of the room is the worst place to put your woofers for a number of different reasons which I can't go into now due to an upcoming mutual coupling.

[dooper]

Thanks for taking time out from your courtin' to clarify for me

Cheers,

dooper

[BassAwdyO]

I dont see any Coupling per say. Anyone can see that two drivers will yeild 3db more for the sheer fact that they will be twice as powerful.

If it were like 6db or something then perhaps that would be something!

Basically what you're asking about is the concept of acoustic phase.

HERE'S FACT

Descrete frequencys have descrete wavelengths.

what you're trying to do is make two radiating bodies act as one.

[dooper]

My understanding, based on information from E-V and other sources, is that when two LF drivers are situated close together, 1 + 1 = 4, and it is indeed a 6 dB increase compared to one driver with the same available power.

Free watts are good, and so are free decibels. Are you saying that a division of Telex Corporation would lie to us?

Cheers,

dooper

[Bill Fitzpatrick]

Originally posted by BassAwdyO

If it were like 6db or something then perhaps that would be something!

When they are in parallel it is 6db. Ain't that something!

[BassAwdyO]

Yeah that really is something...

well I myself would want to test that! What I wonder is how well this works at extremely low frequencys. If its true you will get a boost of 6db at 20hz and below with speakers spread over 25' apart.

Where does that extra 3db come from?

[dooper]

The phenomenon of mutual coupling always
comes to our aid in increasing the power output of
combined subwoofer units. Figure 7-6A shows the
transmission coefficient for a direct radiator as a
function of cone diameter. The solid curve is for a
single unit, and the dotted curve is for two units
positioned very close to each other. In addition to the
double power handling capability afforded by the two
units, the dotted curve shows a 3 dB increase in
transmission coefficient at low frequencies. This is
due basically to the tendency for the two drivers to
behave as a single unit with a larger cone diameter,
and hence higher efficiency. Thus, at B, we see the
relative response of a single woofer (solid curve)
compared to two such radiators (dashed curve). Note
that the upper frequency transition point for the pair
is 0.7 that of the single unit. The more such units we
combine, the lower the effective cut-off frequency
below which mutual coupling is operant.
As an example, let us pick a large cinema with
the following physical parameters:
V = 14,000 m3
S = 3700 m2
T60 = 1.2 seconds
R = 2500 m2
We will use the JBL 2242H LF transducer.
Taking into account its power rating and its dynamic
compression at full power, we note that its power
output in acoustic watts will be:
WA = (WE x reference efficiency)10-dB/10
where WE is the transducer’s continuous power
rating (watts) and -dB is the transducer’s power
compression at full power.
Substituting the values of WE of 800 watts,
reference efficiency of .004, and power compression
of 3.3 dB, we get the value of 15 acoustical watts.
The reverberant level in a space with a room
constant of 2500 is then:

LREV = 126 + 10 log 15 - 10 log 2500 = 104 dB SPL

We can now construct the following table:
Number of Units     Maximum Level     Power Input
      1                104 dB              800 W
      2                110 dB             1600 W
      4                116 dB             3200 W

We cannot continue this process much beyond
that shown here. What happens is that the frequency
below which mutual coupling takes place falls below
the nominal cutoff frequency of the system, and
eventually all we see is a simple 3 dB increase per
doubling of elements.
For multiple subwoofers outdoors, it is best to
assume that levels fall off according to inverse
square law.

I would welcome a simpler explanation. Let's keep this going

dooper

[454Casull]

Originally posted by dooper
My understanding, based on information from E-V and other sources, is that when two LF drivers are situated close together, 1 + 1 = 4, and it is indeed a 6 dB increase compared to one driver with the same available power.

Free watts are good, and so are free decibels. Are you saying that a division of Telex Corporation would lie to us?

Cheers,

dooper

My Linkwitz SPL spreadsheet tells me the same thing (double displacement and you get an increase of 6dB of max SPL). Perhaps the increase in sensitivity is not the same as the increase in max SPL?

[SY]

Don't overcomplicate it. Let's think about a situation where two drivers are acoustically close. Ignore one of them for a moment. Feed a signal voltage to the other- it will push a certain amount of air. Now feed the same signal voltage to the other one. It will push the same amount of air. 1 + 1 = 2. You'll double the acoustic power for the same input signal voltage. That's 3dB.

Now at the same time, for that given signal voltage, you've doubled the current (you're running two motors). That means you've doubled the driving power. Voila! There's the other 3dB! 3 + 3 = 6.