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Old 14th March 2004, 04:35 AM   #1
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Smile Two 8" driver equivalent to one 12"?

The max size of subwoofer driver available in my god-forsaken town is 8". However I was told that 2 of these would be equivalent to a single 12" driver, since the effective cone area is almost the same. Is this true?

No specs available for these speakers except 180W, 8-ohms.

Under these circumstances the best enclosure for this would probably be a sealed one. But would a port offer any advantage even though its not tuned? This assuming the vent is made quite large to avoid port noises.
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Old 14th March 2004, 09:40 AM   #2
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Default Re: Two 8" driver equivalent to one 12"?

Well just using basic math, surface area of a flat circle is A=3.14(r*r)

so 8"

A=(3.14)(16) = 50.24

and 12"

A=(3.14)(36) = 113.04

So you would need at least 2 8" woofers to match a single 12".

However even though they have the same area as a 12" as a system, individually they are still 8", so I don't know if these means your still limited to the frequency response of a 8" driver.

At first I thought it said 8" is the biggest allowed in your town...I was like what kind of rule is that!


PS: If that math is wrong, I'm sorry. I'm stupid
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Old 14th March 2004, 10:20 AM   #3
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well... 8inches are going to need a big, ported box to match 12inch woofers...
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Old 14th March 2004, 10:41 AM   #4
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For a sub driver, as well as Fs being important (quite likely higher for the 8" units compared to 12"), the maximum excersion needs to be considered. It is much easier to design a 12" long excersion driver than an 8", and this will limit the output at low frequencies.

If you don't have any other specs than power (ie. no T/S specs), then you will not be able to design the appropriate cabinet, and will most likely shred the speakers.

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Old 14th March 2004, 10:46 AM   #5
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Well, looking at a few of peerless' data sheets it appears as if 8" drivers have about 1/2 of the equivalent piston area as a 12" driver. However the Xmax is also about half, typically, which would mean that you would need *four* 8" drivers for the same max acoustic output.
You may design a bass-reflex box, but then you would need the T/S parameters for the driver, and it is not sure that the driver is suited for a BR box. If the T/S parameters are not available from the vendor, you can measure them yourself.
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Old 14th March 2004, 10:49 AM   #6
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Default Re: Two 8" driver equivalent to one 12"?

Quote:
Originally posted by roshan101
Under these circumstances the best enclosure for this would probably be a sealed one. But would a port offer any advantage even though its not tuned? This assuming the vent is made quite large to avoid port noises.
My opinion -> make the box as big as conveniently possible and then make the port to tune it to 30-35Hz. Even if it doesn't increase the output much at these low frequencies, it will definitely reduce the cone excursion to maybe a quarter of what it would have been. Gives a lot of scope for winding up the bass...
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Old 14th March 2004, 09:46 PM   #7
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A couple of hints.

When we talk of driver size, we talk about the outside of the basket. So a 12" driver really does not have a 12" cone. It has a 10.5" cone, or thereabouts.

Rule of thumb: for speakers whose baskets are 8" or higher, the cone diameter is 1.5 inches smaller than the basket diameter. So a 15" speaker has a 13.5" cone diameter.

For speakers with baskets under 8 inches, cone diameter is 1 inch less than basket diameter. So a 6.5" speaker has a 5.5" cone diameter. These are approximations, but they will work well enough in any program or calculation you need.

You might be surprised if you measure with a ruler. I just want to point out that these approximations include one half of the surround, so a ruler might decieve you a little. But this is the accepted method.

Although there is a tendency for cone excursion to get larger as the cone diameter increases, this is not always the case. There are long throw 8 inchers. However, I would say the average 8 incher has a cone excursion of ±4.5 mm, while the average 8 incher ±6mm. So it is greater, but not twice as great. And there are higher excursion woofers for both sizes.
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Old 14th March 2004, 10:24 PM   #8
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Quote:
Originally posted by kelticwizard
However, I would say the average 8 incher has a cone excursion of ±4.5 mm, while the average 8 incher ±6mm. So it is greater, but not twice as great.

Do you mean the average 12-incher has an Xmax of +/- 6mm? This seems fairly small if the subject is 12-inch subwoofers.
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Old 14th March 2004, 10:37 PM   #9
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You are right, that would be for the average 12 inch woofer. That would be small for a 12 inch that marketed itseld as a 12 inch sub.

For a long time, ±6mm was indeed as large as most speakers available to the DIYer. Changed about 10 years ago or so.
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Old 14th March 2004, 10:58 PM   #10
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Don't forget that using 2 8" units gives you the opportunity to parallel them.
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