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  #1  
Old 11-01-2010, 01:27 PM
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Default HELP! This is making NO sense!

I am going to have an enclosure built tomorrow and the information that I have on the subwoofer is making absolutely no sense to me. I need someone that really knows what they are talking about to interject. I bought an eclipse SW8010 and the way they present the info is confusing. Rather than giving figures like "displacement" and "net internal volume" they simply give this info -

recommended internal enclosure volume - 2 cuft
tuning frequency - 33 hz
internal height - 18.5
internal width - 12.5
internal depth - 20
port opening area - 2"x12.5" (25 squared)
Port length - 28.5

I'm used to seeing sub requirements in a completely different way and I cannot make sense of it. I need to change the shape of the enclosure and retain the internal volume and tuning frequency but I am unable to extrapolate anything from the info above.

If it helps the TS parameters are
nominal impedance - 4 ohms
Voice coil DCR - 3.3 ohms
FS - 21hz
Qms - 5.25
Qes - 0.39
Qts - 0.36
Vas - 28 L
MMS - 356g
Xmax - 22mm
Dia/Sd - 347 cm2
BL/No - 20 Tm
1 watt SPL - 86 dB
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  #2  
Old 11-01-2010, 01:42 PM
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20*12.5*18.5 = 4625 / 1728 = 2.677 ft^3 gross internal volume before port/woofer displacement

2.75*12.5*28.5 = 979.6875 / 1728 = 0.567 ft^3 port displacement (added 0.75 to width - assuming one wall of port is wall of enclosure and wood thickness is 0.75, this would make width 2.75 total)

2.677-0.567 = 2.11 - 2.0 = 0.11 driver displacement

Someone please correct me if I am wrong but this is what I made out of the figures given. **Fixed after SPY pointed out a mathematical error
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Originally Posted by 80INCHES
just build the box and sit back and let the bass flex the $hit out of ur ride
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Last edited by Hertz5400LincolnLS; 11-01-2010 at 01:52 PM.
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Old 11-01-2010, 01:44 PM
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say your on 3/4 inch mdf
internal height - 18.5 + .75 + .75 = 20H 14W 21.5 deep tune it to 33hz with 25 square inches of port....thats what i get from it anyway
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Old 11-01-2010, 01:44 PM
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Quote:
Originally Posted by Hertz5400LincolnLS View Post
20*12.5*18.5 = 4625 / 1728 = 2.677 ft^3 gross internal volume before port/woofer displacement

2.75*12.5*28.5 = 886.875 / 1728 = 0.513 ft^3 port displacement (added 0.75 to width - assuming one wall of port is wall of enclosure and wood thickness is 0.75, this would make width 2.75 total)

2.677-0.513 = 2.164 - 2.0 = 0.164 driver displacement

Someone please correct me if I am wrong but this is what I made out of the figures given.
Right idea but your math is a little off.

2.75x12.5x28.5= 979.6875
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Old 11-01-2010, 01:46 PM
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Quote:
Originally Posted by SPY View Post
Right idea but your math is a little off.

2.75x12.5x28.5= 979.6875

Ooops, you are right...I was even using a calculator. Not sure how that happened.

**Fixed it
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Quote:
Originally Posted by 80INCHES
just build the box and sit back and let the bass flex the $hit out of ur ride
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Last edited by Hertz5400LincolnLS; 11-01-2010 at 01:51 PM.
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Old 11-01-2010, 01:53 PM
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Thanks guys. For some reason the way they presented it is totally backwards from how I normally look at the info. It was driving me nuts!
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Old 11-01-2010, 02:04 PM
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I am confused as to why you found this difficult to comprehend.
2ft3 net
25in2 port 28.5in long
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Old 11-01-2010, 02:07 PM
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Quote:
Originally Posted by Benny33 View Post
I am confused as to why you found this difficult to comprehend.
2ft3 net
25in2 port 28.5in long
Maybe he didn't know how to do the calculations and work backwards? Last time I checked there are forum members with all different levels of knowledge on here. Ridiculing a new member is not the best way to retain them.
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Quote:
Originally Posted by 80INCHES
just build the box and sit back and let the bass flex the $hit out of ur ride
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