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IDSkoT
04-18-2008, 07:22 PM
Okay, so I've been winging it-- but it's time to get edumakated.

On the Fi website it says "Ported box: 1.8-2.5 cuft @ 28-33Hz" for a 12" Fi Q.

I was wondering, does that include port displacement? I.E. 1.8-2.5 ft^3 including the port?

Don't use 'gross' and 'net' 'cause that ***** honestly confuses me.

Net = Total and gross = Box with out port, right?

SPY
04-18-2008, 07:38 PM
The specs given by Fi are after sub and port displacement. So

1.8-2.5+sub+port = gross

1.8-2.5 after sub and port = net

IDSkoT
04-18-2008, 07:53 PM
The specs given by Fi are after sub and port displacement. So

1.8-2.5+sub+port = gross

1.8-2.5 after sub and port = net

AKA the whole box should be within 1.8-2.5 ft^3?

SPY
04-18-2008, 09:04 PM
No, the whole box will be more like 2.0 to 3.0ish depending on sub displacement and port displacement.

IDSkoT
04-18-2008, 09:13 PM
No, the whole box will be more like 2.0 to 3.0ish depending on sub displacement and port displacement.

I was throwing this around:

Internal: 18"x18"x12" [2.25ft^3], the sub has a .16 displacement...
With two 2.5" (Diameter) Aeroports.

I did the math, and they have to be 7" long. I'm going to double-baffle to make it look pretty and make it strong [like bull], does the 7" include from the double baffle?

I.E. 7" from the first opening to the end, right?

EDIT:: Cache this-- I just realized my trunk's only 17" high at it's highest point :crap:

Back to the drawing board.

But an answer to my question would still be awesome :D

IDSkoT
04-18-2008, 09:47 PM

New internal dimensions: 16.5" x 13.5" x 19.25" = ~2.48 ft^3 tuned to 38hz

DB w/ two 2.5" [Diameter] Aeroports going in 6" starting from the outside of the DB

Aye or nay?

SPY
04-18-2008, 10:11 PM
It's more like 2.28 ft^3 after sub and port displacement.

so (2) 2.5" Aero's 6.47" long will give you 38hz

IDSkoT
04-19-2008, 01:28 AM
How is it 2.28?

(16.5/12) x (13.5/12) x (19.25/12) = 1.375 x 1.125 x 1.6041 = 2.4814 ft^3 - net.

As for the port (I got this formula off a Rockford Fosgate PDF Aeroport manual):

LV = (8466 x R^2 / Vb x Fb^2) - (1.463 x R)

So...

LV = (8466 x 1.25^2 / (1.16) x 38^2) - (1.463 x 1.25)
LV = (13228.125 / 1675.04) - (1.82875)
LV = (7.89719947) - (1.82875)
LV = 6.06...etc.

Am I missing something?

2.48 for the entire box, -.16 for the sub displacement... and then calculate? I mean, I don't see how I can find the area of the port and subtract it with out finding it's length.