New to Ports

[IDSkoT 5,000+ posts]

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 sh!t honestly confuses me.

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

Click to read more...

 

[SPY 5,000+ posts]

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 5,000+ posts]

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 5,000+ posts]

No, the whole box will be more like 2.0 to 3.0ish depending on sub displacement and port displacement.

 

[IDSkoT 5,000+ posts]

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 //content.invisioncic.com/y282845/emoticons/crap.gif.7f4dd41e3e9b23fbd170a1ee6f65cecc.gif

Back to the drawing board.

But an answer to my question would still be awesome //content.invisioncic.com/y282845/emoticons/biggrin.gif.d71a5d36fcbab170f2364c9f2e3946cb.gif

 

[IDSkoT 5,000+ posts]

Reply for bump:

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 5,000+ posts]

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 5,000+ posts]

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.