Question about port length.

I got 23.5" //content.invisioncic.com/y282845/emoticons/eyebrow.gif.fe2c18d8720fe8c7eaed347b21ea05a5.gif so yeah........ hmmm....

Inaccurate programs FTL

 
//content.invisioncic.com/y282845/emoticons/fyi.gif.9f1f679348da7204ce960cfc74bca8e0.gif to change port area, just change Xmax //content.invisioncic.com/y282845/emoticons/fyi.gif.9f1f679348da7204ce960cfc74bca8e0.gif
 
Question: Are you going to use the height as a port opening? If so, it will only be 15.5" and not 16.

As far as dimensions, I used a port of 16x7.5 and a length of 23.75" and that gives a tuning of about 33.3hz and an enclosure net volume of ~8.8 cubes after sub displacement.

 
Box volume in inches using 3/4" wood (H= exterior height; W= exterior width; D= exterior depth):

(H-1.5)(W-1.5)(D-1.5)/12^3

I use Winisd Alpha to calculate port length.

If you use three walls of the enclosure for your port, this is how I get port volume (Ph= interior port height; Pw= interior port width; Pl= interior port length):

(Ph)(Pw+.75)(Pl)/12^3

The +.75 acounts for the width of the wood used to make the port.

Subtracting these two will give you the net volume in the enclosure. After that, I take away about .05 cubic feet per sub but this varies.

 
Subtracting these two will give you the net volume in the enclosure. After that, I take away about .05 cubic feet per sub but this varies.
My 12" sub takes up .09 cu ft and it's a TC-1000 just to give some reference.

 
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