Sup,I believe that this is my 1st posting here at SS.I didnt know that there was even a forum for Stereo Integrity but I'm glad to see there is one.First I like to say that the Magnum is a monster of a sub(I have the 'claw' basket),definitely one of a kind.If something were to happen to this driver I'll either get it fixed or just keep it b/c the basket IMHO is a classic.Props to Nick & the gang @ SI for well built product.As u can see by my sig.(pic in my other reply is 1.4cf) that I am presently running 1 Mag in a 1.0cf sealed box ,the output from this setup is truly amazing.Even though it sounds great sealed,I really the ported sound better.Now to the fun stuff,I have a box that is 32(w) x 17(t) x 17(d) that I would like to recycle.The dimensions mentioned works out to be 5.3cf externally;4.2cf internally.By cutting it down to 32(w)x 17(T) x 11.5(D),this works out to 3.6cf. externally;2.7cf. internally,now I figured out the port will be 2(W) x 15.5(T) x 29(long),giving me a tuning of 30hz. Do these #'s look right?.....thanx in advance 4 ne replies!

For a port that is 2" wide by 15.5" tall I'm coming up with vent lengths around 38" long. A vent of that size displaces ~0.68 cubic feet, which brings your overall enclosure volume down to right at 2 cubic feet, giving you your 30 Hz tuning.

...unless I totally missed something.

Thanks for the comments on the sub! It's actually pretty rare to hear back from people once they've bought something us. ...unless we see them on the internet.

Hey Nick thanx for the reply,I guess I just cant figure out Winsid.That's why I posted here,I just knew I'd get the right answer from either u or ur mods.I'm the one missing something ,anyways now I can proceed with the project.Luckily I waited to hear from u b4 cutting up the box. Thanx again Nick,I'll post pix ASAP....BTW,can u give me a quick tutor on how u got that figure(port length),later

Well just to check WinISD's math, I did some math with a single 4" port tuning a 2 cube enclosure to 30 Hz. Here's what I did:

Vent length = [1.463 * (10^7*R^2) / (fB^2*VB)] - 1.463*R

(58520000 / 3110400) - 2.926 = 15.08 inches.

In other words a 4" diameter vent in a 2 cubic foot vented enclosure (after port and driver displacement) would need to be 15.88" long.

WinISD's port calculation for the same alignment was 15.08". That's .8" off. Who was 0.8" off really doesn't matter (me or WinISD).

So, assuming WinISD is correct, a slot port that is 2" by 15.5" would need to be 38" long. 2x15.5x38 = 1178. 1178 / 1728 = 0.68 cubic feet of vent displacement. Subtract that volume from your total internal volume and you've got how much actual enclosure space you have to work with (which is right at 2 cubes).

I hope this helps.

PS: You can also add end correction by adding 1" to your vent length (since your port is 2" tall).

on a side note, just want to say welcome to SSA

Thanx Nick & denim,I think I got it now