port

I'm building a 3ft^3 box for a 12" l7. Im porting it to 34hz. And everythings okay until I get down to finishing with the port. I'm using the port calculator on here. But I'm not sure exactly how it cAlculates it. Do I stick the port inside those 3^3ft? Or does it need to have the 3^3ft space and then also the port area? Long question short. After using the port calculator will my total airspace be 3^3ft? Or 3^3ft plus port?

If your enclosure is 3 cuft after deducting the volume occupied by the driver and bracing inside the enclosure, then enter 3cuft in the port calculator. Support the12volt.com

I know how to put it in the calculator. Let's say it says I need a 20" port. Do I incorporate that into my airspace? Meaning that'd it'd only really be about 2.4^3ft airspace and the rest port? I apologize. Im not sure exactly how to word this question

No.  Although if you use MDF or something else that is substantial to form the port, you need to deduct the volume of the rigid material from the available volume.Support the12volt.com

So the total would be about 3.6?

acer9876 wrote: So the total would be about 3.6?

??  Adding a port does not make your enclosure larger. 

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DYohn wrote: acer9876 wrote: So the total would be about 3.6? ??  Adding a port does not make your enclosure larger.

Acer,

What he's trying to say is that the structure used to create the port is usually thin enough to be negligible in volumn.  The actual port is open on both ends and does not seal its volumn away from the enclosure.

Rob

b_real45 wrote: [QUOTE=DYohn] Acer, What he's trying to say is that the structure used to create the port is usually thin enough to be negligible in volumn.  The actual port is open on both ends and does not seal its volumn away from the enclosure.

Sorry about the spelling guys.. I'm "new" and cannot edit my post.  I've been typing the work "Column" all day so ...   *sigh*

Rob

b_real45 wrote: b_real45 wrote: [QUOTE=DYohn] Acer, What he's trying to say is that the structure used to create the port is usually thin enough to be negligible in volumn.  The actual port is open on both ends and does not seal its volumn away from the enclosure. Sorry about the spelling guys.. I'm "new" and cannot edit my post.  I've been typing the work "Column" all day so ...   *sigh*

I think I'll just quit typing all together today.

Rob

A box volume specified by a driver manufacturer is a *net* volume! it can be dramatically larger in GROSS volume, however. If you need 3 cubes, you will figure 3 cubes, and then ADD TO IT. You will add to it the displacement of the driver basket and cone, the bracing, and the entire volume of the port, NOT JUST THE PORT WALLS, b_real! As you can see, your overall outside volume can be FAR greater than the 3 cubic feet specified If you are stuck with a given outside volume, then you subtract the volume occupied by all of the above mentioned components AND the wall volume, to arrive at your net volume, which is the volume the driver interacts with. After all that, in answer to the OPs question, you will have 3.3 cubic feet, plus the entire port volume, not wall volume. Calculated this way: (pi) times (radius, squared) times (height) or: (3.1415)(r^2)(port length) If you have a 6" port, 16 inches long, it will calculate like this: 3.1415 X 3^2 X 16 3.1415 X 9 X 16 452.4 cubic inches 452.4/1728 = .26 cubic feet. You can see that if you don't add this, you can seriously and adversely affect the desired volume calculations.It all reminds me of something that Molière once said to Guy de Maupassant at a café in Vienna: "That's nice. You should write it down."