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. 

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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.

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So the total would be about 3.6?

acer9876 wrote:

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So the total would be about 3.6?

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

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DYohn] wrote:

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acer9876 wrote:

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So the total would be about 3.6?

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

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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.

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Rob

b_real45 wrote:

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[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.

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

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Rob

b_real45 wrote:

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b_real45 wrote:

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[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.

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

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

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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.

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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."

Thanks for that info but I still don't understand.

Example: if I have a cardboard box that has an internal volume of 3 ft^3 and I place let's say an empty paper towel cardboard roll, my box's internal volume is increased by the volume of the roll? How could that be if the roll is in the box?

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Rob

The volume did not change.  The airspace has been slightly reduced by the thickness of the cardboard.  I thought it was you who first suggested that adding a port would increase airspace... or perhaps I misunderstood you.  Regardless, you do not change the volume of the enclosure by adding a port.

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DYohn] wrote:

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he volume did not change.  The airspace has been slightly reduced by the thickness of the cardboard.  I thought it was you who first suggested that adding a port would increase airspace... or perhaps I misunderstood you.  Regardless, you do not change the volume of the enclosure by adding a port.

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I agree with you - that's my thought.  Haemphyst suggests differently - unless I read his post wrong.

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Rob

When adding a port to an enclosure, the volume contained within the tube/slot/whatever is not part over the net volume of the enclosure. I'm on my phone right now, and it's a PITA to type. :) DYohn, explain net volume versus gross volume... :]

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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."

I am a layman, so let me try laymen terms <<<<<<<<<<<<delete if wrong

example:

your driver requires 3cuft volume

your port (6X16) take up .26cuft of space

your driver displacement (on mfg spec) takes up .15 cuft of space

3 + .26 + .15 = 3.41

figure your box to be 3.41cuft and after you put the prot and speaker in it will end up at the 3cuft you need.

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2004 Hyundai Sonata
Kenwood KDC-X599 HU
2-Sundown X-18D4
2-SQ Q4500.1
2-SQ Q90.4
1-Massive DBX4
8-Niche 5.25 mids
8-Niche Tweeters
4-Skar 8" mid-bass
OhioGen 220a alt
OhioGen 350a alt
156.0

I would agree that a sub's displacement will cut into the gross volume of an enclosure, but I'm not sure how a port (has both ends open) would cut into the gross volume of the enclosure.  There's the same amount of airspace in the cylinder as there was before the cylinder was placed in the enclosure.  I'm not trying to argue - I'm just trying to understand btw.  Love this forum.

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Rob

04nata wrote:

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I am a layman, so let me try laymen terms <<<<<<<<<<<<delete if wrong

example:

your driver requires 3cuft volume

your port (6X16) take up .26cuft of space

your driver displacement (on mfg spec) takes up .15 cuft of space

3 + .26 + .15 = 3.41

figure your box to be 3.41cuft and after you put the prot and speaker in it will end up at the 3cuft you need.

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see that? your not the only one who cannot type.......

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2004 Hyundai Sonata
Kenwood KDC-X599 HU
2-Sundown X-18D4
2-SQ Q4500.1
2-SQ Q90.4
1-Massive DBX4
8-Niche 5.25 mids
8-Niche Tweeters
4-Skar 8" mid-bass
OhioGen 220a alt
OhioGen 350a alt
156.0

That wasn't a mistake.. you meant it as "prototype port"... 

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Rob

Still on my phone, but here we try...

The port is a mass of air, this mass of air is what resonates within the port to provide the effect that a port provides. Ideally, the column of air within a port will never fully leave the port, in either direction. That mass of air acts against the springiness of the air inside the box, excited by the motion of the woofer.

Your analogy of the paper towel tube within a box isn't correct. In THAT case, where there are no complex acoustic effects, then yes... the port wall volume will affect the net volume of the box only very minimally. In a Helmholtz resonator (a vented enclosure), the vent volume must be subtracted from the whole.

If the air within the vent moved completely in or out of the port, in that case, you would then allow for just the vent wall's displacement, but it doesn't work like that.

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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."

haemphyst wrote:

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Still on my phone, but here we try...

The port is a mass of air, this mass of air is what resonates within the port to provide the effect that a port provides. Ideally, the column of air within a port will never fully leave the port, in either direction. That mass of air acts against the springiness of the air inside the box, excited by the motion of the woofer.

Your analogy of the paper towel tube within a box isn't correct. In THAT case, where there are no complex acoustic effects, then yes... the port wall volume will affect the net volume of the box only very minimally. In a Helmholtz resonator (a vented enclosure), the vent volume must be subtracted from the whole.

If the air within the vent moved completely in or out of the port, in that case, you would then allow for just the vent wall's displacement, but it doesn't work like that.

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Excellent info!  That's what I didn't know.  This makes sense to me now.

Thanks

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Rob

It's what we do.

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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."

haemphyst wrote:

---

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.

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This is the exact answer I was looking for. Thank you so much.
Getting this answer turned out to be far more complicated then I imagined. Haha. But thanks everybody for the input

04nata wrote:

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I am a layman, so let me try laymen terms <<<<<<<<<<<<delete if wrong

example:

your driver requires 3cuft volume

your port (6X16) take up .26cuft of space

your driver displacement (on mfg spec) takes up .15 cuft of space

3 + .26 + .15 = 3.41

figure your box to be 3.41cuft and after you put the prot and speaker in it will end up at the 3cuft you need.

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Also a good example of what I was looking for