port

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?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.Support the12volt.com

DYohn wrote: 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.

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

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... :]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.

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.Rob

04nata wrote: 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.

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

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"...  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.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: 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.

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

Thanks

Rob