Subtracting port volume question...
When building a slot or square ported box using 3/4"mdf; do you subtract the volume of the whole port, say if it was 2" X 8" X 12" long (external dimensions), would you sub 192in^3 ? Or, do you sub just the mdf walls themselves? In this case (3/4 X 8 X 12) + (2 X 3/4 X 12)= 90in^3 ? In this case the port runs along the corner inside utilizing two outside walls.
Does this make sense? [img]graemlins/dunno.gif[/img]
Thanks!
Gene
Does this make sense? [img]graemlins/dunno.gif[/img]
Thanks!
Gene
You subtract everything that the port takes up. The walls, and the air inside the port. Consider everything from the internal mouth of the port outwards as external - not part of the volume of air the speaker is acting on.
For example:
3" ID port x 9"
The port walls are 1/4" thick so consider it to be a 3.5" ID x 9" port
Then, if you will be using 1" MDF you can subtract 1" from the 9" because that 1" was part of the box wall and never part of the box volume
Now we have 3.5" ID x8" port
Simple math Area = pi*2r^2
3.14*3.5*3.5*8=307.8^3 In
Subtract that from the box volume
Side note:
Do you know of JVC's newer minisystems? They have speakers call the "giga-tube." An Esternal port on the top - it is essentially design laziness. They do not have to subtract the port volume because it is an external port.
For example:
3" ID port x 9"
The port walls are 1/4" thick so consider it to be a 3.5" ID x 9" port
Then, if you will be using 1" MDF you can subtract 1" from the 9" because that 1" was part of the box wall and never part of the box volume
Now we have 3.5" ID x8" port
Simple math Area = pi*2r^2
3.14*3.5*3.5*8=307.8^3 In
Subtract that from the box volume
Side note:
Do you know of JVC's newer minisystems? They have speakers call the "giga-tube." An Esternal port on the top - it is essentially design laziness. They do not have to subtract the port volume because it is an external port.
Sorry, I skirted your original question.
2" x 8" x 12" = 192 cu" As you calculated. This is the number you subtract from the box volume.
But, if the port is 12" in total length, remember that 3/4" of that depth will be passing through the wall of your box. So if you are looking for a really accurate measurement, multiply by 11.25 instead = 180 cu"
So, you are using 3/4 mdf? Then the actual size of the port (just the air) is 1.25" x 7.25" x 12" (or 12.75"?)?
I hope I haven't been too confusing.
2" x 8" x 12" = 192 cu" As you calculated. This is the number you subtract from the box volume.
But, if the port is 12" in total length, remember that 3/4" of that depth will be passing through the wall of your box. So if you are looking for a really accurate measurement, multiply by 11.25 instead = 180 cu"
So, you are using 3/4 mdf? Then the actual size of the port (just the air) is 1.25" x 7.25" x 12" (or 12.75"?)?
I hope I haven't been too confusing.
Thanks for the reply maltesechicken. [img]graemlins/thumb.gif[/img]
I kind of figured I would need to sub the whole port volume, walls and inside port airspace, I just needed to make sure.
The dimensions I gave was just an example.
Good reminder of not counting the 3/4" that pass through the face of the box, I won't forget that.
A couple of months ago, a buddy and I were passing through futureshop and I don't know if it was jvc, but we saw a mini system with the external ports; looked kinda like a cross b/w a ram, medusa, and a wmd
Thanks again.
Gene
I kind of figured I would need to sub the whole port volume, walls and inside port airspace, I just needed to make sure.
The dimensions I gave was just an example.
Good reminder of not counting the 3/4" that pass through the face of the box, I won't forget that.
A couple of months ago, a buddy and I were passing through futureshop and I don't know if it was jvc, but we saw a mini system with the external ports; looked kinda like a cross b/w a ram, medusa, and a wmd
Thanks again.
Gene
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