1. ## Do I have my stuff straight

I have tried building boxes using bbp 6 and either I am idiot or something because they don't come out correct. I am trying to construct a box for my friends two L7 15s and I want to be sure my thoughts are indeed facts

a) you tune the box, not the subs

b) you make the box bigger to account for the sub displacement. Ie a box that held two subs that displaced .5 cf would have to be acutually build 1 cf bigger. For instance, if you wanted 2.5 cf per sub, you would make the box 6 cf to account for this used area.

This is actually a question:

1) Do port length calculators, such as the one for precision ports take into account the displacement of the port. The port does displace some area, correct. And that area would affect the volume of the box and therefore affect port length tuning. If it does not, I would have to make the box bigger correct?

2) What is the displacement of a 2006 L7. I learned through the sticky it is Sd x Xmax, neither which are published in the manual.

Thanks,

Josh

2. ## Re: Do I have my stuff straight

Originally Posted by Flipx99
a) you tune the box, not the subs

b) you make the box bigger to account for the sub displacement. Ie a box that held two subs that displaced .5 cf would have to be acutually build 1 cf bigger. For instance, if you wanted 2.5 cf per sub, you would make the box 6 cf to account for this used area.

This is actually a question:

1) Do port length calculators, such as the one for precision ports take into account the displacement of the port. The port does displace some area, correct. And that area would affect the volume of the box and therefore affect port length tuning. If it does not, I would have to make the box bigger correct?

2) What is the displacement of a 2006 L7. I learned through the sticky it is Sd x Xmax, neither which are published in the manual.

Thanks,

Josh
A.) Correct
B.) Correct, if each sub displaced .5ft^3 you would make a 6ft^3 box to get a net volume of 5ft^3. But you also have to account for port displacement...
1.) No it does not calculate port displacement for you. Yes the port displaces internal volume. Net volume is the enclosures internal volume after you deduct the driver and port displacement. You tune with the NET volume. So you need to deduct both port and driver displacement before calculating the tuning.
2.) Sd x Xmax = Vd not displacement. Vd is the peak diaphragm displacement volume . How much air the cone will move.

You find the driver displacement from the manufacturer. 0.09ft^3

Edit - I'm assuming you have 12's...

3. ## Re: Do I have my stuff straight

Originally Posted by tRiGgEr
A.) Correct
B.) Correct, if each sub displaced .5ft^3 you would make a 6ft^3 box to get a net volume of 5ft^3. But you also have to account for port displacement...
1.) No it does not calculate port displacement for you. Yes the port displaces internal volume. Net volume is the enclosures internal volume after you deduct the driver and port displacement. You tune with the NET volume. [bold]So you need to deduct both port and driver displacement before calculating the tuning.[/bold]
2.) Sd x Xmax = Vd not displacement. Vd is the peak diaphragm displacement volume . How much air the cone will move.

You find the driver displacement from the manufacturer. 0.09ft^3

Edit - I'm assuming you have 12's...

Not mine, but they are 15s.

As for the bolded part: you tune to net and port length is determined by net box size. Would you say you increase the box size afterwards to include net.

For instance:

I want each sub to have 5 cf so:

b) then add .16 * 2 = .32
c) then add .305 (volume of each port*4 [four ports])

So the actual box volume I will need to make the dimension to is :10.625 cf
Last edited by Flipx99; 06-28-2006 at 02:34 PM.

4. ## Re: Do I have my stuff straight

Originally Posted by Flipx99
Not mine, but they are 15s.

As for the bolded part: you tune to net and port length is determined by net box size. Would you say you increase the box size afterwards to include net.

For instance:

I want each sub to have 5 cf so:

b) then add .16 * 2 = .32
c) then add .305 (volume of each port*4 [four ports])

So the actual box volume I will need to make the dimension to is :10.625 cf
That is correct. You are getting the hang of it

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