CA, check my math

[maldecido33 10+ year member]

Just double checking my math

First scenario:

2.0cu ft after sub displacement

32" of port

6" diamater round port

==

32.344 Hz

OR

Second Scenario:

2.0cu ft after sub displacement

30" of port (per port)

two 4" diameter round ports

==

32.057 Hz

Both of these work out?

Enclosure is for a 12w7, fyi

Click to read more...

 

[Flipx99 5,000+ posts]

Assuming everything is correct, I would vote for the second one. As far as I tested, 2 ports are louder than 1

 

[maldecido33 10+ year member]

thats what im needing, is for someone to check to make sure the math is correct

 

[Flipx99 5,000+ posts]

according to http://www.psp-inc.com, you are correct. They make the flared-port...I highly reccomend. You can get them from partsexpress

 

[maldecido33 10+ year member]

up

 

[squeak9798 5,000+ posts]

Just double checking my math
First scenario:

2.0cu ft after sub displacement

32" of port

6" diamater round port

==

32.344 Hz

You are correct

Second Scenario:
2.0cu ft after sub displacement

30" of port (per port)

two 4" diameter round ports

==

32.057 Hz

You are correct

 

[tRiGgEr 5,000+ posts]

You are correct. But that port will have to go on the outside of the box. Because as soon as you put the port in the box, it will effect your net volume. Lowering the internal space and raising the effective tuning.

Option 1

2.0ft^3 net volume.

Box needs to be at least 2.52ft^3 for the port displacement. Not including driver displacement.

Option 2

2.0ft^3 net volume

Box needs to be at least 2.44ft^3 for the port displacement. Not including driver displacement.

What ever the driver displacement is add it to the volume I supplied and you're good to go.

 

[maldecido33 10+ year member]

You are correct. But that port will have to go on the outside of the box. Because as soon as you put the port in the box, it will effect your net volume. Lowering the internal space and raising the effective tuning.
Option 1

2.0ft^3 net volume.

Box needs to be at least 2.52ft^3 for the port displacement. Not including driver displacement.

Option 2

2.0ft^3 net volume

Box needs to be at least 2.44ft^3 for the port displacement. Not including driver displacement.

What ever the driver displacement is add it to the volume I supplied and you're good to go.

the actual ideal net volume for my driver is 1.75 cu ft (12w7) I had the .14 driver displacement factored into the volume I used above. i figured you used the net volume before port displacement in the formula. so, i need a box with a volume of 2.186ft^3 before port or sub displacement?

 

[squeak9798 5,000+ posts]

You use the gross volume less all displacements in the formula

If you calculated based on 2cuft, then after port and sub displacement you need a net internal volume of 2cuft.

 

[PV Audio 5,000+ posts]

As far as I tested, 2 ports are louder than 1

//content.invisioncic.com/y282845/emoticons/uhoh.gif.c07307dd22ee7e63e22fc8e9c614d1fd.gif What testing have you done?

 

[tRiGgEr 5,000+ posts]

Like squeak and I said. Your final required volume (net volume) is after all displacement. IE - Bracing, port, driver.

If you need 2ft^3, and you have .14 driver displacement.

Option 1

2.66 total box volume before displacements.

Exaple a box of 16.5 x 16.5 x 22 with .75" build material = 2.67

Add the speaker and the port in there and you have 2.01ft^3

Option 2

2.58 total box volume before displacements.