Port Question

[ChevyBoy]

when i put in my specs into an online port calculator and i want to use 2 ports and it says i need say 17", is that 17" a piece or do i divide that between the 2?

[roscoe1129]

not sure which calculator ur using but I would try a single sub enclosure and see how big of a port it says to use for one sub and then you can build ur double sub enclosure and use two of the ports from the single enclosure.

[Derrick824]

It should be 17" each.

[ChevyBoy]

well ill be building 2 different boxes with 2 10s in each. they call for 1.0 for ported a piece. so when i put in 2 cubes, 2 ports, and 32hz tuning, it gives me ~16" of port length with a 3" port. like i said though, would that be 16" per port, or 8" per port?

http://www.carstereo.com/help/Articles.cfm?id=31

thats the calculator im using.

[Eric B]

Ive always wondered this, it makes sence to divide by 2 since your finding total port length, im not sure though

[andym85]

from jl audio's site... http://mobile.jlaudio.com/support_pages.php?page_id=165

Multiple Ports

There are two widely used methods for calculating multiple ports for a single chamber. Only one method is correct but unfortunately it is the least commonly used.

The first and incorrect method takes it's thinking from the original port formula and says basically that if we take two ports and sum their cross-sectional areas, we can just plug this total into the port formula for Av to get our port length. This would sound reasonable, but it can lead to serious mis-tunings in some cases as we'll see in an example below.

The second and correct way to figure out how long each port should be follows this simple three-step procedure:

• Divide the chamber volume by the number of ports you wish to use for that one chamber.

• Take the quotient and use that as your Vb (box volume) in the port formula

• Do the number crunching and figure out how long each port should be.

Example:

Let's take an arbitrary box volume of 2.5 cubic feet that we want to tune to 25 Hz with a 4" diameter port. If we plug and chug with that big hairy formula (or let our favorite software package churn out the numbers), we'll find that Lv = 18.844 inches.

Now let's decide that we don't want just a single port because it looks boring. Let's put a 2" port in each corner of the box for a total of 4 ports and see what the two methods give us:

Method 1:

Each 2" port has a cross-sectional area of 3.142 square inches so we multiply that by 4 to get 12.57 square inches. Plugging in 12.57 for Av in the port formula yields Lv = 18.844 inches for each port.

Method 2:

We want to use 4 ports so we divide 2.5 cubic feet by 4 and get .625 cubic feet. Vb now becomes .625 cubic feet. We are using 2 inch diameter ports so Av is 3.142 square inches. Plugging these numbers into the equation leads to Lv = 20.302 inches for each port.

Notice that Method 1 produces the same port length as did our single 4" diamter port as it should (after all, we have the same total port cross-sectional area which this school of thought proclaims is correct!). But the first method is incorrect because it neglects the frictional losses encountered by using many smaller ports--there is a higher port wall surface area to cross-sectional area ratio which raises the total amount of frictional losses in the ports and thus shifts the tuning!

Edited May 1, 2007 by amatt85

[sean 371]

Great info man thanks a lot! I'll be using that I'm sure.

[ChevyBoy]

english please. i hate trying to put 4 into x and 6 into y and what not. algebra was a few years back.

[roscoe1129]

Typically whatever a manufacturer recommends for a port is per sub and they will tell u to add one per sub for maximum efficiency that y i said to take whats comes up for one sub and use two...

[ChevyBoy]

so for my box, i should use 2 8" ports correct?