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 its thinking from the original port formula and basically says 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" diameter port, which is to be expected. (After all, we have the same total port cross-sectional area, which this school of thought proclaims is correct!) However, 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, therefore, shifts the tuning!