Does anyone have info on port tuning frequency when you have an enclosure with 2 ports that are the same diameter, but different lengths? I have a .7 ft^3 net volume box with two 2" diameter ports. One is 8.5" long, the other is 4.35" long. Thanks! Brian
How to tune a vented enclosure. There is no way to exactly pre-calculate the exact length for tuning purposes. However I recommend that you download WinISD Pro (Freeware) and do simulations with the Volume of the enclosure and the Port Diameter, then when you get the Port Velocity under 18m/s then take the length given by the program then add about 2 inches to that for your prototype port. Take the Prototype port and install it in the enclosure and hook it up to your system the way it would be used when finished. Then make a CD from WinISD using the built in Tone Generator by recording a 5-10 sec passage of a Sine wave of each Tone from about 15Hz to 45Hz, then arrange these tones on a disk with each tone on a separate track. Now play this CD through the connected Sub with the prototype port, and while holding your finger lightly on the cone feel for the difference in cone movement while switching tones (Tracks). When you get to the corresponding track for the tuning frequency of the box you will notice a great reduction in the cone movement. Hopefully by cutting the port longer than you need the corresponding tone will be lower than you’re target frequency or if your lucky exactly spot on. If the tuning is lower than expected then just cut about 1 inch off per Hz to dial in the exact target frequency. This method worked great for me, and believe me when you hit the Fb of the box you will know, as the drivers cone is barely moving relative to just 1Hz below or above.
Not sure if I believe that one. I calculate tuning frequencies all the time using math formulas on DIYSubwoofers.com and the port calculator on Loudspeakers101. The question is what to do with two ports that are the same diameter but two different lengths in a box. I actually know what the box is tuned to, but I would like to know if there is a formula for it so I can change the tuning by changing the length of one or both ports. Brian
Yes you calculate, that doesn’t prove that the tuning is what the calculated target is, the method I described proves the outcome of the target, that’s why you cut the port purposely long because these programs are not that accurate. Doesn’t matter how many ports you have, using this method will prove it, believe me, when you hit the Fb (tuning frequency) of the enclosure you will know, the cone will be almost still compared to just 1Hz above or below. In addition, if you want to know the tuning with one port open an the other closed, then stuff one port and run the test tones and feel the cone movement to see the difference in Fb. These programs cannot account for the volume of air that is in the port, the port volume must be deducted from the gross volume of the enclosure, when you deduct the port volume from the gross volume it changes the tuning, which changed the port length, which again changed the port volume. This is a circular, undefeatable math anomaly that can’t be solved; you must use the math to get close then fine tune by a real world method if you want to really know the exact Fb. If you don’t really care then no big deal, you probably won’t be off more then a few Hz using the math method.
I usually recalculate three times to account for this. Look at the difference (use my car enclosure): 4.35 ft^3 (net after port and driver displacement) with 7.5" diameter port 23.75" long. tune = 30.6 Hz If the net was 4.0, same port would be 31.9 Hz. If net was 4.35, the port would have to be +/- 2" to make 1 hz difference. Point being you would have to be off pretty good to make a big difference. My box calculates to 30.6, and I have measured it to be 31 Hz using the "cone excursion" method you describe. My last ported home sub calculated to be 17.5 Hz, and measured 17 Hz. . . I think the formulas work great. Biggest error I see with people, is they do not know how to get the correct box volume if it is not a simple L x W x H calculation. Still wondering (right or wrong) if there is a math formula that can predict two ports that are not the same length. . . Brian
It's probably not a good idea. You're going to get lower resistance and so more airflow in the shorter port. Which I think means the shorter port is going to be disproportionately more significant in terms of influencing tuning frequency. Because of this you can't average the two lengths and say that it'll be like two a+b/2 equal length ports. I believe it'll be something like 2 a*x+b*(1/x)/2 ports where x is the percentage of airflow that travels down each port, though I'm probably forgettting a few factors, and I have no idea how you would calculate x. In any case if all of that's true, it'll be like two ports which are closer to the shorter size than the longer size.
I still think this method will work for multiple ports, why wouldn’t it? You said your self you have used this method to check you calculations, and since you’re familiar with how to do it this way, then why not try. The cone excursion method will not lie, regardless of what math and theory say, the only proof is in the physical (lack of ) movement of the cone.
That is what I am going to end up doing. I was hoping to have a way to calculate and predict the results to give me a starting point. . . . Brian