I went and built a ported enclosure today at work and I made the port too big for the enclosure. If I keep the height and length of it alone, then I will need to take off 4 1/2 inches on each side of the port. I dont want to bust apart the box, but was thinking If I would cut pieces the same length of the port and same height, could I just block in the sides till I get the correct width?
Perfect. Ok. I went and put the sub in the box. She has a 10" rockford foasgate H2 dual 4 ohm coils. My volume came out to 1.58????. Now I made my tube 1 7/8 high, 15" wide and 11" deep. In order for it to be properly tuned to 35 HZ, I need to bring each side in 4 1/2". That way I can keep the heigth and length. But I listened to the sub in her car and it sounded great. It used to be in a sealed box. The low bass sounds good. I dont have any port noise. The low bass doesnt sound dead nor "airy-like". But I know I dont have the right port size. So I wanted to know, If it sounds good, all bass notes hit perfectly, can I just leave it go or will it mess up the sub?
The 1.58 cu ft is figured after you deducted port structure volume, correct? If so, leaving the port as it is results in a tuning freq of about 56 Hz. If you block up 4.5" on each side of the port, or 9" total, to reduce port dimensions to 6" X 1 7/8" and 11" length you will lower Fb to about 44 Hz. I think that it would have to sound better with the lower tuning freq.
But since it sounds good the way it is, you might try this: cut your blocking pieces and insert them into the port but don't glue them (just a dab of caulk to hold them from making noise). Make your decision based on the difference that you hear. A single 9" X 2" X 11" piece of shipping styrofoam would do the job quite handily for the test.
And here's a way you can use the port length calculator on this site to help you out. When you know the slot port opening dimensions you can use pi R squared to find what the equivalent round port would be. Use the formula in reverse:
15" X 1.87" = 28
28 / 3.14 (pi) = 8.93
square root of 8.93 = 2.99
2.99 X 2 = 6 (real close)
6" diameter is the equivalent of the port opening size you now have. If you block out 9" off the 15" width, then equivalent 4.6 inch diameter is what you are looking at.
Use the calculator included with your computer to do the math, including the square root step in the equation. Anyway, that's how I did the math to get those results. There are some other port calculators floating around cyberspace but I didn't want to look.
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Build the box so that it performs well in the worst case scenario and, in return, it will reward you at all times.