Ok, I'm trying to design a box for my Alpine Type-R powered by a 500/1. My internal dimensions (D X H X W) are 9 X 16 X 22. That gets me 1.83 cubic feet. Once i take away the sub displacement and port displacement it will be around 1.5 cubic feet. I am useing 3/4 MDF so the actual box will be bigger. Well i want to make the box ported. I get the equation posted on here my CPLKittle. The only part i dont understand is the Vb for the equation. Do i use the Vb not including the port and sub displacement in the equation becuase i cannot find the port lenght to find the final displacement until the equation is over. I'm just a little confused. If i use the 1.83 volume i got without the port i get a length of 13.3387 @ 33Hz. Once i get that displacement and the sub displacement of .075 I get a final volume of 1.5557. Is this math correct and a good set-up? Thanks for all the help! Oh and alpine recomments .75-1.75 for ported.
Replies (2)
cplkittle on 03/18/2008 20:28:15
The port displacement is something you have to work with. Once you get the port size and length, you have to go back and do it all again with the port displacement figured in. The end result will be a minimal change in port length (1/2 - 3/4") in most cases that only changes the tuned frequency by 1 Hz.
your box dimensions are different here than in the other post. (30x20x9.5), so the port size will be different as well.
If the port is too long, making the port size smaller will shorten the port length. Here are some general rules:
larger box = shorter port
larger port area = longer port
longer port = lower frequency
small port area =port noise
Small box + low frequency = really long port (depending on port opening area)
F150Monster on 03/18/2008 22:53:18
Ok well i went back and used the same equation but i figured in for the port and sub displacement and i got a length of 16.2295 instead of 13.3387 for 33 Hz. Is my math correct? I have a relatively small box and i'm going for a low frequency. I dont have 16.2295 inches.
Ok i have done the math. I get a 13.8 inch port for 35 Hz with all the displacement calculated for, so i think i am going to go with a 9 X 16 X 22 Box with a 14 inch 2 X 6 Port for around 34-35 Hz. That will give me 2.75 inches from the bottom of my box to the port opening. Thats acceptable right? Is my math correct on everything? Thanks cplkittle!
Replies (2)
cplkittle on 03/18/2008 20:28:15
The port displacement is something you have to work with. Once you get the port size and length, you have to go back and do it all again with the port displacement figured in. The end result will be a minimal change in port length (1/2 - 3/4") in most cases that only changes the tuned frequency by 1 Hz.
your box dimensions are different here than in the other post. (30x20x9.5), so the port size will be different as well.
If the port is too long, making the port size smaller will shorten the port length. Here are some general rules:
larger box = shorter port
larger port area = longer port
longer port = lower frequency
small port area =port noise
Small box + low frequency = really long port (depending on port opening area)
F150Monster on 03/18/2008 22:53:18
Ok well i went back and used the same equation but i figured in for the port and sub displacement and i got a length of 16.2295 instead of 13.3387 for 33 Hz. Is my math correct? I have a relatively small box and i'm going for a low frequency. I dont have 16.2295 inches.
Ok i have done the math. I get a 13.8 inch port for 35 Hz with all the displacement calculated for, so i think i am going to go with a 9 X 16 X 22 Box with a 14 inch 2 X 6 Port for around 34-35 Hz. That will give me 2.75 inches from the bottom of my box to the port opening. Thats acceptable right? Is my math correct on everything? Thanks cplkittle!
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