lol, don't I feel stupid! I spoke too soon //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gifand why do people buy plans again? it's that easy??? lol.
I think I'm thinking too hard...
lol, don't I feel stupid! I spoke too soon //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gifand why do people buy plans again? it's that easy??? lol.
Now, I didn't know you could assume how much port area a sub needs. Maybe Donn or Jmac can answer that....I'll assume that the box needs about 50 square inches of port (this is suitable for a single 15" or a pair of 10's or 12's in most cases) and about 4 cubic feet after all displacements
Av is your port area.
from a post above,
Now, I didn't know you could assume how much port area a sub needs. Maybe Donn or Jmac can answer that.
But if you already have a set dimensions of the port, it'd just be port_width * that_other_dimension. Like in my case, my box's max height was 16.5, therefore 15 inches internal. Then I just chose a port_width of 1.75 arbitrarily because just from analyzing the equation of Lv (as related to Av, Fb, and Fb), the more port area, the longer the port; which makes sense, cause the bigger the area of the port, the less resistance therefore the length of the port has to make up for that resistance. I'm just guessing though, again, more experienced people like Donn or Jmac would know the real answer/reason.