Posted July 15, 200916 yr Ive been told through another post I need to stick to Atomics enclosure reccommendation to get the most out of my APX 18 the enclosure info says sub and port displacement is already figured in their equation which is APX18-ported 6.0ft3 (40six10”long) ( 32Hz)=5000w (SB) So do I figure 6 cubes as if it was sealed then do the port to fit their req.? and I dont know how to read thier port rec. I see a 10" long port, 6" wide, but whats the 40. or is it 6" in height? Please help so I can get this box built! Or if someone wants to give me an easy out and tell me exact outside measurements (.75 MDF.) I have up to 50" in width, 25" in height and 19" depth. Thanks guys!!!
July 15, 200916 yr Are you going to be doing a slot port? P.S. I think the recommendations are 6 cubes after Displacements (Sub, Port, MDF)
July 16, 200916 yr Author Ok its 6 cubes net. Now I have to figure out the port (slot) Atomic says to do what they say period. The port area is suppose to be 40 sq."x10"long. How do I figure out what equals 40 Sq. ". Do I for example do 8"tall and 5" wide since 8x5 is 40? Then 10" long/deep?
July 16, 200916 yr yes.. it's that simple.Just stick to the 8:1 rule of thumb.Dont make a 20x2 port.. that's 10:1, u see?
July 16, 200916 yr Author Even though the port is 10" long would the 40 sq. in x 10 long still be just 40" of port area or do I multiply the 40" by something to figure out my port area. 40sq. in. seems tiny!
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