AP Zoutes Posted August 28, 2006 Report Share Posted August 28, 2006 Try to come up with my own plans since I can't think of somebody else that can do it right now so here I go.. Dimensions- 48"= W 18"=H 34"= D Multiplied the numbers 46.5 X 16.5 X 32.5= 24,935.625 Did 24,935.625 divided by 1728 and that = 14.430 14.430 Divided by two = 7.215 So that would be aproximetly 7.2 cubic ft per chamber before port displacement. Using four 6" ports, two per chamber, trying to get a tunning of 35hz. I would cut each port at 13" (length). What do you guys think? Do you think by those dimensions it would sound good or no? This is for two 18" BTLs/ MTs. I am really not an "expert" at this, so i might be wrong, hopefully someone can clarify this if i am but... the only thing that doesn't look right to me is that you did not account for the center divider, that seperates the two chambers of the box.. the 48" in width, should have had .75 + .75 +.75 = 2.25" subtracted from it. That accounts for the two ends(.75" each) and the center wall .75".. here are my calculations... 48 - 2.25 = 45" 18 - 1.5 = 16.5" 34 - 1.5 = 32.5" 45 x 16.5 x 32.5 = 24131.25 24131.25/1728 = 13.965 ft^3 13.965/2 (chambers) = 6.9824 ft^3 (2) 6" ports in each chamber @ 35hz = 13.43" the only thing that i did not account for is speaker displacement, since i have no idea what it is... :^ You forgot ot subtract port displacement as well..... Like I mentioned build the enclosure to your outside demensions one it's all done measure internal and subtract displacements then deside you port length Quote Link to comment Share on other sites More sharing options...
Ernesto Posted August 28, 2006 Author Report Share Posted August 28, 2006 It's not that I forgot the subtract port displacement it's that im unsure of it yet :- Quote Link to comment Share on other sites More sharing options...
AP Zoutes Posted August 28, 2006 Report Share Posted August 28, 2006 guestimate Quote Link to comment Share on other sites More sharing options...
AP Zoutes Posted August 28, 2006 Report Share Posted August 28, 2006 j/k Don't stress about it your on the right path. Build first stress later :-[ Quote Link to comment Share on other sites More sharing options...
dmz2711 Posted August 28, 2006 Report Share Posted August 28, 2006 BTW DMZ your right. I forgot to do it at 2.25 instead of 1.5 alright cool, here it is w/ sub displacement, goodluck man better do a worklog with a ton of pics :^ 48 - 2.25 = 45" 18 - 1.5 = 16.5" 34 - 1.5 = 32.5" 45 x 16.5 x 32.5 = 24131.25 24131.25/1728 = 13.965 ft^3 13.965/2 (chambers) = 6.9824 ft^3 so 6.9824 - .26 (sub displacement)= 6.7224 ft^3 (2) 6" ports in each chamber @ 35hz = 14.12" Quote Nobody makes me bleed my own blood, NOBODY!!! Link to comment Share on other sites More sharing options...
AP Zoutes Posted August 28, 2006 Report Share Posted August 28, 2006 do you know how to figure port displacement? Quote Link to comment Share on other sites More sharing options...
dmz2711 Posted August 28, 2006 Report Share Posted August 28, 2006 do you know how to figure port displacement? well, you have to figure out your port measurments first. Like ernesto knows that he wants (2) 6" diameter ports, and he needs them to be 14.12" long... A = pi x r^2 the radius of a 6" diameter is 3" (simply 6/2 = 3) A = 3.14 x 3^2 = 28.26 now i'm pretty sure you take 28.26 x 14.12 (length of the port) = 399.0312 399.0312/1728 = .231 ft^3 which is the displacement of each port .231 x 2 (2 ports in each chamber) = .462 ft^3 total displacement of both ports in each chamber... once again, i am pretty sure thats how its done, hopefully someone can clarify this Quote Nobody makes me bleed my own blood, NOBODY!!! Link to comment Share on other sites More sharing options...
AP Zoutes Posted August 28, 2006 Report Share Posted August 28, 2006 you are right exceptyou need the total width of the port. In other words the ports will be 6" internal but maybe 6 1/2" total across external. But close enough :^ It's also the length of the port inside the enclosure. FOr example if the port is 14.12" and it's flush with the surface af 1" MDF then you would subtract 1" to make it 13.12" Quote Link to comment Share on other sites More sharing options...
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