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  #1  
Old 09-21-2012, 12:59 PM
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Default port length question

Wassup guys i have a question regarding something thats im kinda stuck on right now. I normally use a combination on pen, paper, and win isd to do designs for me and my homeboys but i normally do ports on either side or a square port along in the middle on the bottom. If i do a slot in the middle of the subs would i just take the normal port length and double it? Or how does that work? im sure id have to take into account the port displacement im just not sure where and how everything would factor in with that particular scenario.
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Old 09-21-2012, 05:00 PM
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It would be the same length and cross sectional area as any other port, except you wouldn't compensate for a shared wall like you would if ti was on the side of the enclosure.
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Old 09-22-2012, 01:38 AM
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Originally Posted by SPLEclipse View Post
It would be the same length and cross sectional area as any other port, except you wouldn't compensate for a shared wall like you would if ti was on the side of the enclosure.
Sorry about the delay in response i was at work when i made this thread. So basically youre saying that instead on one port 27 inches long, (just picking a number) i would just have two inside port walls of 27 inches?
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Old 09-22-2012, 02:19 AM
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If the port uses one of the side walls of the enclosure as one of it's sides, you have to account for that. It's like building an internal port with two unevenly long sides. That's why with some online port length calculators it will ask you whether or not you are using a slot port. That's the only difference in length you will have to worry about. For example, in the attached picture both ports are the same tuning. The one on the left has one shorter port wall to account for the fact that the other port "wall" is longer.

Other than that, there's no other change in port design. As long as the port area stays the same the port length will stay the same, regardless of where the port is located on the box.
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Old 09-26-2012, 03:09 PM
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Ok guys so i think i have this figured out. Ill go step by step that way if ive messed it up lemme know so i can straighten it out

Box is 49wx20hx30d (using .75 wood) gives me an internal volume of 14.493 cu ft.

Port is slotted in the middle dimensions are 9.5wx18.5hx31.25d which gives a port displacement of 3.680

This would leave me with an internal volume of 10.813 and just a little or 16 sq inches of port per cu ft tuned at 33 hz.

Does this sound correct to you guys?
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Old 09-26-2012, 07:59 PM
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anyone care to chime in on this one?
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Old 09-27-2012, 02:40 AM
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I ran the numbers and got the same thing...well, it was more like 32.xxx hz, lol.

Here's what I got from your numbers. The inner port being longer than the depth of the box will require a split at the back. Keeping the port area the same means you have to divide the width in half for each side of the "T", so each opening in the back will be 4.75" wide. Accounting for the slot effect at the back of the enclosure (indicated by the effective port length in red), you need to subtract 1/2 of the port width from the calculated port depth, which leaves you with about 28.85" of "real" port length. 28.85" - 23.75" = 5.1" length for the "T" sections. I would just go with 5" or 5.5", whatever is easier to measure.
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Old 09-27-2012, 01:24 PM
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ok man cool deal. I didnt really understand what you were saying when u posted the first time but as i got into it i kinda figured it out. I had the "T" part figured out but im just not good enough at google sketch up to draw it out? Is there a way i could figure out the displacement of 45's?
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Old 09-27-2012, 01:33 PM
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To figure out the displacement of 45s, just measure the full square or the area to be displaced and divide it in half, so say you are working with a corner that is 4"x4" by 18" tall.

4 x 4 x 18 = 288/1728 = .17 cubic feet. Divide that in half for the total displacement.

.17/2 = .08 cubic feet

The red line represents where the 45 will be. the black box around it represents the "square" measurement. I find this easier to do than the volume of a pyramid.

I can't figure out sketchup either, lol. As you can see I prefer Paint. :P
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Old 09-27-2012, 03:33 PM
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Quote:
Originally Posted by SPLEclipse View Post
To figure out the displacement of 45s, just measure the full square or the area to be displaced and divide it in half, so say you are working with a corner that is 4"x4" by 18" tall.

4 x 4 x 18 = 288/1728 = .17 cubic feet. Divide that in half for the total displacement.

.17/2 = .08 cubic feet

The red line represents where the 45 will be. the black box around it represents the "square" measurement. I find this easier to do than the volume of a pyramid.

I can't figure out sketchup either, lol. As you can see I prefer Paint. :P
Lol yea man i appreciate that times a million broham. this should be a sticky as alot could prob benefit from the knowledge you just dropped
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