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    dual ports info

    were can i find info on 2 dual slot ports?

    any help would be great







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    Re: dual ports info

    what do you wanna know about em?



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    Re: dual ports info

    What kind of info do you need? Dual slot ports function no differently than dual round ports. You still calculate for the enclosure volume and target tuning frequency in the same way.




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    Re: dual ports info

    I guess i should be looking for port square area?
    what do you guys do on avg.?
    37.5 square (would that be all right for 2 12's in 3.6" box)?




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    Re: dual ports info

    3.6 inch box?

    guessing you mean 3.6 cubes? If so that is a little on the small side of port area. I would atleast bump it up to 45-50 sq inches of port



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    Re: dual ports info

    *edited for retardedness*
    Last edited by spoonraker; 08-20-2007 at 02:22 PM.



    References : chaunb3400, Team H&K, diminishedpower, pickup1, sublime06, sporty_drew, JRock10, Jando86, sundownz

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    Re: dual ports info

    Quote Originally Posted by spoonraker View Post
    Just do the calculations the same way you would for a box with a single port, only the specs of the single port come from two ports added together. Port area, length, and displacement must be added together from each port to get the total.
    completely and udderly WRONG. you divide your box in half(on paper) and calculate 1 port per 'box'.



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    Re: dual ports info

    Quote Originally Posted by jellyfish420 View Post
    completely and udderly WRONG. you divide your box in half(on paper) and calculate 1 port per 'box'.
    Thank you!



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    Re: dual ports info

    Quote Originally Posted by jellyfish420 View Post
    completely and udderly WRONG. you divide your box in half(on paper) and calculate 1 port per 'box'.
    only if the box is going to be divided... if its all one open area, the ports need to be calculated together, right?

    eaisest way would be to use a port calculator that allows to figure dual ports




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    Re: dual ports info

    Quote Originally Posted by nimbyfaygo View Post
    Thank you!
    one thing that i just cannot stand, is people spreading mis-information. i'm pretty smart...but if i'm not for positive, i wont guess. i'll let you know.



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    Re: dual ports info

    He never said the box had two chambers did he? Or does that not matter? If it does then yeah...I guess I was misinformed along with a lot of people on here because that's where I read it at.

    The way I was told is take the NET volume of the entire box. Take the total port area (both ports) and then figure your port length. Divide it in half and each of your ports should be that long...this is for a single chamber box btw, a dual chamber box is obvious....just design two boxes.
    Last edited by spoonraker; 08-20-2007 at 10:19 AM.



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    Re: dual ports info

    Quote Originally Posted by spoonraker View Post
    The way I was told is take the NET volume of the entire box. Take the total port area (both ports) and then figure your port length. Divide it in half and each of your ports should be that long...this is for a single chamber box btw, a dual chamber box is obvious....just design two boxes.
    lets see if it works.....i don't believe it will....

    for numbers sake, let use a 4cu' tuned to 30Hz. our ports will be 2 ports each at 2" x 12" so 24sq" each...and 48sq" total...

    if you divide the box by the numbers of ports first......

    Lv = [(14630000 x 7.64) / (900 x 3465)] - (1.463 x 2.76)
    Lv = (111773200 / 3118500) - 4.04
    Lv = 35.84 - 4.04
    Lv = 31.0

    now lets add the port area to get our 48sq" and use the total box volume...

    Lv = [(14630000 x 15.28) / (900 x 6912)] - (1.463 x 3.91)
    Lv = (223546400 / 6220800) - 5.72
    Lv = 35.94 - 5.72
    Lv = 30.22
    now you say to divide this by 2 and you get...
    Lv = 15.11

    as you can see the results are very different from each other.

    YOU MUST DIVIDE THE BOX BY THE NUMBER OF PORTS FIRST AND FIGURE IT OUT FOR EACH PORT.



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    Lv = [(14630000 x R^2) / (Fb^2 x Vb)] - 1.463R
    *Vb = vol. in cu"
    **R = sq.rt. A / Pi (for square/slot port)

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    Re: dual ports info

    Quote Originally Posted by Trixter View Post
    lets see if it works.....i don't believe it will....

    for numbers sake, let use a 4cu' tuned to 30Hz. our ports will be 2 ports each at 2" x 12" so 24sq" each...and 48sq" total...

    if you divide the box by the numbers of ports first......

    Lv = [(14630000 x 7.64) / (900 x 3465)] - (1.463 x 2.76)
    Lv = (111773200 / 3118500) - 4.04
    Lv = 35.84 - 4.04
    Lv = 31.0

    now lets add the port area to get our 48sq" and use the total box volume...

    Lv = [(14630000 x 15.28) / (900 x 6912)] - (1.463 x 3.91)
    Lv = (223546400 / 6220800) - 5.72
    Lv = 35.94 - 5.72
    Lv = 30.22
    now you say to divide this by 2 and you get...
    Lv = 15.11

    as you can see the results are very different from each other.

    YOU MUST DIVIDE THE BOX BY THE NUMBER OF PORTS FIRST AND FIGURE IT OUT FOR EACH PORT.
    very well put. i didn't feel like going that in depth. but now maybe people can comprehend...

    if anyone feels like reading heres a good one that tells all about it...
    http://mobile.jlaudio.com/support_pages.php?page_id=165



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    Re: dual ports info

    Wait...maybe I'm missing something. The two results were only different by seventy-eight hundeths of an inch....is that a big difference?



    References : chaunb3400, Team H&K, diminishedpower, pickup1, sublime06, sporty_drew, JRock10, Jando86, sundownz

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    Re: dual ports info

    Well after reading that JL tutorial and the examples they provided there is no doubt that you are correct, which I never doubted. I guess the arbitrary numbers we chose didn't end up being a good example

    Quote Originally Posted by JL tutorial
    Notice that Method 1 produces the same port length as did our single 4" diamter port as it should (after all, we have the same total port cross-sectional area which this school of thought proclaims is correct!). But the first method is incorrect because it neglects the frictional losses encountered by using many smaller ports--there is a higher port wall surface area to cross-sectional area ratio which raises the total amount of frictional losses in the ports and thus shifts the tuning!
    That's the part that explains why this occurs for anybody wondering. I learned something today.



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