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    Ingenuity needed

    Okay my issue is I'm trying to figure out what to do with an 8 inch aeroport. I want to flush mount it firing up along with the subwoofers. It's going to be a tight fit the flare is 12 inches. My problem is I have too much port length for my required tuning and desire to flush mount the port. It's around 18 inches port length. On a 18-inch box height. I need six inches to clear the bottom... putting a right angle on on the port might be an option just not sure if I'm going to have the room with the 13 inch inner flare. I'm thinking if I choke down the inside diameter of the aeroport to 7 or little over 7 inches I could cut that port length down to 11-13inches. I've been going over ideas onto how to do that. Fiberglass perhaps. I wish thinking about making a bunch of wood rings and stacking them up gluing them. Sanding them down routing the edges for easy air flow..... any other ideas.... too bad somebody didn't make Port inserts at different sizes. So you could easily change your tuning at will



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    BOOMINGRANDPA's Avatar
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    Re: Ingenuity needed

    well i'll let others help u with that but cutting some off that will make it louder than 29 hz i'd do it..



    KENWOOD X301 HU , PRECISION POWER Phantom 1000, FU 750 12 (prefab until box is built) , soundqubed 6.5" coax, xover (hu) AGM starting batt. https://www.youtube.com/watch?v=rTcgVGvuFX8 https://www.youtube.com/watch?v=WI4yGDqN27g

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    Re: Ingenuity needed

    I need 18 in to be at around 35 36hz



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    Re: Ingenuity needed

    You don't need more clearance then 2 inch from the rear. There is an equation for the radius of the port to the rear of the aero port. Let me look it up




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    Re: Ingenuity needed

    Here I was wrong not the radius but the port area using the equation there to see how far the rear of the port to the box

    long round ports in small enclosures - Car Audio Classifieds!




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    Re: Ingenuity needed

    Ok. Im not fully understanding the math. Can anyone help?
    Hello!

    View Album, Port by jjzickowski1 | Photobucket



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    Re: Ingenuity needed

    i was always told if port is 6" then needs 6" away but i'm sure less would work..



    KENWOOD X301 HU , PRECISION POWER Phantom 1000, FU 750 12 (prefab until box is built) , soundqubed 6.5" coax, xover (hu) AGM starting batt. https://www.youtube.com/watch?v=rTcgVGvuFX8 https://www.youtube.com/watch?v=WI4yGDqN27g

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    Re: Ingenuity needed

    I thought that it was no less than half the diameter.. but apparently it's not the diameter that matters. I just don't understand the math to calculate in the above link. My boxes going to be. O.D. 18 x 18 x 35... after displacement I'm looking at about 4.44 cubes. Width 18 inches Port all the way inside the Box. Every aspect of car audio is a headache.. soon as you think you have one aspect of something figured out.. it tosses another aspect plan out the window.... FML



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    Re: Ingenuity needed

    WOW... learn something new every day!.. I never even considered the configuration suggested... looks like a round port playing into the rear wall with just enough clearance equaling the port area surrounded by a larger port with the exact same area additionally going the other way... what a brain teaser!.. IDK if trying to explain this is even possible as I had a tough time writing this sentence! I do not envy you the math.




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    Re: Ingenuity needed

    Someone can check my math but unless I did it wrong it looks like your 8" port to the rear wall with 2" clearance to meet your port area and then use an 11.5" ID going the other way to get your overall length... I hope I got it right. I've been in a trance studying that link and It makes sense finally... the #'s do add up... this is something I will try myself eventually... it will be my last option for sure... IDK where to find really large pvc or aeroport anyways... LOL!.. of course you'll need to account for the wall thickness of the inner port unless it's really thin... LOL!
    Last edited by shredder1; 06-19-2017 at 02:24 AM.




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    Re: Ingenuity needed

    Quote Originally Posted by shredder1 View Post
    Someone can check my math but unless I did it wrong it looks like your 8" port to the rear wall with 2" clearance to meet your port area and then use an 11.5" ID going the other way to get your overall length... I hope I got it right. I've been in a trance studying that link and It makes sense finally... the #'s do add up... this is something I will try myself eventually... it will be my last option for sure... IDK where to find really large pvc or aeroport anyways... LOL!.. of course you'll need to account for the wall thickness of the inner port unless it's really thin... LOL!
    Huh? Now im really lost



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    Re: Ingenuity needed

    Quote Originally Posted by My63bug View Post
    Huh? Now im really lost
    Well... After a good nights sleep... I've decided to try this... I was VERY confused as well until I noticed something... Let's try...I am doing this from memory now as the link won't work anymore unless I sign up... OK... how do you find the area of a circle? radius squared X pi = area... correct? so an 8" is 16 x 3.14 = 50.24 1n2 easy. We all know how to figure that. The next part however is what stumped me . The tutorial said to find the minimum clearance to the rear wall that the equation is 2pi x radius x height. There is only one variable here and that is the clearance to the rear wall as 2pi will always be 3.14+3.14... OK so 6.28 x radius is next and we know our radius is 4 so... 6.24 x 4 = 24.96... Now here comes the part that threw me... you need to find the last variable... the height. So... 2pi x radius is 24.96 in2 and what number multiplies by this to get the area of 50.24 in2 ?.. I'm getting 2.0128... now unless your using a micrometer thats 2" clearance. Now you need to find the clearance that you need from the inner port to the outer port and that's easy because we know the area of your 8" port is 50.24 in2 so... I'd account for the port material displacement as well if your using 1/4" wall PVC so the area of the inner port displacement here is now 56.716 in2 but whatever size pipe your using for the outer port must be oversize by 50.24 in2. So port area with port wall is 56.7 in2, port area needed is 50.24 in2 so add them up. 50.24 + 56.7 = 106.94.So the outer port needs to have an inner area of 106.94 in2. After a few tries I came up with 11.75" ID port as the outer. Lets check... the radius is 5.875 so using pi x radius2 = area I get 108.37 in2 now subtract the inner port with wall number which will be 108.37 in2 - 56.7 in2 = 51.67 in2. That's as close as I could get it and a lot more accurate than last time as I DID account for the material. I'm guessing that being off by an square inch won't matter. Hope this helps!
    Last edited by shredder1; 06-21-2017 at 06:15 AM.




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    Re: Ingenuity needed

    Quote Originally Posted by shredder1 View Post
    Well... After a good nights sleep... I've decided to try this... I was VERY confused as well until I noticed something... Let's try...I am doing this from memory now as the link won't work anymore unless I sign up... OK... how do you find the area of a circle? radius squared X pi = area correct? so an 8" is 16 x 3.14 = 50.24 1n2 easy. We all know how to figure that. The next part however is what stumped me . The tutorial said to find the minimum clearance to the rear wall that the equation is 2pi x radius x height. There is only one variable here and that is the clearance to the rear wall as 2pi will always be 3.14+3.14... OK so 6.28 x radius is next and we know our radius is 4 so... 6.24 x 4 = 24.96... Now here comes the part that threw me... you need to find the last variable... the height. So... 2pi x radius is 24.96 in2 and what number multiplies by this to get the area of 50.24 in2 ?.. I'm getting 2.0128... now unless your using a micrometer thats 2" clearance. Now you need to find the clearance that you need from the inner port to the outer port and that's easy because we know the area of your 8" port is 50.24 in2 so... I'd account for the port material displacement as well if your using 1/4" wall PVC so the area of the inner port displacement here is now 56.716 in2 but whatever size pipe your using for the outer port must be oversize by 50.24 in2. So port area with port wall is 56.7 in2, port area needed is 50.24 in2 so add them up. 50.24 + 56.7 = 106.94.So the outer port needs to have an inner area of 106.94 in2. After a few tries I came up with 11.75" ID port as the outer. Lets check... the radius is 5.875 so using pi x radius2 = area I get 108.37 in2 now subtract the inner port with wall number which will be 108.37 in2 - 56.7 in2 = 51.67 in2. That's as close as I could get it and a lot more accurate than last time as I DID account for the material. I'm guessing that being off by an square inch won't matter. Hope this helps!
    Thank you for your time. Your explanation was easy to understand until to the part you said about the size of my outer Port I'm using. I lost everything from that point down LOL again thank you for helping me.



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    Re: Ingenuity needed

    [Img]
    }[/img]
    The only difference between what I have and this picture is I have a 13 inch interport Flair. And the inside dimensions of my box (that this Port will be stretching across) is 16 and a half inches. I'm not sure if that changes anything.



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    Re: Ingenuity needed

    I follow what shredder is saying. Actually makes sense and is logical lol... never tried it though.

    63bug : how did you get such clean flares? I tried with 6" and didn't turn out nearly that good




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