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Thread: Port size change.

  1. #1
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    Port size change.

    I'm using an 8'' JBL woofer that calls for a 2'' diameter X 5.5'' deep port. Using a 3'' diameter port, what would the depth in inches be?
    Gary Miles

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    Moderator / Treasurer/Marketplace Czar boputnam's Avatar
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    Volume of a Right Cylinder

    You just need the equivalent volume.

    V = pi * r^2 * h

    V = 17.279 in3, where r = 1 in (d = 2)

    For r = 1.5 in, h = 2.444 in

    bo

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    donīt want to offend, but thatīs wrong. If you increase the port diameter you need a longer port.

    As a rough starting point you should expect your required new port length to be:

    old length X ratio of port areas, i.e. for your example:

    SQR (3/2) = 2.25 X 5.5 = 12.375 inches.

    Due to port end correction that errs on the short side i.e. the real required length should be a bit more than that probably 13 -14 inches

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    Originally posted by Claus K
    don't want to offend...
    No worry there...

    I've got to re-check some things at home, but regardless, it seems that directionally your method is correct. I just want to verify the result (FWIW). I should delete my prior post for correctness, but the time has passed for such things...
    Last edited by boputnam; 03-12-2004 at 11:33 AM.
    bo

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    Port size change.

    Thanks for your help. Luckily I can install such a port into the pre - built cabinets.
    Gary Miles

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    Moderator / Treasurer/Marketplace Czar boputnam's Avatar
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    Originally posted by Claus K
    old length X ratio of port areas, i.e. for your example:

    SQR (3/2) = 2.25 X 5.5 = 12.375 inches.

    Due to port end correction that errs on the short side i.e. the real required length should be a bit more than that probably 13 -14 inches
    Hey, Claus...

    What you (not "you" - 3rd person...) are really doing is maintaining the ratio of the port area to duct length. Squaring the ratios of the ports as you did gives the same result, but isn't this what's going on?

    Original 2-in diam has area of 3.14159 (pi); ratio to area to duct of 5.5 = 1.7505
    The 3-in diam has area of 7.0686; maintaining the ratio of 1.7505 requires duct of 12.3735.
    Last edited by boputnam; 03-12-2004 at 01:17 PM.
    bo

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    Hi Bo,

    yes, absolutely right, that is just another way of putting it.

    Hereīs some (sorry, lenghty) explanation of this, if anyone should be is interested:

    A bass reflex is basically a system of two coupled resonators. One is the driver the other one
    is the vented bin. This forms a "Helmholtz" resonator. It consists of a kind of "virtual piston"
    formed by the air that is trapped in the vent and a kind of "virtual spring" formed by the air
    that is trapped in the bin. This air is compressed and expanded when the "piston" moves in and
    out. The resonance frequency of this resonator is determined by "piston" mass and "spring" strenght.
    If you increase spring strength by a certain factor you need to increase the piston mass by the
    same factor to end up with the same resonance frequency.

    The piston mass is the mass of the air volume in the vent which obviously is proportional to the
    vent volume. The spring strength is a bit harder to deduct. The counterforce that the "piston"
    sees when it moves inwards is caused by the compression of air in the bin which creates a rising
    pressure. For small movements the pressure increase is proportional to the volume change. As
    "pressure" is a measure of "force per area unit" the total counterforce that the piston "sees"
    is the pressure multipied by the piston area. However this means that for changing the piston
    area by a certain factor the spring strength will be changed by the square of that factor.

    If you e.g. triple the vent area the piston will move three times as much air for a certain
    piston movement. This will create three times the pressure increase. However this pressure
    is now also applied to three times the piston area. So the resulting total counterforce that the
    piston "sees" is nine times the original value. You have basically increaed your spring "strength"
    to nine times the original value. If you want to end up with the same resonance frequency you need
    to have nine times the piston mass. If you would keep the vent length as it was you would end up
    with only three times the original vent mass. So you have to triple the vent length to end up with
    the desired 9 times vent volume to meet the requirement. In the end it means that the vent length
    needs to be increased by the same factor as the vent area.

    In reality this is not 100% exact because there is some difference between the physical vent lenght
    and the "effective" vent lenght. Some amount of air in front of the vent openings will also move, so
    the effective vent lenght is a bit bigger than the physical one.

    Regards

    Claus K

  8. #8
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    "Hereīs some (sorry, lenghty) explanation of this, if anyone should be is interested:"

    Definitely interested Claus, thanks!

    Here's a shot of a portion of my old tuning spreadsheet with D.B. Keele's formula - shown is the port/duct arrangement for a 4435. As you can see, it is highly doubtful the 4435 is really tuned to 26 Hz. I created this spreadsheet when the first version of Quattro Pro came out way back when and it's followed me through various spreadsheet software upgrades . Before I owned a PC, it was programmed into various HP and Ti calculators.
    Attached Images Attached Images  

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    Moderator / Treasurer/Marketplace Czar boputnam's Avatar
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    Thumbs up

    Originally posted by Claus K
    ...In reality this is not 100% exact because there is some difference between the physical vent length and the "effective" vent length.
    I'm guessing this relates to air being a compressive medium. Where this water (non compressible), the effective and actual vent length would be equal.

    Great post, Claus! Thanks for taking the time to put this together so it is on the Forum.

    Hey, Giskard - wanna share that tiny spreadsheet? Sure, I could build it, but then again, it is a Giskard Original.
    bo

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  10. #10
    Senior Member GordonW's Avatar
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    Actually, it's mostly because of "end effects"... the air just outside the mouth of the port, gets "caught up" in the "air piston" from the port, and adds just a little bit of effective length to the port. This, in fact, would also happen, no matter what medium is in use.

    This effect varies, depending on end treatment. IOW, a port mounted flat into a baffle, reacts with a different "end effect" to one with the end in free space, and a "flat" (ie, un-beveled) port end acts different to one with a flared, rounded or befelled end.

    If you want to see specifics on how end effects are calculated, you can go to Dr. Wm. Marshall Leach's Audio Engineering web site at Georgia Tech School of Electrical Engineering; he has references there, to all sorts of formulas for this sort of thing there, that he has calculated over the years.

    Regards,
    Gordon.
    This Is Gordon's Page: www.geocities.com/gordonwaters

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