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Old 20th May 2009, 02:07 AM   #11
gedlee is offline gedlee  United States
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Default Re: unequal multiple ports

Quote:
Originally posted by j.michael droke
Hi there: If unequal multiple ports act "nearly" the same as a single port of the averaged size, why not use use the single average port? .... regards, Michael
The staggered ports act as one as a lumped mass, but have different resonances within the tubes when they have internal standing wave resonances.
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Old 20th May 2009, 03:32 PM   #12
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Default unequal multiple ports, "UMP"

Hi there, Earl and Patrick: Thanks for the explanations regarding unequal multiple ports. Is there a formula or design chart for utilizing UMP's?

As an example: 8cf box, single 5-inch port 13.3-inches long, box tuning Hz= 20hz, F3=20hz, Fs=19.7hz, if two UMP's are selected, would one be tuned below 20hz and the other above 20hz? How is this determined?
....regards, Michael
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Old 20th May 2009, 05:16 PM   #13
gedlee is offline gedlee  United States
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find the correct length for one port and add and subtract some value, like an inch, to each, which need to be 1/2 the area of the single one. Its not complicated.
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Old 22nd May 2009, 08:13 PM   #14
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Sounds like it won't have the effect I expected.

Oh well.

Looks like an interesting thing to try to do, perhaps with an existing subwoofer...

All I need now is an existing subwoofer.
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Old 23rd May 2009, 03:23 PM   #15
xpert is offline xpert  Afghanistan
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Quote:
Originally posted by gedlee



This is precisely what happens both theoretically and in practice. I typically use three, have used five, all different lengths. Just take the average if they are all the same diameter. (Thats not exact, but close enough).
To calculate the effective port length take the equation along which paralleld inductances add:

given are L1, L2, L3 ... port lengths, port lenght correction included

Leff = 1 // ( 1/L1 + 1/ L2 + 1/L3 ... )

It's not exactly the average of L1, L2, L3 ...!

As You ask further paralleld port areas add like capacitors

Aeff = A1 + A2 + A3 + ...

The resonance frequency depends on the relation of area to length

Fr^2 ~ Aeff/Leff = (A1/L1 + A2/L2 + A3/L3 ...

Quite simple. One should keep in mind that narrower && longer ports resonate with a higher Q (on the pipe resonance) than one wider port alone. So there are diminishing returns with that tecnique. The benefit will strongly depend of the special situation.

hope this helps

ps: due to the square in Fr^2 ~ Aeff/Leff small errors in port length or area won't do to much. A deviation of 10% in Leff or Aeff will only cause Fr changeing about 5%. That is neglegible compared to atmospheric influences as air pressure, humidity etc.

The above correction to Mr. Geddes' note was done in case of someone building a double port of lets say 1" length + 10" length. The effective length will be much smaller than the average of both! It will be more around 1.1" instead of 5.5".
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Old 25th May 2009, 02:53 PM   #16
gedlee is offline gedlee  United States
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Quote:
Originally posted by xpert


To calculate the effective port length take the equation along which paralleld inductances add:

given are L1, L2, L3 ... port lengths, port lenght correction included

Leff = 1 // ( 1/L1 + 1/ L2 + 1/L3 ... )

It's not exactly the average of L1, L2, L3 ...!

One should keep in mind that narrower && longer ports resonate with a higher Q (on the pipe resonance) than one wider port alone.



xpert

I said that the average was not exact, but when the deviations are a small part of the total lenght, as they should be, then the average is quite accurate.

And the statement about higher Q for narrower ports is not theoretically correct.
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Old 26th May 2009, 01:38 PM   #17
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"And the statement about higher Q for narrower ports is not theoretically correct."

So why are they noiser than ports with a lower aspect ratio when used at high volume?
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Old 26th May 2009, 03:27 PM   #18
Eva is offline Eva  Spain
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Something that lowers the Q of main port resonance is to place it near a wall or corner.

Lower Q produces less bass boost but reduces cone excursion (and thus distortion) over a wider frequency range, which was probably the initial goal...
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Old 26th May 2009, 04:07 PM   #19
gedlee is offline gedlee  United States
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Quote:
Originally posted by djk
"And the statement about higher Q for narrower ports is not theoretically correct."

So why are they noiser than ports with a lower aspect ratio when used at high volume?
Different effects altogether. The port noise is a nonlinear turbulence effect that depends on velocity which will be higher for a narrow port. The "Q" that is being takled about here is the Q of the first mode resonance of the port, NOT the Q of the box tuning. Again completely different things.

The simple math for the higher mode resonance of a tube does not contain the cross sectional area as a parameter. The damping of the higher modes (and hence the "Q") depends only on the boundary conditions, namely the radiation. Now there is a secondary effect on the radiation resistance with area, but for multiple tubes close together at the end, this will depend only on the total area and NOT of the subdivided areas, meaning that the "Q" of the subdivided ports will be the same as the single port - contrary to xperts claim.
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Old 26th May 2009, 05:10 PM   #20
djk is offline djk
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So the staggered ports act as one as a lumped mass, as far as the box tuning goes.

Say we use four smaller ports vs one large one of the same area. The four smaller ports will have a circumference of twice that of the one large port. Is this a potential source of port compression at high volumes?
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