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port size?

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I am considering a project that would have taking a KLF-10 apart and building two smaller "mid woofers" for under the couch. Just an experiment as I have the speakers and wood on hand. The speaker currently has a single port approx 3" in diameter and 5" in length. Is there a rule of thumb of any type that would dictate what size the port should be if the new box is half the size and containes only one woofer?

Thank you!

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There are equations for this. I don't know them off the top of my head, but I suspect they can be found with an internet search. The port diameter, port length, and enclosure internal volume determine the frequency at which the port it tuned. So you will need to determine that for your stock KLF-10 and keep it the same value for the new enclosure. You can replace the two 10" drivers with four 6.5" drivers because they two setups should move about the same amount of air. I did that when creating a center channel for my KG5.5 mains (they also use a pair of 10" woofers) -- see my avatar picture. When selecting 6.5" drivers you will want to ensure you get ones that are suitable for vented enclosures. That can be determined by looking at the Qts parameter, but again I don't recall the rule of thumb there. You should consider buying yourself a copy of The Loudspeaker Design Cookbook as it would contain all the info you need -- and then some.

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There is software available for these type projects and you should probably invest in it if you're going to do this kind of stuff. You will find that you have a choice: You can maximize the bass response but at the expense of power handling. Port length/diameter will be determined by the T/S parameters of the driver in conjunction with the box size. If you don't know the T/S parameters of the driver you are going to use, there is software for that too. I suggest leaving your 10s as they are and starting from scratch. There are quality 10" available for not very much money, and dealing with known parameters is always easier than shooting in the dark.

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Thanks for the input guys. I had hoped that it was as simple as half the box size, half the drivers so half the port size!

I do not not any of the specs on the woofers so maybe I will just start from scratch.

Thanks again!

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Port length/diameter will be determined by the T/S parameters of the driver in conjunction with the box size.

I am pretty sure that the driver's T/S parameters don't come into play when tuning a port. That said the driver must have a Qts that says it will work in a vented enclosure.

I found this:

Port_Calc.png?1314153234

Where Av is the cross-sectional area of the port (in square inches),

Lv is the length of the port (in inches),

Vb is the enlcosure's internal volume (in cubic inches), and

Fb is the tuning frequency (in hertz).

from http://www.jlaudio.com/header/Support/Tutorials/Enclosure+Ports/Tutorial%3A+Enclosure+Ports/287541 .

Edited by STL

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So you actually want to just use the 10" woofers from a KLF-10 in some new enclosures? That shouldn't all that hard to do, and you do NOT need to know the driver's T/S paramters at all for this. Since the drivers are already in a vented enclosure you know they are suited for one! I did something similar when I make new enclosures (that fit into some built-in bookcases in my old house) for my KG 3.5s.

You just need to use the above equation to determine Fb. Then determine how big your new enlcosures are, pic a port diameter, then shuffle the above equation around to solve for Lv. You can to use the largest diameter port possible (to avoid any port noise) but the larger the diameter the long it will need to be.

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Thanks for the input guys. I had hoped that it was as simple as half the box size, half the drivers so half the port size!

I do not not any of the specs on the woofers so maybe I will just start from scratch.

Thanks again!

Actually, you can just make two enlcosures that are each half the size of the current one, then you would need to make the port in each enclosure be half the crossectional area of the existing port (and then the length would remain the same). For instance, one 3" port has the same crossectional area as two 2.12" ports.

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I think you can halve the port length. It might actually be slightly shorter than half. Try 2.5" lengths of 3" for each enclosure and measure.

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I am pretty sure that the driver's T/S parameters don't come into play when tuning a port.

Funny. Fb, Vb, and Qts are T/S parameters. :)

http://www.troelsgravesen.dk/ports.htm

From http://en.wikipedia.org/wiki/Thiele/Small:

"Thiele/Small" commonly refers to a set of electromechanical parameters that define the specified low frequency performance of a loudspeaker driver.

Fb and Vb are not T/S parameters. Fb is the tuning frequency of the enclosure or box (which I think is where the 'b' comes from) and Vb is the volume of the enclosure or box. As the equation above shows (and what I originally stated) only the port diameter, port length, and enclosure internal volume determine the frequency at which the port is tuned. And not one of those are parameters pertains the driver itself! And I said before, you do need to ensure the Qts of the driver you use in that enclosure has a value that is vent-able. But since he is using the same 10" drivers from the vented KLF-10, we already know that. So he does not need to know the driver's T/S parameters at all here. To be perfectly clear, he can create enclosures -- even ones that are slightly larger or smaller than the volume each driver is using in the stock enclosure -- and port them to the same frequency as the stock enclosure all without knowing any T/S parameters of the K-1056-K (the 10" driver in the KLF-10). And I am not just speaking theorectially either; I have done this myself when I made new enclosures for my KG3.5s.

Edited by STL

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I think you can halve the port length. It might actually be slightly shorter than half. Try 2.5" lengths of 3" for each enclosure and measure.

Actually he would need the port to be twice as long if he keeps the port diameter the same (and makes the enclosure half as big) -- so 10" long. But since each enlcosure would only have one driver -- thus half as much air would be traveling through the port -- it would be reasonible to cut the port's area in half (leaving the length the same) like I suggested above.

Here is an equation solving for port length this time:

Lv = (23562.5*Dv^2/(Fb^2*Vb))-(0.732*Dv)

where:

Dv = port diameter (cm)

Fb = tuning frequency (Hz)

Vb = internal volume (litres)

Lv = length of each port (cm)

from http://www.troelsgravesen.dk/vent_tuning.htm. Do note that port diameter/length are in centimeters and the volume is in litres (instead of cubic inches).

Edited by STL

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I think you can halve the port length. It might actually be slightly shorter than half. Try 2.5" lengths of 3" for each enclosure and measure.

Actually he would need the port to be twice as long if he keeps the port diameter the same (and makes the enclosure half as big) -- so 10" long. But since each enlcosure would only have one driver -- thus half as much air would be traveling through the port -- it would be reasonible to cut the port's area in half (leaving the length the same) like I suggested above.

Here is an equation solving for port length this time:

Lv = (23562.5*Dv^2/(Fb^2*Vb))-(0.614*Dv)

where:

Dv = port diameter (cm)

Fb = tuning frequency (Hz)

Vb = internal volume (litres)

Lv = length of each port (cm)

from http://www.troelsgravesen.dk/vent_tuning.htm. Do note that port diameter/length are in centimeters and the volume is in litres (instead of cubic inches).

I stand corrected, thanks. It seems contrary to common sense, but the equation you provide is right on. Except it's metric, and we are talking about inches etc., and the above would require converting those before tuning for frequency.

For a 3" port who's area is 7.0685834705770345 sq inches the new port area would need to be 3.53429173528851725 which equates to a diameter of precisely 2.1213203435596424.

As for the above question originally asked. The closest fraction would be 2 1/8" diameter. The closest metric diameter would be 5.4cm. A slot could be 1" X 2-1/8". Any of them the same length as the original 3" port.

Basically all stuff STL said.

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Gentlemen, again thank you for all of the info!

mustang guy, I like the idea of the slot vs. port. Free!

Dean G, no issue on the slow typing. I read rather slow.

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Alright, another question! I like the idea of the slot for a port and just want to verify the size.

The recommended size is 1" x 2 1/8". It states that the lengthould be the same as the original 3" port. The original port is approx 5" in length. So the true slot size is???

I have looked at the formulas from the links but am but a deer among headlights.

Thanks again!

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Yes, in theory it doesn't depend if the venting is done with a tube port of a slot port. If the both have the same cross-sectional area, then they'll have the same length (to attain the same tuning frequency). That said, the area of a 2 1/8" circle is 3.54 inches squared -- so you'd really need a 1" x 3.5" slotted port (that is 5" deep). Remember the area of a circle is Pi times the square of the radius.

Edited by STL

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I'd think you'd need 1/2 the original ports VOLUME if the box were half the size. Length or cross sectional area shouldn't matter as long as the volume came out correct. You may have said that already, I didn't do the math. Or, I could be way off base!

Edited by CECAA850

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Yes, in theory it doesn't depend if the venting is done with a tube port of a slot port. If the both have the same cross-sectional area, then they'll have the same length (to attain the same tuning frequency). That said, the area of a 2 1/8" circle is 3.54 inches squared -- so you'd really need a 1" x 3.5" slotted port (that is 5" deep). Remember the area of a circle is Pi times the square of the radius.

Right. Good catch again...

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I'd think you'd need 1/2 the original ports VOLUME if the box were half the size. Length or cross sectional area shouldn't matter as long as the volume came out correct. You may have said that already, I didn't do the math. Or, I could be way off base!

Like I said, it is contrary to common sense. Half the size box and half the drivers and you need half the port area. Oddly if you keep the port area, you near double the length.

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