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5th December 2004, 10:36 AM  #11 
diyAudio Member
Join Date: Jul 2004
Location: Scottish Borders

Hi,
I used Bullock and put all his equations into a spreadsheet. His explanations are very clear & I really understood what I was trying to achieve. Using someone elses software saves a lot of your time but usually you do not learn to understand the mechanisms that are at work. How did you do you calcs? Oh and finally after optimising the box and port I had cut 150mm off the height of the tower and resealed the base. I then found that adding mass to the cone gave a good way to optimise the bass. That must have really upset sensitivity but it worked for me. regards Andrew T. 
3rd February 2005, 02:01 AM  #12 
diyAudio Member
Join Date: Jan 2004
Location: Erwin, Tennessee

Back To The Grind
Ok guys I have constructed an enclosure for a PK BL10 subwoofer. I have a car audio store near buy lets me play with their stuff allot. Anyways I have taken in consideration the volume occupied by the back of woofer and magnet structure. I very carefully added the volume of the duct to the box volume. The port was rectangular not round. After I constructed the enclosure and generated an impedance plot for the enclosure in its sealed state and vented state. I calculated the box compliance and worked backwards using the Vas of the woofer and calculated the box volume. The volume measured 1.53. My target volume was 1.5 five. So this answers my question that the duct volume is included in the overall volume of the enclosure. But the thing that I found that was so interesting was that my calculated tuning frequency was 34Hz but my measured was 26Hz. Why such a huge difference. I was very meticulous with all of the construction and I know the duct cross sectional area and length was exactly correct. From what I have seen on the internet and in literature there is a large amount of error in what is calculated and what is measured in the duct tuning frequency. So my next quest is to find out how to make the duct equation produce better results. I reviewed the derivation of the duct equation and learnt how it was derived. My first thought is that the equation doesn’t take into consideration the air outside of the enclosure directly in front of the port. If the woofer displaces air the amount displaced will exit in or out of the port and then expands. Regardless a certain amount still acts as part of mass of the port. So I figure that the maximum amount of air that can be displaced is the Vd of the driver. Assuming for a peak to peak excursion the volume displacement of the woofer is Vd=2*Xmax*Sd. I re derived the duct equation to consider this. It is given below. Any thoughts on this?
(c^2*R^2)/(4*pi*Fb^2*Vb)(2*Vd)/(pi*R^2)1.463*R=Lv The equation does produce closer results but isn’t entirely correct. 
3rd February 2005, 03:02 AM  #13 
diyAudio Member

you are correct, the effective port length extends beyond the actual duct length. This effect is not entirely consistant however. It varies for flanged and free ends, and also for different flare curvatures and radii. WinISD takes this into consideration somewhat with their standard end correction factors. I have become desensitized to the subtle differences in calcualted tuning and effective tuning. If they're off by less than 1hz you will probably never tell the difference.
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3rd February 2005, 03:10 AM  #14 
diyAudio Member
Join Date: Jan 2004
Location: Erwin, Tennessee

Port Type
I agree it does depend on flared or non flared ports. But regardless the most conservative value would be the greatest possible amount of air displaced. By the way I just downloaded WinISD. I will give it a whirl.

3rd February 2005, 03:17 AM  #15 
diyAudio Member
Join Date: Nov 2004
Location: Québec, Québec

Did you use this formula to calculate the 34 Hz ?
( ( 8466.4 * R^2 ) / ( Vb * Fb^2 ) )  (1.463 * R ) So for a 3" port in your 1.5 cu.ft = ( 19049.4 / 1734 )  2.1945 = 8.79" long Then if it's rectangular : ( ( 8466.4 * ( Surface / Pi ) ) / ( Vb * Fb^2 ) )  (1.463 * sqrt( Surface / Pi ) ) So for a slot port 1" high by 6" long in the same box so Surface = 6 cubic inches... ( 16169.6 / 1734 )  2.0218 = 7,30" long But then my formula might miss a correction factor caused by the way air enter the port in the box by only by one side. My formula for the round port is the usual dual flared port with correction factors included.
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3rd February 2005, 03:29 AM  #16 
diyAudio Member
Join Date: Jan 2004
Location: Erwin, Tennessee

Rectangular Port
The cross sectional area of the port was 19.52 in^2. I did use the equation you have above but the actual port was not flared. Please do the calculations and let me know what length you got so I can compare it to mine. I got 27 inches for 34Hz in a 1.5 ft^3 box.

3rd February 2005, 04:02 AM  #17 
diyAudio Member
Join Date: Nov 2004
Location: Québec, Québec

I got 26,69" long for 34 Hz.
I guess maybe the box volume is not correct, let me check for the correction factor of a slot port. Checked WinISD for fun, it gives 26,56" long for a slot port. I guess the volume is not correct, will continue to search. If you know the internal box dimensions, we could check that to have a rough estimation.
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3rd February 2005, 04:23 AM  #18 
diyAudio Member
Join Date: Nov 2004
Location: Québec, Québec

I found out a good website with all the correction factors needed...
http://www.jlaudio.com/tutorials/ports/ I checked and doublechecked... Are you sure you calculated the port length in the box according to the method listed on the website above? Could you give us internal box measurements? Even with the correct correction factor, you're still at 34 Hz with 27 inches. Maybe your measurements about the actual Fb are wrong? Good luck !!!
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