Which port equation to use?

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Yoda

Member
2001-07-30 1:35 am
I got a port length equation from http://www.diysubwoofers.org, and it seems to be giving me very long port lengths for a given driver. the equation is:

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

Lv = length of each port (cm)

my data are:

Dv = 7.303cm (2 7/8 in.)
Fb = 48hz
Vb = 12L
Np = 1
k = .732

it yeilds 40.106cm = 15.8in.
this is only a 6" woofer, and a 15.8x2.9in port will take up 1.7L , thats almost a 2L pop bottle as a port! it just seems to be very large for what i've got.

more generally, why do i get so many different answers from different sources, i.e. the zoebel calculator at http://www.lalena.com/audio vs. the PE util (press shift-F3)?
 

kelticwizard

diyAudio Moderator Emeritus
2001-09-18 2:33 am
Connecticut, The Nutmeg State
I got 16.2 inches length using the port program on Steve Ekblad's site. The difference is less than 1 Hz anyway, so the calculation is correct. http://www.wssh.net/~wattsup/audio/

The port takes up about 15 percent of your total enclosure volume, which is a lot.

Three inches is a commonly used port for a 10 inch driver, which has twice the surface area as your 6 inch. So feel free to go to a smaller port.

[Edited by kelticwizard on 11-09-2001 at 08:22 PM]
 
more generally, why do i get so many different answers from different sources

.... because of inconsistencies between equations used. Different constants like the speed of sound used in the equation will, of course, yield a slighly different answer. Things like the amount of end-correction also has an effect on the resultant port length. However, small differences aren't going to have a significant (audible) deviation from the target tuning frequency, so it's really not that much a problem.

:)

Isaac
 
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