Which port equation to use? - diyAudio
 Which port equation to use?
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 Multi-Way Conventional loudspeakers with crossovers

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 9th November 2001, 07:42 AM #1 diyAudio Member   Join Date: Jul 2001 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)?
 9th November 2001, 11:23 AM #2 diyAudio Member   Join Date: Nov 2001 Location: Ewersbach port I think a 5cm diameter port will be sufficient for a 6" driver. Possibly you should check the Thiele-Small Parameters of your driver,it´s probably not suitable for bass-reflex. Greetings : Arne
 10th November 2001, 01:20 AM #3 Wizard of Kelts diyAudio Moderator Emeritus   Join Date: Sep 2001 Location: 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]
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Join Date: Sep 2001
Location: Sydney
Quote:
 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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