Is the tuning freqency of a box determined only by the port area, or also by the dimension of a port? For instance, If I had one port that was say 12" high x2" wide x 10" deep that caused a box to be tuned to 36hz that would be 240"^2 of port area. If I had a 10" high x 4" wide x 6" deep port also being 240"^2, would it still be tuning the SAME box to 36hz?

Thanks

the length and area must be consistant along with the volume of the box for it to be tuned the same. so the answer is no.
and 12x2 is not the same area as 10x4

Right. Tuning is a combination of box volume, vent diameter, and vent length. Changing any of the three alters tuning.

Ok lets see if I have this right here....

In addition to enclosure volume, the depth/length of the port MUST be constant too?

Also, if the port is say 10" tall by 2" wide, can I change it to 8" tall by 2.5" wide to keep the same tuning?

So basically the 3 constants here to follow are:
1)Enclosure net volume
2)Port surface area
3)Port Depth

...where every other dimension involved can be altered, correct?

This also puts my first assumption to death:

the AMOUNT of port volume doesn't determine tuning fq

it is the AMOUNT of depth and AMOUNT of surface area

Correct. Area and volume have a direct relationship - that means that as the vent gets bigger in area it also has to get longer.

To figure tuning of your current enclosure
Fb = 0.159*Sqrt[Av*1.84*10^8/(Vb*1728*(Lv+0.823*Sqrt(Av)))]

To figure the length needed to tune your enclosure to the desired frequency
Lv = Av*1.84*10^8/[Vb*1728*(Fb/0.159)^2] - 0.823*sqrt(Av)

Lv = Length of port in inches
Av = Port area in square inches
Vb = Net internal volume of enclosure in cubic feet
Fb = Tuning frequency of enclosure in hertz
Sqrt = Square root

[ August 27, 2004, 05:03 PM: Message edited by: Jmac ]

Thanks guys for the help.


Jmac, I will definitely use that forumula soon before building the box to make sure I am getting the correct Fb.

go here: http://www.carstereo.com/help2/Articles.cfm?id=17

also really helpful: http://www.carstereo.com/help/Articles.cfm?id=31

Dukk I was trying to use kicker's specs on their 2.5 cubic foot box for the 12l7, and I used their Av, Vb and Lv values and plugged it into this equation:

http://www.jlaudio.com/tutorials/ports/images/Fb.gif

Now I have done engineering algebra at a university level, and so computing was not difficult....but I was unable to compute the Fb that kicker stated...the value I computed was much lower than their stated value. Their Vb #'s are always ususally stated as a net value, correct?

Thanks a LOT for those calculators...I think I am just going to use them instead of computing it manually (saves a lot of time and eliminates any quick errors).

Dukk, that 2nd calculator you provided has the option of choosing a "slot port" as an option. Is that what I am making here (the sides of the vent will be created by the enclosure wall except for one side)

ALSO, Do you guys factor in the "end correction factor" ? Is that "end correction factor" only used with vents that are not straight and have bends/corners in them?

^not Dukk here, not even close; but yes if one of the enclosure walls is part of the port, its a slot port.
There's a correction factor applied, about 1/2" added or subtracted (I forget right now, I think sub) in the calcs.

*edit* I think that's the correction factor I mentioned above^.

When you turn corners for the port, you measure your port length in the middle of the port, turn the corner in the middle of the 'around the corner' port, and continue measuring.

I hope that makes sense; and wait for confirmation on that "end factor correction". *edit*

Gene

[ August 28, 2004, 11:22 AM: Message edited by: Fever ]

Kicker takes into account driver displacement but not port displacement ...

However, if you look in the technical manual and scroll down to page 25, there are enclosure plans for the S12L7 ...

I have no idea how Kicker measures their boxes [img]graemlins/dunno.gif[/img]

End correction for a slot vent is to shorten the vent by half the width of it. So if the slot is 2" wide you would shorten the length by one inch. Incidentally, it wouldn't be audible either way. ;)

Ok so when creating a slot port, do I create the inner wall of the slot port (the wall which isn't part of the outer enclosure) exactly equal to Lv(eff)?

Sorry about all the newbie port questions

You measure down the middle of the port ... So the inner wall will be shorter than the length ...

Are you making an L-Shaped port ?

Yes I am making an L-shaped port.

Ok so the inner wall will be shorter by 1/2 lf the width of the port.

Port wall 1 = Internal depth - Port width
Port wall 2 = Total port length - External depth

http://www.soundsolutionsaudio.com/f...1093205548.jpg

Example :

Port length = 50"
External depth = 20"
Internal depth = 18.5"
Port width = 3"

Port wall 1 = 18.5" - 3" = 15.5"
Port wall 2 = 50" - 20" = 30"

This will give you a length of 50" measuring down the center of the port.

Port displacement (in cubic inches) = (Port wall 2 length x Internal height x Port width) + (Internal depth x Internal height x Port width)

More info

[ August 29, 2004, 11:09 AM: Message edited by: Jmac ]

hey, hwta did i tell you about asking St*pid questions?!?

ahhah J/K. glad you're gettingto wok on the box!