View Full Version : 2 subs, do you double the port size?

JonnyBlaze

05-08-2004, 03:14 PM

a friend of mine just put 2 xtant m412's in serperat sealed boxes. they are just under 1 cubic foot each. hes pushing them with a punch 360. it sounds decent the way it is but i want to build him a ported box hoping it will be louder.

the spec sheet for the speaker recomends a 2.50cf box with a 4"x8.75" port and it says that would be tuned to 37hz.

for 2 subs, i would make the box 5 cubic feet but what do i do with the port dimension? does the length double? does the diameter double?

i probably was going to use a rectangular port, which would be 1.5"x8.38" if that matters.

there are also specs for a bigger box (3cu each) with a lower tuning frequency (35hz). would i be better off using the biggest box they say?

thanks alot

JB

Casserole

05-10-2004, 04:31 PM

http://www.jlaudio.com/tutorials/ports/index.html

"Notice that Method 1 produces the same port length as did our single 4" diamter port as it should (after all, we have the same total port cross-sectional area which this school of thought proclaims is correct!). But the first method is incorrect because it neglects the fricitional losses encountered by using many smaller ports--there is a higher port wall surface area to cross-sectional area ratio which raises the total amount of frictional losses in the ports and thus shifts the tuning!"

EDIT: just realized it was two subs vs two ports. so what i just posted may have nothing to do with this situation.

Acidburn

05-10-2004, 09:15 PM

use the JL Audio equation with the increased box volume and the same tuning frequency and that should tell you

JonnyBlaze

05-12-2004, 11:37 AM

does anyone have a clue how to get that equation into excel? and where does the number of drivers come into play?

JB

JnEsPappa

05-12-2004, 06:14 PM

to get that equation into excel?

JB

=.159*((AV*(1.84*10^8))/(VB*(LV + .823(AV^2))))

Instead of AV, type in the cell where AV is located and do the same for VB and LV

The parentheses are important to get right.

Benny212

05-12-2004, 11:49 PM

does anyone have a clue how to get that equation into excel? and where does the number of drivers come into play?

JB

number of drivers doesn't come into play with that formula. Only thing that more drivers would change is the optimum port area, because more air being moved means it might be a good idea to go with a larger port

JonnyBlaze

05-13-2004, 11:10 AM

=.159*((AV*(1.84*10^8))/(VB*(LV + .823(AV^2))))

Instead of AV, type in the cell where AV is located and do the same for VB and LV

The parentheses are important to get right.

thanks alot.

this is my first ported box design and theres alot to learn. i want to get it right the first time.

i think im going to port it high, maybe around 45hz and make a plug that will change it to 27hz. ill post my designs once i get it drawn up.

JB

edit: its my second ported box actually. my first was a 5cubic foot for my stroker, but i wouldnt bet i did it correctly :)

Omarvelous

05-14-2004, 06:34 AM

=.159*((AV*(1.84*10^8))/(VB*(LV + .823(AV^2))))

Instead of AV, type in the cell where AV is located and do the same for VB and LV

The parentheses are important to get right.

I'm not seeing how you implemented the Square Roots in the equation.

I'm sure excel can do Square root.

First get the Square Root of AV then plug it in here * <- (indicated by asterik *)

=((AV*(1.84*10^8))/(VB*(LV + .823(AV*))))

Then take the value from the equation above and Square Root it wid a Calculator then x that by .159

Unless i'm missing something...

Also might wanna check my thread on the Swappable port - might have a method that will work for u

Omarvelous

05-14-2004, 07:00 AM

Fb = 0.159*[Av*1.84*10^8/Vb/(Lv+0.823*Av^0.5)]^0.5

I think ... It's late ...

Yes ^.5 will work also! That is better!

JonnyBlaze

05-14-2004, 09:47 AM

thanks for the help guys. but is there a way to modify the equation so instead of giving you the tuning frequency (Fb), it gives the port length (Lv)?

as it is i have to give the length to get the frequency.

and i used to think i was good at math ;)

thanks again

JB

Omarvelous

05-14-2004, 11:04 AM

Well.... since you are the one who is giving the LV in the formula i would imagine you already have the length of the LV.

I know what you mean by flipping it, but if you have LV set to a cell, you could jus keep changing the length until you get your measure.

But....

I guess you could ^2 the equation and then work from there. I'm not attempting it however! lol

Acidburn

05-14-2004, 05:36 PM

i got: Lv = ((4651704 * Av)/(Fb^2 * Vb)) - 0.823 * Av^.5

Acidburn

05-14-2004, 06:03 PM

oh well, just plug it back into the original to make sure whatever you get works

JnEsPappa

05-14-2004, 06:50 PM

=.159*((AV*(1.84*10^8))/(VB*(LV + .823(AV^2))))

Instead of AV, type in the cell where AV is located and do the same for VB and LV

The parentheses are important to get right.

Thanks for poiting out the sq. root guys.

=.159*(((AV*(1.84*10^8))/(VB*(LV + .823*(AV^.5))))^.5)

You can never have enough parentheses I say. That way there is no confusion over the rules of order.