How to get tunning, square ports

[Texas Tech]

so you are saying that the port length given from WinISD does not give you the actual port length? You have to calculate in displacement? what displacement is there in the port? unless you are using outside dimensions (are you using outside dimensions and thats why you are calculating displacement?)

[babyjoker]

try downloading it from Limewire, that were i got my bass box 6 pro, and it works fine.

[Texas Tech]

but do you have to calculate displacement from the port length that is given to you from WinISD?

i dont see why you would have to calculate any displacement unless you were giving outside measurement values

[alpine]

whats the website to bass box pro like the exact url if not just the eneral website url?

                                                              thnks,joe

[Guest Nismofreak]

http://www.ht-audio.com/bassbox.htm

[andym85]

i could never figure winisd out for some reason...but here is the formula i use, even threw in the excel setup i have

Fb = .159*((Av(1.84*10^8)/(Vb(Lv+.823*Av^.5))^.5

Fb = tuning frequency

Av = vent area

Vb = net volume in cubic inches

Lv = effective length of vent

       effective length = physical length + end correction <---1/2 of the ports width, end correction is only used if one of your walls of the box doubles as a port wall

here it is in excel:

Av, insert value in b10

Lv, insert value in b11

Vb, insert value in b12

Fb, insert formula in b13

formula =0.159*((B10*(1.84*10^)/(B12*(B11+0.823*B10^0.5)))^0.5

[BiggyDon]

I got this formula off caraudioforum long time ago

the equation for the tuning frequency is like this

L=(4651704*Pa)/(Fb^2*Vb)-.823*sqrt(Pa)

where

L is the port length in inches

Pa is the port area in square inches

Fb is the tuning frequency of the box

Vb is the internal volume of the box in in^3

if you wanted to make a 1.5cf (2592 in^3) box that had a 30hz tuning frequency and 20in^2 of port area, you would need the port to be

L=(4651704*Pa)/(Fb^2*Vb)-.823*sqrt(Pa)

where

L is the port length in inches

Pa is 20 in^2

Fb is 30hz

Vb is 2592 in^3

so

L=(4651704*20)/(30^2*2592)-.823*sqrt(20) = 36.2 in

so with a 1.5cf box with 20 in^2 of port area, you would need a 36.2 in long port to get a tuning frequency of 30hz

[Texas Tech]

do all these equations work? because these formulas seem so simple compared to the formula given on the jl audio website.

[Texas Tech]

i have a question about measuring port length when it is a square port but there is no "elbow" or bend in the port. So basically there is only one board that creates the port assuming you used the side of the box to make the port.

I will just throw some numbers out there they are of no significance so:

lets say a box is

24 inches wide, 12 inches in height, and 14 inches in depth. Then you put a board (of wood) 4 inches away from the side (so the port is 4 inches wide, internally this is after displacement is calculated (not that it matters), and the board is 10 inches in depth. Would the port length be 10 inches, because thats how long the board is that actually creates teh port? would the port be 12 inches, 14 inches. The depth of the box is 14 inches and the length of the board that creates the port is 10 inches, so there is 4 inches between the two.

i hope you can understand what i am saying, sorry i cant draw a picture, i am on a new computer and i didnt see a program.

I have another question regarding port building, when you put your dimensions into WinISD for the ports width and height, is that the actual width and actual height, so it wouldnt include the wood or anything, tis the actual measurements of teh port (internally you could say)

thanks for all the help guys and all teh formulas

[David]

yes those are the proper mathmatical  formulas to tune your port.