ports, displacement

if i am designing a box on win isd alpha and it shows the port lenght like this

 number of ports: 2

port size: 21.00" x 4.00"

port length: 34.52

does that mean that i need two ports @ 34.52

or 34.52 combined total, split equal between the two

and if i put a single port in the center of the box 21"x4" the length is 13.47"

can i put a wall on each side of the port

and if so do they both have to be 13.47 or a combine total of the two split equal?

thanks for the help if any is possible

number of ports: 2           port size: 21.00" x 4.00"             port length: 34.52         does that mean that i need two ports @ 34.52

Yes

The program is giving you actual inside dimensions of each port.  To keep the same tuning frequency, using two ports of the same opening size (such as 4 X 21) will result in each port being more than twice as long as just a single port of that same opening size.  This is because you have dramatically increased port opening area by using multiple ports.

Use your program to look at port noise with one 4 X 21 inch port opening.  That's a big opening and you should notice that air speed will be quite low.  If you want to use two ports, such as for aesthetic design reasons, make each port about half of that opening size...for example 2 X 21...or better, 4 X 10.5   You will now see that port length remains about the same as if you were using one port.

and if i put a single port in the center of the box 21"x4" the length is 13.47"    can i put a wall on each side of the port

Yes,  you will have to.  Otherwise it wouldn't be a port.  This is, by the way, a good design for a dual subwoofer arrangement, as it strengthens the middle section of the box.  When you are figuring displacement of the port structure, you will count both walls as part of the port size.  Port airspace, on the other hand, is just the airspace between the walls and corresponds to the measurements you get in the program.

how do you configure port displacement?

what is the formula to figure out mass of the port so much port takes up so much space right well how do figure how much if you are using 3/4" wood?

Figure port displacement the same way you figure box displacement.  If the port is slotted, figure any wall that separates the port from the rest of the box airspace as part of the port structure.  The port structure has cubic dimensions (width, length and heigth) and will result in cubic inches when multiplied.

Figure cu inches for the total box, deduct port structure...result is net airspace.  Net airspace is the volume of air the woofer has to work with.

If using a round port, use a volume calculator for this:  https://www.the12volt.com/caraudio/boxcalcs.asp#cyl  Measure the overall outside diameter of the port tube and use that number for diameter.  Enter 0 for thickness of wall, as this is already incorporated in the measurements.  Answer is in cubic feet.

One cubic foot is 1,728 cu inches.  To remember this number, think in terms of 12 X 12 X 12, the dimensions of a cubic foot.  Multiply the dimensions.

Measure in inches.  Convert each volume of space to cubic feet ( divide by 1728).  Subtract displacements after the conversions, so that you are always subtracting cu ft from cu ft.  Woofer displacement is given in cu ft, for example.