Okay, I've got a question for expert box builders. I am new by the way, and I've been lurking quite a bit up until now. When designing a box with a center port between the two subwoofers, is the port width equal when it separates to each section of the sub? A picture is attached to make what I'm talking about clear.

Thanks guys!

 

if the center of the port is 4" in width, then to two ports going to each chamber will be 2" in width. you are basically build two seperate enclosure, without the center port having a divider

 

So using your example, would I have to calculate the port length of a 2" width or a 4" width, or a combination or both...?

 

I thought port tuning had to do with the surface area of the port... you may be in for some serious number crunching.

I would just do two separate 2 inch ports, or one big 4 inch port on the side.

 

Port tuning has to do with how large the port is (Surface Area) which affects how long the port length will be.

Oh yeah, but I can do it. If anyone could input their expertise please, it would be really appreciated. I really don't like the idea of two separate ports, I like a cleaner look of one big port, and I don't like how the box is not symmetrical with the port on the side. Little things that make your system look a lot better, and like a expert built it. That's what counts for me

 

Ok, if you want to do the math.

4 inch in the middle, two inches down the back. Do your calculations for a 4 inch port getting WINisd or another program to calculate the length for you.

1. Figure out the surface area for the one 4 inch port (M1)
2. Calculate the surface area for the middle 4 inch piece of your box (M2)
3. Calculate the length of the two 2 inch pieces so the surface area of the two 2 inch pieces (M3) plus the middle equals your first calculation.

In the end, the surface area of your 4 inch middle plus the surface area of your two 2 inch ports, must equal the surface area of the original port (given by WINisd).

M1 = M2 + M3 + M3

Good luck. It will take a little bit of algebra but you can do it.

EDIT: Crap, you quoted me before I fixed my spelling mistake... I used to be a Math/English teacher.

 

Wait, could you explain that a bit more? You kinda confused me when you said figure out the surface area of the one 4 inch port, and the figure out the surface area of the middle four inch port...? And I'm guessing the surface area we're talking about is the port surface area.

 

Yes, port surface area.

Your computer program is going to give you the length for a single 4 inch port. Calculate the surface area of it (do the math). (M1) Example: 50 sq inches

Then you need to decide how long you are going to make that middle 4 inch port. You can make it as short as you want. Calculate the surface area for it (do the math). (M2) Example: 20 sq inches

M1-M2 will give you the remaining surface area. Example: 50 - 20 = 30 sq inches divided by two (2 ports) = 15 sq inches each (M3)

The surface area of your two 2 inch ports (M3) plus your 4 inch port (M2) must equal the original surface area. This is the tough part because you need to work backwards. You know the surface area you want, but you need to work backwards to find the dimensions for your two 2 inch ports. Using my Example: you would need to figure out the length of a 2 inch port so that when you calculated its surface area it would be 15 sq inches.

M1 = M2 + M3 + M3 or using the example above
50 = 20 + 15 + 15

This will probably be easiest to do with rectangular ports.... I am assuming that is what you are doing...

Hope that made sense.

 

4" width.

You should have a flow splitter/blocker at the intersection of the two tops to that T as well. Make it a triangle and chamfer the corners of the vent to match.

 

End correction with a 4" wide port - the physical port length should be 2" shorter because it will act like it's 2" longer. Whatever the calculation gives, you make the port around that number. The rule is - the port will act like it's +half of the port width, it doesn't just stop at the port, it runs along the wall. It would be easier to measure the port length of each box separately, and then "take out the divider".