Need help with slot port

[C-Bass 0]

I need some help with the dimensions of my slot port before I go ahead and make a disaster.

I have a 15" SSD in a 3.8 cuft box. I'd like to tune it to 33Hz as per Fi's recommendation

Using one of those online calculators I got the following dimensions 3"H x 6"W x 5.22 L

Does that look right? How would that be affected if I wanted to make 2 ports?

Any help would be appreciated

[ANeonRider 3]

18in^2 is not even close to enough port area........

Plus, I can't really help much, since I don't know what kind of space you have to fit it in.

[C-Bass 0]

18in^2 is not even close to enough port area........

Plus, I can't really help much, since I don't know what kind of space you have to fit it in.

I'm pretty tight on space.

The front baffle is approx 36" W x 18" H with the sub in the middle. That leaves about 8" and change on either side of the sub for the port. the box is an odd shape and the front baffle will be made out of fiberglass

This might give you a better idea

[C-Bass 0]

so nobody has ever built a vented box for a 15 SSD?

What about round ports??

[C-Bass 0]

??????????

Any port calculator I use comes up with the same figures (obviously) so why is 18 sq in of port too small? What basis do you have for that comment?

[ANeonRider 3]

umm, port velocity, therefore causing port noise... that's why.

I would not go with less than 30.

[C-Bass 0]

umm, port velocity, therefore causing port noise... that's why.

I would not go with less than 30.

Thank you

makes sense...I'll come back with some hopefully better figures

[MTA 0]

12-16 sq in of port are per cubic foot of air space works pretty good

at 3.8 cubes youd be lookin around 36 sq in as a good low noise port now just figure out how long it needs to be... dont be afraid to fold the port if need be to accomodate the length needed

now is that total volume or minus sub displacement?

assuming you have total 3.8 your port at 36 sq.in. will need to be aprox 27" long thats if u build the port out of 3/4" mdf and sub displacement im guessing at 0.2 cu.ft.

anyone can correct this as its just a really quick design

hopefully this helps ya a bit

[C-Bass 0]

12-16 sq in of port are per cubic foot of air space works pretty good

at 3.8 cubes youd be lookin around 36 sq in as a good low noise port now just figure out how long it needs to be... dont be afraid to fold the port if need be to accomodate the length needed

now is that total volume or minus sub displacement?

assuming you have total 3.8 your port at 36 sq.in. will need to be aprox 27" long thats if u build the port out of 3/4" mdf and sub displacement im guessing at 0.2 cu.ft.

anyone can correct this as its just a really quick design

hopefully this helps ya a bit

That helps a ton, I'll play around with the numbers and see what I can come up with

[C-Bass 0]

After some more screwing around with numbers I'm even more confused.

You mentioned 12-16 sq.in. port area. How did you get 36 sq.in for my box volume?

If I have 3.8 cubes * 12 sq.in = 45.6 sq.in.

Let's say I use the 36 sq.in. figure. The way I would come up with that is 6"x6" = 36 sq.in. Assuming my math is correct, that would mean I will have a port with the opening of 6" x 6" right?. If I plug that into my handy online port length calculator it gives me the length of 15.49"

Let's assume for the moment all that is correct. In my scenario, using a 6" High port would come within 2" of the sub. From what I have read, that is too close for comfort. True or false?

Let's assume again that it is too close. Say I use 6"x4", my port length (according to my online calculator) would have to be 8.59". Would that be correct?

Again assuming that my calculations (more so the web site calculator) are correct, 6" x 4" (24 sq.in.) isn't enough sq. in of port. So let's say I want to put another 6" x 4" port to increase the sq. in.

That should give me 48 sq.in. of port no? 2 x 24 sq.in.

Again, assuming everything is correct so far, the two ports would give me near the recommended port area. So now I would just have to figure the length right?

Dividing the box volume by 2 (2 ports) I get 1.9 cu.ft. That means I would have to have to ports at 4" x 6" x 24.22" Length??

I don't have that much room to work with these ports. My original calculations used much smaller ports, but as someone mentioned they don't look to have enough port surface.

So I'm now more confused....

[amatt85 0]

once again, assumin that is 3.8cubes before any displacement and that the sub displaces around .2cubes...u could try aeros. two 4"aeros 15" long would put u right aroung 33hz and still have almost 3.5cubes net...

someone can check my math if they want...im like 95% sure im pretty close though

[C-Bass 0]

I don't have any place I can get flared ports around here and it would take at least a couple weeks for someone to bring them to me. I'm trying to avoid waiting so long. Plus the wood is already paid for

The total box dimensions is approx 4 cu.ft. sub is .17

Edited October 3, 2007 by C-Bass

[MTA 0]

if u do a wooden port that is 3x12 27long it is actually 4.5x13.5 x27long external dimensions..that is what u use for the amount of air volume it will displace inside ur box ( 0.949 cubic feet)

so that makes ur box just under 3 cubic feet after displacement... now 2.8x12=33.6 now does that make sense more?

as judging by the pictures it doesnt look like u have enough depth to do a pair of 4" aero ports 15" long anyways... with wood..you can actually build them L shaped to accomodate the length