Why not talk about slot ports, since they are common.

[TheOtherSide]

Here's a doozie for some of ya'll...maybe I can get a straight answer from one of the knowledge gurus here.

I was thinking a while back about building a slot ported enclosure for my BM when I first got it. Problem is, in an Ext Cab truck, there are very few straight, square edges to use as 2-3 sides to the port. Which got me to thinking, is a port on a ported enclosure tied directly to length and surface area or volume total? Such as if the external opening were 9"x4" (36 square inches) and the internal opening was 7"x5 5/32" (36.09 square inches), port length would be calculated the same as if it were, say, a tube with a surface area of 36 square inches? Or would it be figured differently?

I ask because it still kinda bugs me. I'm not losing sleep over it but it still makes me wonder if I could've done a slot port instead of a tube port.

[ku40]

It's basically just surface area. In the same box, a 36 sq. inch slot port would be the same length as a 36 sq. inch round port for the same tune.

I'm not quite sure I follow your example. Are you saying if the outer and inner openings of the port were different dimensions? I'm not sure why you would do that? Unless you can taper it to the inner size with a smooth wall, it would be bad. You don't want to have excessive corners within the port.

[Electrodynamic]

If I followed your post correctly (I'm still not on the same page about the differences in the port opening vs. the inside of the port if we're talking a slot port), ku40 is correct. It's all about surface area and volume.

In the same box, a 36 sq. inch slot port would be the same length as a 36 sq. inch round port for the same tune.

is right on the money.

[TheOtherSide]

I understand the length/area thing, but let me see if I can rephrase/draw this for you. That is typically where I lose people, when they just say, why do that?

Imagine a wedge shaped box roughly 14 inches tall at the front and 9 inches tall at the back. The length of the box is 14 inches, front to rear, and width is 54 inches (the top is level, the bottom is angled {see picture below}). If total volume (just a random #) was 2.2 cubic feet net and you were porting at 34 hz (old BM specs) then you'd need a round port of diameter 6 inches and length roughly 25 37/64". If one wanted to slot load the box and use the 3 walls of the box for sides of the port, then at the front of the box you'd have a height of 12.5" and a width of 2 17/64" (assuming 3/4" MDF) for total port opening area of 28.27 square inches. At the rear of the box, the port can only be 7.5 inches tall and would change the width required. So here's the question, what would the width need to be because if you turned the port in the corner, to get the required length, the port shape would be like this:

because of the turn like this:


Would you set the area at the end of the port inside the box to that on the outside? If no, then how would you do it. I just want to understand the theory behind it, not actually build it. I think of it as a sort of T-Line but the fold in the corner doesn't make the opening smaller or larger.
Did I lose you yet? Basically it probably is more work than its worth but just figured I could ask if anyone knew about it.

[Tirefryr]

Go with the smaller of the two openings for figuring your area, so use the back in this instance. You won't notice a difference in tuning by ear anyways. I've done this several times before with no negative effect. If you are truly worried about it, carry the area all the way throughout the enclosure.

[ku40]

You'd also have to remember that as your first port section went towards the back, the top of the box would be collapsing, lowering the cross sectional area of the port. So you'd have to angle that port board in towards the center of the enclosure as it went backwards to make up for the height lost.

In that situation I'd just make a port along the bottom on the side. It's flat and even there.