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.