Whoa, I was looking up instrument loudspeakers on google and came by this:

http://www.instituteofbass.com/product_spotlight/

This makes me wonder if something tricky can be done with the WO as well.

At first glance, this makes no sense to me.  This theory may work in cylinder heads, where 4+ smaller valves amount to more flow than a single large valve, but I thought one of the necessities of creating lower frequencies was starting with a cone surface that could create them to begin with!  Low frequencies seem lower pitches when played very loudly, but... well you get my drift.  What is really going on in those cabinets with all the 5" drivers?

All I can figure is they're using multiple drivers to create larger SPL since there is as they said more effective cone area, but making up for the lack of actual low frequency reproduction from the driver cones by using a transmission line or folded horn of sorts.

You can't compare speaker output to valves in an engine.

Valves in an engine optimize air velocity and flow. For SPL you want acoustic pressure.

On that page, I see a setup that has an 8x4 array of what I assume are these 5" 'woofers'. That's roughly 628 square inches of cone area in about 960 square inches of space. We could put 6 12" speakers in that same amount of space but have about 678.5 square inches of cone area. That's 50 square inches more cone with less wiring headache.

20hz from an 18" speaker is a lot different than 20hz from a 10" speaker.

Well-

Lets look at the 4 x 8 array.  It says it uses 5 inch drivers.  We can assume these drivers ar emounted as close as possible in lines, meaning each 5 inch speaker is mounted in a 5 x 5 square.  This means the 32 5 inch speakers use up 800 sq inches of space, and create 628 sq inches of radiating area.

Similarly, we could fit at most, 4 12" drivers (in the 800 sq in and in lines.)  These 4 12s would have a radiating area of 452 inches.  

That means the 5 inch speakers have more radiating surface then the 12s.  

The 5s will play higher frequencies easier and sound better doing it, plus they will displace more air then the 12s at the same excursion.

Problem is, the 12s will have a higher excursion, therefore equal or exceed the displacement of the 5"s.  The only real benefit is a somewhat similar output with higher more controlled frequency response.

[quote author=colinreynolds  link=1165668742/0#5 date=1165811224]And as said, the louder you play low frequencies, the lower they sound.
[/quote]
i tought it was the lower the frequency, the louder it seems to the ear

[quote author=J_Rock  link=1165668742/0#4 date=1165727407]Well-

Lets look at the 4 x 8 array.  It says it uses 5 inch drivers.  We can assume these drivers ar emounted as close as possible in lines, meaning each 5 inch speaker is mounted in a 5 x 5 square.  This means the 32 5 inch speakers use up 800 sq inches of space, and create 628 sq inches of radiating area.

Similarly, we could fit at most, 4 12" drivers (in the 800 sq in and in lines.)  These 4 12s would have a radiating area of 452 inches.  

That means the 5 inch speakers have more radiating surface then the 12s.  

The 5s will play higher frequencies easier and sound better doing it, plus they will displace more air then the 12s at the same excursion.

Problem is, the 12s will have a higher excursion, therefore equal or exceed the displacement of the 5"s.  The only real benefit is a somewhat similar output with higher more controlled frequency response. [/quote]
Lets say these 5" speakers have a peak-peak xmax of 1 inch. That means on one single peak to peak stroke (- to +) of all speakers at the same time moves roughly 418.46 cubic inches of air with the 1 inch stroke and 1 inch deep cone. Now those four 12's would move roughly 597.15 cubic inches of space if they had a 2inch xmax and 2inch deep cone.

Now, my math could be wrong, as it's 1:30am, but the same formula (wrong or right) was used on both examples, so it still stands that the 12's would move more air.