Is there a formula to calculate SPL loss in relation to frequency?

Is there a formula to calculate SPL loss in relation to frequency?

Q: Is there a formula to calculate SPL loss in relation to frequency? For example, if a subwoofer can produce 140 dB SPL at 30 Hz, how many dB would it produce at 23 Hz? Alternatively, is there a formula to determine how much lower you can go in Hz frequency by pairing multiple subwoofers?

A: This is a subject that requires far more information than is made available through spec sheets. It is a subject that I’d love to pursue in a video or blog but it would be possible to test against only the speakers available at the test at the time. 
The limiting factors in loudspeaker performance are amplifier power, driver displacement (ie, cone area multiplied by excursion distance, aka vD) and the relative efficiency of the cabinet design at given frequencies. There are trade-offs. The phrase, “all other things being equal” comes up a lot, but between products, that’s essentially never going to happen. One can compare the same amplifier and driver in different cabinets but changing more than one element introduces too many variables to make the theoretical or even simulated comparisons relevant.  They must be tested in situ, and comparing any two would provide very limited benefit, so many need to be compared, and that requires the commitment of a lot of time and money. 

Considering the initial meaning of volume, it’s essentially a synonym for displacement, and frequency defines of segments of time. So SPL is equated with volume, aka displacement. Frequency is the duration of time over which that volumetric displacement must be maintained. In digital terms, that’s on, off. In sinusoidal terms its rise-fall. Nevertheless, positive (and then negative) pressure must be maintained for twice as long to deliver 20Hz than 40Hz. (Or any 2:1 frequency ratio, of course) In order to provide the same pressure for twice the duration demands twice the displacement, but not just in one direction. It demands twice the displacement in both directions. Thus the ability of a piston of a given diameter (woofer cone) to deliver the same pressure an octave lower requires 4x the displacement, all other things being equal… The ability for that piston to achieve that displacement requires more power and more acceleration as well, so there’s more to add. 

This is where cabinet design comes into effect. If one is to design for maximum efficiency at a lower frequency, higher SPL can be achieved at that target frequency, usually with an increase in physical size, but inevitably there will be a loss of efficiency at higher frequencies. It’s possible to compensate for those upper frequency losses, but that also adds to the physical size, the complexity, the weight and the cost.

Essentially, it comes down to personal preference as much as anything else. If you want deep bass, and you’re willing to pay the price to get it in terms of physical size, cost, or whatever the price may be, then you pay it. When you do, you have something rare and special because so few are willing to pay that price. "Good enough" is cheaper, and only those who appreciate the value are willing to pay the price. 

Interestingly, there is research being done about the tangible value of deeper bass now, and it seems to offer advantages in perceived value of items sold, in time spent in venues, in time spent dancing and in other areas such as perceived quality. That’s another subject, but I believe one worthy of discussion with any potential customers.

I realize this doesn't exactly answer your question, but I hope it provides a… deeper understanding… of the subject. ;-) 
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