Here is a question to our tech experts: How do I calculate total efficiency of two cabs with different efficiency ratings, for example a 115 with 98dB and a 210 with 100dB? I just know that you have to add 3dB if you have 2 equally loud sources and that you gain another 3dB if they are close together (stacked cabs) due to coupling effects or something like that. Thanks, Matthias

I dont know any formula, but my thinking got me to 102dB 103 dB = Is not achievable. 101 dB is Achiavable with 98 +98.. So.. (98 + 100) /2 = 99 99 + 3 = 102 dB Mayble Im completely wrong, but that´s my empirical aproach.

Hi, Matthias, First, I must point out that sensitivity and efficiency are not the same. Sensitivity *can* indicate efficiency if all else is equal (e.g., directivity). The reason is that efficiency is the total acoustic output power of the cabinet in 3-space (i.e., a complete envelope around the cabinet) divided by the total electrical wattage input. Sensitivity is the sound pressure level measured at *one* point in space. Second, the mutual coupling only works at the lowest frequencies, and diminishes to nothing eventually (typically in the upper bass region). Others have indicated that sensitivity is usually measured anechoicly at 1 kHz, so it is doubtful that mutual coupling contributes anything to the sensitivity at that frequency. However, at lower frequencies, you can get some benefit from the coupling. On your question, if you can put equal wattage into both cabinets, the 115 will give you 95 dB at 1/2-watt, and the 210 will give 97 dB at the same wattage. If you combine these, you get 99.1 dB (you have to add their individual acoustic power intensities, which involves using logarithms). This is the likely number at 1 kHz. If the drivers are placed within 2 feet of each other, mutual coupling can assist starting at below ~140 Hz. But putting your cabinets on a floor, or in a corner, etc. boosts the lows as well. - Mike

Luis: Yes, that's what I thought about, too. But as I have learned from this forum that acoustics is not so simple, I thought I'd better ask... Mike: Thanks a lot for this detailed answer (again). I appreciate your contributions to this forum very much - I've learned a lot in the past year, and a big share of it came from your posts. (not to forget Joris, Throbbinnut, Bruce L and Bruce G, Dave B and many others of cause) Matthias

Oh, one more question please: How do I do this? I tried to figure out, but well... (maybe that's the wrong moment to tell you, but I've studied mechanical engineering - as it seems I've forgotten some of the theory...) Matthias

Acoustic power intensity, I, is the acoustic power divided by area, and can be referenced to Iref = 10^-12 W/m^2 (a picowatt per square meter). This intensity would correspond to 0 dB. To get the intensity for a given SPL value, you use I = Iref*10^(L/10), which is the inversion of the normal logarithm, L = 10*log(I/Iref). L is the sound power level, expressed in dB, and is interchangeable with SPL (sound pressure level). So, you take each SPL, calculate the associated power intensity, I, then add each I. Take Itotal and divide by Iref, take the log, etc. and you have the effective L. Even if Iref doesn't exactly correspond to SPL's Pref (depending on the acoustic impedance), the difference washes out because of the ratio. - Mike