I am sorry, you guys are correct on the calculations of the math.

I had written 104dB and changed it for reasons that elude me right now.

in any case, the principle is still valid.

SPL is equal to 20 log pRMS / 2x10^-5

one equation I have relating RMS air pressure to a moving coil loudspeaker is:

pRMS = p0/(2*pi) * BL*EgRMS / (Sd*RE*MAS) *G(s) * Tu1(s)

BL is the BL product, RE is voice coil resistance, Egrms is the generator/input voltage, Sd is cone area, MAS is acoustical mass of diaphram and air load in infinite baffle, G(s) and T(s) represent the 2nd order high pass and 1st order low pass transfer functions and dont affect pRMS in the midband.

If you double EgRMS, pRMS doubles, giving rise to a 6dB increase in SPL.

however, when you double the power, EgRMS only increases by sqrt(2) = 1.41, and thus pRMS increases by 1.41 and SPL increases by 3dB. 20 * log(1.41) = ~3dB vs 20 * log(2) = 6dB.

You can also use the pressure equation above to explain why a doubling of the cone area will produce a 6dB increase in SPL.

MAS = Mass of Diaphram / (Sd^2) + 2*MA1

Since MAS is inversely related to the square of Cone Area, we end up with Sd also being directly related to Prms. MA1 is the mass of the air load impedance on the piston.

When you connect two drivers you are doubling the cone area which calls for a 6dB increase, and dividing the power between them in half, which only calls for a 3dB decrease.

Suppose you had two 92dB/1w/1m 8 ohm speakers with the amp I referred to above ( capable of 11.32V(16W) into 8 ohms). If you wire the coils in series, you'll be left with a 16 ohm load for the amp. 5.66V will appear across the coil of each speaker which is one doubling of 2.83V, a 6dB increase.

92+6dB = 98dB per speaker. Adding the 6dB for the increase in cone area brings us to 104dB. Note that the amplifier is only supplying 8 watts total or 4 watts to each speaker.

I will redo the math for the next one since i wrote the wrong numbers down in my other post. 2 speakers each on their own 8W amplifier(8Vrms output)

The voltage is doubled from 2.83 to 5.66 and then 8/5.66 = 1.41

2*1.41 = 2.83...20 * Log(2.83) = 9dB

so you get 92+9+6 = 107dB

a 3dB increase in efficiency compared to one speaker with 16 watts input at 104 dB.

Now consider two 8 ohm speakers with their coils wired in parallel. Here each speaker will be receiving 16 watts each.

The voltage doubles twice from 2.83V to 11.32V. 20*log(4) = 12dB + 6 more because there are two of them.

92+12+6 = 110dB assuming your amplifier is capable of delivering the extra current required.

Sorry if i made this a lot more confusing than it really is, I hope I'm never a teacher...

jt