How do you determine if one cab is "more efficient" than another and how much more efficiency is needed for a given application? Also, how do you know what "class" (A, B, C, D) an amp is?
In most instances companies tell you the efficiency of their cabs. It's usually expressed as how load a cab plays with 1 watt applied at a distance of 1 meter. For example the rating would look something like this: 97dB 1W/1m
Sometimes sensitivity rating are expressed in terms of 2.83V with the impedance specified. So you might see something like 99dB 2.83V/1m at 4 ohms.
The tricky part of this is 2.83V at 4 ohms = 2W and 2.83V at 8 ohms = 1W. To get the 1W/1m rating for the 4 ohm cab, subtract 3dB. So in this case 99dB 2.83V/1m at 4 ohms is the same as 96dB 1W/1m.
My feeling is the 2.83V rating is often a way to try and hide the fact that the 4 ohm variant of a driver is slightly less efficient than the 8 ohm variant.
When comparing I suggest using the 1W/1m rating.
If you know the 1W/1m sensitivity rating and how much power the amp is rated for, you can estimate the max SPL. This is because we know that SPL increases by approximately 3dB every time you double the power.
Let's assume a 100W amp and count the doublings of power from 1 to 100.
1,
2,4,8,16,32,64,128.
As you see, the power doubles 6 times (count the numbers in bold).
Now multiply by 3dB for each complete doubling. 3x6=18dB. This is the decibel change from 1 to 64W.
Add this to the 97dB 1W/1m sensitivity rating. 18X97=115dB.
I suggest you also consider how close the available power is to the next doubling. 100W is closer to 128dB than 64dB, so let's add 2dB.
115+2=117dB. This is our estimated max SPL