I REALIZED I MADE A MISTAKE IN MY MATH ABOVE: the Atom can do 53mW at 1400Ω, not 5.3mW, which will produce more than 114dB. A30 can do 61mW, which should be just south of 117dB. Both painfully loud.

By the way, if you want to figure this stuff out on your own it's quite simple with a bit of basic algebra.

Two things to remember: 1) Voltage (for the most part) remains constant no matter the load (impedance); and 2) The relationship of power to volume is logarithmic– every additional 3dB of volume requires a twice as much power.

Watts = Power = P

Volts = Voltage = V

Amps = Current = I

Ohms (Ω) = Restistance/Impedance = R

P = V*I

I = V/R

Therefore:

P = V^{2}/R

and:

V = √(P*R)

Using the Atom as an example, we know that at 600Ω that it's capable of 125mW (0.125W), so:

0.125W = V^{2}/600Ω

V = √(0.125W * 600Ω)

V = 8.66V

Then, since voltage should remain constant, we can plug that voltage in to figure out how much power it can do into 1400Ω:

P = 8.66V^{2}/1400Ω

P = 0.053W = 53mW

The T 1 has a sensitivity of 102dB at 1mW, so it requires 4mW to reach 105dB, 8mW for 108dB, 16mW for 111dB, 32mW for 114dB, etc... I'm much too lazy to remember how to do logarithms to figure out precisely how loud it'll be at 53mW, but you're certainly welcome to do that.