Don't suppose anyone has a table showing amperage output at sub ohm levels do they?watts/volts
The following may be a bit complicated for some - and not absolutely necessary to know. But when playing with low resistance coils and high battery currents, it is nice to be able to verify that everything is fine before risking blowing up a battery.
To expand on what Dan wrote a bit:
You want to vape at a maximum of 25 watts. What is the circuit resistance you need?
A fully charged battery is 4.2 volts. Dividing 25 by 4.2 gives 5.95 amps.
You can now find the necessary resistance - since Resistance = voltage divided by amps.
R = 4.2 / 5.95 or approximately 0.7 ohms
Hopefully, the coil will be most of this resistance - with the internal battery and connectors contributing a negligible amount.
If you have an accurate voltage meter, you can fire the coil with a fully charged battery and see what the voltage measurement is across the coil. It should be very close to 4.2 volts. If it isn't, you can figure how many watts you are losing in the battery/connectors.
Assume that the voltage across the coil is 4.0 volts. You know the coil is 0.7 ohms, since you carefully wound it and measured it at the binding posts.
power dissipated by the coil = voltage(across the coil) squared divided by the coil resistance
So power (watts) = 16/0.7 = 22.9 watts
This means you are losing 2.1 watts in the battery and connectors.
You don't have 5.95 amps flowing through the coil as was calculated for an ideal circuit - you only have 4/0.7 = 5.7 amps.
There are other ways to get to the same answer, especially if you know all the relationships between watts, volts, ohms and amps.