I don't know what amount of volts and watts a mech mod can put out with a freshly charged battery. But can a regulated mod at 8.5 v compete? I don't quite understand how a mech mod works in regards to resistance because when you lower the resistance you lower the volts. Surely the more volts the better?
A mechanical mod will never put out more than what the fully charged battery has -- 4.2 volts. The battery output declines as the battery is drained to approximately 3.4 volts, then needs to be recharged.
The lower one goes in coil resistance, the more current (amps) is drawn from the battery. This is why sub-ohm coils require higher amp batteries (20 - 30 amps). Since a mechanical mod doesn't have a processor to adjust the power output, one must depend solely upon changing the coil resistance to change the vaping experience.
1.0 ohm = 4.2 amp draw
0.9 ohm = 4.6 amp draw
0.8 ohm = 5.2 amp draw
0.7 ohms = 6 amp draw
0.6 ohms = 7 amp draw
0.5 ohms = 8.4 amp draw
0.4 ohms = 10.5 amp draw
0.3 ohms = 14.0 amp draw
0.2 ohms = 21.0 amp draw
0.1 ohms = 42.0 amp draw
0.0 ohms = dead short = battery goes into thermal runaway
Regulated mods still use Ohm's Law, but instead of the relationship of coil resistance & battery amperage, the mod's processor chip determines the current (amps) limits. Since the newer high wattage regulated mods have enough power to fire standard resistance coils with a lot of power, there's no real need to build sub-ohm coils any more.
Ohm's Law for Dummies and Vapers