Just for an FYI...
The reason higher 3.6 voltage is used, as opposed to 1.5v, is to reduce the amperage running through the device. Lower voltages create higher amps, which heats-up the components between the battery and the high wattage coil. Higher voltage allows for thinner wires between the battery and the coil, since the parts stay cooler, and there is less power lost from less amps flowing through the device. (This is also why power-lines use high voltage over long distances. It is cooler on the wires with less amps flowing through them.)
There is one missing element for any real calculations here. We don't know the wattage of the coil itself. However, we can estimate the wattage by puffs and puff-lengths between charges. (Got a pen? LOL.)
There are three things to determine power consumption.
Volts * Amps = Watts
Watts / Volts = Amps
Watts / Amps = Volts
Uber power pack!
http://www.batteryspace.com/index.asp?PageAction=VIEWPROD&ProdID=2357
Also note...
A 120w device rated for 120v will draw 1A at 120v, but will draw 10A at 12v.
A 12w device rated for 12v will draw 1A at 12v, but will draw 0.1A at 120v.
You know your volts, 3.6v, and you know 1200mAh or 1.2Ah.
(V * A = W) {V * Ah = Wh} Same thing {Hours are assumed}
(3.6v * 1.2A = 4.32W)
At 3.6v, with that 1.2Ah battery, you can power a 4.32W device for an hour.
Reality check...
Most devices require a range of operating voltage. A 3.6v 1200mAh battery, should provide around 3.6v @ 1.2A of power for an hour. However, if the device you are powering is half the wattage (4.32w / 2 = 2.16w), it will operate at 3.6v for 2 hours.
The wattage is the same, but, depending on the voltage, your amps will rise and fall, reverse of the voltage.
Higher voltage will consume less amps, at the same wattage.
Lower voltage will consume more amps, at the same wattage.
A voltage regulator will result in constant voltage that draws more amps from the battery, as the voltage in the battery falls below the output voltage. On the opposite side, the regulator will consume less amps from the battery if the battery voltage is higher than the output voltage.
EG,
A 12v 10Ah battery can handle 120Wh.
If that were regulated to 1.2v output voltage. The 120Wh device would be drawing 100A. Since the battery only has 10Ah, but now draws 1/10 the amps since it is 1/10 the voltage, it can run that 120Wh device for 1 hour.
Battery sees 10Ah load. (Battery stays cool)
Circuit sees 100Ah load. (Circuit is hot, and device is hot.)
Another situation, with a smaller wattage device.
A 12v 10Ah battery can handle 120Wh.
If that were regulated to 1.2v output voltage. A 12Wh device would be drawing 10A. Since the battery has 10Ah, it can only run that 12Wh device for an hour.
Battery sees 10Ah load. (Battery is warm)
Circuit sees 10Ah load. (Circuit is warm, and device is warm.)
Another situation, using higher voltage.
A 12v 10Ah battery can handle 120Wh.
If that were regulated to 120v output voltage. The 120Wh device would be drawing 1A. Since the battery only has 10Ah but now draws 10x more amps in the conversion which is 10x up from 12v, it can only run that 120Wh device for 1 hours.
(12v SLA battery, upped to 120v, will power a 120w light for 1 hour.)
Battery sees 10Ah load. (Battery is hot)
Circuit sees 1Ah load. (Circuit is cool, and device is hot.)
If you add a micro-controller, like this...
http://www.electronicspecifier.com/...eading-Position-in-6-pin-Microcontrollers.asp
You could fully regulate power and control of the device, like the battery that comes with the vaporizers. (Since they use a similar micro-computer controller in the battery.)
That same micro-controller can also be used to charge the battery, or regulate the charge power going into the battery. But you may get the same cruddy results as we get from our battery chargers. It has amperage limitations, so it can not provide the high-amp charge that these batteries like, when they are low on power.
There is another chip that can handle that. (Same chip they use in our cell-phones.) No programming required.