Indeed I have. I've been doing the 3.7v/1.5 ohm thing until now. I have around twenty new 1.5 ohm Cisco 306s left in my stockpile.
Is LR the culprit? Seems counterintuitive to me, like having to produce higher voltage would shorten the battery life, but I guess I don't know how the internals work to make such an assumption. If LR is the culprit, I guess that gives me an excuse to try out an HH357.
making it produce a voltage higher (or lower) than the battery is going to use up the battery. Basically its a buck/boost converter, which basically pumps charge from a switching transistor (the switching frequency corresponds to the voltage you want outputted I believe but could be wrong) through a small capacitor which is then connected to a inductor, making a "constant" voltage. through this process of increasing or decreasing the voltage compared to the reference (the battery, 3.V), there are losses due to heat, from switching a transistor on and off quickly (the faster the "clock" rate, the more energy used by the power transistor) to charge the cap (which has internal resistances and heat losses). All elements, from resistors, caps, inductors all have a non-ideal impedence. For example, a cap has some inductance and resistance, a resistor has some inductance and capacitance, etc. Not that these things are particularly lossy (except for resistance, which creates heat), its just that no element is ideal. The coils in our atomizers probably have a relatively substantial inductance since it is coiled (aka a inductor is coiled, but also connected to a ferrous core to increase inductance). Also, the chip used in the Darwin, the DNA12 (amirite?) has an efficiency of 87%, and is always on, so it is discharging ever so slightly as it's "off". The buck/boost converter I described may not be exactly what the darwin uses, but it's a pretty typical circuit and charge pumps are used in all kinds of applications.
ANYWAYS, back to the topic. LR is intuitively going to drain more power than HR or SR. a lower resistance path from the + terminal (Vdd) and - ("ground") of a battery means that there is almost nothing "resisting" it from discharging at it's full current output. A higher resistance, like 3 ohms, resists twice as much as 1.5, so the current being pulled from the battery is half.
Simple V=IR
I= 3.7V/1.5ohm=2.4666 Amps
I= 3.7/3ohm = 1.233 Amps of instantaneous and sustained current as you drag.