I was being an idiot and remembered ohm's law wrong. Guess that's empirical evidence that drinking makes you stupid. Taking the simplist view above, I=V/R, or R=V/I, or R=0.4/1.5, or ~0.25ohm, or ~0.6w. Likely 0.4v is an overestimate, according to Cree's spec sheet.

Except that's really not the proper way to solve it. The system is nonlinear, and you need to take--the graphs for Vf vs I for the emitter, I=V/R for the resistor and additional resistive loss in the system, and V vs I for the battery--and find the point at which they all meet.

As an example for the above, assume 1.5A through everything, which will take the battery from about 3.8-3.4v full to discharge (through it's not going to be 1.5A fairly quick in a resistive circuit). Assume you'd only want 1.5A max, we'd take the 3.8v. Also at 1.5A, the emitter according to cree has ~3.5Vf. At 0.3v through everything else, that's only 0.2ohm. There's going to be probably that much resistance through everything in the system (including the battery), so chances are you may be alright without anything.

However to illustrate the problem with these kind of estimates is for instance, this guy only measured 3.36 vf at 1.5a for his sample http://media.photobucket.com/image/xp-g%20vf/jtr1962/CPF/Cree_XP-G_bin_R5_Lumens_Chart.gif

if that's true for the specific emitter you use, the "extraneous" voltage drop will be 3.8-3.36 = .44v. If indeed you only had 0.2ohm in the system, ~1.8A through the system would be likely given the same assumptions (say 3.78v in battery - 3.40 from chart = 0.38v, 0.38/1.8 ~ 0.2ohm).

By the same token, when the battery starts getting drained, it'll get to be around 3.5v at ~1a. Using the graph above, that'll be around 1.1A @ 3.28v for the emitter. However, again, if we use the cree sample data, it's around 900ma.