hmm let me see if i got that straight...
the electronics are in fact switching resistors as the voltage gets lower....so in effect the headlamp is a resistored drive.
I think it switches in different resistances for different power settings, but for a given power setting, there's only one fixed resistance.
I do have trouble visualizing what you say about the graph....
Basically I made a graph from your figures of current and battery voltage. For each battery voltage (full alkalines, used alkalines and NiMH cells), I plotted points for the boost, high and medium power levels, which gave three points for each of the power levels.
For the high power level, the thee points were pretty much on a straight line. Drawing a line through the points, and carrying it on, it crossed the voltage axis at around 2.8V
That's pretty much what would be expected for a fixed resistor drive. Since even the highest current (high power, full alkaline battery) was only 0.18A, the actual Vf of the LED at that current would have been very little different from the Vf at the lowest power level (0.12A), and would also not be massively different from the Vf of the LED when it was just barely turning on.
As a rough approximation, if we pretend that the Vf of your LED was close to constant, having a resistor in series with it would give a graph which started at Vf on the voltage axis, and then went in a straight line with increasing current with increasing battery voltage, with the slope depending on the resistance.
Things were very similar for the medium level, just with the crossing point being very slightly lower, though it's hard to be completely accurate with lower currents measured in 10mA steps
For the boost level, the three points weren't really in a straight line, but that might well have something to do with voltage measurements being made without a load. If the alkaline voltage levels had each been a little lower, a very similar picture would have emerged to that at the high and medium levels.
Doing some *very* rough calculations, for the high level the results are consistent with a total resistance of about 10 ohms, and for the medium level, about 19.
Bearing in mind the roughness of the calculations, that could be something like the high level using two 16R resistors in parallel (giving 8R) and the medium level using one (giving 16R), with possibly a little more resistance from elsewhere in the light (or one of the 1R5 resistors?), and/or just from error in the rough calculations and lack of precision in the numbers.
With even rougher calculations, in the case of the boost level, that seems like a resistance of something like 2 ohms, which could be a 1.5R resistor and extra resistance in the circuit or LED, though at lower resistances, it's a lot harder to be sure since all kinds of things become relatively more important.
you measn that in both cases the peak is at 2.8Volts (is that under load?) and amp draw will not get any higher as voltage increases, because the extra voltage is consumed by the resitor that is selected each time?
Very roughly yes, at least at the medium and high settings.
At those relatively low currents, the Vf won't vary much, so as a first approximation, an LED and resistor could be visualised as a 'constant voltage sink' with the current flowing in the circuit varying linearly with the 'surplus' battery voltage above the level of Vf.
