Wiring question

ALthough this is for a bike light, I hope it'll be more appropriate to put it here...

I am looking to 'manually' (i.e. with my own wiring, as it will be to custom lengths etc) a length of wiring from the battery pack (which is 4x18650) to the headlight of my bike light.

What's the thinnest possible wire I can use for this task? I was hoping for something <2mm.... I think the light is rated at 3 amp,are there any other things I should be looking for?

It dependes upon what you are willing to live with, and the type of driver you are using for the LED's. 3 Amps at the led is ~10 watts.

If the 4 x 18650's are in parallel, that is one problem, and if they are in series, that is an entirely different problem. The rule of thumb is that you don't want to lose more than 5% of the power in the wiring. The fomula for losses is I^2 x R. So if you have a Buck Driver, the voltage to the driver from the battery pack will be fairly high, ~15 volts, and the LED @ 3 Amps will be seeing around 10 watts. From the 15 volt battery pack, that is about 700ma, so the resistence in the wiring should be no more than about 1 ohms (.7x.7=.49 x 1= .49 watt. If the Batteries are in parallel, then 3 amps at the LED is going to require about 3 amps from the battery pack. Using the I^2 x R formula again, 3 x3=9 x .055=.5watt, or the wiring will need to be about .55 ohm. Those two scenarios dictate very different wiring selection.
So let's say you need a total of 5 feet of wire (total). For the 15 Volt scenario, it would need to be less than .2 ohms per foot, and that corresponds about to 30 gauge copper. I'd probably use 22 gauge, it is a bit less fragile, and easier to work with. The parallel battery configuration needs to be on the order of .01 ohms per foot. That in turn dictates a wire of at least 18 gauge copper, and 16 gauge would
probably be preferable.

I suppose you could also be looking at a series/parallel configuration. That would be about 7.5 volts, and 10 watts would be about 1.33 amps. I^2=1.3x1.3=1.8 x .28 ohms=.5 watts. That works out to about
.052 ohms per foot, of about 26 gauge copper.

Basically the losses go up as the square of the current, so relatively high voltage feeding a remote buck regulator/LED is almost always preferable to a Boost regulator driving the same configuration.

Although a bit more expensive, I'd probably go with Teflon insulation. It is pretty impervious to oil/grease, grim, sunlight and heat. If the wiring length changes, the numbers will change, but the formulaes do not.

Hope this is all useful l to you.

I wouldn't recommend using super thin wire.

The reason being that the smaller the wire the more resistance it provides, you could easily be burning up 30-40% of the voltage you started with at the battery by the time it makes it to the driver. If you care about efficiency at all I wouldn't use any less than 14AWG for a three foot run at 3 amps, doing some VERY rough math in my head that with burn around 2% of your starting voltage.

I'll let others correct me and add on, as it is well past my bed time and I have been sitting here rambling off nonsensical paragraphs that thankfully none of you will never see as I still have the good sense to delete them.

Thick wire good, thin wire bad.