oorrrr... just for the heck of argument, how about a REO / BF hybrid? an atty like the RM2 built onto the body itself and the only contact point is the bar to the positive of the 510 (the negative being one part with the body)
just tinkering with ideas. I'm already at 4 volts on a .6 ohm, I have only one or twice gone below .5 so I'm at a perfect spot for my device
on a 4.15 V battery off charger, how much better can it get? 4.1?
4.1^2/.6 = 28 watts
4^2/.6 = 26.66
28 - 26.66 = 1.34 watts
100 * 1.34/26.66 = 5% improvement???!?
Time invested on such small gain is not worth it for me at the moment
Unless you keep doing it just for the heck of exploring alternative designs / fun of it it's not worth it, a battery after 10 min / 20 min of use would have already dropped to 4 volts if you are doing SLR, which I understand completely when one has time
Reason why I don't look at dual coils that much, battery life is the next step for me
Now, a blem REO is $115, regular $145, with a little modding I can get that puppy to 4 volts on .6 ohm, which is better than many many high end devices out there
that's why i told you about my plans for my next two REOs, one for DNA20Ding (which I know you don't care much for) and one without the Bottom feeding feature but with 2x 18650 in parallel -> on a sub-ohm you may go 24 hours + with these.
I mean, 2x these:
$12.77 Panasonic NCR18650A 3.7V 3100mAh 18650 Rechargeable Batteries 2-pack - button top / 2-pack at FastTech - Worldwide Free Shipping
or these:
$11.66 Panasonic NCR18650 Rechargeable 2900mAh 3.7V 18650 Li-ion Batteries 2-pack - authentic cells / 2-pack at FastTech - Worldwide Free Shipping
and you are talking about a mechanical device with 5800 / 6200 mAh of charge!
that mod would use two screws at the bottom (one where the feeding bottle is) and a custom flat bar at the top, below the black plastic and to which the firing bar connect, something that I can place and remove at will (if I really miss BF on a second, third REO?)
Batteries in parallel last twice as much because the circuit demands (both being equal) half the amps from each
I don't have hard numbers of equations to back this up, but with less amps being demanded from each battery the drops maybe just a tad smaller. You still have the same amps at
most contacts, but will have half of it at each battery + / - contact. EDIT: I may be wrong here.
And then you have the internal resistance of the battery, but when you are measuring 0.0x ohms that should make no real difference that I could imagine