Is my maths right?

I was curious how much more power an 18650 has than a typical AA battery, being not that much bigger in size.

A duracell AA at 1A output (typical for a small flashlight) returns 506mAh (source: 1.5V AA Duracell Alkaline Battery Tests - RightBattery.com)

so that’s 506mA at 1.5V. Adjust that up to 3.6V - 506 / (3.6 / 1.5) = 210.8mAh

So a 3500mAh 18650 has the same amount of power as 16.6 duracell AAs at 1A output. That’s staggering. It does halve to 8.3 AAs at 0.5A current but even still.

Using HJK’s data, a Keep Power 3500 mAh 18650 has 11.881 Watt hrs and a Duracell AA has 1.297 Watt hrs, both at 1 Amp.

18650

AA

11.881/1.297 = 9.16 so this 18650 has a little over 9 times the power of the AA cell.

negev, as xxo already used, the comparison for total energy of a battery is not amp-hours but watt-hours. This is calculated by multiplying the capacity in amp-hours by the nominal voltage of the battery. So for a 3500 mAh (=3.5 Ah) 18650, we multiply by 3.6 (or 3.7 for some cheeky manufacturers) for a value of 12.6 rated watt-hours. Of course xxo referred to HKJ’s test (and at a specific discharge current, too!) so the value is different, but perhaps better since it’s based on an actual test rather than rated capacity.

If we were to compare the amount of energy the cells are able to deliver in bursts, the 18650s pull even further ahead! Check 10, 15, or even 20 amp discharge curves for 18650s (maybe 30Q or VTC6). If you want to be more generous to AAs you could use Energizer Lithium or Eneloop NiMh to compare.

Error; Confusing watts with watt hours

Power can be expressed in watts, not watts / hours.

Power equals work divided by time. Power = Watts. Watts = amps times volts.

Watts equal work divided by hours.

Work equals power times time.

Work equals watts times hours.

Time equals work divided by power.

Hours equal work divided by watts.

Whatt?

I don’t know how many watt hours I worked shoveling snow today, but I could feel the BTU’s dripping from my forehead

Energy = Watt Hours

‘xxo’ never mentioned power.

Pepinfaxera is right I should have said energy instead of power.

Math’s good but the English is lacking

the BTU is much to large to be used as a headlamp. Wrong tool for the job .

As my dad used to say… “the heat is in the work”

First of all, you are speaking about energy, not power.

Alkaline cells, along with any sort of expendable cells, are not designed for heavy loads.

And this calculation, while seemingly valid is far from correct. This is because you are considering “nominal” voltages, and nominal voltages are sort of average voltage values at typical loads. A high energy 18650 li-ion cell is expected to provide an average of 3.6V during discharge at 1A load current, maybe a little bit more. However, the expected typical load for an alkaline 14500 cell is much lower than 1A, a lot lower. This means that the average output voltage from a 14500 cell at 1A load is quite a bit lower. In fact, in your above review the average output voltage from the akaline cell down to 1V at 1A is just ≈1.12V.

Let's see a more proper Duracell alkaline 14500 test: Duracell Plus Power AA

The guy in the test you linked obtained 506mAh at 1A, cutting-off at 1V. This matters, too.

HKJ (see above graph) obtained ≈687mAh at 1A with cut-off at 1V, and 1097mAh with cut-off at 0.7V. Cut-off must be considered.

You should compare cells in terms of energy, which is average load voltage (V) times capacity (Ah): E = V × Ah = Wh.