Hey, Mamabear, good explanation! I didn't check your math, but your method is good.

Technically, W= VxA is only accurate for DC and it takes a bit more math to figure wattage for AC supplies, but for the purpose of figuring this sort of thing out, how you demostrated it is practical and good enough.

My inverter knowledge is sadly out of date (other than owning some), but for other reasons I've been asking some questions lately. The modern inverters are pretty much all true sine wave now (or close enough to not matter) and efficiences are supposedly 90% or better. (hmmm - I am confident of the sine wave, but need to see real figures to believe the efficiency claim - maybe. Dunno.)

All the inverters we have are well-protected against overcurrent. All will deliver 2x rating (or 3x in one case) for a brief period to account for starting loads. If any of them are overloaded, they simply shut down, no fuss. The surge capability can be misleading if one doesn't do the math like you did. Example: I can fire up a Bosch reciprocating saw with one of the 600 watt inverters and it runs great, just like on line power - the surge capacity is more than ample to fire it up. And the NO LOAD current draw once running is well within the rating, taking only about 3 amps (maybe 360 watts, more or less). Try to cut something with it, and poof! no juice after a couple of seconds. That's because the working power requirement is much higher than that inverter can deliver for very long - it takes about 900w to actually USE that saw.

Battery capacity is a whole 'nother topic, but your advice was pretty good. Like you said, do the math!

Regards,

Tom