Bill your statement:
#3) Provides more AH from the same setup as the inverter is inefficient (90% ish) but more importantly the high currents an inverter needs cause the batteries to actually operate in a mode where they have less AH. Basically a batt used for a high amp load will deliver fewer AH than the same batt powering a low amp load.
Do you mean that if I draw some battery (X) down at a 5 amp per hour rate for 9 hours thereby reaching 50% state of charge, then if the draw down had instead been a 15 amp per hour rate then a 50% SOC would exist before 45 amps had been used? What formula do you use to determine how much?
thanks Lloyd
Lloyd,
This is called Peukert Effect...
A 100Ah battery at the 20 hour rate is rated to deliver a 5 amp load, for 20 hours, before dropping to 10.5V/dead.
The formula is; 20 hour Ah capacity divided by 20.
100Ah Batter / 20 = 5A
or
125Ah Battery / 20 = 7.5A
Any draw above 5A will result in less than 100Ah out of the battery before hitting 10.5V. The higher the load, further away from 5A, the less Ah's you'll have...
Any draw below 5A will result in slightly more than 100Ah's from the battery before hitting 10.5V...
This is why it is sooooo critical to properly program a battery monitor with your manufacturers Peukert value, if you want it to be close to acurate..