>.Not trying to argue, just want to understand.
no argument here either, just trying to explain few things.
first, let's make an assumption that the UA-5 under normal operating conditions has a current consumption of 500mA.
>>If I know how much power my UA-5 will draw in one hour,
this is incorrect. your UA-5 continuously draws 500mA, whether you operate for 1minute or 10hours. a current consumption is not a rate.
>>why can't I divide that by the rate of charge to the battery to determine how long I need to charge the battery for each hour I intend to use it?
1. you're not charging your UA-5. you're charging your battery to power your UA-5. 2 totally independent and different operations.
2/ because your battery's "mAH" rating is based on a full charge. if you want to partially charge your battery, it will affect your voltage. it's quite difficult to know how much a partially charged battery will supply the needed current an keep the voltage up.
let's take a look at this example based on your scenario:
you want to operate your UA-5 for 1hr.
calculation: 500mA X 1 hr = 500mAH
your charger's output: 280mA. divide the 500mAH by 280mA = 1.78hrs.
so, if you take your fully discharged SLA and charge it for 1.78 hours and hope that it will operate your UA-5 for 1 hr.
can you see where the logic fails to make sense? your battery's capacity is missing in the scenario. depending on the capacity of your battery, you'd have no idea whether you'll have enough voltage to operate the UA-5.
marc