ORIGINAL: rctech2k7
BTW, the actual current is based on the actual voltage of the charger less initial battery voltage then devided by their combine resistance.
Depending on the condition of your battery, considering healthy battery at fully drain, devide the mAh capacity by the charging rate of mA will give you time in hours then multiply it with factor that base on your charging rate. Usually 1.25 would give you 20% of energy to compensate loss on charging. Make sure that rate after full charge or charging current would not heat up your battery for a long time.
I'm not quite sure anyone knows the "combined resistance". Well, someone knows, but probably not the average "joe" in the hobby. Ummm... ok. Suffice to say that charging with about 20% overhead is probably sufficient.
CGr.