What percentage of power taken from a batteries pack is going to the load ?
Why are your batteries getting so hot so fast , here is a possible answer
Very suprising !
Was testing some very high capacity NiHm "D" cells 12-15 A/H from china and little socked at 12-18 milliohms readings
averaging about 15 milliohms.
If when assembled into a 48 volts pack (40 cells) would give 600 millohms add another 10% for interconnections
and we are up to 660 milliohms or .66 ohms .Doing a 36 volts or 24 volts pack , the same math applied .
Taking a 48 volts 500 watts motor has a internal resistance of 4.6 ohms .
When the power pack is driving the motor at standart 500 watts load the motor is only getting 87% of
the power coming out of the battery the remaining 13% is going to heat the batteries up .
Now picture yourself accelerating and the load goes up to 1500 peak power load impedance dropped to .833 ohms
[i]Now the percentage to motor is 55% and battery internal loss is 45% so effectively you are using 1/2 the power
coming out of the batteries and the other half to cook your batteries .[/i]Just thoughts that it may be of interest , the low impedance of cell is not only beneficial for peak power surge
but will also prevent damage to the batteries by heat generation .
As mentioned before impedance is king even before amp/hour or volts ,since you can have all the volts you want , but if the internal impedance is hight the net result is cooked batteries and little power to load
The math is the same regardless of batteries chemistries , just the number of cells are differents , ounce you have a impedance reading of a cell the rest is very clear as to efficiency , peak power etc..
Do not understand the tremedous emphasis on amp/hr , cost , weight , size , # cycles etc ... and the complete disregards on impedance measurement , it is the whole key to efficient power transfer both charging and discharging .
Just my two cents worth ! ! !