Capacity is a highly nonlinear function, so the capacity itself is a function of the current draw (and the temperature and the duty cycle and the history of the battery and ....)

This is true, but.... Is this really what toffee is trying to determine? It seems to me he is struggling with the time element.
To determine the exact number will be extremely difficult if not impossible.

http://batteryuniversity.com/learn/article/calculating_the_battery_runtime

 

This is true, but.... Is this really what toffee is trying to determine? It seems to me he is struggling with the time element.
To determine the exact number will be extremely difficult if not impossible.

http://batteryuniversity.com/learn/article/calculating_the_battery_runtime

To be honest, I'm not sure. I also thought the main difficulty was grasping the time element, but when he mentioned practical experience with real batteries, that opens up the possibility that he's also seeing the effect of current-dependency on the capacity.

 

thanks guys

well in a nutshell say i have a battery here with xyz mAh capacity [xyz]mAh

I run some tasks that need 5Watts etc etc over such a time, and maybe 2watts over another time period.

how much of these can i fit into a battery of said xyz capacity, of course i have depth of discharge and etc to figure out but first its what the tasks themselves take up, how much would 1 of them consume in mAh.

 

Just remember that an amp-hour is nothing more than an amount of charge -- 3600 coulombs to be precise. So you just need to integrate the current over the time to determine how many coulombs of charge have been extracted.

But do be sure to look at the battery specs to make sure that you use the effective capacity that is appropriate for your current draw. And keep in mind that the curves are based on a uniform (though often pulsed) current draw and not a multi-level current schedule. In general, a period of a high current draw followed by a low current draw will drain the battery a different amount that the same low current draw followed by the same high current draw -- and the data almost certainly doesn't exist to predict the result with any level of accuracy. So leave yourself plenty of margin.

 

As I noted in #5, watts are volts x amps, and watt-hours are volts x amp-hours. If you're battery is more-or-less at 3.6V, then the wattage it is producing is 3.6V x mA/1000. Or the current is amps = watts/3.6. Just multiply both sides by hours of operation if you want the total energy instead of the instantaneous power.

The slight wrench in all of this is that the nominal battery voltage is not really constant. It will drop a little with time (state of charge) and also with the current draw, due to its internal resistance. Cold temperature as well. So for something critical, you'd want to have more nominal battery capacity that required.