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I have found many references in my reading about battery life being affected by many factors. I am curious if anyone has attempted to design a battery life prediction algorithm which would take these factors into account and perhaps use the data from the FNDC/CC logs?

- okkman
- Forum Whiz
**Posts:**38**Joined:**Sat Mar 14, 2015 6:25 am**Location:**Seymour, CT**My RE system:**

Assuming you are considering charge available only down to some voltage threshold, the actual charge available from the battery depends on temperature and the rate at which you discharge it. Lower temperatures slow the chemical reaction in the battery, making it harder to extract charge. Higher rates of discharge increase losses in the battery, decreasing the voltage, thus hitting the "discharged" voltage threshold limit sooner.

The electric potential difference provided by the chemicals in the battery is actually constant; what makes the voltage decrease is the depletion of the chemicals around the electrodes and degradation of the electrodes and electrolyte. This is why battery voltage can recover after a period without use. So, the point at which the threshold voltage is reached can actually be quite complex to determine.

If you can find a good datasheet for your battery, it may give some insight into the parameters under which these calculations were made.

I hope the information was helpful : )

The electric potential difference provided by the chemicals in the battery is actually constant; what makes the voltage decrease is the depletion of the chemicals around the electrodes and degradation of the electrodes and electrolyte. This is why battery voltage can recover after a period without use. So, the point at which the threshold voltage is reached can actually be quite complex to determine.

If you can find a good datasheet for your battery, it may give some insight into the parameters under which these calculations were made.

I hope the information was helpful : )

- Chris
- Forum Junior Member
**Posts:**2**Joined:**Thu Feb 07, 2019 3:17 am**Location:**Baltimore

I went to some notes, here's the best to my knowledge

Look for the reserve capacity, which usually shows the approx time where it still power without the recharge. This can be found on user's manual or battery label.

The value often takes into account some conditions, like 25 amperes of current at a voltage of 15 volts (an example, look for it in the label, manuals)

If the circuit uses more/less power than this given circuit, then that determines whether the battery life would be shorter/longer.

You can predict the battery life by calculating the battery's overall capacity and then divide that by the circuit's power.

First, multiply the battery's reserve capacity by 60. E.g. a reserve capacity of 220, you have 220 x 60 = 13,200.

Then, multiply 13,200 by the battery's rated wattage, 280 for instance, you have 13,200 x 280 = 3,696,000 (the energy, megajoules)

You then divide the result by the battery's voltage. For example, it has 15 volts: 3,696,000 / 15 = 246,400.

After that, divide it by the current of the circuit. Example, the circuit has 25 amps: 246,400 / 25 = 9,856. This now mean that circuit runs for 9,856 sec.

To convert to hours, divide it by 3,600. You get 9,856 / 3,600 = 2.73 hours, or around 2 hours 45 minutes.

Look for the reserve capacity, which usually shows the approx time where it still power without the recharge. This can be found on user's manual or battery label.

The value often takes into account some conditions, like 25 amperes of current at a voltage of 15 volts (an example, look for it in the label, manuals)

If the circuit uses more/less power than this given circuit, then that determines whether the battery life would be shorter/longer.

You can predict the battery life by calculating the battery's overall capacity and then divide that by the circuit's power.

First, multiply the battery's reserve capacity by 60. E.g. a reserve capacity of 220, you have 220 x 60 = 13,200.

Then, multiply 13,200 by the battery's rated wattage, 280 for instance, you have 13,200 x 280 = 3,696,000 (the energy, megajoules)

You then divide the result by the battery's voltage. For example, it has 15 volts: 3,696,000 / 15 = 246,400.

After that, divide it by the current of the circuit. Example, the circuit has 25 amps: 246,400 / 25 = 9,856. This now mean that circuit runs for 9,856 sec.

To convert to hours, divide it by 3,600. You get 9,856 / 3,600 = 2.73 hours, or around 2 hours 45 minutes.

- HonestCalf
- Forum Junior Member
**Posts:**2**Joined:**Wed Oct 10, 2018 8:56 am

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