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# how do you get from Ampere to Ampere hours?

i know, basically it's not too diffy. I checked wikipedia and in the end i get to Coulomb, but the information i need is: i wanna build these cat-ears:

https://www.instructables.com/id/Animatronic-Cat-Ears

and as material one needs a LiPo battery - the guy says "anything that will be able to supply 3A worth of current" and links to a 800mAh-battery at hobbyking. I first bought a battery with 4000mAh, say a battery that gives 4A per hour, but the one i got is quite heavy and i doubt that it's a good idea to use it in this light-weighted project - so how the hell did that guy get from a 800mAH-battery to the 3A worth of current? How do i calculate this way?

## Discussions

800mAH is .8AH. To get rid of amps we just divide:

.8AH/3A=.267H, or 16 minutes.

An 800mAH battery will put out 3A for sixteen minutes.

3 amps is a lot of current, especially for a controller circuit. Although I guess the motors will pull a lot of current.

BTW, a 4AH batter means that you can pull 4 amps for an hour, not per hour. if you wanted to pull 16 amps out of that battery, it would last for 15 minutes.

Good Luck.

yea thanx, guess that does it, thanx for the calculation.

an amp HOUR, is the total number of electrons that would flow over a given amount of time (an hour).

Like how speed is distance/time, and if you multiply by an amount of time, you get just distance...

The water analogy falls apart for lots of electronics but is good for explaining amps and volts when referring to DC:

Think of a river flowing downstream. The flow rate or current (cubic meters per second) is like amps. The total amount of water that goes by in an hour (just plain old cubic meters) is like amp-hours.

Then consider a slow moving river, Not much vertical difference between start and end. not much 'potential energy' difference...This is much like volts, specifically LOW voltage. how much the electrons (or water in this case) wants to get from point a to point b is good to represent voltage. Again, the width and depth of the river gives a cross section, good for representing the flowrate, or amps.

Now consider Niagara falls. Large potential-energy difference between the top and the bottom (height)...the water at the top once it falls off the cliff is VERY excited to get to the bottom. This is a good representation of HIGH voltage. When it's a rainy season and the river is bursting its banks, the current will be high and the voltage will remain high. When it's a drought, and only a trickle remains, there is still high voltage (potential), but not a lot of current.

Roundabout way of saying it:The problem arises when people use short-form or skip words when referring to different units.

A battery has:

- a capacity (total amount of energy) in Amp-hours
- A voltage (total voltage potential of cells) in Volts
- a discharge rate 'C' - the amount of amps you can pull out of the battery at any given time
- various other specs like temperature tolerances, weight, etc that don't affect this math *much.

So, a 4000mAh 7.2v battery with a rating of 20Csupplies 7.2 volts, and is capable of supplying 4A*20 = 80 Amps

If you draw 80 amps out of a 4000mAh battery (4Amp hours) it will last for 1 hour * (4amp hour/80amps)

= 1 hour * (1/20)

= 3 minutes.

At absolute maximum discharge rate, the battery will last 3 minutes