Calculating battery power
 

So i know a bit about the voltage and the mAh but i may have my info a bit wrong. Volts are the "strength" of the battery and mHa are the amount of energy the battery holds. But in testing the battery i have i found some crazy differences.



8.4v-1100
8.4v-1400

Both these are almost the same.

8.4v-4000 Question 1: Will more mHa effect the life of the internals?

In theory should this battery just last longer? Its so much stronger, and a huge ROF increase. Question 2: Why is more mHa making this change?

9.6v-1100

I get a better ROF with this guy.

I'm a bit rusty on my physics but essentially Voltage is the ability to "push" current. Higher voltage means that you get more "umph". So as it stands higher voltage translates into a higher ROF.

Amperes on the other hand (in this case milliampere*hours) is the amount of charge or "juice" you have. The unit derives from some weird measure that I forget, it has something to do with "Coulombs" though (a unit of electric charge). Like I said it's been a while since I last took physics.

So take these examples:

8.4V-1400 mAh battery: you shoot 1200 BB's (assuming that you get 1BB per mAh) that means you only have about 200 mAh (worth of "juice") in the battery at the end of the day. When you have no more (or very little) what happens? Lets say you do one last round and shoot your last 200 BB's. You have no "juice", there isn't any charge or "juice" in the battery pack that can be used to propel a BB.

Now take that same battery but instead add another cell to it turning it into a 9.6V-1400 mAh battery. Notice how it's still 1400 mAh? You didn't increase the amount of juice you have in your battery pack BUT you increased the voltage by adding another cell. Why? Well, the battery is setup in "series". Adding cells adds to voltage, a disadvantage being if one cell fails (not very likely) the pack is unusable until you replace or repair the pack or cell, just like older christmas lights where you had to find which bulb burnt out or it wouldn't work. These days christmas lights are setup in parallel. Anyways, you didn't increase the capacity of the pack to hold a charge rather you increased the "umph", the "push" of electricity to the motor/internals. As a result you notice the trigger response is "faster" as well you get a ROF increase because there is now more juice to pull the spring back faster.

Since we are already on the subject of batteries. Does the same thing apply to lipo's? I only ask because I bougbht a charger thats capable of charging lipo's too, and I'm considering making the jump.

AFAIK from what I learnt in physics a battery is a battery is a battery. They are all pretty much "generic". We don't talk about specific formula when talking about batteries unless we need to calculate the voltage or something. The only difference being that different chemistries produce differing amounts of "nominal voltage" (a LiPo cell is like 3.7 or 3.8, NiCd 1.2, NiMH also 1.2).

There is also a new type of electrochem cell I've heard being developed called "LiFe" batteries made of Lithium Ferrite with some other additive (to keep it stable or something) I forget exactly what's so good about it but apparently RC hobbyists are treating it like it's some crazy new development that's going to change the hobby as they know it. It's supposed to have better power density than LiPo's or something.

Think of it as a river
voltage is how fast the river is moving
amperage is how big the river is
and resistance is the rocks slowing down the water

NiMH cells don't actually provide all the amperage needed to operate airsoft gun motors, so we up the voltage to get higher ROF. more voltage means the same amount of amperage delivered faster.
It's also why an 8.4v 4200mah battery will fire significantly faster than an 8.4v 1600mah battery. The larger surface area in the cell offers more amperage, and therefore a higher ROF. Same voltage, more amperage.

A LiPo offers all the amperage your motor needs, therefore less voltage is required. A 7.4v 20C 1600mah LiPo will have the same ROF as a 9.6v 1500mah NiMH battery.
Now since the mini LiPo already offers all the amperage you need, it will have the same ROF as a 7.4v 20C 4900mah battery.

And I think the batteries your talking about are lithium iron phosphate, supposed to have higher energy density and be 'safer' lol

Got it thanks guys!!!

It also depends on the quality of the cells, not just voltage and mAh.

And also, the total power is equal to the product of the Voltage and Amperage

P = UI

So you either have a higher voltage or a higher amperage to get the same power. This is why they usualy go with higher voltage on Ni-Mh cells, because they are not made to provide a high Amperage, as opposed to Li-Po cells.

The problem with higher amperage is that Heat dissipation is only affected by either Resistance or Amperage, not voltage. This is why the hydro company transports all the electricity at very high voltages, to minimize the loss in power (heat).

P = R*I^2

The mAh means milli-ampere by hour. If you have 1000 mAh, it means that the battery can provide 1 ampere (1000 mA) for 1 hour before being drained out. So a 4000 mAh is not a battery that is more powerful, but it is a battery that will last longer that the other one. For exemple either 4 amperes for 1 hour, or 1 ampere for 4 hours. I think you got it, it's real simple.

I'm talking about different cell sizes, a 4000mah battery is more powerful because the cells are physically larger.
A 1400mah sub-c cell and a 1000mah sub-c cell have the same amperage, but a 4000mah C cell provides more amperage.
If you want hard proof, hook up an 8.4v mini to your gun, and then hook up an 8.4v 4000mah battery. There's a very clear audible difference in ROF, and they have the exact same output voltage, therefore the large battery provides more amperage.

Yeah thats what i was talking about with the high mah battery. It will drain the same mag an easy 2-3 secs faster. But what i was really wondering was using a high mah 9.6v would be more harmful for my internals.

I explained what mAh is for. It's not because of the mAh rating that one 9.6V is "stronger" than the other one, it's beacause of it's discharge rate.

It's like if you had a car, and said that it was more powerful because you had a bigger gas tank.

Yeah i was referring to this statement. My original question had more to do with ROF and wear on my gun. TC hit the nail on the head. Thanks!!!

ThunderCactus: As much as I love your name, I've never heard of mAh giving a faster ROF.

I think of it this way. Its a pool of water.

Volatage is the size of the pool
mAh is how much water is in the pool

Voltage increase gives ROF increase. mAh increase gives longer play time.

mAh can give RoF increases: I agree with thundercactus. I find that generally the batteries with big mAh can dump the current better. This is probably due to what thundercactus said: physical size and/or quality of the battery. I had a 9.6V ~2000mAh and 9.6V ~3900mAh battery, both sub-C cells. The 3900mAh had an edge on the RoF (albeit by a pretty small margin, but certainly significant I would say)

Understandable, I wasnt catching onto what cactus was referring too.