The first point you make is excellent. And with the numbers you have chosen to use, the point is well illustrated. It should be made clear that the discharge curves are not linear (batteries typically drop quickly initially, roughly plateau in voltage over a given output of total capacity, and then tank at the the end). But even with this in mind, I think this first point you make is key. I am glad you posted it.

Regarding your second point, I should make it clear I still don't think it is the case that "a higher capacity pack also has a higher discharge rate with the same C rating." Sure, you can say a higher discharge rate is "available," but then you have to construct the battery in such a way to accommodate it.

And the third point is good. It should just be clear that it is not the higher capacity that equates to power. It is the higher voltage (and all things being equal, it is higher longer into the discharge time because of the higher mAh capacity).

Edit: cut and paste below to save others from having to go back and look. Still new to the mechanics of posting and quoting in this forum.

Originally Posted by drolmaeye View Post
I'm not an expert on batteries, but just to clarify (or start the discussion, if it can help me or others), I think I am with Keddy on this one. The capacity of a battery, x-thousand mAh, is really just like the size of your gas tank. An R/C with a 6000mAh battery will run longer than with 3000 mAh just like a car with 10 gallons of gas will run longer than it would if you only put 5 gallons of gas in it.

Right for the most part. The key is that the voltage is constantly dropping the whole time. Lets say your car uses 2500 mah in a 5 minute race. With 30 seconds to go this puts your 3000 mah battery at ~7 volts nominal while the 6000 mah pack will be around 7.5. Assuming that the voltage doesn't dip under load at all (infinite C rating), the 6000 mah pack will provide almost 10% more power. It's not just that, your lines would have to change as the race went on to compensate for your continually dropping power and consistency is what wins races.

I don't think this is correct. The "safe" discharge rate of a battery is listed as a C rating. The actual discharge rate will depend on a number of factors including motor, esc (and settings on both), driving style, etc. A higher capacity does not mean a higher discharge rate.

A higher capacity absolutely means a higher discharge rate is available. What we are really talking about with C ratings is how much current can the pack deliver without the individual cell voltage dropping below 6 volts. A 10c pack can deliver 10 times the capacity (in mah) without damaging itself. A 3000 mah 10c pack will give you 30 amps continuous, but a 6000mah 10c pack will give you 60 amps without hitting the lvc prematurely. A 6000 mah 10c pack will be holding a higher voltage than the 3000 mah 10c pack at 30 amps, so the 6000 mah pack will provide more current, voltage, and 'punch'.

Yes, P=IV. But if you want to calculate the "instantaneous" P you are putting into your car, you have to take the battery voltage (let's call it 7.4V) and multiply it by the amount of current you are drawing, at that moment, from the battery (not the battery's overall capacity).

See above as to why the voltage will be higher with a larger capacity pack.

You do see guys get excited about high C ratings, and insisting that they get more punch with higher C rating. I suppose this is possible (and this would also reduce instances of brownouts, as the battery can always keep up with the demand). This makes sense if battery construction for high C ratings ultimately lets more current flow during bursts.

That is the whole point of a "C" rating in the first place...Its a measurement of pack capacity over time. The C rating establishes how much current I can take out of the cell over a given period of time or how much I can put in over a given amount of time without damaging it. Because the formula uses pack capacity as one of the factors changing the capacity of the pack changes the safe charge and discharge rate of the pack.

Lets say I have two batteries, they are the same brand and both have a 40c continuous rating and a 80c burst rating. The only difference in the packs is that one is 5000mah and the other is 6000mah. If we do the math it is obvious that one can supply more current than the other.

5000mah
1C=5A
40C=200A (continuous rating)
80C=400A (burst rating)

6000mah
1C=6A
40C=240A (continuous rating)
80C=480A (burst rating)

Now typically we arent drawing that much current out of the packs anyways but a pack that will deliver more amperage will also have less voltage sag and will maintain voltage longer than the pack that will not deliver as much current.

+1

This happens because the surface area (inside the pack that is relative to the reaction) increases linearly with capacity assuming the chemistry and layer thickness hasn't changed.

Wow, I hate it when I'm so slow I have to be told the same thing two or three times before I get it. Right; for two battery packs with the same C-rating and different capacities, the higher capacity battery will have a higher safe continuous discharge rate.

Does this mean our cars/trucks actually draw more current from these packs (provided the V is high enough that there is not an issue with the low-V cutoff)?

Put in the exotek front slipper eliminator weds, ran 4-5 packs thru it last night. Super planted and never squirrely under acceleration. Point, squeeze, and SHOOT! Low speed turns suffered, but I know my front diff is thick (50k?) and my rear bar is soft. Maybe just thickening the rear bar will help the back end kick out a little more in the slow stuff. Overall, I loved it. Instant stab of the throttle out of the turns had me winning corner to corner drag races every time. I need a little more work with it, but I was happy...

They do draw more current, but only because a higher voltage is maintained under the load. I'm typing up a more detailed answer, but wanted to get this one in quick.

Example as to why a higher capacity pack will give more current:

Going with a 3000 mah 10c and 6000 mah 10c as example packs and assuming at 10mph the resistance (and impedance) of the electrical system is .1 Ohms and the back emf is averaging 4 volts.

With this setup and assuming a pack voltage of 8 volts (almost full and an even number to work with. lol), the system will pull 40 amps if this voltage is maintained. The math I'm using for that is very simplified for what the whole system is, but it is close enough for this discussion:
Delta(V) = IR
VBattery - Vmotor = I(0.1ohms)
8volts - 4 volts = I(0.1ohms)
4volts/0.1ohms = 40 amps

This is higher than the 3000 mah pack can safely give, so the demand for current pulls it's voltage down to 7 volts. At this point the equation works out to 30 amps:
7volts - 4volts = I(.1ohms)
3volts = I(0.1ohms)
3volts/0.1ohms = 30 amps

30 amps x 7 volts gives you 210 watts.

Same idea, but with the 6000 mah pack. The setup wants 40 amps. The voltage will dip a touch under load, but not as much as the smaller pack. This is a function of the packs internal voltage, but to make it easy lets just say it drops to 7.5volts.
7.5volts - 4volts = I(0.1ohm)
3.5volts/(0.1ohms) = 35 amps

35 amps x 7.5 volts = 262.5 watts

That's 25% more power!

Obviously no one is running packs only capable of 30 amps, but the idea still applies. It's just a diminishing point of return as the voltage drop under load becomes smaller and smaller with each capacity or C rating increase. The biggest arguments about packs are centered around this... sure you can buy those amazing new cells that just came out for $100, but what percentage of power are you gaining over a "sportsman" pack for $60? What about even cheaper packs straight from asia? With no standards when it comes to labeling how a pack will perform (near term as well as long term) it all comes down to the arguments you see every other week on various forums about the pack of the week and personal experience.

Personally I get "enough" power from my inexpensive nano-tech batteries. Would an orion or smc pack provide more punch? Probably, but the small percentage in power gains with large percentage in cost increase can't be justified against my diminishing racing budget. As always, that last 10% of performance costs as much as the initial 90%.

Nice, thanks for the report...Seems like a good product. Damn man, 50k? Slap some 10k in there, that is what I was running with the stock slipper and vts should liven the steering back up.

Wow I am glad I have a battery sponsor. Run them hard, Run out front, Replace when necessary. No problems and no math.

Gasp Murfdog he just said your batteries did not last forever. Take away his sponsorship and give it to me lol
Lucky

Thanks for writing this up. Of course, I am confused.

For the first case, the system "wants" 40 amps, but a 3000 mAh pack with 10C rating can only give 30A, so there is a voltage dip (I'll avoid the word "drop" here). For the second case, the system wants 40 amps, but a 6000mAh, 10C battery should be able to deliver, no? Which leads to no voltage dip? This would give deltaV of 4V, leading to 40A X 8V to 320 Watts!

I think I prolly should not take the numbers too literally. I guess your point is, there will always be a momentary voltage dip, but the higher the capacity (for the same given C rating) the dip will be lower, leading to a larger deltaV, leading to larger I.

I may need to learn a little more about batteries and the overall electrical system, and particularly, how it all works as a function of time, to be fully on board.

That said, I am learning a fair bit while at work. Maybe I should tell my boss about my time well-spent . . . or maybe not.

I just need one Reedy Pack & I get her done ....

Sweet!

Leaving the specific math aside, this info could still be of use to you when you explain to people why you run your sponsor's batteries. I can't figure you would simply tell a fellow R/C'er the only reason you use them is because you get them for free (although at my driving level and budget, it would be reason enough).

True, but I also let people run my packs if they want to try before they buy. Never hurts to let someone feel the difference first hand. Running a slipper last year my truck would pull a wheelie going down the straight. Tested a Venom pack of the same "C" rating and no wheelie. Not sure about the math but I could see and feel the difference.

If you work at a Duracell they'd probably be impressed! Otherwise, probably not. lol

The voltage will always drop for batteries under load. When we say the battery should be able to deliver, we mean that it won't drop to 6 volts while providing "X" amount of current. Here's another example to help clear it up:

For this example, I'm simplifying the "C" rating to the idea that the 6000 mah 10C pack can deliver 60 amps continuous before it drops below 6 volts. This doesn't mean that when it's fully charged it will supply 60 amps at 8.4 volts and then all of a sudden drop to 6 volts when you try to pull 61 amps. The voltage drops roughly proportional to the percentage of power demanded from it. If you demand 30 amps, a full pack will probably be around 7.5 volts. If you demand 50, it will probably be close to 6.5 volts. 60 amps will put it right at 6 volts. If you remove any current demands, the pack will jump back up close to 8.4 volts.

The "drops quickly, levels out, and then dumps" voltage curves you are thinking of are for constant current discharge rates. You take a pack, discharge it and plot the voltage. This just tells you how the pack will react under constant currents, not varying ones. If you discharge one pack at 30 amps and another at 60, the voltage curve for the 60 amp run will always be below the 30 amp run when plotting it not only against time, but against how many mah have been discharged.

Voltage dips that you see from in car telemetry devices or other on track recording systems are caused by high current demands at low rpm. The voltage quickly recovers primarily because the demand diminishes as the rpm of the motor increases. At low rpm there is very little impedance and the back emf is also very low. This creates a high delta voltage and a very low overall resistance, so high current spikes and battery voltage drops are the result.