Electric Stunt > Gettin all AMP'ed up!

Battery capacity vs. voltage

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Mike Palko:
Randy Smith asked if I could share some information about batteries and what happens when we change the voltage and capacity. Randy says customers often ask questions about batteries and he thought it would be a good idea to have an answer online to point them to. His idea was to use StuntHangers electric forum for those answers.

DISCLAIMER: The calculations shown here are purely estimates not taking into account heat, motor efficiency, resistance, voltage drop or any other factor. With more information about a specific power system more accurate results can be achieved.

The voltage and capacity determine the power potential of the battery and how long it can produce this power. Working through a few examples will make it easy to understand.

The first example will use a 15C 4S (14.8volt) 4000mah pack. I will show how to calculate the maximum power this battery can deliver and how to calculate an estimated flight time at different power levels.

4000mah / 1000 = 4ah

15C x 4ah = 60amps (maximum)

60amps x 14.8volts = 888watts (maximum) or 888 / 746 = 1.19 hp (maximum)

To calculate the flight time at this power level you need to know the following, capacity in ah, % discharge, and in flight current draw. If you have those three pieces of information plug them into the following formula.

((capacity in ah) x (% discharge) x (60)) / (current draw) = flight time

((4ah) x ( .8 ) x (60)) / (60amps) = 3.2 minutes of flight time at the maximum power level.

You can see if you know your desired flight time you can determine the maximum current draw to achieve that time. You can find any variable by substitution. I use 80% because this is the deepest discharge a lipo is safely capable of, anything less is also safe, but you will be leaving fuel in the tank. There is absolutely nothing wrong with leaving fuel in the tank, and if you would like, you can split the flight time and get two shorter flights.

A second example of a more usable current level for C/L aerobatics would be 32amps. To predict the power and flight time just substitute the information.

32amps x 14.8volts = 474watts or 474 / 746 = .64 hp

((4ah) x ( .8 ) x (60)) / (32amps) = 6 minutes of flight time

To show the difference between a 15C 3S (11.1volt) 4000mah pack and a 15C 4S (14.8volt) 4000mah pack I will redo the calculations.

60amps x 11.1volts = 666watts (maximum) or 666 / 746 = .89 hp

32amps x 11.1volts = 355watts or 355 / 746 = .48 hp

The flight time will remain the same because the capacity has not changed. However, you can see the maximum power output has dropped by 25%.

You can play around with the formulas and see that if you increase the voltage you can reduce the current draw and still maintain the same power output. By doing this you can adjust your power system to better fit the efficiency of the motor you are using. Some motors are more efficient at 25amps and others may be more efficient at 35amps. The closer you are to the max efficiency the less power you will turn into heat and the more power that will be put towards turning the prop. You can also use the voltage/current draw to adjust your flight times at different power levels. Randy has a few other questions customers often ask, so this thread will be followed by a series of threads to help entry level electric fliers get started.

Mike

Pinecone:
Being an R/C electric flyer, some things to consider.

1)  A 3S pack under heavy load runs more like 10 volts rather than 11.1 volts.  A 4S pack, nominal 14.8 volts, run more like 13.2 - 13.4 volts.

2)  Your max amp draw for long pack life should be no more than 80% of the max C rating.  But anyway, if you want to run 8 minutes, you cannot draw more than 7.5 C.  1/C x 60 = Run Time in minutes.

3)  The rule of thumb for power that seems to work for R/C is Displacement in CI x 1000 = Desired Watts.  So to replace a .40 CI engine, you want about 400 watts.  It seems low, but the efficiency of electric seems to make it work.  Probably need to fine tune for CL work.  So you would want about 40 amps on a 3S or 30 - 31 amps on a 4S setup.  That would require a 5400 mAH 3S pack or a 4200 mAH 4S pack.  In the high power and helicopter world, things are going more to higher voltage, lower current setups.  It requires more S, but is offset by the small capacity required.

Jeffrey Olijar:
Okay so I have a 20c  3s  11.1v  1800mah pack

1800mah / 1000 = 1.8ah

20c x 1.8ah = 36amps (max)

36amps x 11.1volts = 399.6 watts or  0.54 hp Max

1.8 x .8 x 60 / 36 = 2.4 min

but my motor has a max amp of 14.5  w / 8x4 prop so does that mean that if I set my timer to 100% that I get
1.8 x .8 x 60 / 14.5 = 5.95 min

and am I correct in saying that the max amps I can pull is the lowest max amp between the motor/esc/batt.
so when I set my z-tron timer to 90% I should get

1.8 x .8 x 60 / (14.5 x .9)
1.8 x .8 x 60 / 13.05 = 6.62 min

please tell me if my math is correct of if I am understanding this wrong.

Mike Anderson:

--- Quote from: Jeffrey Olijar on September 26, 2008, 02:43:14 AM ---Okay so I have a 20c  3s  11.1v  1800mah pack

1800mah / 1000 = 1.8ah

20c x 1.8ah = 36amps (max)

36amps x 11.1volts = 399.6 watts or  0.54 hp Max

1.8 x .8 x 60 / 36 = 2.4 min

but my motor has a max amp of 14.5  w / 8x4 prop so does that mean that if I set my timer to 100% that I get
1.8 x .8 x 60 / 14.5 = 5.95 min

and am I correct in saying that the max amps I can pull is the lowest max amp between the motor/esc/batt.
so when I set my z-tron timer to 90% I should get

1.8 x .8 x 60 / (14.5 x .9)
1.8 x .8 x 60 / 13.05 = 6.62 min

please tell me if my math is correct of if I am understanding this wrong.

--- End quote ---

You do not want to discharge the battery more than 80% of full capacity
(ie: 1.8 * .8 = 1.44 A-h = 86.4 A-minutes)

I don't know what current your motor pulls, at the voltage you are using it and with the prop you have on it, but (since you have previously mentioned a 2212/10 on 3 cells (?)) let's estimate that you are pulling around 15 amps in flight with the 8-4, based on the charts for that motor at Lightflight RC.  Therefore, you can safely fly 86.4/15 or about 5:45 with that setup.  You can get an estimate of average amp draw in flight if your charger gives you a readout of total charge returned to the battery -

My FMA charger typically reads 850 - 950 mah, after charging my little 2-cell batteries on my "1/2E" Skyray.  Since I typically have the timer set for 2:30, and fly twice on each charge, I can estimate that I am using 900 mah (= 54 A-minutes) per 5 minutes of flight.   54/5 is then, about 11 Amps average.

My cells are 1300 and 1500 mah, so on the 1300's I would not turn my timer more than

1.3 * .8 = 1.08 A-h = 62.4 A-minutes / 11 Amps = 5 min. 40 sec  (for the 1.3 A-h cells)

and

1.5 * .8 = 1.2 A-h = 72 A-minutes / 11 A. = about 6 min. 30 sec (for the 1.5 A-h cells).

So it is not over-discharging unless I fly for a greater than that amount of time (total) between charges.

Incidentally, I'm using the cousin of your motor - 2212/6 (which has a Kv of 2300), and a prop that is cut from a 7-6 APCe to about 5.75 in. diameter, and set my timer for about 11000 RPM, which is plenty of grunt to get the beginner pattern on 40 foot lines.

Mike A

Jeffrey Olijar:
ok now im confused I did a 4 min run on my system and it only drained the battery 35%...  so I had 65% left on the battery  a 4min run should have taken much more juice from the battery...  also my lap times were around 5 seconds on 35' lines.  what % power should I use to get 6 second lap times... I was never good at math.  HB~>