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

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.

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. 

Hi Motorman,
   Your number for Amps, 82.5 is the maximum amps possible to pull from your 25c battery.
You should use the actual current you are using to work out duration.
Or if you don't have the kit to measure current use your duration and the amount of charge
you have to put back into battery to work out your actual average current for the flight.

John