BMFA have issued some sound advice on how best to look after and use lipo batteries.
LipoSafety
by Mike Pirie
When buying a battery for your electric model, one of the things you have to consider is the capacity.
What exactly is meant by capacity?
The capacity is usually stated in milli-amp hours (mAh) and is an indication of how much energy is stored in the pack. Put another way, it indicates how long the fully charged pack will supply a particular current until it is discharged. The higher the capacity, the more energy is stored and the longer the flight will be (for any motor/prop/ model combination). Of course you have to remember that the penalty for increasing capacity is increasing size, and more importantly, weight.
Okay, suppose for argument’s sake, you have chosen a 2000 mAh battery pack for your electric glider. What does this figure mean and how can you make best use of it?
First of all, let’s sort out the terminology.
The capacity in mAh is the product of mA (milli-amps) and h (hours). So mA x h = mAh.
A milli-amp is a thousandth of an amp. In theory a 2000mAh battery would provide:
2000mA for 1 hour (2000 x 1 = 2000)
200 mA for 10 hours (200 x 10 = 2000)
20 A for 1/10 hour (20,000 x 1/10 = 2000)
I say in theory because in practice the usable capacity actually declines with increasing current. The nominal capacity stated on the cell refers to a discharge rate of 0.1C (where C is the nominal capacity).
Charging and discharging rates can also be calculated and are usually defined as a multiple of the nominal capacity (C). For example, for the 2000mAh battery, a rate of 2C would correspond to a current of 4 amps and a rate of 0.1C would correspond to a current of 200mA.
You can also see that the time taken to charge your battery from flat at a rate of 1C would, in theory, be 1 hour. Charging at 2C, it would be ½ hour.
A useful formula for calculating your motor run time is:-
Run Time (minutes) = Cap.(Ah)/Current (A) x 60
For example, a motor drawing 20A from a 2000mAh battery will run for: 2(Ah)/20 x 60 = 6 minutes
Conversely, the average current during the flight can be calculated from:-
Current (A) = Bat cap (Ah)/Run time(mins) x 60
For example, the average current used for a motor run time of 6 minutes would be: 2(Ah) x 60/6 = 20A
RCME
by Mike Pirie
Do you really understand the terms, VOLTAGE (V), RESISTANCE (R) and CURRENT (I) - (?????) , and the relationship that exists between them (as defined by Ohm’s Law). If this seems like school stuff, then you’re right, but no harm in a little refresher course. I came across a cunning little analogy in the ‘Guide to Electric Flight’ (the beginner’s booklet issued by BEFA, the British Electric Flight Association). It goes like this …..
Imagine that you have in your back garden, a tall water tower of specific dimensions, with an outlet pipe at ground level which can be turned on and off and whose diameter can be adjusted at will. Now suppose the tank is half filled with water, then it is clear that the water at ground level will be under considerable pressure due to the weight of water above it. The outlet pipe is now opened, and water flows through the pipe at a certain ‘rate of flow’. Filling the tank to the top will result in a rate of flow twice as great as it was before due to the increased pressure. Alternatively, the same result can be achieved by increasing the diameter of the outlet pipe thus reducing the resistance to the free flow of water. In this case, the tank will empty more rapidly.
In our analogy then, the water pressure can be likened to VOLTS, so that when the water level in the tower is low, we have a low voltage, and when it is high, we have high voltage.
The constriction, or resistance, to the free flow of water imposed by the variable diameter outlet pipe, can be likened to RESISTANCE. Electrical resistance is measured in OHMS.
The rate of flow of the water, of course, is the CURRENT and is expressed in AMPS.
So we can see that, in a given arrangement, the CURRENT in AMPS may be altered by changing the voltage (filling up the water tower or partially draining it), or by changing the RESISTANCE (altering the diameter of the outlet pipe). Increasing the VOLTAGE will increase the CURRENT and vice versa. Increasing the resistance will reduce the CURRENT, and vice versa.
This takes us on to OHM’s LAW which states that: One AMP is that current which will flow through a resistance of one ohm under the pressure of one volt. The basic formula in Ohm’s Law is that the current is equal to the voltage divided by the resistance, or I = V/R. From this we can evolve two other equations: V = IR and R = V/I.
Note: The ‘Guide to Electric Flight’ can be downloaded free from the BEFA web site and is a very worthwhile booklet for anyone starting out in electric flight.