Choosing a power system - simple version

For a long time, I struggled with understanding how to optimize multirotor propulsion components.  My research led me to many dead ends.  Most advice consited of simply, "try e-calc", but this is nothing more than a trial and error approach.  I also found a curious lack of what the motor parameters mean, such as kV.  To aid other newbies, I'd like to share my simplified method for choosing motors, ESCs, and propellers, for multirotors:

1.  Determine how much thrust you require from one of the props on the craft.  

This requires an estimate for the total mass of the multirotor, with battery.  For example, if you have a quad that weighs a total of 1000g, each prop would need to produce 250g of thrust at hover.  In order to do more than hover, the prop must produce more thrust than this.   Let's use a multiplier of 2x, thus in our example the prop must produce 500 g of thrust.

2. Decide what battery voltage your'e going to use.

Most crafts use 3S or 4S.  This is because that's a normal voltage level for which there are cheap power electronics available, from which ESCs are constructed.   There is no fundamental reason that a 6S or 8S pack could not be used on a multirotor.  For our example, let's decide on a 3S pack, which produces an average voltage of 11.1 V.  

3. Choose a propeller that is the largest diameter you can fit on your craft

I say this because as propeller diameter increases, thrust efficiency increases.  If your goal is not maximum flight time, then this rule is less important.  For our example, let's use a 10x4.7 prop, which is a good size for an average quadcopter.

4.  Calculate the rpm required for your propeller to produce this amount of thrust

This information must be found from manufacturer data, or from independent thrust stand testing.  A good place to go for it is here: http://m-selig.ae.illinois.edu/props/propDB.html.  These grad students spent time testing propellers in a wind tunnel.  To calculate thrust from those data points use the equation Thrust = Ct *air density * rotation speed ^2 * diameter ^4.  Use SI units for all parameters and you'll get Newtons of thrust as the answer.  However for rotational speed, there are two options for which units to use.  The paper that accompanies that site is lacking in its technical capacity and doesn't clarify this, but they used rev/s for rotational speed to come up with this thrust and power coefficients.  In our example, a 10x4.7 prop needs to spin at 6000 rpm to produce 500g of thrust.

5.  Calculate the required kV to drive the propeller at this speed

This critical motor constant tells you how fast it will spin for the input voltage, according to the equation : rpm=kV*V, where V is the input voltage.  If you've chosen a battery, you can calculate your requied kV.  If I have a 3S battery, the input voltage will be (on average) 11.1 V.  If I need 500 g from a 10-inch prop, I need to spin the prop at ~6000 rpm.  Thus my required kV is rpm/V=6000/11.1=540.  In reality the motor will not spin as fast as the equation predicts, due the internal resistance of the motor, and the current flowing through this resistance.  So at this point you should add 20% to the kV number. Thus our requried kV in our example is 648.

6.  Calculate the required current to drive the propeller at this thrust level

Manufacturer data is needed here again.  The data should tell you how much power is required to drive the propller at the rpm needed.  Once you know the power required, you can calculate the required current using the equation : I=P/V, where P is the required power, and V is the input voltage.  In our example, the 10-inch prop producing about 500g of thrust would require 70 W of power.  With a 3S pack, that means the motor would require 70/11.1 = 6.3A of current.  Some may notice that I've substituted mechanical and electrical power liberally here.  The prop requires 70 W of mechanical power, and I've use this number to calculate the required electrical power.  In reality there are inefficinecies in the motor when it converts electrical power to mechanical power.  Thus the real electrical power required will be higher than the mechanical power required by the prop.  Thus we need to add margin to the required current.  A margin of 20% here is good as well, thus in our example the required current would be 7.5A.

7.  Choose a motor that has the required kV, and is capable of at least the required current

In our example we'd be looking for a motor with a kV of 650, and capable of 7.5A.  Using a motor with higher current capability would only add unnecessary mass, and flight time would suffer.

8.  Choose an ESC that can handle the chosen battery voltage, and the required current.

Similarly, using an over-sized ESC would only add unnecessary mass.

Other remarks:

The goal here is to use components of minimum mass to do the job.  One could chose a motor capable of 40 A, but you'd never use it's full capability, and it would be heavier than necessary.  Thus you'd be simply carrying around dead weight.  You'll notice that there are an infinite number of combinations of battery voltage and kV.  I could use a 3S pack with a 1000 kV motor and have it spin a propeller exactly the same as a 6S pack coupled to a 500 kV motor.  So which is better?  Well if you use a higher supply voltage, the power system will require less current, and the overall efficiency will be higher.  The only thing limiting voltage is the ESCs.  Few ESCs are made to handle 6S or more.  Those that do are usually of unnecessarily large size, for example helicopter ESCs that can take 12S but are large enough for 150 A.  Mathematically though, current is the only thing responsible for inefficiency in motors.  So mathematically I could keep driving supply voltage up, which would reduce the required current.  Theoretically I could have a power system with huge voltage and only needs 0.001 A.  In such a case, the efficiency of the motors would approach 100 %.  The same tactic is used to transmit power across transmission lines.  These lines are usually operating at many kilovolts.  The voltage is only reduced to our 120 Vac at a transformer close to our homes.  They do this because it minimizes the current flowing in the long-distance wires that we see along the highways and roads.  Thus it minimizes the loss of power during transmission.