## Multi-Rotor Efficiency

One topic that seems to be on the mind of every Multirotor pilot is efficiency. One of the most commonly asked questions is “How long can I fly per charge” or “How big of a battery do I need to fly for 15 minutes.” In this installment of Multirotor Flight, we will look at how to optimize the efficiency of your multirotor, and how to calculate flight times from a given battery size.

There are several key components that contribute to the efficiency of multirotors, but the most important ones are motor efficiency, prop efficiency and overall weight. The weight part of the equation is easy, the lighter the craft is, the longer it will fly. No real mystery there, a lighter machine takes less energy to keep aloft, and will draw less power from the batteries, so naturally flight times will be longer. The other two components, Motor efficiency and Prop efficiency are a bit more detailed, so we will spend more time looking at them.

Motor efficiency can be looked at as the ratio of the amount of power you get out of a motor versus the amount of power that you put into the motor. Every motor has internal losses, and these are due to several causes. First are the frictional losses, which result from drag in the motor bearings and the drag caused by the air as it is pulled through the motor by the cooling fan or cooling holes. In a modern motor with ball bearings these losses are small, but do account for some of the motor’s inefficiency.

Next are the heat losses, which show up in the form of resistive, or I-Squared-R losses, and hysteresis losses in the core of the motor. Whenever you run current through a wire there will be heat losses. Copper is an excellent conductor of electricity, but it does have some resistance. The number watts of power lost to heat in a motor can be calculated by taking the number of amps of current running through the motor, multiplying that value by itself, and then multiplying the result by the resistance of the motor windings. Mathematically this can be expressed as I x I x R and is commonly called I-Squared-R losses. If you have a motor that is running 20 amps of current through it, and the resistance, or Rm value, of the motor is 0.1 ohms, the power lost in the motor is equal to 20 x 20 x .1 or 40 watts. One thing to notice is that this value is independent of voltage. If you are running a motor on 3 cells, 4 cells or 5 cells, and you prop the motor so that the current is the same in all 3 cases, the heat loss in the motor will be 40 watts, regardless of the voltage. In the case of 3 Li-Po cells, if you have 12 volts at 20 amps you are pushing 240 watts of power into the motor. With 40 watts of energy lost to heat, the heat losses are 40/240 or 16.7% of the total power. If you use 4 cells and have 16 volts, then you are putting in 16 x 20 or 320 watts of power. In this case, the losses to heat are 40/320 or only 12.5%. If you go up to 5 cells, then the total power into the motor is 20 x 20 or 400 watts, and the 40 watts of heat loss only represents 10% of the total input. That is why higher voltage set-ups are more efficient than lower voltage set-ups.

Hysteresis loss is the energy dissipated in the stator core of the motor. The coils of wire in a motor turn the sections of the stator into alternating electro-magnets that go back and forth from a positive charge to a negative charge, over and over again. Ideally, when the current stops flowing, the stator pole should go to a completely neutral charge. In actual practice however, there is a small amount of residual magnetic force that stays in the stator and this must be overcome in the next transition. This causes heating in the core of the motor, and contributes to the overall losses. As the motor warms up during use, this heat also causes the resistance value of the copper wire to go up, so the I-Squared-R losses go up as well. Larger motors that are being run at a fraction of their full capability will run cooler and be more efficient. At this point you are probably wondering “What can be done about these losses?”

The best answer is to buy better quality motors. Brushless motors vary widely in the materials used, and in the quality of construction. Cheap motors will use thicker stator plates and cheaper grades of magnets, and this will reduce the overall efficiency. By using thinner stator plates, the hysteresis losses can be greatly reduced. Figure 1 shows close-up views two different stators. The one on the left has stator plates that are 0.5mm thick, and is fairly inefficient. The stator on the right has stator plates that are only 0.2mm thick and have a lot higher efficiency. Some of the worst quality motors out there will have an overall efficiency of as little as 65%, and some of the top quality premium motors can have efficiencies as high as 90% . In the cheap motor, 35% of the power you put into the motor gets wasted as heat, and in the high quality motor only 10% of the energy goes to waste heat. In real world numbers, if the good motors got you 9 minutes of flight from a charge, the cheap motors would only get you 6.5 minutes of flight. That is a huge difference!

The next big thing for efficiency is the propeller. The propellers job is to convert the rotational energy of the motor into thrust, and then use this thrust to lift the multirotor off the ground. With all propellers, the maximum efficiency point occurs at relatively low thrust, around 10% of the thrust that is available at full throttle. The efficiency of the prop drops off in a fairly linear manner as the thrust increases to full power. Figure 2 shows a typical efficiency curve for several props on a brushless motor. From this graph you can see how much the efficiency of a propeller varies over the throttle range of a motor. In this example, the best prop reaches an efficiency level of a little over 12 grams of thrust per watt of input power at 20% throttle and falls of to only 5 grams of thrust per watt at full throttle. At 50% power, which is where a properly set up multirotor spends most of its time, the thrust efficiency is around 9 grams per watt.

What this graph shows you is that as a prop spins faster and faster, its thrust efficiency goes down. This is why running a larger prop at a lower RPM on a multirotor is better than running a smaller prop at a higher RPM from an efficiency standpoint. For example, on a smaller machine, you might have a choice of running 10 inch props or 8 inch props. The 10 inch props will probably generate the required thrust at about 40% throttle, and have an efficiency close to 10 grams per watt, while the 8 inch props might have to spin up to 70% throttle to generate the same amount of thrust, and at that speed the prop efficiency will be down to about 7 grams per watt of power.

If you have a multirotor that weighs 1000 grams, and you are running 10 inch props that produce 10 grams of thrust per watt of input power, it will take 100 watts of power to make the craft hover. On the other hand, with the 8 inch props that only produce 7 grams of thrust per watt of power, to get 1000 grams of thrust it would take 143 watts of input power! Again, a huge difference in efficiency just by changing the props you use. There are also props being developed specifically for multirotors now that have blade profiles optimized for hovering flight, and this can increase the prop efficiency even further. Figure 3 shows examples of propellers that are specifically designed for multirotors.

Looking at the two extremes presented here so far, you can see how the flight time you get from a battery can vary dramatically depending on the motors and props that are used. If you took low quality motors and matched them up with the smaller props, you will get a certain amount of flight time. If you took the same multirotor and put on the high quality motors, and matched it with the larger, more efficient props, you can increase your flight time by as much as 50% or more! Now let’s take this newly acquired knowledge about motor and propeller efficiency and put it to good use to calculate actual flight times for a multirotor.

Whenever I make rough calculations about flight times, I like to use a prop efficiency number of 8 grams per watt of input power. This assumes a decent quality motor with an efficiency of around 80% and a properly sized power system that requires 50% throttle to maintain a hover. To properly size your motors, you want to see how much thrust each motor makes at full throttle with the prop you intend to use, and multiply that value by the total number of motors on your multirotor. Then, take half of that number and make sure the weight of your multirotor does not exceed that value. For example, let’s assume that we have a motor that spins a 10×4.7 prop and produces 35 ounces of thrust at full throttle. If we are building a quad-rotor machine we need 4 motors, so if we take 35 ounces times 4, we get a total of 140 ounces of available thrust at full throttle. Based on this thrust, we want our multirotor to weigh no more than 70 ounces, ready to fly, with batteries and any cameras or FPV gear. With these assumptions, we will now go through the step by step procedure for calculating flight time or required battery size for a given multirotor airframe.

In this example our craft will weigh 70 ounces including a 3-cell 5000mah Li-Po battery and a GoPro camera. Converting this weight to grams we take 70 ounces x 28.35 grams per ounce, and get 1985 grams. If we use the prop efficiency number of 8 grams of thrust per watt then we divide 1985 by 8 to get the required number of watts, and this equals 248 watts. Since we are using a 3-cell Li-Po battery we will have around 11.1 volts on average for the flight. To calculate the number of amps we need to pull from the battery you take 248 watts and divide it by 11.1 volts to get 22.3 amps.

Now that we know the current draw we can convert that to the C-rate of discharge. A 5000mah battery can also be called a 5 Amp-Hour battery because there are 1000 milliamps to the amp. If we are pulling 22.3 amps from the battery, and have a 5 Amp-Hour battery, then the C-rate of discharge is 22.3 divided by 5, which equals 4.46. At a discharge rate of 1-C a battery will take 1 hour or 60 minutes to fully discharge. To calculate the number of minutes of flight time, you simply take 60 minutes and divide that by the actual C-rate of discharge. In this example that would be 60 divided by 4.46 which equals roughly 13.5 minutes.

With the calculations we just made, we now know that our 70 ounce quad-copter that is powered with a 3-cell 5000mah battery pack will fly for 13.5 minutes to completely discharge the battery. In actual use however, you never want to fully drain the battery pack. You always want to leave about 20% of the energy in the battery at the end of the flight. Doing this will greatly extend the life of your batteries and prevent a situation where you actually run out of power and fall from the sky. If we abide by this “80% Rule” and leave 20% of the energy in the battery pack at the end of the flight, our actual flight time would be 13.5 minutes times 0.8 which equals 10.8 minutes.

This number gives you a starting point to work from, but your power usage will vary depending on how aggressively you fly. If you are doing aerial photography and fly smoothly in a hover the entire flight, then you will get a nice long flight. If you fly aerobatics the entire flight, looping, rolling and performing high speed passes, this can cut your flight time in half. The calculations above are for estimating purposes only and get you in the ball park for your size multirotor. If you want to find out how much flight time you will actually get for your specific flying style you can use the following method.