# Does the required torque to spin a propeller increase with RPM?

I‘ve currently been experimenting with different propellers: 12x12 prop, 16x12 prop (both two blades) and a 22x14 (6 blades) prop. They were propelled by a 2.4 kW electric motor with 260 kV on a 6S/5200mAh/12C battery setup.

The 12x12 prop produced 2 kg of thrust at full throttle, the 16x12 produced 4.5 kg of thrust. So I thought the 22x14 would produce around 6 kg because its bigger blade area would be more efficient. However, I got only 0.8 kg of thrust and recognized that the motor would slow down at a certain RPM.

I have to say the big prop is pretty heavy so could it be that the motor would have enough power to spin the prop faster but it lacks enough torque to get it there?

• "could it be that the motor would have enough power to spin the prop faster but it lacks enough torque to get it there ?" - I don't think power and torque are separate like that. Failure to produce enough power is failure to produce enough torque, and failure to produce enough torque is failure to produce enough power. Though it is true that the maximum power output of a motor depends on the torque. – Terran Swett Sep 29 '18 at 0:04
• @TannerSwett: At any given speed of rotation, power and torque are proportional, but the proportionality is different at different speeds. Motors are generally torque limited at low speed (they can produce plenty of torque but not much effective power) and power limited at high speeds (i.e. delivering maximal power at only a fraction of the torque they can reach at low speed). The behavior in between depends on the details of the engine. – Henning Makholm Sep 29 '18 at 13:03

Of course. More and bigger blades is more drag. Drag is also proportional to the velocity squared. Even maintaining the RPM, a bigger propeller (diameter) will have the additional diameter flying faster.

To maintain a higher RPM (or to use a bigger prop) the force to cancel the increased drag is higher, and thus the torque (force times distance) that drives the propeller.

Place the propeller in vacuum, and the engine will have no trouble with the propeller apart from any loads on the shaft (but there won't be any thrust).

• So you think it could be that my motor just can‘t deliver its 2.3 kW because the required torque is too high so it doesn‘t reach its max RPM at which its power is at its maximum? That would basically mean that I need a motor that produces its peak power at lower speeds right ? – Simon Henn Sep 29 '18 at 16:19
• @SimonHenn - Correct, bigger torque-hungry propellers typically run slower. – ymb1 Sep 29 '18 at 18:43

Mechanical power is always the product of an effort (torque) times a flow (RPM) variable. Similarly, electrical power is always the product of an effort (voltage) times a flow (current) variable.

The design problem of optimizing a motor-and-propeller combination for maximum power then always boils down to this: 1) determining the RPM at which the motor produces peak power (torque x RPM), 2) knowing the voltage and current required to produce that power, and 3) specifying the propeller which can absorb that amount of power at that specific RPM.

The analysis is complicated by the fact that from a dynamical systems modeling standpoint, a DC motor is a gyrator, in which the input effort variable (voltage) is proportional to the output flow variable (RPM).

• I added a link to the Wikipedia article for "gyrator," because I had never heard the term before. Although after reading the article, I can't say I really understand it any better. – Wayne Conrad Sep 29 '18 at 2:11

If the other factors remain constant, the torque required to rotate a propellor will increase with RPM. The power requirement is a function of torque and RPM.

Yes the required torque increases with RPM. With simple impulse theory, the torque Q on a rotor or propeller is:

$$Q = C_Q \cdot \rho \cdot A \cdot \Omega^2 \cdot R^3$$

Where $$\Omega$$ is the angular velocity = RPM. So the torque required goes up with the square of RPM, and since power = $$Q \cdot \Omega$$, the power increases with RPM$$^3$$. All other factors remaining as they were of course.

That is simple impulse theory however, which does not consider disk solidity and blade profile drag. For a given thrust and blade radius, torque and power increase with number of blades since there is more blade profile drag to overcome.

Firstly, take the six blade and throw it away. 2 blades are best until the arc hits the ground in front of your plane, then go with 3. Curious as to the application here, is it a microlight? Might help also to consider pitch too, but you are definitely on the right track with the 16x12. If I am not mistaken, it is longer than the 12x12 with the same pitch. You can get a rpm measuring device from a hobby shop to measure peak rpms, but as you are already measuring thrust it is not absolutely needed. Now you can try 15x12, 17x12, maybe 18x10 (longer with less pitch) until you get it perfect. Pitch is just like gears on a car, lower means better acceleration from standstill and better at climbing, higher will give lower rpms at cruising. I would keep trying, and make sure the battery/engine can handle the load without going up in flames. Definitely check with an expert as you have some serious prop there.

• Thanks for your answer. I know that 2 blades are usually most efficient. However, the purpose of this project is to maximize the thrust per area so 6 or even 12 blades should be better here. My problem was that with the 6 blade 20 inch prop I got like a third of the thrust of the 12 inch prop with the same motor. Now I know it´s because of the higher torque requirement of the bigger prop. – Simon Henn Oct 2 '18 at 19:37
• OK. This is a great one to continue for three reasons. First, we are also seeing blade turbulence interference on the 6 blade compared with 2. Second, because of 12 pitch, you are probably spinning stalled blades. Third, seeing maximum load on your battery/motor, which you can monitor a watt meter. So maybe get a 10 inch 6 blader with pitch range of 4 to 12 (several of them). Notice this all changes once the plane is moving. With a 150 mph headwind, higher pitch is better. With static bench testing, the AOA is more, just like a wing. This is why variable pitch props were developed. – Robert DiGiovanni Oct 2 '18 at 21:23
• So, not to spoil the ending, after safely continuing with smaller blades, you will find, just like with jets, you get more thrust per area by increasing power of engine and adding blades, at the expense of efficiency. Jets have more thrust, but are not as fuel efficient as piston engines with 2 blade props. But they are much faster indeed. – Robert DiGiovanni Oct 2 '18 at 21:34
• So after you pick your prop diameter limit, spin a low pitch 2 blader right up near the speed of sound. That is your RPM limit. Then try pitch for max thrust, that is your best AOA. If RPM is not dropping yet, motor has more to give. Widen the 2 blade or move to 3 blade. Very cool experiment. Might be good to approach from small/fast side. Don't think you will go past 4 or 5 blades, I could be wrong! – Robert DiGiovanni Oct 3 '18 at 0:33
• Ok, what would you suggest as a limit for blade tip speed ? Mach 0.6 ? – Simon Henn Oct 4 '18 at 12:26