# Does the continuous power-to-weight ratio of electric motors decrease with size?

Background: According to the DEP overview section of this NASA paper, aviation companies are interested in distributed electric propulsion (DEP) because the scale-agnostic power-to-weight ratio of electric motors enables aerodynamic advantages from distributed propulsion.

However, I'm struggling to make sense of that claim in the context of commercially available electric motors. Small air-cooled Hobby motors, supposedly have continuous power/weight ratios of >7 kW/kg while Siemens' aerospace-optimized AC motor has a "record-breaking" 5 kW/kg ratio with liquid cooling.

What gives?!

Question: Does the power-to-weight ratio of electric motors not change with size, or does power/weight vs size just not change as quickly compared to combustion engines?

EDITS/Understanding thus far: According to Neil_UK, Brian Drummond, and Charles Cowie, a motor's ability to dissipate heat is proportional to surface area ($Q \sim DL$) while its peak power is proportional to volume ($P_{max} = T_{max}\omega \sim(D^2L)\omega$). Assuming weight is linearly proportional to volume ($W \sim D^2L$), then the continuous power/weight ratio of electric motors actually decreases with size because

$P_{cont}/W \sim DL/D^2L \sim 1/D$

Correct?

• Welcome to EE.SE. Maybe someone here has a few answers, but asking so many questions at once tends to chase people away. With the wide use of once-too-expensive NIB magnets, the electric motor industry is changing. – Sparky256 Feb 12 '18 at 3:15
• distributed propulsion .... maybe it means 100's of small motors with propellers attached all over the aircraft – jsotola Feb 12 '18 at 3:16
• @Sparky256 Thanks for the feedback. How does my question look now? – DEEPquery Feb 12 '18 at 3:27

Continuous power output is usually limited by cooling, the ability to get rid of the waste heat mainly produced by $I^2R$ losses in the coils.

In a conventional un-ventilated motor, the peak power output and the weight vary as the volume, a dimension cubed, however the ability of the motor to get rid of heat varies as the surface area, a dimension squared. If you allowed this trend to continue as motors were scaled up in size, then the continuous weight to power ratio (note the way up that ratio is) would vary as a dimension.

This is why large motors are invariably ventilated, or even water-cooled, to reduce the dependence on surface area for cooling. Adding water cooling to a motor adds weight, that isn't itself generating power.

There is another more subtle effect which exacerbates the scaling problem. Thin sections of material are not as stiff as thick sections. Stiffness in a motor is needed not simply for mechanical strength, but also to push mechanical resonances up in frequency to well above the motor operating speed. This means wires and supporting members will tend to be thicker in larger motors for mechanical reasons than they would be for thermal reasons. This means the continuous power output scales at less than dimension cubed, even with active cooling.

• Are you saying $P_{cont} \approx 1/D$ without water-cooling assuming $P_{max} \approx D^2L$ and $Q \approx DL$? – DEEPquery Feb 12 '18 at 14:57
• I don't quite understand the "the continuous weight to power ratio (note the way up that ratio is) would vary as a dimension" part of your answer. For an unventilated motor, you're saying that $P_{cont}/W$ decreases with size because of surface area cooling limits while $P_{max}/W$ increases with size? – DEEPquery Feb 12 '18 at 15:07
• Both $P_{max}$ and W tend to increase with $d^3$, so the ratio stays constant(ish). $P_{cont}/W$ tends to decrease with size because of cooling. – Neil_UK Feb 12 '18 at 16:31

Hobby motors are normally run from battery packs that last a few minutes, so they aren't often given honest continuous power ratings.

Note the "continuous" rating under "specifications" in your linked example is qualified by (180s). That is, they admit it can generate that rated power for 3 minutes max. It's only starting to warm up at that point; a real continuous power rating will be lower, to avoid burning out the motor..

Both points in the other answers are good too, though.

Neil is absolutely correct that smaller motors are more easily air cooled thanks to the increased surface area/volume.

Charles is correct to point out that increased speed increases power for "free" in an electric motor (from the point of view of efficiency and wasted heat; up to the practical limits of bearing speeds and material strength) while increased torque cost efficiency, through increased current and thus heat loss in the winding resistance. So running smaller motors fast increases efficiency. (However, large slow propellors produce thrust more efficiently, so expect to see fast motors geared down).

• So the high cont. P/W ratio of small motors is due in part to 1) more efficient air cooling 2) low torque at really high speeds? Is that why direct drive torquers have such a lower cont. P/W? They can produce much higher torque but up to lower speeds. – DEEPquery Feb 12 '18 at 15:32
• Quick question about those direct drive motors: they have insanely high $k_t$ ratings. Is this why they have such a low RPM range compared to BLDCs? IIRC, $T \sim -{k_t}^2\omega$ from the standard motor model. I'm sure the structure of the motor can handle higher speeds! – DEEPquery Feb 12 '18 at 15:36
• Kt * Kv = 1 for every (ideal) motor (with everything in SI units). That's simpy another way of expressing the law of conservation of energy; V * I = Power = Torque * rotational speed. So high Kt implies very low Kv (and I saw about 2rpm/V for one of those motors). In such a case, the limitation on speed is imposed by insulation breakdown limiting the max voltage. Also note, you can rewind any motor for any Kv (or Kt) you want by changing the number of turns. – Brian Drummond Feb 12 '18 at 15:43

Small hobby propellor motors have a high power to weight ratio because small motor and small propellors can easily operate at high RPMs, in the 10,000 RPM order of magnitude. The cited paper is about motors in the area of 2000 to 3000 RPM.

Motor weight and volume is somewhat proportional to torque and not so much related to power. A motor of a given size will provide the same torque over wide range of speeds. The higher the speed, the higher the power for the same size motor. The same thing is generally true for heat engines also.

See if the cited paper makes sense with that in mind.

I still don't understand the connection of these 2 arguments WRT the NASA paper.

The paper is lengthly and complex. It cites several references to other papers. This forum is intended to deal only with relatively narrow questions. I only scanned the paper briefly.

If heat engines also follow this trend, then why are people pursuing DEP all of a sudden?

Serious large electric aircraft design is a new field. We should expect to see a lot of approaches explored. There is significant history in determining optimum number if engines for aircraft as illustrated below. ... the continuous power/weight ratio of electric motors actually decreases with size...

Since power = torque X rotational speed, a power/weight ratio is only meaningful if either torque or speed is constant for a given comparison.

In all aspects of this question, the balance of system (BOS) is an important factor. The BOS includes the control and monitoring system, fuel storage and delivery system, lubrication system, cooling system, structural support and enclosure system and perhaps others. Some parts of these may be integral to the motor.

• I understand your point on torque and volume, and I understand your point about $P_{cont} \approx T\omega$ for the rated torque, but I still don't understand the connection of these 2 arguments WRT the NASA paper. If heat engines also follow this trend, then why are people pursuing DEP all of a sudden? – DEEPquery Feb 12 '18 at 15:16