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How does motor and regen efficiency scale with power?


Dec 9, 2013
New York
I'm curious about the scaling of efficiency of a as you increase power, volume, and mass.

I've been thinking about it since Elon mentioned a Tesla truck. Trucks tend to have a significant amount of mass, and when towing or carrying a heavy payload, the mass is increased even more. But with regen, mass isn't the issue, motor/regen efficiency is.

The energy required to move an object one location to another depends on the potential energy at each location, not the horizontal distance between the locations. In the ideal frictionless vacuum loved by the writers of physics test questions, altitude change is the only thing that changes the energy requirement for a given vehicle. If you travel to a destination at the same altitude, no energy is required, if you travel to a higher destination, energy is lost, if you travel lower, energy is gained.

But we don't live in an ideal world. When you run a motor, or run it in reverse to gain electricity, some energy is lost. If you run it hard, more is lost. What I am looking for is information on how this inefficiency scales. If you were to create a motor with relaxed mass and volume constraints that could provide twice as much torque, how much more energy would you have to provide?

If you were to get linear scaling, or near linear scaling, creating a bigger vehicle with more towing capability and more cargo capacity wouldn't decrease range too much. The dollar cost would increase, but range wouldn't drop significantly.

The ways energy exits the system are by air resistance and drag, rolling resistance on the tires. and loss due to the inefficiency of the motor and due to the regenerative brakes.
Increasing volume increases the cross section, which increases drag, but the increase is small relative to the change in power. Drag also depends on speed, not mass, which means that a heavier vehicle will lose less energy to drag as a percentage, because if you assume the same shape, drag is the same, but kinetic energy at speed has increased. Increasing the weight of the vehicle may increase the rolling resistance of the tires, but again, I don't think that is a major problem. The biggest loss as I see it is due to the motor's imperfect efficiency.

In the model S, and other EVs, if you accelerate aggressively, you lose efficiency. The maximum output of the S motor is less efficient than half of its output. Imagine you had a motor that had twice the maximum output power. If you used half of it's capability (the maximum output of the S motor), what energy would it require? Would it be less than the S? more? the same? If it is less, then large vehicles could be more efficient, and creating something like a pickup truck, or even a freight vehicle would come down to choosing the right size for the application. If bigger tasks are accomplished with greater efficiency (though they take more energy), then any application is approachable.

If I am missing an important factor, please chime in.

In short, it takes more energy to accelerate a heavy vehicle, but more energy is available in the heavy vehicle when you slow with regenerative braking, so large BEVs should be feasible, at least from a physics perspective.


Ex got M3 in the divorce, waiting for EU Model Y!
Feb 9, 2012
Drammen, Norway
Great question. I don't have the answer to all of it but here are my thoughts:

- There are many examples of very large electric motors (large trains, factories) and very large electric generators - regenerators if you will - (power plant turbines), all operating at very high efficiency so scaling of the motor including regeneration should be possible without loss of efficiency?

- When it comes to building a truck I think the whole technology will be nicely scalable and as you say the vehicle will be bigger, but with bigger battery and motor it will just cost more otherwise be as efficient and have the same range.

- When you start to think about large trailers, big rigs etc. weight with regards to rolling resistance becomes a bigger issue than perhaps you have considered? I know that wear on the road surface increases i believe with a factor of ^4 proportional to weight - meaning in effect that even though there may be one hundred personal vehicles per heavy trailer on the roads more or less all wear of the road surface comes from heavy transport. If you read this article: http://www.theicct.org/sites/default/files/publications/ICCT_work_rollingresistance_nov2012.pdf this is well explained:
...an almost linear relation can be found between CO 2 and rolling resistance, with only the vehiclemass as a parameter (Barrand & Bokar, 2008). This surprising conclusion is the result of the following:

• The effect of rolling resistance on CO 2 is linear. When comparing two tires with different rolling resistance coefficients on the same vehicle over the same driving cycle, the absolute difference in road load will be proportional to the difference in rolling resistance coefficients. Although the driving cycle can change the contribution shares of the different road load components, the difference in the amount of fuel consumed will be independent of the cycle.

• For typical engine operation points, fuel consumption maps show that a decrease in engine load translates into an almost linear decrease in fuel flow rate. So even though a change in road load is affecting the engine efficiency, the resulting effect on fuel consumption is nearly linear. Because engine efficiency decreases at lower engine loads, the linear correlation coefficient between road load and fuel consumption will be less than 1.

In other words, it can be concluded that even though the fuel consumption over a driving cycle is dependent on parameters related to speed and dynamics of the cycle, the difference in the amount of fuel consumed between two tires on the same vehicle and over the same test cycle will be proportional to their rolling resistance coefficients and independent from that cycle.

In Barrand and Bokar (2008), this is represented by an empirical relation with the following formula:
ΔFC = α · ΔCRR · M
where ΔFC is the change in fuel consumption (which may also be expressed as ΔCO2), α is the linear correlation coefficient (which takes into account engine efficiency among other parameters), ΔCRR is change in rolling resistance, and M is vehicle mass.

The above quote though relates to a regular ICE engine the efficiency which of varies more with different load than an electric motors, so perhaps the correlation does not exist in EVs?


Dec 9, 2013
New York
I suspect it does exist in EVs. A linear scaling isn't a huge problem, it isn't ideal, but it isn't prohibitive. Adding more batteries should increase range, but with diminishing returns. Going from 60 to 80 kWh gives more range than going from 80 to 100, at least as far as rolling resistance is concerned, because adding battery increases the energy density of the vehicle as a whole. Of course, to move that weight around you need more power from the motor, and at high load, the motor loses efficiency.

So we have drag which scales with velocity, we have rolling resistance which scales with mass, and motor inefficiency which scales with power ouptut. Adding mass doesn't change the scale factor for rolling resistance, but adding mass in the motor may decrease the scaling factor of motor and regen efficiency.


(S85-3/2/13 traded in) X LR: F2611##-3/27/20
Supporting Member
Mar 8, 2012
- When you start to think about large trailers, big rigs etc. weight with regards to rolling resistance becomes a bigger issue than perhaps you have considered? I know that wear on the road surface increases i believe with a factor of ^4 proportional to weight - meaning in effect that even though there may be one hundred personal vehicles per heavy trailer on the roads more or less all wear of the road surface comes from heavy transport.

There are two things that contribute to the road damage heavy vehicles cause:

1. The shear slope of the road base. The downward force from the contact area spreads out along a cone shape, the angle of the cone's slope being determined by the subgrade material (the pavement's slope is practically vertical). So for an 18,000 lbs (4080 kg) load on the front axle, the contact area is about 75 sq.in (485 cm[SUP]2[/SUP]) per tire. The shallower the slope of the cone is the lower the pressure on the road base. This works fine for the front.

2. The use of dual tires. The problem with duals is that no one tire ever carries 25% of the axle load (can be as high as 50%-0%). Besides the unequal ground loading from each tire, where the cones overlap the force is doubled, which makes some areas of the road base have more force applied than others--not a good thing. If duals were replaced with large single tires, road damage would decrease but in most areas the tire manufacturers haven't made a lot of headway with the government. And because most trucks drive through multiple jurisdictions, they have to install tires that meet the regulations everywhere they drive.

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