I'm trying to understand the physics of why accelerating rapidly uses more energy than a more gentle acceleration? Is the motor/inverter less efficient at high current?

Written for gasoline cars, but the same principles apply....See #2 http://www.pepperdine.edu/sustainability/services/EcoDrivingUSA_EcoDriving_Practices.pdf

Good read; tips that I follow both with my wife's ICE and some with the MS. Our resident hypermilers, @jerry33 and @Sparrow should author something like this for Model S owners.

The energy applied to the car is the force (lb for Newtons) times the distance traveled under that force. Faster acceleration requires higher force far a given mass. This is somewhat of a simplification, but more energy is required. Therefore, even if the motor efficiency is exactly the same it will still take more energy from the battery for faster acceleration. Motor efficiency will also have an effect, and most motors will be most efficient at medium torques and speeds. GSP

Resistive power losses in the battery pack, drive inverter, motor, and associated wiring are proportional to the square of the current. So doubling the motor current doubles the torque but quadruples the losses.

I think an EV will be less effected than a gas car. A gas engine will adjust things like fuel/air ratio for maximum power when accelerating quickly, and that's never the same as for maximum efficiency. I don't think an EV would have any analogous behavior. I would be interested in seeing some scientific tests of this for some gas cars and EVs to measure the difference under different acceleration scenarios. I once tried to do the calculations to determine how much energy was required to accelerate at different rates to a certain speed and was surprised to discover no significant different at different acceleration rates. But perhaps much of the losses are due to resistance/friction which I didn't account for, or maybe I just did it wrong. I don't know.

It depends on what your goal is. If your goal is to accelerate to a target speed (for example during a freeway on ramp) then the energy needed is just the kinetic energy (1/2*mv^2). Using the analysis of force, yes, the acceleration and force is higher, but the distance required for reaching that speed is lower. Of course when you throw in other increased frictional and heat losses from higher acceleration, you still turn up worse (how much worse is unknown, esp. for EVs). Also those that accelerate fast tend to either overshoot their target speed (causing unnecessary energy loss by having to slow down again) or it leads to excessive braking (like between stoplights).

But at higher acceleration won't it be for a shorter duration? i.e. won't the total area under the curve still be the same? Slow acceleration will be a long, slowly increasing curve. Fast acceleration will be a steep, but short curve. Total area the same. This is something I've wondered about as well. Assuming the motor is equally efficient at all power inputs, it should take the same amount of energy to accelerate a mass to a given speed, regardless of how fast. Faster requires more energy, but is done over a shorter period of time, so the net is the same. I think the only difference then is attributed to lower motor efficiency at greater power inputs, where more energy is lost to heat. That, and also the real world tendency to overshoot the target speed when accelerating fast, which will be wasted energy. An ICE has considerably greater inefficiencies at higher power inputs due to difficulties in maintaining adequate air intake as well as increased friction losses at the pistons and crank, resulting in lower efficiency at higher acceleration. An EV has none of these problems.

You both nailed it. I tend to disregard these factors and go just with the the theoretical physics and hence always accelerate as fast as possible to my target speed

I tried to use a simplification, but I think if you integrate force over the same distance you still get more energy required for the more aggressive drive cycle. GSP

No. Simple kinetic energy change. Doesn't matter how fast/slow you accelerate. The amount of energy required (assuming no friction, etc) is identical if you accelerate slowly or quickly. The ICE example given earlier in the thread does not apply to EVs. An ICE has a most efficient RPM to produce energy. You would want to accelerate such that the RPM stays within the most efficient band including gear shifts. Since the Tesla is single speed, this is irrelevant since you have no control over RPMs. There are other factors in play. What is the most efficient power range for the Tesla ... e.g., does the inverter/motor operate more efficiently at a given power/current? There is air resistance to take into account and friction losses in tires, bearings, etc. At the end of the day, the engine efficiency has the greatest impact in the ICE; I'm not sure what would be the over riding factor in the EV.

Wow. This is very interesting. You get so used to ICE driving that you think the same rules would apply to your Tesla. It would be really interesting to see what the efficiency ratings for the Tesla drive train are at various loads. Clearly there are greater losses at high loads, but is it really significant? I would suspect that the impact of various acceleration rates on a Tesla drive train would be tremendously smaller than an ICE.

This is one of the reasons. Here is a chart for the NCR18650A battery cell. Voltage slumps at higher discharge currents, requiring more current to compensate. Total discharged mah also decreases in one instance. At higher currents you get a lot less energy out and a lot more lost to heat in the batteries.

No. Work = integral(F(t)v(t) dt, 0 to T) where F(t) is Force, v(t) is velocity at time t, T is total time of acceleration. But under constant acceleration, v(t)=Ft/m where m is mass and force F is constant. = integral(F^2 * t/m dt, 0 to T) = F^2 * T^2 / 2m But, F=ma, so T is inversely proportional to F, T = V/a = Vm/F where V is target velocity. = F^2 * (Vm/F)^2/ 2m = F^2 * V^2 * m^2 / (F^2 * 2m ) = mV^2/2 This is independent of rate of acceleration. Hmm, now where have I seen that last equation again? Yes, it's kinetic energy. The simpler way of saying it is that in a perfect system the work is all coverted into kinetic energy, so rate of rate of acceleration would be irrelevant and we can leave the calculus as an exercise for the reader.

Qhmic resistance is 4 times as high at twice the current. This means losses everywhere in the electrical system are quadruple. Higher load on the battery causes voltage to drop lower. The same amount of capacity (in Ah) is drawn from the battery, but the voltage is lower at higher discharge rates, thus you get less energy out compared to lower discharge rates. I assume the losses in friction are also not linear

The energy required to accelerate the mass in both scenarios is the same; however, the energy from the battery required for a higher rate of acceleration is higher due to the lowered conversion rate, as discussed above.

And thus do the electrical engineers continue to trump the mechanical engineers in this new era ... I've got no skin in the game, I was nuclear. Thanks for the info, I didn't know the efficiency curve.

Does anyone know what the most efficient acceleration is for the Model S? I've read that for ICEs at low speeds, say 30 mph, a very efficient (presumably somewhat unpleasant) style of driving is pulse and glide - rapid acceleration at the ICE's optimal acceleration followed by gliding. I doubt this technique would buy one much for the Model S, but it does make me wonder what the most efficient acceleration is.

How big would the battery have to be to make this of low significance? The bigger the battery, the less effect it will have on each individual cell. 85 kWh is a pretty big battery.