So that leaves the battery, and I think I have a model that explains why the battery causes the car to "waste" energy during heavy acceleration (or even deceleration?)
Batteries are often modeled as an ideal voltage source and a "internal resistance". This isn't perfect, but works pretty well. Here's an article discussing the internal resistance and other factors in 3 types of EV batteries: www.evs24.org/wevajournal/php/download.php?f=vol3/WEVJ3-5340444.pdf
In this model, the "waste energy" from the battery is I^2 * Ri, where I is the current, and Ri is the internal resistance of the battery. I think this waste energy essentially becomes heat. Here is some information about these Lithium 18650 batteries Tesla uses: http://laserpointerforums.com/f67/how-healthy-your-batteries-how-measure.... It says a "healthy battery" has an internal resistance of ~0.1ohm, and a depleted battery more like 0.250ohms. At 0.400ohms internal resistance, the battery is toast.
The power meter on the Model S goes from -60kW (regen, pedal up) to +320kW (pedal down), so lets run throw some numbers on how much "waste energy" is used at 2 different accelerations. Lets accelerate at 40kW and 320kW.
I'll do a lot of rounding here, because I don't know exact numbers--like how many cells there are, or how many are in use at the same time--and hoping that it kind of averages out to the right answer.
I believe there are in the range of 7500 lithium batteries in the 85kWh models. Lets assume they are all in use all the time. These cells deliver ~4V.
So how much current is each cell delivering when they are delivering 40kW of power?
P = I * V or I = P/V
40000W / 4V = 10000Amps
10000A/7500 cells = 1.3Amps per cell
"Waste energy" per cell is I^2 * Ri = 1.3^2 * 0.1ohm == 0.169W
0.169 W * 7500 cells = 1267W
1.267kW / (40+1.267kW) = 3% wasted energy (battery loss) at 40kW
I would say 3% "waste" is small enough that there may be other causes of loss that are much bigger, so it might even be "negligible".
Now lets try it with the pedal floored: 320kW of useful power.
320000W/4v = 80000Amps
80000A/7500 cells = 10.7Amps per cell
"Waste energy" per cell is I^2 * Ri = 10.7^2 * 0.1ohm = 11.4W
11.4 W * 7500 cells = 86kW
86kW / (320+86)kW = 21% wasted energy (battery loss) at 320kW
At the point that you are producing 86kW of heat, presumably you need to start spending some more energy to cool the batteries as well...
My conclusion is that heavy acceleration uses more energy. I think it's also likely that heavy regen is less efficient than light regen. This could explain another post I read that claimed using the "reduced regen" mode was giving better mileage.
The analysis of how the internal resistance changes as the battery degrades over time also has repercussions on how much power the cells can deliver as they age.
If we go forward in time to the point when the batteries have an internal resistance of 0.2ohms (5 years? 8 years?), then that roughly doubles the waste energy (P = I^2 * Ri and Ri doubled)
So if the car could still output 320kW of useful energy, now the waste energy would be:
172kW / (320+172kW) == 35% wasted energy (battery loss)
But note it's still fairly efficient at low power
2.5 / (40 + 2.5) == 6%
Heavy power consumption comes at a price.
As the batteries age, this cost increases.
It seems likely that top acceleration will degrade as the batteries age.