Just the Aero fraction of the total consumption.
Aero force is 0.5 * rho * Cd * A * V * V
Newtons / 3.6 = Wh/Km
0.5 * Rho * Cd * A / 3.6 = 0.5 * 1.225 * 0.23 * 2.34 / 3.6 = 0.09156875
The C column function then is 0.09156875 * v * v
Do you see an error here ?
Left unsaid, but perhaps I should clarify that I was only interested in the additional energy/Km the car consumes as the speed increases. I presumed that all forces and fixed power consumptions stay the same except the Aero drag.
Actually, any power use that is directly proportional to the speed has no effect on the optimal driving speed. Imagine your car has such a power use and you cut your speed in half. Then you will cut that power use in half. But since you have also just doubled your driving time, the energy consumption is unchanged.
The car also has some power consumption that is independent of the speed (e.g. lights, infotainment, seat heating). It may seem counter-intuitive at first, but an increase in this type of power use actually increases your optimal driving speed. This is so because increasing your speed reduces your driving time, thus reducing the amount of time the speed-independent power usage occurs. However, unless the charging power is very small, the effect of the speed-independent power use on the optimal driving speed is negligible.
What remains is the power use that increases with the square and cube (and possibly higher orders) of the speed.
The optimal speed given a specific charging power in the above plot seems a little high to me.
I went to
EV Trip Planner to get their consumption numbers for the LR Model 3. I chose a flat road at sea level, with ambient and cabin temperature equal (at 21C), no wind, no payload (although that should only have an effect if altitude is gained or lost, or many starts and stops are made). Using different speed multipliers, I received its estimate of the consumption per distance (which is really a force, namely the force that the car needs to exert on the road in order to maintain its speed). With speed in unit km/h and driving force in unit Wh/km I got these numbers:
v F
55 79
66 87
77 97
88 110
99 126
110 144
121 164
132 186
143 211
154 238
165 267
176 298
182 315
187 332
193 349
204 386
209 405
220 445
Since the drag power (from the drag equation) grows with the speed cubed, I fitted the above values to a cubic polynomial, with these coefficients (using SI-units, i.e. m/s for the speed and W for the power):
0 328.93
1 211.893
2 -3.2902
3 0.424175
The first coefficient is the constant term, i.e. a speed-independent power use of 329 W (negligible with even a small charging power), and the whole polynomial approximates the driving power, i.e. the power required to maintain the given speed.
For various charging speeds I then set out to solve equation (1) in my optispeed-paper:
https://www.eso.org/~llundin/optispeed.pdf
With the driving power being a cubic polynomial, the equation (1) also becomes a cubic polynomial.
For varying charging speeds I found not only the optimal driving speed but also the effective speed (i.e. the speed when including also the charging time), the driving power and the driving force.
Charging Power 3 kW: Vcar= 61.8 km/h Veff= 22.8 km/h Pcar= 5.1 kW F= 83 Wh/km
Charging Power 11 kW: Vcar= 90.3 km/h Veff= 46.7 km/h Pcar=10.3 kW F=114 Wh/km
Charging Power 22 kW: Vcar=111.9 km/h Veff= 64.0 km/h Pcar=16.5 kW F=147 Wh/km
Charging Power 50 kW: Vcar=145.2 km/h Veff= 89.2 km/h Pcar=31.4 kW F=216 Wh/km
Charging Power 117 kW: Vcar=190.9 km/h Veff=122.3 km/h Pcar=65.6 kW F=344 Wh/km
Charging Power 120 kW: Vcar=192.5 km/h Veff=123.5 km/h Pcar=67.1 kW F=349 Wh/km
Charging Power 180 kW: Vcar=219.6 km/h Veff=142.6 km/h Pcar=97.3 kW F=443 Wh/km
I verified the approach by going back to
EV Trip Planner and by trying out various driving speeds, I found that the optimal driving speed for charging at 120 kW as determined by the above is consistent with the results found manually on
EV Trip Planner.
As can be seen, using the numbers from evtripplanner.com, the optimal driving speed while charging at 50 kW is about 145 km/h. The above plot has that speed closer to 155 km/h.
The speeds I calculated above already seem quite high, but if realistic then the supercharging at 180 kW of the Model 3 will be very handy here in Germany (once the Superchargers become upgraded). But even the charging speed of 117 kW that current Model 3 users are reporting will allow for a useful effective traveling speed.
PS. The fitting of the polynomial and the solving of the differential equation was done with
www.eso.org/cpl/,
a tax-payer funded open-source library for astronomy related tasks.