Calculating 0-60 mph time for an electric vehicle

Easy:

Mechanical energy: E = 1/2 * m * v^2
Constant power: P = E/t
Solve for time: t = 1/2 * m * v^2 / P

For a Model S (m=2100 kg, v = 27 m/s, P = 310000 W) that yields t=2.47 sec.

Caveats:
- Full, constant power from standstill is not possible because that would imply infinite acceleration at standstill
- Also, you won't be able to transfer the full power to the road on low speeds because tire/road friction is not high enough.

This is why real life 0-60 mph speeds are higher.

Kaspar

Torque of the motor is a worthless number, unless you also have the final drive ratio.

If you have the final drive ratio, you would use the torque (assume 100% torque at 0 rpm) and the mass to determine acceleration.

You would need to check with total power to make sure that torque is constant all the way to 60 mph (or whatever speed) and rpms.


I would imagine that rolling losses, and aerodynamic losses are very small 0-60 compared with the amount of energy required to accelerate the mass of the vehicle.

jumpjack said:

Thanks, but unfotunately this is only the easy version.
Actually we must take into account both rolling friction (m*g*Crr) and air drag (0.5 * rho * Cd * A), and I am not able to get any suitable equation: all equations I got give impossible results like t>60 seconds.
Even considering only air drag does not help.

Click to expand...

Have you actually tried using my formula? It will give you reasonable times.

Yes, my formula doesn't include rolling friction or air drag, but as ElSupreme has said that's next to irrelevant because the majority of the energy goes into moving the car, i.e., the mechanical energy stored in the heavy, fast-moving car.

First of all, the data you have is a great starting point, but it’s insufficient.
As ElSupreme notes, the gear ratio is important if you want to use torque. If you’re using power, you need a graph of power vs. RPM.

In ICE cars, it’s even more complicated as those cars have variable torque curves and they change gears at least once.

Let’s look at Model S P85, where we have some more information:
The car is constant torque at 600 N-m to about 19 m/s (42 mph), at which point it changes to constant power at 310 kW.
600 N-m is 310 kW at 4934 RPM (310000/600/2/pi), which matches the published 5000 rpm spec.
The published gear ratio is 9.73:1, and the tires are about 2.21 m in circumference.
5000 rpm / 60 s/min * 2.21 m gets us to 18.9 m/s, so the math is all working.

Let’s start by assuming no losses because that’s easier and is close to accurate (as El Supreme notes).

From 0-19 m/s, the car is linearly accelerating with constant torque (or force).
We can use the torque with the gear ratio and tire radius:
600 N-m * 9.73 / 0.35 m tire radius = 16.7 kN of force. 16.7 kN / 2108 kg = 7.9 m/s²
19 m/s / 7.9 m/s² = 2.4 s.

We could also use the power to calculate force:
P = F . v, so F = 310 kW / 19 m/s = 16.3 kN, or 7.7 m/s².
19 m/s / 7.7 m/s² = 2.5s.

Now we need to use constant power to get us to 26.8 m/s.
This is done best by looking at the energy still needed, which is:
½ * 2108 * (26.8² - 19²) = 377 kJ.
377 kJ / 310 kW = 1.2 s.

Therefore, the total time to get to 60 mph without any losses is 3.7 s.

...

Now you wanted to add loss into the equation.
The only loss that matters is air resistance, whose force varies as velocity squared.

Let’s get an idea about the drag of Model S.
At 80 mph, range is 200 miles. 85 kWh / 200 miles = 425 Wh/mile or 34 kW or 952 N
At 60 mph, range is 285 miles. 85 kWh / 285 miles = 300 Wh/mile or 18 kW or 672 N
That leaves the air drag constant (0.5 * rho * Cd * A) around .8.
Note that even at 60 mph, the 672 N of drag is much smaller than the 11,567 N that the motor is putting out.
Note also that this means that the drag would pull 310 kW at around 160 mph, putting a ceiling on the car’s speed if a transmission was used.

Velocity = integral of acceleration dt.
Acceleration = F/m
Below 19 m/s
F = (16.3kN – 0.8 * v²)
I don’t know how to solve that differential equation, so I’ll just do it numerically with delta_t = 0.01:
I get t = 2.5 s to get to 19 m/s.

Above 19 m/s,
F = (310 kW / v – 0.8 * v²)
Again, I solve numerically, and I get:
1.3 s to get from 19 m/s to 26.8 m/s.

In total, the air resistance adds on a bit under 0.1 s, and the 0-60 time is about 3.8 s.

Hope this helps,
Derek

Amazing reply!

Yeah. Amazing post Derek.

Theoretically how quick could an AWD Model S get with proper tires? Maybe assume whatever you think is reasonable for front motor in terms or torque and HP added.

dsm363 said:

Yeah. Amazing post Derek.

Theoretically how quick could an AWD Model S get with proper tires? Maybe assume whatever you think is reasonable for front motor in terms or torque and HP added.

Click to expand...

Assuming the battery limits you to 310 kW, then snooper77 put a lower bound of 2.5 s on it.

If you put in two identical motors/inverters and didn't change the gear ratio or 310 kW cap, you could get twice as much torque to the wheels.
Running the numbers this way, it would take you 1/4th the time to get to 1/2 the speed, or 0.6 s to 9.5 m/s.
Then you use the 310 kW to get you from 9.5 m/s to 26.8 m/s in 2.1 s, and now you're at 2.7 s with perfectly sticky tires. That's quite remarkable.
In reality, we need to add a little time for the extra weight (maybe 0.2 s), and the front tires won't stick as well as the rear tires when the huge torque removes some of the down force from the front wheels.

Actually, let's look at the weight ratio more closely:
The rear wheels are exerting 5820 N-m of torque.
I think this is a decent approximation, but I'm not sure: I assume all of the car's weight is 1.5 m in front of the rear wheels. Then 5820/1.5 = 3880 N of weight is shifting from the front wheels to the rear wheels. I assume that before stomping the pedal, the front wheels and rear wheels each supported 10 kN. After the weight shift, the rear wheels have 14 kN and the front wheels 6 kN. That's less than half, and the existing tires are currently at the edge of slipping. Let's assume somewhat better tires, so that the front wheels can add 50% more to the torque currently provided by the rear wheels.

If I say that the front motor adds half the torque that the rear motor does, and that the car weighs 150 kg more, we get force of 25 kN, and acceleration of 11.1 m/s² (1.1 g) up to 12.4 m/s in 1.1 s, and 2.1s more to get to 26.8 m/s.
So maybe a 0-60 of 3.2 s? That would be incredible.

derekt75 said:

Assuming the battery limits you to 310 kW, then snooper77 put a lower bound of 2.5 s on it.

If you put in two identical motors/inverters and didn't change the gear ratio or 310 kW cap, you could get twice as much torque to the wheels.
Running the numbers this way, it would take you 1/4th the time to get to 1/2 the speed, or 0.6 s to 9.5 m/s.
Then you use the 310 kW to get you from 9.5 m/s to 26.8 m/s in 2.1 s, and now you're at 2.7 s with perfectly sticky tires. That's quite remarkable.
In reality, we need to add a little time for the extra weight (maybe 0.2 s), and the front tires won't stick as well as the rear tires when the huge torque removes some of the down force from the front wheels.

Actually, let's look at the weight ratio more closely:
The rear wheels are exerting 5820 N-m of torque.
I think this is a decent approximation, but I'm not sure: I assume all of the car's weight is 1.5 m in front of the rear wheels. Then 5820/1.5 = 3880 N of weight is shifting from the front wheels to the rear wheels. I assume that before stomping the pedal, the front wheels and rear wheels each supported 10 kN. After the weight shift, the rear wheels have 14 kN and the front wheels 6 kN. That's less than half, and the existing tires are currently at the edge of slipping. Let's assume somewhat better tires, so that the front wheels can add 50% more to the torque currently provided by the rear wheels.

If I say that the front motor adds half the torque that the rear motor does, and that the car weighs 150 kg more, we get force of 25 kN, and acceleration of 11.1 m/s² (1.1 g) up to 12.4 m/s in 1.1 s, and 2.1s more to get to 26.8 m/s.
So maybe a 0-60 of 3.2 s? That would be incredible.

Click to expand...

Giving this some "real life" perspective...

Current Formula 1 race cars: 0-60... 2.1-2.7 sec
Ferrari Enzo (basically a street legal F1 race car--cost > $1M if you can find one for sale): 0-60... 3.1 sec

So, a dual motor Model S 0-60... 3.2 sec? OMG---all I can say is WOW. And, I want one!

At some point this kind of acceleration gets a little nuts but is exciting what they could offer if they wanted to. And if I understand if correctly they would get even more power with a larger battery pack so the wish list P110+ AWD would be quite the beast if a P85+ AWD gets down to 3.2 seconds. Amazing.