Comparing EVs? Be aware of a factor that’s missing

 

Call it the “drivability factor.”

By John Higham: Legislative Director of the Electric Auto AssociatioN, Electric Auto Association Board Member 

 
The author in his Jaguar I-Pace.

The author in his Jaguar I-Pace.

 

There are endless factors used to compare electric vehicles (EVs) from various manufacturers. There are 0-60 times. All-electric range (AER). Usable kilowatt-hour (kWh) in the battery. Luggage space. Number of seatbelts. Charge rates for DC fast charging. Efficiency.

But something is missing. Something important. For the lack of a better term, I’ll call it “drivability.” Drivability is related to efficiency, but it isn’t exactly that. It’s also related to charge rate, but it isn’t exactly that, either.

I would define drivability simply as how long I have to charge my car in order to get back on the road. In terms of basic arithmetic, it could be defined as the ratio of how quickly I can add energy to the car, to how quickly energy is used while driving the car.

To make drivability a useful metric to compare EVs, some standardizing assumptions are required. So let’s assume that we drive the car for one hour at 62 miles-per-hour (MPH). How long will it take to replenish the energy used? The less time it takes to charge the car to replenish the energy used, the more drivable the car becomes.

From personal experience…

My first EV was a Miata conversion. While I loved driving that car, it had a horrible drivability factor. At 62 miles per hour (MPH), the Miata traveled approximately 2.2 miles for every kWh consumed. Worse, its maximum charge rate was merely 2.5 kilowatts (kW). Expressed mathematically, that is 2.5*2.2 = 5.5, which has units of MPH. Taking it a step further, and since the 2.2 mi/kWh efficiency assumed 62 MPH, for every hour charging, I could drive for only 5.5/62 = 0.09 hours! As I’m defining it, my e.Miata had a drivability of 0.09. Not good for long trips.

Now I drive a Jaguar I-Pace. Its efficiency isn’t much better at 2.4 mi/kWh at a constant 62 MPH, but its maximum charge rate is far superior at 100 kW from a capable DC fast charger. Running through the same arithmetic gives a driveability factor of 3.8. What this means is that for every hour I charge my I-Pace, I can drive for 3.8 hours at 62 MPH. Those with an eye for numbers will cry foul about now because most EVs will not sustain the maximum charge rate on a DC fast charge for very long. But we’re going to ignore this reality for a moment.

Let’s look at one more example. I convinced my younger brother to buy a Tesla Model 3, which has an efficiency at 62 MPH of 4.5 mi/kWh, and will charge at 120 kW. Running through the same arithmetic gives it a driveability of 9. Ergo, for every hour of charging one could drive for nine hours, assuming the car’s battery and the driver’s bladder have that type of capacity. This can also be interpreted to mean that every minute charged enables nine minutes of driving. 

The numbers discussed above reflect a maximum drivability using the maximum DC fast charging rate for a specific model. Understanding that no EV can sustain maximum charge rate when using a DC fast charger, it makes sense to be more focused on the average drivability. Over the course of a full charge, the I-Pace average charge rate at up to 90% state of charge is about 70 kW. The Model 3 is about 100 kW at up to 90% state of charge. This yields an average drivability factor of 2.7 for the I-Pace and 7.1 for the Model 3.

So, what should we do with this information?

I am suggesting that manufacturers start using a standardized metric similar to drivability, enabling consumers to compare different models among various brands. Until that happy day, I’d like to see EV enthusiasts use the drivability factor that they’ve calculated themselves while comparing EVs.

How to do this? Simply determine the efficiency of an EV in miles/kWh at 62 MPH--the results should be between two and five. Most EVs will express this number either in a connected app or in the driver’s user interface display. Some manufacturers express this number in watt hours per mile, or even in kWh/100 mi. To go from watt hours per mile, simply divide into 1,000. For example, an EV that has an efficiency of 333 watt hours per mile is 1,000/333 = 3.33 mi/kWh. To go from kWh/100 mi, just flip it around. For example, an EV that uses 33.3 kWh/100 miles is 100/33.3 = 3.0 mi/kWh.

Once you have efficiency in miles/kWh, find out the average charging speed for a DC fast charge session in kW. This can be easily determined by charging from a near 0% state of charge for an hour. The DC fast charger will tell the driver how many kWh have been added to the car. Once you have efficiency in mi/kWh and the average charge rate in kW, simply multiply the two numbers; 62 divided by that result yields the drivability factor. Stated more plainly, for a drivability factor of five, charging for one minute yields five minutes of driving. Charging for five minutes yields 25 minutes of driving. Charging for one hour yields five hours of driving.

What is the drivability factor of your electric car?