Tesla Model 3 Is 166% More Efficient Than Gasoline Cars: Jason

A Tesla Model 3 is 166% more efficient than a gasoline-powered car, which returns 40 Miles Per Gallon (MPG), according to Jason Fenske.

Jason is one of the most brilliant car guys on YouTube and runs his channel by the name of Engineering Explained. He owns a Tesla Model 3 Performance and drives it daily.

Speaking of the Model 3, we all know that it delivers a fuel economy of 116 miles per gallon equivalent (mpge). However, what we do not know is how much energy is lost from the battery to the wheels.

Tesla does not merely consist of a steering wheel, tires, or a battery. It has a ton of electronics and automotive equipment to carry as well. We will be looking at how much energy the car consumes besides making the tires roll.

For example, a gasoline vehicle delivers 40 MPG from 1 gallon of fuel. A gallon of fuel has 33.4 kW energy, so the final efficiency of such a car is around 26%.

Similarly, for Tesla, fuel efficiency is around 70%. Here’s how Jason got this number. Warning: Complex maths ahead.

How A Tesla Model 3 Is 166% More Efficient?

To calculate the total energy consumed, we need a bunch of data. Namely, we need rolling resistance of the tires, aerodynamic drag, air density of the wind, area of the front of the car, load of the AC, and atmospheric pressure.

Keep in mind that the numbers are not accurate to their final digits, but they do represent a fair state of things.

The main task here was the collection of data. So Jason performed a series of tests and made a bunch of calls to get as accurate of data as possible.

To calculate the energy required to overcome aerodynamic drag, the frontal area of the Model 3 was needed. Jason figured the frontal area of the Model 3 by making a few estimates from the height and width data of the Model S.

The frontal area of the Tesla Model 3 came out to be 23.69 ftÂ².

Then, Jason needed the consumption data of the Model 3 Performance. He obtained the consumption data by finding a straight patch of the road and driving up and down it several times. He drove at successively increasing speeds from 50 to 70 mph to get data at several consumption points.

After getting the consumption data, he calculated the aero energy losses by multiplying the Model 3 frontal area 23.69 ftÂ² with air density, drag-coefficient (0.25), mass, atmospheric pressure, and distance.

E (aero loss) = ATM x Mass x Frontal Area x Distance

After assuming coefficients of friction for the winter tires on his Tesla Model 3 Performance, Jason calculated energy losses due to rolling resistance.

E (rolling resistance loss) = Rolling resistance x total weight on tires x Distance traveled

After obtaining these figures at each speed, he subtracted them from the overall energy consumed. It gave us the final remaining energy used by air-conditioning, speakers, center consoles, and all the other electrics on board the Model 3 Performance.

The final efficiency percentage for the Tesla Model 3 Performance, driving between 50-70 mph with no air conditioning, came out to be 70.66%.

Meaning if a Tesla Model 3 is using 300 wh/mile to go 70 mph, then 212 wh/mile is used to drive the car. And the remaining 29% is utilized by everything else from power transfer between the inverter and the battery for running all the sensors, cameras, AC, center console, etc.

Remember the 26% I mentioned earlier? Now consider this: a gasoline vehicle with 40 mpg uses 33.4 kW of energy at a rate of 26%, meaning 73% of total energy is lost in a gasoline engine.

Power transfer loss is so significant for conventional cars that it takes almost 2.6 times as much energy to move them the same distance as compared to electric vehicles.

What do you think of this experiment? Let us know in the comments below.

Yetnesh Dubey

Associate Editor at Fossbytes. Yetnesh manages the everyday editorial duties and oversees the writing staff. He occasionally covers news related to electric vehicles and tech.
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