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Here's An Engineering Problem For You

Here's An Engineering Problem For You

Everyone knows that excessive acceleration accompanied by corresponding braking is an inefficient way to use stored energy, whether it be for ICE or EV's. However, what about the situation where you accelerate from one speed to another at a constant rate, and hold the final speed (like pulling onto a highway)? I contend that it doesn't use any more energy whether you creep up to speed or make it fun.

Here's my reasoning: the work done to accelerate a rigid body just depends on the initial and final velocities, not the acceleration rate. If you accelerate twice as fast it takes half the time to reach your final speed, and you use the same amount of energy.

Now cars are not rigid bodies, and energy is lost due to friction, heat and the inefficiencies of the conversion of stored chemical energy to mechanical power. However, if these losses are linearly related to the acceleration rate, then again the energy used to accelerate from one speed to another is the same no matter how fast you accelerate.

Now here's my final curve: the aerodynamic drag might be negligible for the few seconds of going from 30 to 65 mph, but it becomes more significant the longer it takes to do that. So if you accelerated at a very low rate, and it took you an hour (or many hours) to go from 30 to 65 mph, then the aerodynamic drag will become a more significant factor in the energy used, and you will use MORE energy than if you accelerated faster.

So I contend that putting your foot into it to pull onto a highway (as long as you don't exceed your desired final speed) doesn't use any more kW-Hrs than poking along. So (safely) enjoy yourself! Would be great to hear from a Tesla engineer on this one!

evanstumpges | October 26, 2012

Just came across this Tesla blog post. It's addressing the Roadster, but I suspect the Model S would have similar characteristics. There are several helpful graphs in the post. I'm attempting to display the most relevant graphic here, but in case it fails, check out the blog...

http://www.teslamotors.com/blog/roadster-efficiency-and-range

evanstumpges | October 26, 2012

To add some detail to the graph above, it is showing the losses of various aspects of the car. The red line is the sum of all losses. Looks like the most efficient speed for the Roadster is 15MPH...

Aerodynamic Losses: From air drag forces on the car/radiator.
Tire Losses: Rolling resistance of the tires
Drivetrain Losses: Motor, inverter, bearings, and gearbox
Ancillary Losses: 12V equipment, cooling, pumps, lights, aux power use

As expected at very low speeds the motor efficiency drops significantly and ancillary power consumes a greater percentage of overall vehicle power. Aerodynamics losses increase with the square of velocity. Rolling resistance losses are more or less constant throughout the speed range.

Getting Amped Again | October 26, 2012

Hey nickjhowe, I think your 821 kJ/mile figure for 60 mph is equal to 228 W-hr. Looks like it's 250 W-hr/mile for a Roadster. Nice job!

I think you should just add the drivetrain loss from the graph above to your calc's for a Model S, and I think you'd be pretty damn close!

Getting Amped Again | October 26, 2012

Also from evanstumpges's link (thanks):

"Drivetrain losses include those that the user doesn’t typically control: the efficiency of the motor controller, the motor itself, the gearbox and generally all losses in converting the DC electricity from the battery pack into useful torque at the wheels of the car. This is proportional to speed due to spinning losses in the gearbox and motor and also proportional to power output due to conversion losses in the various subsystems."

Maybe the drive system efficiency ain't that nonlinear after all! That would make my original proposition more likely - that it doesn't matter that much from a range standpoint whether you poke up to speed or get there with some fun, as long as you don't exceed your desired end speed.

nickjhowe | October 27, 2012

@GettingAmpedSoon - in your 3s and 6 s scenarios, the speeds are the same and the distances are the same, so Ek and Er are identical. The only difference therefore is the slightly higher aero drag over 3s seconds the first car is at 65 while the other car is still accelerating.

My calcs suggest the difference is about 6kJ (out of a total energy draw of 1505kJ)

Getting Amped Again | October 27, 2012

@nickjhow - thanks man, that's 0.4% for a one mile trip. For trips longer than a mile the difference becomes even less.

Conclusion - drive safely, but enjoy the great acceleration of a Model S, whether you get a 40 kWh or a Performance 85. Unless you needlessly over speed-up and brake constantly, it really doesn't matter if you have some fun pulling away from a stoplight or onto a highway- your range isn't affected hardly at all.

Science wins out over folklore.

nickjhowe | October 27, 2012

Two big caveats:
1) I may have made a huge error in my math (not unknown!)
2) We know that the drive train efficiency suffers under high load - lap times fall off a lot on track after a few laps so there may well be losses associated with a heavy right foot that I haven't taken into account. These losses might not be trivial

But until 1 and/or 2 are proved to be true - have fun!

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