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!

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@Tiebreaker

Yes, I can see now that your comment was :-P (brain was still stuck in "serious" mode after my previous post and took your words at face value).

It took me a good few seconds to work out what :-P mean too and I only saw your definition below after I'd figured it out - guess it pays to look ahead :)

@Getting Amped Soon

I'm confused by what I see as contradictions in your original statements and contention, but it could also just be that I've misunderstood your intended meaning...?

You say that you are only interested in the energy needed to accelerate to your final speed, but the purpose of your question seems to be to understand how this will affect overall energy efficiency.

This is two different questions:

1) Energy to accelerate to final speed, versus

2) Energy to accelerate over fixed distance

If you want to know how different acceleration rates will affect the efficiency of your energy usage when you "pull onto a highway", then you must measure the energy losses over a fixed distance.

From the fact that you want to disregard the constant-speed component of the total losses, it seems that you are considering the first question...

Carl Barlev - yes I'm only interested in the acceleration portion of the problem (your #1). Battery usage at constant speed is well understood: it's rolling friction and aerodynamic drag, and I think the aero drag dominates and increases with the square of the speed.

I confused things in my original post with my statement about extremely slow acceleration rates as my dumbed-down "limit analysis". Should have just left that out.

This all stems from my frustration when I get behind a slow-poke when pulling onto a highway or away from a stoplight. They think they are saving gas (or in this case charge), but what really hurts your gas mileage is excessive acceleration and braking - like zooming up to a stoplight and jamming the brakes and always driving like that.

People are brainwashed into thinking that puttering onto the highway is saving them gas, but I contend that the savings is insignificant. As us engineer-types understand, if the system losses are linearly related to the acceleration rate, it makes NO difference what the acceleration rate is. If you want to accelerate at 1.0X and you have 10% losses, you have to use 1.1X of power. If you want to go 2.0X you have to use 2.2X of power, but you use the higher power for 1/2 the time, and the total energy consumed is the same in both cases.

Some posters have pointed out that the electrical losses are nonlinear, which I totally get. (Some posters have also talked about aerodynamic drag - not significant in my problem.) I was just hoping that someone had some real data on just how nonlinear the electrical losses (or inefficiencies) were. If I had to guess, there is a sweet-spot that the motor operates at, and it's somewhere between creeping onto the highway and flooring it. I also think that "pretty decent" acceleration is not any more wasteful than puttering, and might even be less so due to the fact that the motor might not be very efficient at low RPMs or loads (I don't know anything about electric motors).

It's aggressive stop and go driving with excessive braking, and driving at high highway speeds that kill your range. I contend that in situations where you want to increase your speed to some "cruise" level, it doesn't make that much difference if you poke up to speed to go 3/4 throttle to get there. I was hoping for some facts to prove or disprove my theory (as I knew I had rigid-body physics on my side).

Some engineer at TM has all this built into a spreadsheet and could answer my question in two minutes. I was hoping he or she might read my post.

Thanks.

Getting Amped Soon

"Some posters have also talked about aerodynamic drag - not significant in my problem."

Not sure why you keep saying that aerodynamic drag is not significant. You haven't documented that, other than to say that it is well known. However, it is the one factor that is definitely known to be non- linear.

I am too tired tonight, but I will try to put a reasonable aerodynamic study of this together this weekend.

GoTeslaChicago and kalikgod - I'm only talking about accelerating from 30 mph to 65 mph, and nothing more. I agree my original post went off into "very slow acceleration" as an FYI, but I'm just talking about the acceleration portion. Aerodynamic drag is miniscule compared to the force to accelerate a 4647 lb weight! It comes into play only when traveling at a constant speed, and all you're overcoming is friction and drag.

Here's my previous post, which you missed. Feel free to check my calculations.

Here are some more facts (did the calculations in a hurry, hope they're right):

Assuming it takes 3 seconds to accelerate from 30 to 65 mph, the constant force required on a 4647 lb weight is 2469 lbs.

Assuming a frontal area of 20 sq. ft., and a drag coefficient of 0.25, the drag force at 65 mph is 106 lbs. At 30 mph it's 23 lbs.

So the aerodynamic drag is only between 1-4% of the force required to accelerate the car in my scenario - not that significant.

At a guess, I would say accelerating smoothly in an ICE IS more fuel efficient, but I think it's because the ICE has better efficiency in a temperature range and an RPM range. Don't have facts to back this up, but I think power varies in an ICE based RPM per unit of fuel. I think that might explain the hesitation of hypermilers to floor it.

@Getting Amped Soon, you seem to have mixed statements.

This one

People are brainwashed into thinking that puttering onto the highway is saving them gas, but I contend that the savings is insignificant.Talks about gas savings in general, and people saying that are quite correct, it saves them fuel.

however this one

yes I'm only interested in the acceleration portion of the problemis talking about half of the problem, not the whole picture. You are comparing different results using different endpoints, which gives you twisted results.

Energy you put into car is independent of the acceleration rate, you are correct about that (you end up with same kinetic energy, but then the difference comes from that point forward where your faster car has already accelerated to end speed (which is the point where you stopped your calcs) and is going in constant speed, while the other car is still accelerating. That gives you another set of energy to consider: how much energy you use to travel the same distance. That's the one where the savings in fuel comes from.

Slower acceleration gives you better result because you spend less time in higher loss speeds. It's not big difference if both cars accelerate in reasonable rates but it is there.

Comparing two different sets of losses using same endpoint in time or in distance (instead of speed) is kinda nasty integral and I'm way too tired to make calculations in my head for that.

+1 Timo

@Getting Amped Soon,

I see now why you were saying that drag is insignificant. You were refering to the fact that drag losses will be very small compared to the total energy expended during acceleration.

This is correct, but this doesn't mean that you can disclude drag losses in an energy efficiency comparison of different acceleration rates. You must compare apples with apples... or in this case, energy losses over any fixed distance (must be the same in each case).

Overall, you will use less energy if you accelerate slowly and steadily. Perhaps not during the actual acceleration, but certainly once you factor in the additional losses from traveling further at a higher-loss speed.

Timo and Carl Barlev - I'm only talking about the energy used to accelerate from one speed to another - an event that lasts 3-4 seconds. I confused the issue with next-to-last paragraph, which I wish I could edit out. I was just using an extreme example of a very slow acceleration, and it took the post off into a different direction and more complicated problem.

Focusing on the acceleration problem alone: most people think that slowly accelerating to highway speed saves them energy, and 2X acceleration uses 2X more energy (gas or electrical). In fact it's not anywhere close to that. If losses were linear, it wouldn't matter what your acceleration rate was. I was hoping someone could quantify the battery and motor behavior throughout its torque range with real data - like "3/4 accel uses only 10% more energy than 1/4 accel when the change in speed is the same in both cases." I don't think I'm going to get that information from this post, but someone at TM has it in a spreadsheet.

Work (Energy) = integral of Force over a distance

Let's ignore rolling resistance. At any speed the force required merely to overcome drag and maintain velocity increases cubically with velocity. Accelerating force is applied on top of that and is greater for a car accelerating faster.

Since s = (v+u)t/2 and t is inversely proportional to the constant acceleration, distance traveled is inversely proportional to the rate of acceleration.

So, accelerating from a stop, just overcoming drag, a car accelerating twice as fast will reach target speed in the first half of the distance of the car accelerating more slowly while having to apply forces to overcome drag that are at least 8 times as great at each point.

Assuming a perfect motor I'd suggest it'd need to do 4 times the work to overcome the drag and then using the constant F=ma the acceleration work would be the same.

Of course there's rolling resistance to overcome, so you could conceivably use more energy for a given _period of acceleration_ but because, unlike ICEs a simple electric drivetrain has a faster=worse efficiency rule, for a given _distance_ slow acceleration is better. I don't know how rolling resistance works but I'd guess that's why traveling really, really slowly is inefficient too.

Oh damn, there's no delete. They'll each be traveling the same speeds at some point, so the slow car will end up doing double the drag work?

But it's more efficient to accelerate slowly because to cover the same distance the fast car will be traveling at a higher average speed so with higher average drag.

OK - here we go. I've dusted off my calculus and noodled on this for a while. If you want to see the calculations check out this link. can't guarantee they are correct, but the best I can do.

Imagine three scenarios:

Scenario a: accelerate from 35-60mph in 2s;

Scenario b: accelerate from 35-60mph in 10s;

Scenario v: accelerate from 35-60mph in 2s then drive at 60mph for 8s.

Assuming motor efficiency/losses are constant regardless of load (!!), then

Energy (a) = 518kJ; Energy (b) = 594kJ

So slow acceleration takes 14.5%

moreenergy (because it takes longer!)Energy (c) is 628kJ (5.6% more than (b))

So if you are accelerating onto a freeway, you will use less energy over the same time if you accelerate slower (but you'll also get there a little later)

FYI In scenario (a), 96% of energy is acceleration; 4% is overcoming drag/rolling resistance. In Scenario (b) it is 86/14 (but the absolute energy to accelerate is the same)

If motor losses vary by more than 5% under heavy load then the calc may go the other way.

Thanks nickjhowe! I knew I was right of course, but people will point out that the motor losses are nonlinear with current/rpm/load. How nonlinear they are is really the crux of the problem.

Your scenario C is interesting and correct, but misleading. Trips are based on distance, not time. Can you re-run for trips of varying distance, say 1, 5 and 10 miles? You don't have anything better to do, right?

Thanks for your work!

At 60mph, aero drag uses about 322kJ per mile, rolling resistance about 499kJ per mile for a total of 821.

It takes about 750kJ to go 0-60 in pure kinetic energy. With fast acceleration, drag is small (25kJ) as the car only covers 60m.

So - 0-60 in 4.4 is about 1 mile of driving at 60mph, give or take. (at least according to my possibly flawed calculations)

Interestingly, with such low Cd, it is only above 60mph that aero drag is more than rolling resistance (based on a coefficient of resistance of 0.015 for the tyres). If anyone has a better estimate I can update the graph:

(y Axis is in N; x axis is MPH)

That aero stuff is really a drag!

Hence the appeal of vacuum tunnels ... ;)

nickjhowe - great work! I have one more request:

on-ramp acceleration from 30-65 in 3.0 seconds, total trip distance 1.0 miles.

Same as above but the on-ramp accel period is 6.0 seconds.

I think you are saying that, as your trip gets longer than a few miles, what you did to get on the highway becomes less important with respect to the range used up. I'm interested if it matters in a one mile trip.

Also - pulling away from a stoplight in a one-mile trip, same accelerations scenarios as above (except 0-35 mph). Drag is much less significant in this case (as you know).

I think these two scenarios are going to show that puttering around in a Model S doesn't improve your range, at least when accelerating to a "cruise" speed.

Thanks man!

I meant "what you did on the on-ramp becomes less important..."

Too much fun at a Haloween party tonight, I'll cruch the numbers tomorrow. :-)

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

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.

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!

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.

@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)

@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.

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!