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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Can someone please find the thread where this has been discussed at length?

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.Only if you time the time used in acceleration. If you just time the same time to both cars, or the distance, then the one that accelerated faster use more time in higher air drag speeds, which means it uses more energy.

I had a friend at school and in the age of 11 he had a good idea: If You go double fast, You need half the time... so you need less gasoline, or at least the same amount.

That's -for sure- nonsense.

From 0-60 in 4 Seconds uses much more energy then from 0-60 in 20 seconds...

Assuming without losses 0-60 in 4s and same in 20s you end up with same kinetic energy for both cars, so there should not be any difference between the two. Difference comes from the losses in efficiency and time used in higher loss speeds. For BEV those two should be a lot closer to each other than with ICE because of very high efficiency of the motor regardless of accelerator pedal position.

Volker's right, discussed at length before. Summary:

Serious calc's require accurately modeling many variables to honestly quantify this. But here are some topline observations:

1. In ideal (unreal) conditions ... Perfect battery impedance, no drag, etc., yes, the energy it takes to accelerate the car is the same whether done fast or slow.

But in the real world, the losses are significant. Battery electrochemistry results in higher, nonlinear thermal loss at very high current (i.e. rapid acceleration). And as Timo points out, the time/miles spent at the higher aerodynamic drag speed will affect the budget too.

2. Aerodynamic drag dominates above 35 mph. (so it's the major loss on freeways, even at steady state speed without acceleration).

3. Despite the losses from nonidealities, an EV is far less lossy on any given acceleration profile than an ICE.

So in that sense, yes, live large and gun it (relatively) guilt-free.

Further to that, aerodynamic drag increases as the square of velocity so think of it as a miles thing rather than time thing.

If you drive 10 miles at 30mph, your drag loss per mile will be just 1/4 the loss at 60mph.

Regardless of the longer journey time, the actual total energy loss due to drag is 4 times greater on the same trip distance.

So aero drag is, uh, a real drag.

That's why TM put so much engineering effort into getting the lowest Cd figure on the market.

Apropos Cd:

http://www.teslamotors.com/forum/forums/aerodynamics

I wonder if you could actually save energy with rapid acceleration to about 20-25mph, and then slow the acceleration down. That's the sweet spot for Model S, at 20mph your range is over 450miles with 85kWh battery.

Heh. "rapid acceleration to 20mph" sounds amusing. How would you distinguish it from slow acceleration? ;)

Timo - Sorry I don't follow this:

"Only if you time the time used in acceleration. If you just time the same time to both cars, or the distance, then the one that accelerated faster use more time in higher air drag speeds, which means it uses more energy."

"Both" cars are going the same speed at the beginning and the end. The one that accelerates slower has more losses due to aerodynamic drag than the one the accelerates faster because it takes longer to achieve that final speed. The aerodynamic drag is most-likely insignificant compared to the energy required to accelerate the vehicle however - it's more important at constant highway speeds, where you are just overcoming friction, drag and chemical-to-mechanical inefficiences.

Whity Whiteman "That's -for sure- nonsense.

From 0-60 in 4 Seconds uses much more energy then from 0-60 in 20 seconds..."

Actually Whity I would be 100% correct if all the losses where linearly related to the acceleration rate. Mark K points out that they are not, and that's the only hole in my hypothesis.

It would be interesting to know how nonlinear the losses are. I would bet that most people (like Whity) would assume that 2X acceleration uses 2X more energy, and I would also bet it's no where near that - probably 10-20% more. Anyone know the facts on this?

@Getting Amped Soon

You seem to think drag is proportional to acceleration rate. You suggest that drag losses will be greater if you spend more time accelerating between two speeds.

I think this is where your logic has taken a wrong turn. Drag losses are not proportional to rate of acceleration, but rather to car speed at each point along the acceleration curve (roughly speaking). Higher speed = more drag.

A simple example:

Car A takes 1000 m to reach final speed.

Car B reaches final speed in 500 m and then maintains final speed for the next 500 m (same total distance as Car A).

Car B has a higher average speed over the 1000 m and will thus experience higher drag losses over this distance.

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.

Also, your electrical losses increase at I^2 (current squared). This is defineatly not linear. The faster you accellerate the more current is used at once and you have much more power lost to heat in all the wires. That heat by the way also increases the resistance of the wires which makes your losses worse yet.

Other losses include air drag (as mentioned) and friction (tires on the road and all the moving parts of the car) and NONE of these are linear in nature.

All the losses do add up, and that is why you are hearing about how "spirited" driving and drag racing with your Model S can drastically reduce your range.

The only thing I agree with you on is that the efficiency of doing this in an EV is better than doing it in an ICE.

you're making this more complicated than necessary.

Although there is no acceleration penalty in the model S like there is in an ICE car, slower acceleration will still result is less energy consumption simply because the car spends more time at the slower speeds where aerodynamic drag is less.

Let's perform a thought experiment.

Suppose your choice is 0 to 60 in 4.4 seconds and then drive at 60 mph for one hour, or take an entire hour to go from 0 to 60. In the first case your energy consumption is basically that of a car going 60 mph plus 4.4 seconds at maybe 6(?) times that rate.

In the second case you energy consumption will be close to that of a car averaging 30 mph.

However, if you're going to spend an hour driving at freeway speeds, the difference between getting there in 5 seconds, vs 30 seconds will have only a very tiny effect on the average energy consumption for the hour as a whole.

I probably should have added that the measured parameter is energy consumption per total travelled distance, not per time period.

Can anyone quantify their opinions with engineering data or specs? That's what I'm really after and I'm only getting qualitative responses, and they range from I'm essentially right to I'm dead wrong.

I thought that when you stomp down on the accelerator, it squirts a whole lot of electrons into the carburetor, which then don't burn as efficiently.

GoTeslaChicago - I'm only talking about the energy used to change speed from 30 to 65 mph, nothing after that. Aerodynamic drag is insignificant in my scenario - see my post and calculations explaining why.

There are two different questions here I think. One is about total energy consumed over a period of time or over a distance under a combination of acceleration and constant speed. The other is about energy consumed at different accelerations. They are related but not the same.

If the question is: is it better to mash the throttle or take it easy (with easy being some level less than "mash") then the end point needs to be defined. A better question is "what is the lowest energy way of getting to xx mph" (a follow up question might be around constant or variable acceleration)

There are two different energy components - (A) using chemical energy to increase kinetic energy (acceleration) and (B) using chemical energy to maintain kinetic energy (speed). The former is most effected by efficiency of conversion. The latter is most affected by duration and drag (CD, rolling resistance). Neither measures are linear with time.

If we accelerate twice as hard, (A) may be higher or lower depending on the efficiency of energy conversion. If it takes twice as long to get from one speed to another, then the car will spend twice as long at each incremental speed, and therefore (B) will be some multiple as large.

So...if it is a 0-60 question, sum (A) and (B), integrating over time. Anyone want to throw some numbers into Matlab?

If an apple a day keeps the doctor away, does two apples keep two doctors away for one day or one doctor away for two day?

Nickjhowe - really I'm just after your second question. At a constant speed it's really all about how fast you are going. My question is about accelerating from one speed to the next, like pulling onto a highway. Most people would assume that accelerating twice as fast uses twice as much battery energy. I contend that it's no where near that but I'm looking for hard numbers.

It definitely wouldn't use twice as much ENERGY to accelerate twice as fast. It would require twice as much FORCE (assuming no losses). In a lossless environment it would use the same amount of energy. The losses associated with accelerating twice as fast are not only twice as much, more like 4 times as much since electrical losses and air drag losses will tend to increase as a squared function. No one here is going to do real hard numbers for this cause you would need to know exactly what all of these losses are. Do you know the impedance of the motor and all wiring or the frictional losses of the gear and tires? What about losses associated with drawing power from the battery? Nope. Not to mention it will depend on the temperature outside and how long and how hard you have been driving prior to this one particular acceleration.

But a good estimate is probably accelerating less than half peddle to the floor probably uses 10-50% less energy for the acceleration period as opposed to flooring it.

When you get your car you can run a couple hundred field tests and record all the data and then you'll know. Let me know how close my wild guess is.

Great!!! What other scientific excuses can we find to justify the pure joy of insane acceleration?

Getting Amped Soon,

I am an electrical engineer. I majored in power systems and power electronics and also took papers in aerodynamics and mechanical systems.

Unless I have misunderstood the context of your original question, I can assure you that crunching hard numbers won't increase my certainty of the fact that accelerating faster will use MORE energy for any given trip.

As you correctly noted, the gain in kinetic energy will be the same regardless of how you get up to speed. The only difference then will be in which acceleration path has the higher losses.

There are three major loss factors and they all follow the same relationship, namely that faster acceleration will increase losses.

* Higher drag losses over acceleration distance

* Higher resistive losses in wiring

* Higher losses in battery

We don't need to know how much greater each of these will be, to know that the total losses will be greater when you accelerate at a faster rate.

Sorry Tiebreaker, but the science doesn't justify insane acceleration (if you goal is energy efficiency). Doesn't mean you can't enjoy though :)

Another variable to throw into the mix is motor efficiency, which is not constant across the torque/speed range. Motors tend to be less efficient at low speed and low torque. I haven't seen an efficiency map for Tesla's AC Induction motor, so I can't quantify this or even know to what extent efficiency drops off for their motor at low speed/torque.

This said, I agree with others that faster acceleration results in greater overall losses in the real world.

Carl Barlev - you're not correct about the drag losses. In the scenario I presented they are insignificant (see my previous post) plus their effect is more pronounced the longer the acceleration period. The resistive drag force is proportional to speed, not acceleration, and the speeds are the same in both scenarios.

As others have pointed out, it really only matters if the mechanical and electrical losses are nonlinearly related to the acceleration rate, and a few have proposed that indeed they are. I don't doubt that, but nobody can quantify it. Everyone just says "more" and "higher", just as you have. That's not what I'm after but thanks for your reply.

@Carl Barlev: I am an EE too, that was :-P . Coz that what the thread really does - looking for excuses... :-D

:-P -> tongue in cheek

The "terminal" speed is the same, but the higher accel version spends more time at high speed. That's what high accel is for.

@Getting Amped Soon,

I did mean exactly what I said. If you time only the time used to accelerate then the faster acceleration rate gives you less loss, because you spend less time in higher loss speeds,

butthat car has also a) gone less distance compared to other car IE. has done less work before it reaches the same speed, and b) at that same timepoint the faster acceleration gets to the high speed the other car has used less energy because it has not yet got to the high loss speeds.To really compare the two cars you need to give them either same distance or same timeframe where

bothcars have reached the same speed. In both of these cases the car that accelerates slower wins. You get twisted results because you stop the measurements in different points for both cars.