Hard Acceleration / Energy Consumption

Hard Acceleration / Energy Consumption

Imagine the following two scenarios:

1. Accelerate HARD to 60 mph, drive for a short while and then decelerate to a stop for an average speed of (say) 40 mph.

2. Accelerate very gently to slightly above 60 mph such that the average speed for the entire run is also 40 mph, then decelerate to a stop as in scenario 1.

So, apart from the (slight?) difference due to additional tire distortion in 1 vs. 2, the total energy consumed should be the same, right? But I'd bet that the car's energy consumption in 1 is likely to be greater than in 2. If I'm right, where did the energy go?

Blueshift | 5 octobre 2013

I don't know if your conclusion is correct. But if it is, then a possible explanation could be that the efficiency of energy transfer is inversely related to the magnitude of energy transfer. That is, as more energy per time is transfered from battery to wheels, the efficiency drops, because more heat is generated, etc.

Ignoring the battery, the work done by the wheels in either scenario would be the same, at lest in vacuum. Since there is air resistance however, I would expect #2 to require more work (thus energy). [Consider this thought experiment: accelerate from 1 to 60 over 100 days...the energy used to counter air resistance on just the 99th day, say, will easily exceed the energy used in #1].

jat | 5 octobre 2013

Higher acceleration means higher current, and many of the losses are proportional to the square of the current, so #1 has higher losses.

MacDaddyDude | 5 octobre 2013

I actually know the answer to this:

The higher amp drawn in scenario 1 creates greater resistance/loss.

jat | 5 octobre 2013

More specifically, resistive losses are I^2 * R where I is the current in amps and R is the resistance in ohms. Accelerating at 318kW compared to 40kW dissipates ~63 times as much waste heat in the battery, wiring, inverter, and motor. Even after allowing for accelerating longer at 40kW, slower acceleration wastes far less energy (to a point - below a threshold, other inefficiencies start to dominate).

soma | 5 octobre 2013

That is somewhat correct (only regarding the idea of resistive losses), but I don't think resistive losses in the electrical system could expend that much energy, or be the primary factor. If that were true, we would regularly be wasting astounding amounts of energy at high speeds.

I think the majority of energy loss in #1 compared to #2 is probably due to the same thing that affects energy efficiency vs. speed: the wind resistance.

jat | 5 octobre 2013

Steady-state high speeds don't require high power outputs, and most of the loss there is due to drag (which is proportional to the square of velocity). Also consider that assuming you could maintain 318kW output you would deplete the battery in about 15 minutes.

But the OP's question was assuming the same speeds, just different acceleration rates to get there. So, aside from taking longer to get to the higher speed, the drag at steady-state is exactly the same.

jjb94941 | 5 octobre 2013

So the answer, for someone who remembers the law of conservation of energy but not much more about thermodynamics from college physics is that scenario 1 consumes more energy than 2 and the difference is dissipated in heat energy. Right?

jat | 5 octobre 2013

Yes, virtually all loss eventually winds up as waste heat (for example, drag heats the air and the car, friction heats the tires and the road, internal resistance heats the battery, etc).

triss1 | 6 octobre 2013

Agree almost completely with @jat, but power loss is proportional to the cube of velocity (even though jat is correct that drag itself is proportional to the square). Per Wikipedia's Drag (physics) article:

"Note that the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome air drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW). With a doubling of speed the drag (force) quadruples per the formula. Exerting four times the force over a fixed distance produces four times as much work. At twice the speed the work (resulting in displacement over a fixed distance) is done twice as fast. Since power is the rate of doing work, four times the work done in half the time requires eight times the power."

jjb94941 | 6 octobre 2013

No question that all other things being equal, the quicker the vehicle is moving, the greater the drag and hence the energy wastage (proportional to the cube of the velocity). That was the reason that I carefully set up the two scenarios to achieve the same average speed for their runs: in order to eliminate the drag component in order to try to isolate the effect of acceleration.

thranx | 6 octobre 2013

If there is a pop quiz on this next week, I'm outta here.

Brian H | 6 octobre 2013

It went into the road, of course, which is what you pushed against to gain speed.

ye | 7 octobre 2013

Brian H, I can't tell whether you're joking. If so, I don't get it. If not, you're mistaken.

jat | 7 octobre 2013

@ye - actually, part of the lost energy does go into heating the road from friction, though it isn't clear it will be much different depending on how fast you get there.

ye | 8 octobre 2013

Any energy from the battery that doesn't increase the car's speed gets turned into heat. At higher speeds, most of that energy is used to overcome air resistance, thus heating the air. Some of the energy is used to deform the tires against internal friction within them, as they roll along the road, thus heating the tires. I suppose the road gets heated via contact with the hot tires, but I don't believe there is significant direct heating of the road by friction, because it hardly deforms at all as the car rolls over it.

To gain speed, the car pushes against the road, but the road gains no kinetic energy thereby. Brian H seemed to me to be saying that it does. I probably should have asked him to clarify what he meant, instead of simply saying that he was mistaken. Sorry, Brian H.

Brian H | 8 octobre 2013

The tires are pushing the planet on a vector tangent to the planar circle intersecting the core, oriented directly opposite to its acceleration. Laws of motion. Energy is being expended to alter the Earth's rotation.

aviationfw | 8 octobre 2013

Put simply the MS does not make instant torque convertion to tire rotation. it takes 5 to 6 seconds to get to 60 mph . When you apply full accelerator the motor has a large current applied to it. During the time this current is applied and the motor is trying to accelerate to a steady speed a lot of heat is generated in the motor windings the heat is proportional to the diiference between the speed requested and the actual speed at any instant in time. For example Try taking a small motor and apply electricity( kids toy motor) to it while stopping it turning using a vice. It will produce so much heat it will melt the windings and fail within seconds resulting in lots of smoke and heat etc. My point being that this MS motor is a loaded motor accelerating with a high load ( heavy car) and this produces a lot of heat losses. Take the small toy motor example again and accelerate it slowly and it will not get warm or burrn out. The same theory applies to the MS, lots of heat is generated by demanding a speed and the lag time for the motor to get to that speed . Accelerate slowly and gap between demanded speed and actual speed is much less resulting in less heat generated and less losses. Secondly we all know rapid acceleration produces more road friction wearing out our tires quickly. There is alot of heat energy lost here. Try acelerating from 0-60 with full acceleration and then stop feel your tires. Do the same again for slow acceleration on a different day starting with cold tires. There is a difference how much? Well think of it like this. We loss a few microns of rubber to the road evertime we accelerate quickly. Imagine if you took all 4 tires and tried manually sandpapering that amount of rubber off all four tires thats a lot of sanding of all 4 tire surfaces, I bet you would be tired after the task and would have expended a lot of heat energy in the process. Just food for thought.

ye | 8 octobre 2013

Well, now that you've clarified what you meant, I can definitely say you're mistaken. :-)

Let's get specific. Suppose a Model S accelerates from a standstill to 100 km/h. How much kinetic energy does the car gain? How much kinetic energy does the Earth gain?

(A Model S weighs 2100 kg. The Earth weighs 6,000,000,000,000,000,000,000,000 kg.)

triss1 | 8 octobre 2013


The amount of force is the same. For every action there is an equal and opposite reaction. The resultant velocity of the car and the Earth will, of course, be very different.

ye | 9 octobre 2013

triss1, I agree with you about the forces and velocities. What about energy?

BrassGuy | 9 octobre 2013

One point that seems to be overlooked - in order to result in the same average speed for both scenarios, scenario 2 will have to drive much farther maintaining highway speed. Depending on how "very gently", one could drive the entire distance of scenario 1 before reaching 60.

Geert.Snijders | 9 octobre 2013

Good example with the toy motor in a vice. What would happen in a 300KW Tesla motor if standing still, full brakes and full accelleration are applied at the same time...

Anyone tried this?! Is there an accelleration-by-wire protection for the possible overheating of the motor? Brake XOR Accellerate.

July10Models | 9 octobre 2013

The fact that we are even having this discussion is a testament to the great engineering achieved all around the MS. The average energy in the OP scenario is about the same. In the first scenario there is a spike of energy usage with the car quickly settling into reasonable energy usage. The second example, the energy usage is at a moderate level for a longer period of time. In both cases the area under the curve is the same. Which is why the city and highway rating of the MS is about the same.

July10Models | 9 octobre 2013

Yes Brianh if you point your car Eastward and stand on the accelerator, time will stand still.

redacted | 9 octobre 2013


I think momentum (mass x velocity, or mv) is conserved, so the earth's mv change will be the same as the car's. Kinetic energy is mv^2, so because most of the v change is in the car, it will acquire the lion's share of the kinetic energy.

I've never quite grokked the mv^2 thing, why it should be that it should take 4 times as much energy to double your speed. The regen curve on the car makes it pretty obvious that you take out a lot more energy at high speeds than at low speeds, but I don't get why

ye | 9 octobre 2013

redacted, you are correct.

The difference between momentum and kinetic energy is the difference between time and space. The amount of momentum you give to an object depends on how much time you spend pushing it. The amount of kinetic energy you give to an object depends on how far it moves as you push it.

Suppose that applying a certain force to a car for one second increases its speed from standstill to 10 miles per hour. Applying the same force for another second will further increase its speed by the same amount, from 10 to 20 miles per hour. But the car travels farther during this subsequent application of force than during the first one, because it's moving faster, and so it gains more kinetic energy. Which means that you have to supply more energy to it, because the energy has to come from somewhere.

During the first second, the car speeds up uniformly from 0 to 10 miles per hour, so its average speed during this second is 5 miles per hour. During the next second, it speeds up uniformly from 10 to 20 miles per hour, so its average speed during this second is 15 miles per hour, which is three times as much. So, the car travels three times as far during that second as it did during the first second. Therefore, it gains three times as much kinetic energy.

jjb94941 | 9 octobre 2013

As fascinating as the latest interchange has been, in both scenarios no kinetic energy is exchanged between the car and the earth since in both cases, the car begins and ends at a rest. Right?

So, based on the excellent engineering explanations and the excellent intuitive examples, it seems that the main reason for the additional energy consumption in 1 vs. 2, is that the transfer of energy from the battery to the wheels is less efficient in scenario 1 due to the fact that the battery generates more wasted heat in scenario 1.

mal42north | 9 octobre 2013

From the various panasonic data sheets floating around it seems that the internal resistance of the battery cell is about 0.06 ohm, and the maximum discharge current is about 4C or 12 A. So at max acceleration you dump about 8.6W of heat into the battery, while total power from the cell is about 48W (assuming 4V open circuit cell voltage). So the end result is you waste about 20% of your energy heating the battery at maximum acceleration.

riceuguy | 9 octobre 2013

On a side note (since I got a "C" in physics and this entire discussion reminds me why!), for those who haven't been wasting too much of their time on this forum for at least a year, please don't be fooled into taking Brian H seriously! Brian H serves two purposes on these forums: (1) He saves me from having to be the grammar and spelling police by playing that role with far greater frequency and humor than I could, and (2) to post very subtly mocking responses to posts (arguably to see if anyone gets his often subtle humor). I greatly appreciate both of these roles, but have to laugh when people think he's being serious (about role #2...he's quite serious about role #1!).

Brian, if you're reading this, I feel the forum members should all chip in and get you this shirt:

chrisdl | 9 octobre 2013

Absolutely correct.

Like many have said: heat loss is the main culprit.

I wrote up a calculation of the power available to each different Model S here:

If anybody wants to calculate the exact difference between slow vs fast acceleration based on this information, please feel free to do so. I'd be very interested to see how much more power is used when accelerating faster. It's probably easier to do a real life test and look at the consumption, though :-)

This is the part of my calculation which talks about the maximum power loss (heat dissipation in the battery alone, without counting other heat losses):

8) Power loss in battery cells (dissipated heat)
The internal resistance of a fresh battery cell is about 100 mOhm (0.1 Ohm)
=> S60: 750^2 x 0.1 = 56 kW
=> S85: (285 kW / 346 V )^2 x 0.1 = 68 kW
=> P85: 925^2 x 0.1 = 86 kW

9) Total energy usage in battery (power output + heat loss):
=> S60: 227 + 56 = 283 kW
=> S85: 285 + 68 = 353 kW
=> P85: 320 + 86 = 406 kW

Note that I took 100 mOhm as internal battery cell resistance versus the 60 mOhm of the Panasonic spec sheet. This is based on a battery forum where they measured the actual resistance of the cell. Also note that the internal resistance of the cells goes up (and so do the heat losses) as the battery ages.

And thank you jjb94941 for asking this very interesting question.

chrisdl | 10 octobre 2013

Alright, I did an attempt myself. Couldn't resist :-)

Energy loss calculation 0-60 mph

Assumptions for hard acceleration test:
1) Pedal is pressed to the metal until 60 mph is reached.
2) Power output stays at maximum until 60 mph is reached (and hence battery energy loss as well)
3) No other losses are taken into account.

Assumptions for gentle acceleration test:
1) Pedal is depressed slowly so that energy losses are neglectable (this will be quite accurate, since the power loss is the square of the current)

P85 (0-60 in 4.2 seconds at full power, pedal to the metal):
=> energy loss in battery = 86 kW * 4.2 s = 361.2 kJ = 100 Wh

S85 (0-60 in 54 seconds):
=> energy loss in battery = 68 kW * 5.2 s = 353.6 kJ = 98 Wh

S60 (0-60 in 5.9 seconds):
=> energy loss in battery = 56 kW * 5.9 s = 330.4 kJ = 92 Wh

Going flat out 0-60 mph uses less than 0.1 kWh of energy from your battery MORE than going slowly 0-60 mph.

In other words: have at it! (If you can afford the tires.)

chrisdl | 10 octobre 2013

It would be great if someone could do the test in real life to see how accurate the numbers are. Anybody? :)

triss1 | 10 octobre 2013


"triss1, I agree with you about the forces and velocities. What about energy?"

Force is measured in Newtons and energy in Joules. The two are closely related:

One Newton = kg*m/s^2 while One Joule = kg*m^2/s^2 so 1 Joule = 1 Newton * m (m here is one metre).

So the problem with calculating the energy is we would have to know how far the Tesla moved the Earth. Given the Earth's mass and the relatively small force, the distance will be vanishingly small, but not zero. It's easier to deal with force.

ye | 12 octobre 2013

Right, we need to know how far the Earth moves. But that's not so hard to calculate. The Earth moves less than the car in the same proportion as its mass is greater. Dividing the mass of a Model S (2100 kg) by the mass of the Earth (6 000 000 000 000 000 000 000 000 kg) yields a ratio of 0.000 000 000 000 000 000 000 35. That's how many joules of energy are spent moving the Earth for every joule spent moving the car.

Everyone can decide for himself whether that's small enough not to worry about. :-)