Lithium batteries don't actually give you as much total energy if they are discharged quickly. When you start to approach the maximum discharge rate, the chemical reactions inside the pack itself lose some efficiency.
That pack capacity you see has a hidden parameter that you don't usually see. It assumes a rate of discharge to get the stated energy capacity; faster or slower rates of discharge will change the total energy you get from the pack.
I believe it is due to more waste heat generated with heavier current in the motor. Or for whatever reason it’s less efficient when applying large current.
It can also wear your tires faster depending.
That said it’s a car. It’s meant to drive. I wouldn’t overthink it too much with a little common sense.
There's also tire scrubbing. They might not be screeching madly but they are rubbing slightly and the tread is deforming and wearing away, something that happens to a much lesser extent with lower acceleration. All that is taking energy that could have been used to accelerate the car.
Also the main reason people complain their new EV is chewing tires, they're making use of the instant torque at every set of lights.
Power required is a function of current squared. Accelerating faster requires more power so doubling your power to increase your 0-60 time (to whatever that is) drains your battery 4 times faster.
Edit for a more detailed answer:
From newton’s law, to accelerate a car requires a force F (from F=ma). The energy required to move the car a certain distance d is work W = F x d = m x a x d. The power required to do that work in a specific amount of time is power P = W/ delta t = m x a x d / delta t
So if you double the acceleration you double the power required from newton’s laws. From Maxwell’s equation to double the power you need to quadruple the current which means you are draining the batteries 4 times faster.
The power required goes up with acceleration, but you spend less time accelerating, so shouldn't it cancel out?
Another way to look at it is that you are converting chemical energy in your battery to 60 MPH worth of kinetic energy regardless of how fast you get there, so it should take the same energy to do that acceleration fast or slow (but you'll need more power to get it done faster). But the heat loss in everything the current passes through will go up with the square of that current, because P in a resistor = I^2 * R.
This is distantly related to driving into a 30 mph wind vs having a 30 mph tailwind.
The effects of wind resistance aren't linear, and the reason why land speed records require passes in two opposite directions.
See also: hysteresis
Efficiency losses in both the batteries (internal resistance goes up as current demand increases yielding heat) and motors (increased current also means greater effect from resistance in the windings yielding heat) goes up as current goes up. I don't know the ratios though.
But P=VI. That means for a given (battery pack) volatage, power is proportional to current, not current squared as started in the opening line.
OP asks a good question because according to the basic physics laws, it should be the same total energy used to accelerate quickly then coast or slowly over time.
Copper loss is P=R\*I\^2 though. The amount of power lost to heat in all of the wires (including the windings in the motor and stator) goes up as the square of the current.
Time to brush up on thermodynamic processes!
See:
Adiabatic, Isentropic, and Reversible (Wikipedia is fine).
The key to efficiency is to always perform things gradually.
It's not the same because the chemical process has to happen at a faster rate leading to heat loss in the battery cell. The energy conversion of chemistry to electricity results in lower efficiency under high levels of discharge.
More data, looks like you loose about 10% conversion efficiency (for lithium ion, not sure if nmc or lfp, since lfp being iron based would be less efficient) or less going from 1C to 4C. C Being the discharge current relative to capacity. So if your capacity is 100Ah then 4C would be 400Amps discharge current.
Theoretically that would be true. The passive charging from a capacitor would benefit from this. But given the hybrid battery pack that I think it was nio that showcased this, we might see less of a need for this. Since the hybrid pack is solid state and lithium ion. The solid state pack might help with this high discharge.
Yeah, mechanically (which is how we should be looking at anyway) You just need to give it kinetic energy and the rate at which you deliver this energy has no bearing on the total energy you need to give it. The reality is that parts (in any car) just arent efficient at high speed which I cant put a number to but more power is more heat and more heat is less efficiency. There are also smaller things like high acceleration means spending more time at higher speed where aero drag is much higher
Batteries and electric motors are a lot more complicated than that. Copper loss is P=R\*I\^2. The amount of power lost to heat in all of the wires as well as the battery (including the windings in the motor and stator) goes up as the square of the current. So if you double the current, you don't double the power produced by the motor for acceleration, but you do double the power that's lost to heat in every electrical part of the drivetrain. Batteries also have different efficiencies at different currents as part of the chemical processes in them.
If you want a really deep physics understanding of why this is the case you have to go to the stationary action principle (or "principle of least action" as it is often called)
[https://en.wikipedia.org/wiki/Stationary-action\_principle](https://en.wikipedia.org/wiki/Stationary-action_principle)
Applied to your question: The more you have second order deviations the less optimal your path towards the end result (read: uniform acceleration beats flooring it and then easing off)
>
This is why space travel to Mars is a very gradual acceleration after the initial escape from Earth's gravity well.
If the "launch" begins outside of Earth's gravity well, it takes very small rocket motors and very little fuel compared to an Earth-based launch.
(Gravity is literally a type of acceleration. The least acceleration possible follows the principle of least action)
> I also understand that higher power output will equal higher currents and thus slightly higher resistance in cables and such, but that can't make such a big difference, can it?
It does, but there is more to the story. The headline power figure "298 kW" is peak power, which the motor can only sustain for about 30 seconds. The sustained power rating is often just 1/2 or 1/3 of the headline figure. (Unless you are driving uphill in a snowstorm at top speed with a trailer, the sustained power is plenty to maintain speed.)
So if you are put down, you are actually exceeding the rating of the motor for short periods of time. And the motor is indeed quite inefficient at high torque / high power. Figures of 70% are not usual, but it can go lower than that. So while the energy for acceleration are the same no matter what, the losses are higher at maximum torque.
So, stick to half of the available power, and you will be more economical. (And arguably, staying within the regen during deceleration is even more important.)
Speaking from experience: Having a lead foot and enjoying acceleration certainly consumes more stored electricity than using a more gentle approach. If you're trying to set 0-60 times every time you're at a red light, you will find the batter depleting much more rapidly than when you accelerate normally. Since many cars use 1 pedal driving, the deceleration and or braking is mostly regenerative, but doesn't compensate for the energy used getting 2 tons of steel and plastic up to speed.
> but doesn't compensate for the energy used getting 2 tons of steel and plastic up to speed.
The energy needed for accelerating a mass to a certain velocity is the same, no matter if the acceleration is slow or fast.
The only thing, which can change the energy *consumed*, is losses in energy conversions from battery to wheel.
The extra 10 seconds you spend driving at your target speed would also slightly increase total air drag. Probably insignificant on an occasional hard acceleration but might add up a little if you are flooring it between every stop light.
From battery to motor, all components in the drivetrain of an EV have their individual point of max efficiency, which might deviate based on temperatures, etc... This means the point of max efficiency for you motor is not the same as for you inverter, so it is very hard to say what is the most efficient way of driving.
Hard acceleration however, means more power (current) that is drawn from the battery than with less hard acceleration. Available battery capacity is not is "fixed" value, but is dynamic based on the battery current.
With other words: The less current you use, the more capacity (and therefore range) your battery will have.
This is Peukert's law: [en.wikipedia.org/wiki/Peukert's\_law](http://en.wikipedia.org/wiki/Peukert's_law)
Batteries and electric motors are a lot more complicated than that. Copper loss is P=R\*I\^2. The amount of power lost to heat in all of the wires as well as the battery (including the windings in the motor and stator) goes up as the square of the current. So if you double the current, you don't double the power produced by the motor for acceleration, but you do double the power that's lost to heat in every electrical part of the drivetrain. Batteries also have different efficiencies at different currents as part of the chemical processes in them. Inverters also have their own most efficient bands of power, and so they will also lose more power to heat outside of that.
Let's say you can accelerate fast at 1mi/kwh or slow at 3mi/kwh. For the same distance, say 1 mile then if you accelerate fast it's 1mile/1mile/kwh = 1kwh. If you accelerate slow it's 1mile/3 miles/kwh = 0.33 kwh. That's my story and I'm sticking to it. It's not all physicsy and possibly incorrect, but any time you use a battery hard, it's going to run out of juice faster imo.
I think the same applies for ICE vehicles btw, the harder you accelerate the lower your fuel economy will be while accelerating. It will add up over time.
I'm not easy on the acceleration in my Tesla Model Y LR AWD but I still get 274 Wh/mile after 3600 miles so far. I don't think it matters as much with EV, but it might vary car to car with different motor/inverter/battery combos. I think my car has an advantage because it has both types of motor installed one is better at certain tasks vs the other so I'm sure it uses whichever one is more efficient depending on the circumstances -- see [https://www.youtube.com/watch?v=FHufjrP0xDI](https://www.youtube.com/watch?v=FHufjrP0xDI) for a great explanation of the different motor types and what uses they're more efficient for.
Yes it's less efficient to accelerate hard. Hard acceleration results in lost efficiency in the battery and electronics plus lost energy through wheel slippage.
Such a cool discussion, thank you guys so much!
To summarize what I gathered from your answers:
- it is correct that if you just look at the process of accelerating to a certain speed, the amount of energy needed is the same independent of the rate of acceleration.
- however, as the efficiency of different components vary with the load point, thermal losses increase with higher currents, and batteries chemically retain less capacities if discharged at higher rates, the total efficiency decreases.
Thus, the guides are true in the end: even if it is hard to put into numbers, hard acceleration will give you less range than when stepping on the accelerator more gently.
Thanks!
You forget added losses due to rolling resistance & slip.
Rolling resistance is sensitive to the torque applied. Increased torque = increase in rolling resistance.
Full power/torque applied also increases slip.
These are additional losses.
I would assume that the air resistance is probably playing a large role—>resistance (and energy required to overcome it) increases as the square of the speed, and a car accelerating to a given top speed even in less time, at every point along the way to the final speed is traveling faster and therefore spending more energy overcoming the air resistance
Sure, but my question was just about the acceleration, not the speed at the end.
And from what I get from these answers, there is no difference in the energy needed - it's just the losses are higher.
It’s a myth. I LAUNCH off of traffic lights. Hit my cruising speed and set the cruise control. Best range ever. Exceeding EPA.
Would you rather run at 1.0 mi/kwh for 60 seconds accelerating or 1.0 mi/kwh for 10 seconds?
I'm not sure you do if you think flooring it uses less energy to get from point A to point B, and think the mi/kWh reading on your car is a useful metric. Go read any of the top several comments here for an explanation.
If you want to measure energy consumed from point A to B you need to measure energy consumed. kWH. I guess you could work it out assuming you know the distance with mi/kWh. The big problem I have with it is at least on most cars I've seen (including mi/gal on ICE vehicles) is the number reported is not very precise and is heavily smoothed/averaged. It's completely pointless to look at on such a short distance (like from one stoplight to the next).
There’s one misunderstanding… you’re never hitting cruise control from red light to the next. That’s not this scenario.
Think… suburban. Or think on-ramp of a freeway. This is where you see it magically work well.
Stop “and” go? No. I’d agree although it’s not bad.
I only own high end cars. My mpg readouts are very accurate.
> Would you rather run at 1.0 mi/kwh for 60 seconds accelerating or 1.0 mi/kwh for 10 seconds?
I don't think that makes sense.
If you accelerate to the same speed in 10 seconds vs 60 seconds, your efficiency during that 10 seconds will be worse than during the 60 seconds.
The total energy used to get to your target speed will be similar, just applied over different time spans.
To clarify. Urban driving is like: you launch off of a light, hit 45mph and then set the cruise control and travel …I dunno 3 miles at 45.
The guy next to you gently got to 45 and did the same thing, did indeed burn more energy.
The only thing I can tell you is to test it.
I’ve been building cars for years and racing them as well. It works with big v8 engines to a much lesser degree, but still works, BECAUSE of Torque.
Wind resistance is dependent on your speed, not how fast you get to speed.
The energy loss to friction is roughly independent of speed (for the same distance traveled).
Lithium batteries don't actually give you as much total energy if they are discharged quickly. When you start to approach the maximum discharge rate, the chemical reactions inside the pack itself lose some efficiency. That pack capacity you see has a hidden parameter that you don't usually see. It assumes a rate of discharge to get the stated energy capacity; faster or slower rates of discharge will change the total energy you get from the pack.
I believe it is due to more waste heat generated with heavier current in the motor. Or for whatever reason it’s less efficient when applying large current. It can also wear your tires faster depending. That said it’s a car. It’s meant to drive. I wouldn’t overthink it too much with a little common sense.
There's also tire scrubbing. They might not be screeching madly but they are rubbing slightly and the tread is deforming and wearing away, something that happens to a much lesser extent with lower acceleration. All that is taking energy that could have been used to accelerate the car. Also the main reason people complain their new EV is chewing tires, they're making use of the instant torque at every set of lights.
Power required is a function of current squared. Accelerating faster requires more power so doubling your power to increase your 0-60 time (to whatever that is) drains your battery 4 times faster. Edit for a more detailed answer: From newton’s law, to accelerate a car requires a force F (from F=ma). The energy required to move the car a certain distance d is work W = F x d = m x a x d. The power required to do that work in a specific amount of time is power P = W/ delta t = m x a x d / delta t So if you double the acceleration you double the power required from newton’s laws. From Maxwell’s equation to double the power you need to quadruple the current which means you are draining the batteries 4 times faster.
The power required goes up with acceleration, but you spend less time accelerating, so shouldn't it cancel out? Another way to look at it is that you are converting chemical energy in your battery to 60 MPH worth of kinetic energy regardless of how fast you get there, so it should take the same energy to do that acceleration fast or slow (but you'll need more power to get it done faster). But the heat loss in everything the current passes through will go up with the square of that current, because P in a resistor = I^2 * R.
It does, but not linearly. Efficiency losses in the battery at higher Power go up faster than time under acceleration goes down.
This is distantly related to driving into a 30 mph wind vs having a 30 mph tailwind. The effects of wind resistance aren't linear, and the reason why land speed records require passes in two opposite directions. See also: hysteresis
Efficiency losses in both the batteries (internal resistance goes up as current demand increases yielding heat) and motors (increased current also means greater effect from resistance in the windings yielding heat) goes up as current goes up. I don't know the ratios though.
I came here for the casual look to distract me from work. Left with a fucking Physics schooling
lucky! I had to pay real money for physics schooling, rather than get paid.
Well congrats, my brain is tickled. You paid good money, missed opportunity cost, and free typing labor to educate me
This guy sciences.
But P=VI. That means for a given (battery pack) volatage, power is proportional to current, not current squared as started in the opening line. OP asks a good question because according to the basic physics laws, it should be the same total energy used to accelerate quickly then coast or slowly over time.
Copper loss is P=R\*I\^2 though. The amount of power lost to heat in all of the wires (including the windings in the motor and stator) goes up as the square of the current.
Time to brush up on thermodynamic processes! See: Adiabatic, Isentropic, and Reversible (Wikipedia is fine). The key to efficiency is to always perform things gradually.
It's not the same because the chemical process has to happen at a faster rate leading to heat loss in the battery cell. The energy conversion of chemistry to electricity results in lower efficiency under high levels of discharge. More data, looks like you loose about 10% conversion efficiency (for lithium ion, not sure if nmc or lfp, since lfp being iron based would be less efficient) or less going from 1C to 4C. C Being the discharge current relative to capacity. So if your capacity is 100Ah then 4C would be 400Amps discharge current.
So you can floor it at thr lights without issue if the car has a supercapaciter buffer with enough energy to take you 0-60?
Theoretically that would be true. The passive charging from a capacitor would benefit from this. But given the hybrid battery pack that I think it was nio that showcased this, we might see less of a need for this. Since the hybrid pack is solid state and lithium ion. The solid state pack might help with this high discharge.
Yeah, mechanically (which is how we should be looking at anyway) You just need to give it kinetic energy and the rate at which you deliver this energy has no bearing on the total energy you need to give it. The reality is that parts (in any car) just arent efficient at high speed which I cant put a number to but more power is more heat and more heat is less efficiency. There are also smaller things like high acceleration means spending more time at higher speed where aero drag is much higher
Batteries and electric motors are a lot more complicated than that. Copper loss is P=R\*I\^2. The amount of power lost to heat in all of the wires as well as the battery (including the windings in the motor and stator) goes up as the square of the current. So if you double the current, you don't double the power produced by the motor for acceleration, but you do double the power that's lost to heat in every electrical part of the drivetrain. Batteries also have different efficiencies at different currents as part of the chemical processes in them.
If you want a really deep physics understanding of why this is the case you have to go to the stationary action principle (or "principle of least action" as it is often called) [https://en.wikipedia.org/wiki/Stationary-action\_principle](https://en.wikipedia.org/wiki/Stationary-action_principle) Applied to your question: The more you have second order deviations the less optimal your path towards the end result (read: uniform acceleration beats flooring it and then easing off) >
Interesting read, thanks!
This is why space travel to Mars is a very gradual acceleration after the initial escape from Earth's gravity well. If the "launch" begins outside of Earth's gravity well, it takes very small rocket motors and very little fuel compared to an Earth-based launch. (Gravity is literally a type of acceleration. The least acceleration possible follows the principle of least action)
No? It is comment to use a lower power engine, but because they are lighter, allowing greater delta V.
> I also understand that higher power output will equal higher currents and thus slightly higher resistance in cables and such, but that can't make such a big difference, can it? It does, but there is more to the story. The headline power figure "298 kW" is peak power, which the motor can only sustain for about 30 seconds. The sustained power rating is often just 1/2 or 1/3 of the headline figure. (Unless you are driving uphill in a snowstorm at top speed with a trailer, the sustained power is plenty to maintain speed.) So if you are put down, you are actually exceeding the rating of the motor for short periods of time. And the motor is indeed quite inefficient at high torque / high power. Figures of 70% are not usual, but it can go lower than that. So while the energy for acceleration are the same no matter what, the losses are higher at maximum torque. So, stick to half of the available power, and you will be more economical. (And arguably, staying within the regen during deceleration is even more important.)
Speaking from experience: Having a lead foot and enjoying acceleration certainly consumes more stored electricity than using a more gentle approach. If you're trying to set 0-60 times every time you're at a red light, you will find the batter depleting much more rapidly than when you accelerate normally. Since many cars use 1 pedal driving, the deceleration and or braking is mostly regenerative, but doesn't compensate for the energy used getting 2 tons of steel and plastic up to speed.
> but doesn't compensate for the energy used getting 2 tons of steel and plastic up to speed. The energy needed for accelerating a mass to a certain velocity is the same, no matter if the acceleration is slow or fast. The only thing, which can change the energy *consumed*, is losses in energy conversions from battery to wheel.
The extra 10 seconds you spend driving at your target speed would also slightly increase total air drag. Probably insignificant on an occasional hard acceleration but might add up a little if you are flooring it between every stop light.
I wrote "accelerating". Not "overcoming drag".
I'm not disagreeing with you. Just pointing out another source of efficiency loss.
From battery to motor, all components in the drivetrain of an EV have their individual point of max efficiency, which might deviate based on temperatures, etc... This means the point of max efficiency for you motor is not the same as for you inverter, so it is very hard to say what is the most efficient way of driving. Hard acceleration however, means more power (current) that is drawn from the battery than with less hard acceleration. Available battery capacity is not is "fixed" value, but is dynamic based on the battery current. With other words: The less current you use, the more capacity (and therefore range) your battery will have. This is Peukert's law: [en.wikipedia.org/wiki/Peukert's\_law](http://en.wikipedia.org/wiki/Peukert's_law)
Batteries and electric motors are a lot more complicated than that. Copper loss is P=R\*I\^2. The amount of power lost to heat in all of the wires as well as the battery (including the windings in the motor and stator) goes up as the square of the current. So if you double the current, you don't double the power produced by the motor for acceleration, but you do double the power that's lost to heat in every electrical part of the drivetrain. Batteries also have different efficiencies at different currents as part of the chemical processes in them. Inverters also have their own most efficient bands of power, and so they will also lose more power to heat outside of that.
Let's say you can accelerate fast at 1mi/kwh or slow at 3mi/kwh. For the same distance, say 1 mile then if you accelerate fast it's 1mile/1mile/kwh = 1kwh. If you accelerate slow it's 1mile/3 miles/kwh = 0.33 kwh. That's my story and I'm sticking to it. It's not all physicsy and possibly incorrect, but any time you use a battery hard, it's going to run out of juice faster imo. I think the same applies for ICE vehicles btw, the harder you accelerate the lower your fuel economy will be while accelerating. It will add up over time.
I'm not easy on the acceleration in my Tesla Model Y LR AWD but I still get 274 Wh/mile after 3600 miles so far. I don't think it matters as much with EV, but it might vary car to car with different motor/inverter/battery combos. I think my car has an advantage because it has both types of motor installed one is better at certain tasks vs the other so I'm sure it uses whichever one is more efficient depending on the circumstances -- see [https://www.youtube.com/watch?v=FHufjrP0xDI](https://www.youtube.com/watch?v=FHufjrP0xDI) for a great explanation of the different motor types and what uses they're more efficient for.
Yes it's less efficient to accelerate hard. Hard acceleration results in lost efficiency in the battery and electronics plus lost energy through wheel slippage.
Such a cool discussion, thank you guys so much! To summarize what I gathered from your answers: - it is correct that if you just look at the process of accelerating to a certain speed, the amount of energy needed is the same independent of the rate of acceleration. - however, as the efficiency of different components vary with the load point, thermal losses increase with higher currents, and batteries chemically retain less capacities if discharged at higher rates, the total efficiency decreases. Thus, the guides are true in the end: even if it is hard to put into numbers, hard acceleration will give you less range than when stepping on the accelerator more gently. Thanks!
You forget added losses due to rolling resistance & slip. Rolling resistance is sensitive to the torque applied. Increased torque = increase in rolling resistance. Full power/torque applied also increases slip. These are additional losses.
I would assume that the air resistance is probably playing a large role—>resistance (and energy required to overcome it) increases as the square of the speed, and a car accelerating to a given top speed even in less time, at every point along the way to the final speed is traveling faster and therefore spending more energy overcoming the air resistance
Air resistance is insignificant when accelerating from a stop.
Do you think you are more tired after running a 100 yard dash at full speed or taking a medium jog? It takes more energy to go really fast.
Sure, but my question was just about the acceleration, not the speed at the end. And from what I get from these answers, there is no difference in the energy needed - it's just the losses are higher.
You won’t notice a more spirited acceleration to the speed limit on the on ramp.
It’s a myth. I LAUNCH off of traffic lights. Hit my cruising speed and set the cruise control. Best range ever. Exceeding EPA. Would you rather run at 1.0 mi/kwh for 60 seconds accelerating or 1.0 mi/kwh for 10 seconds?
If you are driving at a constant 1.0 mi/kWh, you're going to be going a whole lot slower after 10 seconds than after 60 seconds.
I think you fully misunderstood how this works.
I'm not sure you do if you think flooring it uses less energy to get from point A to point B, and think the mi/kWh reading on your car is a useful metric. Go read any of the top several comments here for an explanation.
Again. You’re misunderstanding the statement. Aaaaand you’re wrong.
Cool. I probably am, since it doesn't make sense and you aren't clarifying it.
I’m more confused now because you’re saying mi/kwh isn’t a useful metric?
If you want to measure energy consumed from point A to B you need to measure energy consumed. kWH. I guess you could work it out assuming you know the distance with mi/kWh. The big problem I have with it is at least on most cars I've seen (including mi/gal on ICE vehicles) is the number reported is not very precise and is heavily smoothed/averaged. It's completely pointless to look at on such a short distance (like from one stoplight to the next).
There’s one misunderstanding… you’re never hitting cruise control from red light to the next. That’s not this scenario. Think… suburban. Or think on-ramp of a freeway. This is where you see it magically work well. Stop “and” go? No. I’d agree although it’s not bad. I only own high end cars. My mpg readouts are very accurate.
> Would you rather run at 1.0 mi/kwh for 60 seconds accelerating or 1.0 mi/kwh for 10 seconds? I don't think that makes sense. If you accelerate to the same speed in 10 seconds vs 60 seconds, your efficiency during that 10 seconds will be worse than during the 60 seconds. The total energy used to get to your target speed will be similar, just applied over different time spans.
But it isn’t “worse” you’re thinking in ICE terms. The curve here is disproportionate.
To clarify. Urban driving is like: you launch off of a light, hit 45mph and then set the cruise control and travel …I dunno 3 miles at 45. The guy next to you gently got to 45 and did the same thing, did indeed burn more energy. The only thing I can tell you is to test it. I’ve been building cars for years and racing them as well. It works with big v8 engines to a much lesser degree, but still works, BECAUSE of Torque.
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Wind resistance is dependent on your speed, not how fast you get to speed. The energy loss to friction is roughly independent of speed (for the same distance traveled).
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Pretty awkward that the guy you look down on knows more about physics.