Why does accelerating fast use more energy than accelerating slowly?

[Rashomon]

This is consistent with ICE findings from BMW testing done some years back when they were evaluating optimal fuel efficiency in acceleration in connection with meeting fuel economy standards. I do not have the data they released anymore but i clearly recall the results: accelerating at about 2/3 throttle produced the best fuel efficiency. At the time teh test was made with both then-current production diesel and gasoline models, which astonishingly (to me, anyway) showed the identical 2/3 throttle for best acceleration efficiency. They also demonstrated another ICE point made here that the most efficient thermodynamic cruise is maximum torque. The latter point totally ignores aerodynamics, which are more important than are thermodynamics above, IIRC, 80 kph.

I never saw any BEV comparable data, but based on posts here and elsewhere, I suspect that battery temperature rise and absolute level is more of a factor than the actual acceleration rate in acceleration efficiency. However, overall BEV efficiency has all the passive losses, transmission/conversion losses as larger factors than pure acceleration rate.

Lacking any specific data, and not being an EE I have no technical qualification for my last point, the underlined part. Am I missing the point, or are BEV's inherently less sensitive to acceleration rate than are any ICE?

No, you are exactly right. ICE engines are most efficient unthrottled (i.e. full throttle), except most engines have their mixture go rich at full throttle for cooling, both to reduce the chance of detonation and to protect exhaust valves, etc. So the most efficient power is usually made at a throttle opening approaching full throttle, but not close enough to to get into the rich part of the fuel injection map. (This part of the map is almost never engaged in emissions tests, or it wouldn't be able to exist!) Best brake specific fuel consumption (BSFC) is almost always near peak torque rpm; go faster and power increases but so do friction losses. There can readily be a factor of 5 difference in efficiency between peak BSFC and BSFC at, say, 5 percent engine power. This is why best range with an ICE vehicle is usually obtained somewhere around 50 mph. That's where the improving efficiency curve of the engine meets the V-cubed nature of aerodynamic drag. Go faster than that in an ICE vehicle and range drops, but because efficiency is still improving with increasing load, it's nothing like the drop in range of an EV as it goes faster.

Because EVs have such a high and flat efficiency, acceleration doesn't matter much, but aerodynamics and speed matter tremendously. You're seeing the actual energy requirements of moving the car down the road, not a reflection of the engine efficiency map.

 

[Larry Hutchinson]

Probably been mentioned before but the main issue is resistive losses. The power to the motor gos as V*I while restive loss goes as R*I^2. It is that square term that causes losses to go up faster than power out.

 

[sorka]

I think you missed my point. You should use regen. If you're going some speed and decide to coast to a stop, you'll coast a certain distance in a certain amount of time. If at the point in time when you decided to coast, you instead regen down to the average of the coasting speed and maintain that speed, you'll come to the stopping point at the same place at the same time as if you had coasted. But because of the lower speed and cubic decrease in power you will use less energy than if you had coasted. You can then regen the remaining kinetic energy down to zero speed. So you are wasting energy if you just coast to a stop.

No, I understood your point and I disagree with it(in some cases). You're making the assumption that all drag is air resistance when in fact at moderate to lower speeds, most of the drag is rolling resistance which is linear.

You'd need to show the energy you capture with regen exceeds the energy you'd lose from for the speed x down to speed y that gets you to distance d in t time vs the the constant speed c that gets you to distance (d) in the same time (t).

What you say is true at a high enough speed. I can tell you for sure that once you're your looking at average speeds of 75 to 80 MPH, your assertion is true and that at average speeds of 50 MPH it isn't. There's a crossover at some but you'll need to do some math to figure it out.

To solve this you'd need to integrate the formula for drag (air + rolling) over the velocity difference over the same distance at the same time vs a constant speed that gets you the same distance in the same time.

The more practical approach would be to calculate this from a table generated from a coast down run using something like a vbox. I have two runs split over several speed ranges that I did when I was generating a correction curve to apply towards horsepower measurements using the vbox to subtract out air and rolling resistance so that power at say 80 MPH could be compared with power at 40 MPH without having drag bringing down the measured power as speed goes up.

I could go out this weekend and do a complete 100 to 0 coast down and give you the file. It would be split up over several runs as the long stretch of deserted flat highway that I live by isn't long enough for a full 100 to 0 in one shot.

The advantage of that approach is that you're using real data from an actual Model S where the standard drag formulas may not exactly apply. Even then, the cross over point will vary slightly depending on air density, tire pressure, tire type, etc but it should be pretty close.

 

[TIppy]

No, I understood your point and I disagree with it(in some cases). You're making the assumption that all drag is air resistance when in fact at moderate to lower speeds, most of the drag is rolling resistance which is linear.

You'd need to show the energy you capture with regen exceeds the energy you'd lose from for the speed x down to speed y that gets you to distance d in t time vs the the constant speed c that gets you to distance (d) in the same time (t).

What you say is true at a high enough speed. I can tell you for sure that once you're your looking at average speeds of 75 to 80 MPH, your assertion is true and that at average speeds of 50 MPH it isn't. There's a crossover at some but you'll need to do some math to figure it out.

To solve this you'd need to integrate the formula for drag (air + rolling) over the velocity difference over the same distance at the same time vs a constant speed that gets you the same distance in the same time.

The more practical approach would be to calculate this from a table generated from a coast down run using something like a vbox. I have two runs split over several speed ranges that I did when I was generating a correction curve to apply towards horsepower measurements using the vbox to subtract out air and rolling resistance so that power at say 80 MPH could be compared with power at 40 MPH without having drag bringing down the measured power as speed goes up.

I could go out this weekend and do a complete 100 to 0 coast down and give you the file. It would be split up over several runs as the long stretch of deserted flat highway that I live by isn't long enough for a full 100 to 0 in one shot.

The advantage of that approach is that you're using real data from an actual Model S where the standard drag formulas may not exactly apply. Even then, the cross over point will vary slightly depending on air density, tire pressure, tire type, etc but it should be pretty close.

I'd be happy to look at your coast down data, but you'll always be able to find a constant speed who's integrated power is less than the coast down integrated area. If you regen down to 2 mph immediately, maintain that speed and cover the same distance as coasting, you'll use some small amount of energy. If you regen down to 1 mph, it will take you twice as long to cover the same distance but you'll use 1/4 the energy. If you keep cutting the speed in half in the limit it will take you forever to get there, but you'll capture all the available kinetic energy. Even if 1/2 the speed results in 1/2 the energy, the end result is the same.
I really would like to see the data.

 

[sorka]

I'd be happy to look at your coast down data, but you'll always be able to find a constant speed who's integrated power is less than the coast down integrated area. If you regen down to 2 mph immediately, maintain that speed and cover the same distance as coasting, you'll use some small amount of energy. If you regen down to 1 mph, it will take you twice as long to cover the same distance but you'll use 1/4 the energy. If you keep cutting the speed in half in the limit it will take you forever to get there, but you'll capture all the available kinetic energy. Even if 1/2 the speed results in 1/2 the energy, the end result is the same.
I really would like to see the data.

Cool. I'll generate a file for you this weekend and PM you with a link.

 

[ohmman]

Cool. I'll generate a file for you this weekend and PM you with a link.

Why not keep it public? It's pertinent to the topic (and interesting to boot). It's also a nice change from the typical rumor mill or drama threads.

 

[Rashomon]

http://www.dimnp.unipi.it/guiggiani-m/Michelin_Tire_Rolling_Resistance.pdf This is a Michelin presentation nominally about tire rolling friction, but it's the best one source paper on loads on a vehicle/fuel economy I've seen. There's a nice ICE efficiency map, but note it's for a low-power engine (51 kW) and only goes down to about 10 percent load.

 

[sorka]

Why not keep it public? It's pertinent to the topic (and interesting to boot). It's also a nice change from the typical rumor mill or drama threads.

Sure. Can I upload the file here?

 

[TIppy]

Cool. I'll generate a file for you this weekend and PM you with a link.

Have you tried using neutral rather than modulating the accelerator? I'm not sure what neutral is. It could be zero torque, zero kw or inverter off. If it's the last one, that could save some energy. In any case it could reduce the noise on the data.

 

[sorka]

Have you tried using neutral rather than modulating the accelerator? I'm not sure what neutral is. It could be zero torque, zero kw or inverter off. If it's the last one, that could save some energy. In any case it could reduce the noise on the data.

Neutral is equivalent to perfectly modulating the pedal. 0kw in or out assuming AC and heat are off.

 

[TIppy]

Neutral is equivalent to perfectly modulating the pedal. 0kw in or out assuming AC and heat are off.

Does 0 kw mean you're doing some regen? It takes some power to keep the rotating stator field synchronous with the rotor rpm for zero torque. So for zero power some kinetic energy must be converted to supply the power to run the inverter and magnetize the motor. Right?

 

[sorka]

Does 0 kw mean you're doing some regen? It takes some power to keep the rotating stator field synchronous with the rotor rpm for zero torque. So for zero power some kinetic energy must be converted to supply the power to run the inverter and magnetize the motor. Right?

Not sure how much standby power is used to keep the inverter energized if any. I can confirm that the rate of deceleration is the same for a carefully modulated pedal and neutral.

Not sure why any power would be needed while coasting. Powertools shows 0 as long as you have the climate control off.

At a wedding now. Will get you data tomorrow.

 

[TIppy]

Not sure how much standby power is used to keep the inverter energized if any. I can confirm that the rate of deceleration is the same for a carefully modulated pedal and neutral.

Not sure why any power would be needed while coasting. Powertools shows 0 as long as you have the climate control off.

At a wedding now. Will get you data tomorrow.

In JB's description of coming to a stop with regen, he states that below a certain speed they stop regen because the energy to run the inverter is more than the remaining kinetic energy. This is why you have to use brakes to slow below the last few mph. The current required to maintain the magnetizing flux is what causes the efficiency of the motor to drop when the motor operates below 25% of full load. They could mitigate the losses by reducing the voltage to the motor essentially turning the inverter off.

 

[Stiction]

The regen fade-out has the handy side effect of getting you to put your foot on the brake at the traffic light......at least until the hold is engaged.

 

[TIppy]

I checked the power and torque on the can bus when coasting in neutral. The torque is zero and the power drawn from the battery at speed is 200 watts more than it is when stopped. The inverter is basically off.

 

[sorka]

In JB's description of coming to a stop with regen, he states that below a certain speed they stop regen because the energy to run the inverter is more than the remaining kinetic energy. This is why you have to use brakes to slow below the last few mph. The current required to maintain the magnetizing flux is what causes the efficiency of the motor to drop when the motor operates below 25% of full load. They could mitigate the losses by reducing the voltage to the motor essentially turning the inverter off.

Yes, I know this. I thought you were asking how much energy is needed for the inverter when you you're coasting or in neutral which I don't have an answer for but can't imagine that you'd need any energy at all while coasting.

BTW, I'll have to try again tomorrow. I'm in California's central valley and the wind is nuts today because of the storm.

 

[sorka]

I checked the power and torque on the can bus when coasting in neutral. The torque is zero and the power drawn from the battery at speed is 200 watts more than it is when stopped. The inverter is basically off.

Cool. That's what I would suspect. I'm guessing the 200 watts is for the display and other on board systems while the ms is powered up. While powered down, if you can call it that, it draws something like 70 watts.

 

[TIppy]

Cool. That's what I would suspect. I'm guessing the 200 watts is for the display and other on board systems while the ms is powered up. While powered down, if you can call it that, it draws something like 70 watts.

260 watts everything powered up but not moving. 460 watts while coasting in neutral.

 

[sorka]

260 watts everything powered up but not moving. 460 watts while coasting in neutral.

So do you think the 200 watts is keeping the inverter energized while coasting in neutral? Could it be that the this is required to bootstrap the AC induction to eliminate delay in when activating regen?

Does that extra 200 watts continue as you coast to 0 in neutral?

 

[TIppy]

So do you think the 200 watts is keeping the inverter energized while coasting in neutral? Could it be that the this is required to bootstrap the AC induction to eliminate delay in when activating regen?

Does that extra 200 watts continue as you coast to 0 in neutral?

I think they keep the stator magnetic field rotating at the same rpm as the rotor. This means there is no slip and therefor no torque.

At rated loads the voltage to the motor is increased as the rpm increases to maintain a constant ratio of volts to hertz. If at a fixed rpm you decrease the load from rated down to zero the rotating stator field is still consuming significant power. But as the load decreases you don't need the full voltage that the constant volts/hz would dictate. By lowering the voltage the power required to maintain the rotating stator field is decreased. At zero torque the voltage to the motor is probably close to zero, but the inverter is still maintaining the rotating magnetic field at the same rpm as the rotor prepared to either make or take torque as required. When the car comes to a stop the inverter probably shuts down completely.

I was in traffic so I couldn't coast to a stop. I'll try that today.