# No new petrol or diesel cars by 2040

Back into the battery. .... Yes - it is not a realistic model, but theoretically you accelerate a mass to x m/s, and then you accelerate it to 0 m/s. The mass is irrelevant.
I think the confusion has probably arisen because you appear to be addressing a totally different question from the one being discussed.

You appear to be talking about the overall 'energy balance' when a car accelerates (due to the electric motor) and then decelerates (due to regenerative braking) back to the initial (pre-acceleration) speed (be that zero or whatever). As I think you are implying, if the car were in a vacuum, there was no friction or other losses and everything was '100% efficient' (such that it would require no energy to keep the car travelling at constant speed) then, with regenerative braking, there would be no net usage of energy - the amount used to accelerate the vehicle would be entirely recovered during the deceleration - and that would, indeed, remain true regardless of the mass of the vehicle.

However, that's not what we are discussing. We are simply looking at the difference between having, and not having, regenerative braking when one reduces the vehicle's speed (by regenerative or conventional braking, respectively). DaveHerns simply asked how much energy would be recovered by regenerative braking if a cars speed was reduced from 30mph (presumably to zero), and how far the car could drive with that recovered energy. That is simply a question of converting the car's kinetic energy when travelling at 30mph into electrical energy, and that amount of energy obviously is dependent on (in fact, proportional to) the mass of the vehicle.

If the vehicle is heavier, such that more energy is recovered during braking from a certain speed, it is indeed true that more energy would have been expended in accelerating the car to its speed in the first place - but that is not what we have been discussing.
However: We do not think it strange, or against the laws of physics, if an EV can be fast charged in less time than it can be fast discharged, i.e. used. So I don't see why, per se, a car could not recover kinetic energy at a faster rate than it creates it.
No reason at all - but the rate of creating or recovering the kinetic energy is not something we have been discussing - we've only been discussing the total amount of energy recovered during regenerative braking, regardless of what time interval that recovery occurs over.

Kind Regards, John

And while (tangentially) talking abut where all this extra lecky will come from ...
Energy companies in the US have cancelled plans to build a pair of nuclear power plants – after Toshiba's Westinghouse Electric Company collapsed.
https://www.theregister.co.uk/2017/08/01/nuclear_power_plant_cancelled/

Doesn't bode well for Moorside (3.4GW) which (IIRC) was to significantly involve Toshiba.

Coming down boxhill top to bottom on regen all the way gains me back about 3 miles on the flat - you can see the battery gauge going up. Obviously I don't get back all the energy that was used to get me up there, but it's not negligible by any means.

I think the confusion has probably arisen because you appear to be addressing a totally different question from the one being discussed.
No, I wasn't - I was just pointing out that the mass of the object is irrelevant.

You appear to be talking about the overall 'energy balance' when a car accelerates (due to the electric motor) and then decelerates (due to regenerative braking) back to the initial (pre-acceleration) speed (be that zero or whatever). As I think you are implying, if the car were in a vacuum, there was no friction or other losses and everything was '100% efficient' (such that it would require no energy to keep the car travelling at constant speed) then, with regenerative braking, there would be no net usage of energy
Yes - as I said, it's not a realistic model.

the amount used to accelerate the vehicle would be entirely recovered during the deceleration - and that would, indeed, remain true regardless of the mass of the vehicle.
Which is what I said.

However, that's not what we are discussing. We are simply looking at the difference between having, and not having, regenerative braking when one reduces the vehicle's speed (by regenerative or conventional braking, respectively). DaveHerns simply asked how much energy would be recovered by regenerative braking if a cars speed was reduced from 30mph (presumably to zero), and how far the car could drive with that recovered energy. That is simply a question of converting the car's kinetic energy when travelling at 30mph into electrical energy, and that amount of energy obviously is dependent on (in fact, proportional to) the mass of the vehicle.
But so is the energy used in accelerating the vehicle.

If the vehicle is heavier, such that more energy is recovered during braking from a certain speed, it is indeed true that more energy would have been expended in accelerating the car to its speed in the first place - but that is not what we have been discussing.
Yes it is.

If you get back 50% via regenerative braking, you get back 50% no matter what the vehicle weighs.

No reason at all - but the rate of creating or recovering the kinetic energy is not something we have been discussing - we've only been discussing the total amount of energy recovered during regenerative braking, regardless of what time interval that recovery occurs over.
If we wish to maximise the recovery, then regenerative braking has to recover at a faster rate or the vehicle will have unacceptable braking performance.

The point is the recovered energy will be either extending the range, which is only mildly related to the vehicle mass, or saving charge at the wall, which is not dependent on the mass. The breaking energy is mostly proportional to the mass.
You seem to be assuming that regenerate breaking energy will be used to accelerate the car back to speed (which would be mostly proportional), but you would have set off the same that regardless of whether it had regenerative breaking, so it's not the saved energy.

No, I wasn't - I was just pointing out that the mass of the object is irrelevant.
It's irrelevant to what you were talking about, but crucially relevant to the question I was addressing/answering (how much energy {what absolute amount} is recovered by slowing a vehicle from 30mph to 0mph by regenerative braking).
If you get back 50% via regenerative braking, you get back 50% no matter what the vehicle weighs.
Indeed so, but neither I nor DaveHerns were talking about that percentage - we were talking about the absolute amount of energy recovered.
If we wish to maximise the recovery, then regenerative braking has to recover at a faster rate or the vehicle will have unacceptable braking performance.
I don't understand what you're saying - a "faster rate" than what?

Kind Regards, John

Coming down boxhill top to bottom on regen all the way gains me back about 3 miles on the flat
How long is that downhill stretch?

The point is the recovered energy will be either extending the range, which is only mildly related to the vehicle mass ....
Why do you say that? Are you perhaps assuming, like BAS appears to be, that the recovered energy will be used for accelerating the car, rather than for travelling at a constant speed?
... or saving charge at the wall, which is not dependent on the mass.
Why do you say that? The amount of energy put back into the battery (hence charging not required from the wall) will, for a given degree of braking (speed reduction), be proportional to the mass of the vehicle.
You seem to be assuming that regenerate breaking energy will be used to accelerate the car back to speed (which would be mostly proportional), but you would have set off the same that regardless of whether it had regenerative breaking, so it's not the saved energy.
Exactly. As I've said to him (even if he denies it), I think BAS is discussing a different question from the one which the rest of us are discussing!

Kind Regards, John

John D v2.0 said:
The point is the recovered energy will be either extending the range, which is only mildly related to the vehicle mass ....

=======
Why do you say that? Are you perhaps assuming, like BAS appears to be, that the recovered energy will be used for accelerating the car, rather than for travelling at a constant speed?
No I'm assuming neither, I'm assuming the extra range will be used for normal driving, which will include some accelerating, some going up and down hills, some cornering and some breaking. On average I deemed their use of energy to only be mildly related to the mass.
John D v2.0 said:
... or saving charge at the wall, which is not dependent on the mass.

=======
Why do you say that? The amount of energy put back into the battery (hence charging not required from the wall) will, for a given degree of braking (speed reduction), be proportional to the mass of the vehicle.
Charging any stationary battery I can't see being related to the mass of the attached vehicle! I'm talking about charging at the wall ie plugged in parked at the end of the journey.

No I'm assuming neither, I'm assuming the extra range will be used for normal driving, which will include some accelerating, some going up and down hills, some cornering and some breaking. On average I deemed their use of energy to only be mildly related to the mass.
Yes, but that's just 'normal driving', and I've never suggested that the amount of energy required for constant-speed driving is related to mass (acceleration obviously is). I have been talking about the amount of energy recovered during braking (and put back into the battery) - which IS dependent on (directly proportional to) the mass of the vehicle whose speed was being decreased by the braking
Charging any stationary battery I can't see being related to the mass of the attached vehicle! I'm talking about charging at the wall ie plugged in parked at the end of the journey.
Yes, I know what you're talking about. If, say 1 kWh has been recovered during braking and fed back into the battery, then (compared to the situation without regenerative braking) that's 1 kWh that you will not need to draw from the wall. If the car had had twice the mass, and had been braked (speed reduction) to the same extent, then there would be 2 kWh recovered that you would not need to draw from the wall.

Kind Regards, John

I think I agreed with you the whole time, I was just trying to point to BAS where he was differing.
He said mass is irrelevant (which is only true when the choice is between stop and start from a given speed as opposed to just carrying on at the original speed), however under the alternative assumption that the car would have stopped anyway:
braking energy recovered is proportional to mass
the energy to start the car again would be used with or without regenerative braking
that saved energy will actually be used to extend the range of the car or save on charging, neither of which are proportional to mass
therefore mass is relevant.

I think I agreed with you the whole time, I was just trying to point to BAS where he was differing.
Oh, fair enough - it just didn't seem like that!

Kind Regards, John

I was interested in the electric bike, I find with a standard push bike as you speed up so wind resistance increases, so a down hill stretch will only allow one to speed up to around 20 MPH and so when one gets to the flat the energy will only take one so far. If however one used regenerative braking then the wind resistance is reduced as your going slower, so the distance on the flat increases.

The big question is will the energy stored on the down hill runs be enough to remove the extra energy required to peddle up the hills due to extra weight of the bike. As we add batteries to the bike, so we also need to design the bike for the extra weight, spokes are thicker, we need suspension to stop damage to bike as it hits pot holes because of extra weight, so the whole unit becomes heavier not only the weight of motor and battery.

So this has to be considered with the car as well, the Citroen 2CV is very light, it would do 60 miles per gallon even with the old carburettor, it is the base vehicle, carries 4 people and cheap to run, OK UK we have Morris Minor not quite so good, but again 4 people from A to B with low vehicle weight, the vauxhall agila was another more resent vehicle, they were designed for narrow Japanese roads there is a range of these small vehicles.

If you look at the yaris hybrid it is so much larger than the original yaris, so hard to compare. Once one moves to the larger cars they are no longer cars to get up to 5 people from A to B.

Quite efficient energy recovery then. That's 2.6 miles of downhill coasting giving you 'about 3 miles' flat driving.

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