High acceleration reduces range. but Why?

[RegGuheert]

Presumably there's also no reason to go abnormally slow when climbing a large hill (unless remaining charge is SO low that every aerodynamic advantage needs to be harnessed), correct?

Basically, the capacity of a battery is lower at high discharge rates than at low ones, so you will not go as far if you run the car at high power levels. In addition, there are higher aerodynamic losses, as you mentioned.

I think Tony has determined through testing that the optimum speed for maximum range on level ground is 13 MPH. Driving any faster or slower than this will result in reduced total range.

 

[gbshaun]

Presumably there's also no reason to go abnormally slow when climbing a large hill (unless remaining charge is SO low that every aerodynamic advantage needs to be harnessed), correct?

Basically, the capacity of a battery is lower at high discharge rates than at low ones, so you will not go as far if you run the car at high power levels. In addition, there are higher aerodynamic losses, as you mentioned.

I think Tony has determined through testing that the optimum speed for maximum range on level ground is 13 MPH. Driving any faster or slower than this will result in reduced total range.

Thanks
I think we have all discovered that reducing freeway speed by 10mph makes a HUGE difference (65mph c/w 75mph), but my question was more about climbing a long drag at say 30 rather than 40. We're well down the curve of aerodynamic drag, and I'm not expecting even 40mph to put the car in an inefficient zone for rpm,torque, or at an excessive battery discharge rate.

Shaun

 

[GregH]

Presumably there's also no reason to go abnormally slow when climbing a large hill (unless remaining charge is SO low that every aerodynamic advantage needs to be harnessed), correct?

Basically, the capacity of a battery is lower at high discharge rates than at low ones, so you will not go as far if you run the car at high power levels. In addition, there are higher aerodynamic losses, as you mentioned.

I think Tony has determined through testing that the optimum speed for maximum range on level ground is 13 MPH. Driving any faster or slower than this will result in reduced total range.

If you accelerate faster (draw more current) the IR losses on the battery mean you're pulling out those same Ah at a lower voltage thus less power (netting you fewer kWh to the wheels). Not that this knowledge slows me down at stoplights Personally I take some uphills a bit slower not so much to improve my efficiency but reduce the amount of battery heating (an additional unwanted feature of IR losses in the battery).

 

[edatoakrun]

...
I think we have all discovered that reducing freeway speed by 10mph makes a HUGE difference (65mph c/w 75mph), but my question was more about climbing a long drag at say 30 rather than 40. We're well down the curve of aerodynamic drag, and I'm not expecting even 40mph to put the car in an inefficient zone for rpm,torque, or at an excessive battery discharge rate.

Shaun

In general, I'd suggest you use whatever kW is required to drive at your optimum speed. I use the 30-40 kW range every day I drive home, (and somewhat lower regen rates when I descend from my home).

The exception (other than general the battery heat/resistance factor mentioned) might be near the bottom of the charge, where some have suggested rapid discharge rates may particularly accelerate battery aging. I have (arbitrarily) tried to both minimize and limit my discharge rates to ~ 20-30 kW past the LBW, and ~10-15 kW past the VLBW.

As others have noted, this threads title: High acceleration reduces range, is incorrect.

Furthermore, IMO, the most efficient travel, over a given distance over a given period of time, is generally by maintaining the most constant speed as road conditions permit, and by varying the kW added and recovered by regen as necessary.

Over a fixed travel time, the loss of energy to regen inefficiency is generally less than the loss of energy to atmospheric drag caused by driving outside of a very small speed range.

So unless you are on a constant grade where you can maintain (within a few mph of?) the optimum (constant) speed by coasting in N or ECO, apply or withdraw kW as necessary to maintain optimum (near-constant) speed.

I suspect the optimum efficiency speed differential for descending (faster, to avoid regen) over ascending (slower) on the same grade is quite small. In fact, I often drive more like five mph faster in descent than in ascents, but this is largely due to the prevailing speeds maintained by ICEVs, whose efficiency is often challenged by steep high-speed ascents, and which (of course) can recover no energy (beyond kinetic) in descent, at any speed.

 

[Sven]

There is a similar thread in the Tesla forum,
http://www.teslamotors.com/forum/fo...iciency-different-power-outputs-accelerations

It is using the following article as a reference regarding internal battery resistance,
http://www.evs24.org/wevajournal/php/download.php?f=vol3/WEVJ3-5340444.pdf

The last comment in the thread wraps it up nicely,

cliff@hannelcon... | April 18, 2013

OK, let me try this:

(1) resistive losses are proportional to the square of the current [ri^2]
(2) torque in electrical motors is proportional to current
(3) acceleration is proportional to torque (and thus current)
(4) distance covered to achieve a certain speed is proportional to acceleration][.5*a*t^2)

From (1)-(3), losses are proportional to the square of acceleration. So, when you double acceleration, you have 4x the losses but from (4) you've covered only 2x the distance, so you have 2x the losses per distance, thus reducing your efficiency.

So, for example, if you have 10% electrical losses (90% efficiency) when doing 0-60 in 8.4 seconds, you will have 20% electrical losses (80% efficiency) when doing 0-60 in 4.2 seconds.

 

[RegGuheert]

The last comment in the thread wraps it up nicely,

Except that ignores all loss mechanisms except for battery resistance, including:

- Aerodynamic losses.
- Drivetrain losses.

Note that the resistance of the battery is VERY low, so it's efficiency is extremely high. The resistance of the Model S battery is even lower. So the fact that the LEAF drivetrain efficiency increases from ~88% at low torque to ~95% at maximum torque basically cancels out the additional losses in the battery. Then you are left with mainly the aerodynamic effiects.

 

[TickTock]

There is a similar thread in the Tesla forum,
http://www.teslamotors.com/forum/fo...iciency-different-power-outputs-accelerations

It is using the following article as a reference regarding internal battery resistance,
http://www.evs24.org/wevajournal/php/download.php?f=vol3/WEVJ3-5340444.pdf

The last comment in the thread wraps it up nicely,

cliff@hannelcon... | April 18, 2013

OK, let me try this:

(1) resistive losses are proportional to the square of the current [ri^2]
(2) torque in electrical motors is proportional to current
(3) acceleration is proportional to torque (and thus current)
(4) distance covered to achieve a certain speed is proportional to acceleration][.5*a*t^2)

From (1)-(3), losses are proportional to the square of acceleration. So, when you double acceleration, you have 4x the losses but from (4) you've covered only 2x the distance, so you have 2x the losses per distance, thus reducing your efficiency.

So, for example, if you have 10% electrical losses (90% efficiency) when doing 0-60 in 8.4 seconds, you will have 20% electrical losses (80% efficiency) when doing 0-60 in 4.2 seconds.

While this is all very true, it doesn't accurately answer the question in the OP (Why high acceleration reduces range). In fact if the excerpt you shared is all you go by, you would be mislead into thinking that you should accelerate slowly to get more range when in reality:

1) it has negligible impact compared to speed
2) in the case of constant average speed, slower acceleration will actually reduce your range even assuming constant drivetrain efficiency
3) as RegGuheert pointed out, the drivetrain efficiency profile negates some of the extra losses during high acceleration

However since you are the OP'er, I assume your real question has been answered which appears to be "Why does accelerating faster use more energy per distance during acceleration assuming constant drivetrain efficiency and ignoring aerodynamic losses?" and the excerpt you shared does explain this very well. However, that question won't fit in the subject line.

 

[LTLFTcomposite]

Is it safe to say ev drivers pay a much smaller penalty for jackrabbit starts than ice drivers?

 

[RegGuheert]

Is it safe to say ev drivers pay a much smaller penalty for jackrabbit starts than ice drivers?

Perhaps true on efficiency. As far as longevity goes, I would say only if it doesn't degrade the battery faster. Unfortunately, most research I see in LiIon battery cycling says that cycling at a higher rate causes faster degradation. (Yes, I know, others here dispute that, but you have to decide whether you are willing to bet some battery life on that possibility.)

The LEAF does seem to eat tires rather quickly and jackrabbit starts are indicated here as well. Perhaps with the extra weight it eats them faster than lighter ICE cars? I don't know...

 

[Sublime]

Is it safe to say ev drivers pay a much smaller penalty for jackrabbit starts than ice drivers?

Yes, by a large margin. We're talking about less than a 10% efficiency loss in an EV (possibly way less). For an ICE, fuel efficiency can go down 50% or more at WOT.