Regen vs. Coasting

[Jeff Miller]

Todd,
I thought about this a bit more. If one assumes a constant amount of power going into regenerative breaking, say 30kw as in your initial post, that implies a force Fregen = Power/v . So the equation of motion becomes

mdv/dt = -av^2 -b -c/v

where a and b are as in my previous post and c = 30,000 (corresponding to a constant regenerative power - probably an OK assumption between 75 and 55).

Integrating mdv/(av2 + b + c/v) numerically gives 11.2 seconds.

So what you "gain" from the regenerative breaking is some fraction z of 30,000*11.2/3600 = 93 wh.

What you loose because you are driving at the higher initial speed for longer is ...

[... I need to think about this part as I'm not sure my original calculation was right ... ]

 

[hcsharp]

The analytic solution for the time to decelerate from 75 to 55 by coasting with the frictional forces you specified
( mdv/dt = -av^2 - b ) is a difference of arctangents:

Time = (m/sqrt(a*b)) *(arctan(vi *sqrt(a/b)) - arctan(vf*sqrt(a/b)))

Nice work! What is your background that you were able to see this was a difference of arctangents?

It's an interesting exercise, and I'll be interested to see what you find on the regen side of things, but I think I agree with some of the other commentators that the "fair" comparison of the efficiency of different ways of slowing down requires that one also impose the constraint that
the total time to go from point A to point B be the same. From your remarks you seem to agree with this as well...

How can you impose that constraint (time be the same) when the way the problem is defined the time can't possibly be the same in both cases? He can't start slowing down before he gets on the off-ramp, and the way the problem is presented he starts regen right away to cut wind drag losses. I can't see how regen would ever be more efficient if he had to continue at 75 for several seconds longer than the coasting case. The whole problem goes away. No?

 

[Jeff Miller]

Nice work! What is your background that you were able to see this was a difference of arctangents?

How can you impose that constraint (time be the same) when the way the problem is defined the time can't possibly be the same in both cases? He can't start slowing down before he gets on the off-ramp, and the way the problem is presented he starts regen right away to cut wind drag losses. I can't see how regen would ever be more efficient if he had to continue at 75 for several seconds longer than the coasting case. The whole problem goes away. No?

Too many wasted years in my youth spent as a physics TA ... You just have to integrate the equation dv/(av^2 +b) which is an arctan.

I think you may be right, but I have to think it through again.

 

[JohnQ]

It makes sense to compare the energy usage across a constant distance. Specifically, the distance to reduce speed from 75 to 55 by coasting. Which makes the problem a little more challenging since one needs to solve for distance rather than time.

 

[Jeff Miller]

One more try. For definiteness suppose you are driving a total distance D.

Case 1. You drive at speed vi the whole way and then at the very end slam on the breaks at the
exit ramp bringing your speed to vf. This is obviously very inefficient. Your total energy expenditure
for the trip is

E1 = Force * distance = (av^2 +b ) * D = 771 * D

Case 2. You coast the last part of the trip to the ramp. Todd already showed the coasting time is 29.6 s.
The distance traveled in this time is just the integral of the velocity which numerically turns out to be
852 meters (if I haven't made an error). Since you supply no power during this last distance, your total energy expenditure is

E2 = 771 * (D - 852)

So you save an energy of 771 * 852/3600 = 182 wh by coasting.

Case 3. You use regen breakinging.

Integrating the velocity numerically like Todd was doing, dv = -dt/m (a v^2 + b + c/v) where c = 30,000 as in my previous note,
you find that the regeneration breaking distance is 326 meters. Total energy spent is therefore

E3 = 771 * (D - 326)

Ignoring the gain from regen breaking, you clearly spend more energy than in case 3 relative to case 2
because you are driving further under power. The difference in energy spent is

E3 - E2 = 771 * (852 - 326)/3600 = 113 wh.


From the previous post, the absolute most you can get out of regenerative breaking is 30000* 11.2/3600 = 93 wh.
That assumes 100% efficient regeneration. In reality of course you will get much less.

So in this example, assuming no errors on my part, it seems that that coasting beats regen breaking.
Still need to think about the general case, and the equal time case.

- - - Updated - - -

It makes sense to compare the energy usage across a constant distance. Specifically, the distance to reduce speed from 75 to 55 by coasting. Which makes the problem a little more challenging since one needs to solve for distance rather than time.

Totally agree

 

[hcsharp]

...
E3 = 771 * (D - 326)

Ignoring the gain from regen breaking, you clearly spend more energy than in case 3 relative to case 2
because you are driving further under power. The difference in energy spent is

E3 - E2 = 771 * (852 - 326)/3600 = 113 wh.

I'm confused why you would use 771 in the E3 calculation since the car has reached vf after 326 m. 771 is calculated in case 1 so we have a convenient number to use in the other cases because in each case the car reaches the same energy state at the end of the off-ramp (ignoring higher battery SOC). And you're saying in case 3 there is no energy used in the first 326 meters so all of it has to be used in the last (852 - 326) m?

 

[Todd Burch]

I'll look more into this tomorrow, but we probably ought to be using 60 kW for regen. At higher speeds (~45 mph and above) regen pegs at 60kW.

 

[ItsNotAboutTheMoney]

When you regen instead of glide, the firstthing you are actually doing is maintaining speed, and regen just recovers a proportion of the additional energy used to maintain speed.

The only winning case for regen would be if energy saved through reduced journey time outweighed the losses in regen, or if having the recaotured energy for use in place of grid chargig would improve efficiency.

 

[Hometheatremaven]

All this is very nice, but for those of us who live in the real world, what the heck are we supposed to do to be the most efficient?
Seems there are several possibilities:
1. Set regen to low
2. Shift into neutral
3. Activate our 3rd eye so we know when the stoplight is going to change.
4. Simply look ahead as far as possible so we arrive at the light just as it turns green (kinda like #3 but less sarcastic).

 

[Todd Burch]

All this is very nice, but for those of us who live in the real world, what the heck are we supposed to do to be the most efficient?
Seems there are several possibilities:
1. Set regen to low
2. Shift into neutral
3. Activate our 3rd eye so we know when the stoplight is going to change.
4. Simply look ahead as far as possible so we arrive at the light just as it turns green (kinda like #3 but less sarcastic).

I still need to calculate the final answer (coming today I hope) for my specific original post, but in general, 2, 3, and 4. Leave regen on high so you can recapture as much energy as you can when you DO need to slow down.

Aa for my case, I'm suspecting now that the threshold in which regen is better is higher than the speed we normally drive at, but will post results today.

 

[Jeff Miller]

I'm confused why you would use 771 in the E3 calculation since the car has reached vf after 326 m. 771 is calculated in case 1 so we have a convenient number to use in the other cases because in each case the car reaches the same energy state at the end of the off-ramp (ignoring higher battery SOC). And you're saying in case 3 there is no energy used in the first 326 meters so all of it has to be used in the last (852 - 326) m?

I probably should have a plot, but the idea is basically

case 1. start ppppppppppppppppppppppppppppppppppppppp end (p means under power, ~25kw, Todd's assumptions)
case 2. start pppppppppppppppppppppppccccccccccccccccccc end (c means coast)
case 3. start ppppppppppppppppppppppppppppppppppprrrrrr end (r means regen)
.......................................................|<--526m-->|


In case 3, you end up driving around 526m further at your original speed vi than you do in case 2. To maintain that speed over this distance you need to overcome the frictional forces -a*vi^2 - b = 770 N over this distance. In case 2, you're just coasting over this distance so not spending any extra energy from the battery.

 

[jerry33]

1. Set regen to low

This never works because you always end up throwing away a lot of energy as heat.

2. Shift into neutral

This has safety ramifications, isn't recommended, and is illegal in some jurisdictions.

3. Activate our 3rd eye so we know when the stoplight is going to change.

This actually works well if you have a lot of practice on a well known route.

4. Simply look ahead as far as possible so we arrive at the light just as it turns green

Stated another way. When you see a red light slow down and then drive slowly up to the light. Often it will turn green before you come to a full stop.

 

[stevezzzz]

...Down a long steep hill, speed increases to the point where aerodynamic resistance takes away more distance than the glide gives you. Using regen to hold the speed just below that point is optimal. (Sorry, I haven't done the math on what that speed is. It's possible the optimal speed varies based on the slope of the hill. I've been waiting for someone else to do the math

I've been following this thread closely, and I'm still waiting for someone to tackle the "down a long steep hill" scenario. I'm convinced that for the flatland cases, coasting beats regen at all normal speeds. But what about the huge descents we get here in the Rockies when crossing the major mountain passes? Assume two identical Model S start at the top already traveling the speed limit with cruise control on. Car A leaves cruise control on at that speed and Car B puts the car in neutral. Car B coasts down in neutral reaching terminal velocity---this assumes that you don't a) fly off the road or b) get pulled over for speeding---and arrives at the bottom of the hill going faster than the speed limit, and rolls up the other side until the speed drops to the posted limit. Let's call that spot the destination. Car A comes down the hill at the posted limit using regen to hold the speed, gets to the bottom and starts using power to maintain speed until it reaches the destination. Which car arrives at the destination with a higher SOC?

Case 1: If the hill is precisely the right grade so that Car B's terminal velocity equals Car A's set speed, the two cars reach the destination (the bottom of the descent, in this case) at exactly the same time, at the posted limit, having expended no energy and using no regen. In this case, it doesn't matter how long the hill is: the results are identical.

Case 2: If hill is shallower than in Case 1, Car B's terminal velocity is less than Car A's set point, so car B arrives at the bottom of the descent later than Car A but has used no energy to get there, while Car A has had to use battery power to maintain speed and so arrives at the bottom with a lower SOC.

Case 3: This is the real world case I experience regularly crossing the Divide westbound on I-70, on the way to the Western Slope: the descent from the Eisenhower Tunnel to the bottom of the valley at Silverthorne is 14 miles long and relatively uniform in steepness; terminal velocity is unknown, but significantly higher than the speed limit. I normally choose to descend with the cruise control set to the 65mph speed limit and gain about 2kWh of regen by the time I reach the bottom (I'm Car A): that gives me roughly seven miles of Rated range to expend climbing out of the valley on the other side, before the battery's SOC drops to what it was at the tunnel. It seems clear that, no matter how fast terminal velocity is on the descent, Car B will slow to the 65mph speed limit as it coasts up the far side significantly short of the spot Car A reaches when its SOC drops to the starting SOC. It will, however, get there sooner; but if what we care about is range over other considerations, Car A comes out ahead, without risking an accident or driver B's license.

I also believe that that reducing the cruise control set speed on the descent would add to the amount of regen gained, and thus increase range even more; this is one time when the rule of thumb "avoid using regen" doesn't hold.

Does someone want to do the math to corroborate or disprove this claim?

 

[ItsNotAboutTheMoney]

All this is very nice, but for those of us who live in the real world, what the heck are we supposed to do to be the most efficient?
Seems there are several possibilities:
1. Set regen to low
2. Shift into neutral
3. Activate our 3rd eye so we know when the stoplight is going to change.
4. Simply look ahead as far as possible so we arrive at the light just as it turns green (kinda like #3 but less sarcastic).

2 (maybe), 3, 4.

These are all used by hypermilers.
2: in some places (eg CA) it is illegal to coast in neutral; in many places (e.g. ME) it is illegal to coast in neutral downhill. How well a vehicle tolerates neutral varies, so you need to know the limits of your car. However, in a BEV ideally, you find the glide point instead and use that.
3, 4:
Lights are stop signs that appear and disappear. You want to hit the intersection when the stop sign has disappeared. So you learn sequences and timings (if appropriate) and use anticipatory focus (look ahead, think ahead) to try to arrive as close to normal speed as possible, with as little braking as possible. Remember that it takes time for traffic to get moving, allowing 2s for each stopped car is reasonable, although watch out for lights at dips and crests because you may find that people may not adjust acceleration appropriately with terrain.

Light timing and smart braking (applying minimum necessary braking) are part of the hypermiling principle of Driving Without Brakes (DWB), which seeks to eliminate the deliberate discarding of kinetic energy.

But, hypermiling really keys on what I call the Golden Rules:
- Look ahead (and all around) and think ahead
- Leave a good buffer (not just 2s!) to the car in front
Those give you information and plenty of time and space to act on them and those allow you to eliminate harsh maneuvers. In the case of the original post, the information is that you need to slow down by 20mph on a off ramp, so you try to figure out where to begin your glide such that your car will naturally slow down 20mph by the time you reach the new speed limit sign. That, of course assumes that you don't mind taking a little extra time or potentially holding up people behind you.

 

[Jeff Miller]

I've been following this thread closely, and I'm still waiting for someone to tackle the "down a long steep hill" scenario. I'm convinced that for the flatland cases, coasting beats regen at all normal speeds.

The downhill/uphill question is interesting, I'd like to think about it.

I think you (and some others above) are right about the coasting vs regen though. I see coasting winning in Todd's original scenario at all speeds even assuming a 100% efficient regen. But it's comparing apples to oranges in a way because the times traveled are different. You would expect the shorter total travel time (regen) to be less efficient and it is.

So to be precise about what I'm comparing, I'm assuming that the time spent regening is the minimum time necessary to reach the final speed using regen braking only (otherwise you could regen anywhere along the route and putter along at some slow speed and get any efficiency you want - the problem becomes ill-posed).

Under this assumption, I see that coasting always wins mostly because when you regen you travel further under power and that's very costly. Here is a plot of energy saved relative to hard breaking assuming 60kw regen and 100% efficient. Black line is coasting, Blue is total regen both relative to hard breaking.

​

 

[cantdecide]

Stevezzzz... Agree completely that this is an interesting case I keep coming across myself although usually in the sub case of a smaller hill. Given a hill to come down and a minimum speed, what is the optimal strategy?

Once you reach terminal velocity while coating you clearly gain no kinetic energy or battery energy after that so the rest of the hill is wasted... So regen is obviously better on a long enough hill if only those two strategies are available.
However, I believe a different strategy is optimal... Use the regen strategy until you are near the bottom, then switch to the coating strategy. If you are just coasting for a short time and for a small speed increase then you don't have much additional coasting losses from the additional speed (which are proportional to the time you spend at the higher speed) compared to the battery losses of regen which are about the energy conversion and not the time. Of course the point at which you should switch strategies (as measured by time or distance from the bottom or by soured to coast to) will depend on factors including the exact slope... But won't be dependent on the total length of the hill. Making the hill really long simply assures that you begin with the regen strategy.

 

[SeattleSid]

I must say I'm absolutely impressed, flabbergasted, really, at the brilliance in this thread, and the elegance of various people's solutions. Not to mention depressed by realizing there was a time, around fifty years ago, when I probably would have understood the equations. It stands to reason that Tesla owners are, in general, better educated that average drivers.

Still, like hometheatermaven, I'm thinking specifically and simply. Coming home I go down a long hill, about a mile and a half, with a stop sign at the bottom. So there's no timing to avoid stopping. In that case, it seems intuitive (because, as implied, I no longer have the gray matter to behave otherwise) that in that case it makes sense to use regen. Given a set distance and a predictable stop, getting from top to bottom without regen will leave the battery with less energy than using it. My question as been whether it makes more sense to have low regen vs high: because with high regen there's a need to apply some go pedal to avoid slowing to nearly a stop. I don't want to exceed the 40 mph limit. So, for a constant speed, constant slope, constant distance, predictable stop at the end, is it not the case that coasting is the least efficient? And wouldn't it also be the case that if low regen allowed traveling at the constant speed till the bottom and if high regen required applying a few electrons to maintain speed, that the best scenario would be low regen?

 

[Jeff Miller]

I must say I'm absolutely impressed, flabbergasted, really, at the brilliance in this thread, and the elegance of various people's solutions. Not to mention depressed by realizing there was a time, around fifty years ago, when I probably would have understood the equations. It stands to reason that Tesla owners are, in general, better educated that average drivers.

Still, like hometheatermaven, I'm thinking specifically and simply. Coming home I go down a long hill, about a mile and a half, with a stop sign at the bottom. So there's no timing to avoid stopping. In that case, it seems intuitive (because, as implied, I no longer have the gray matter to behave otherwise) that in that case it makes sense to use regen. Given a set distance and a predictable stop, getting from top to bottom without regen will leave the battery with less energy than using it. My question as been whether it makes more sense to have low regen vs high: because with high regen there's a need to apply some go pedal to avoid slowing to nearly a stop. I don't want to exceed the 40 mph limit. So, for a constant speed, constant slope, constant distance, predictable stop at the end, is it not the case that coasting is the least efficient? And wouldn't it also be the case that if low regen allowed traveling at the constant speed till the bottom and if high regen required applying a few electrons to maintain speed, that the best scenario would be low regen?

Without having thought about it a lot, your point makes sense to me in your particular scenario. Going down the hill, your power source is gravity, not the battery. Plus you have to stop at the bottom anyway. So you should try to collect all the extra gravitational energy (above the amount you need to overcome friction) you can by using regen breaking. Coasting doesn't allow you to collect anything, so it is least efficient in this case. I'm not sure exactly what the tradeoff in your case is between high and low regen breaking. I imagine it depends on the particulars of your speed and the slope of the hill.

 

[SeattleSid]

Thanks, Jeff. I've played with it on a couple of test drives, but won't have my own car till (holy cow!!) this Thursday. Then I guess I can start to figure it out for my particular situations. I'll be driving from Seattle to Portland a lot, too, and there are several long downgrades on I-5 on which I can compare (once I figure out all the available data displays) coasting vs the 2 regens.

 

[wycolo]

> Energy Saved by Coast/Regen vs Braking (chart) [Jeff Miller]

This chart is a bit confusing since an un-careful reading would yield a takeaway: "Coasting is the most efficient mode for an EV, therefore I will always try to maximize COASTING whenever possible." Ok, actually this IS a pretty good goal to have, but reality is that safety & laws limit coasting to unusual circumstances. So most times we end up using the orange/green 'coasting point' to invoke COASTING rather than actually shifting to NEUTRAL. Hovering around this Coasting Point gives us instant POWER or instant Regen/Braking by simple use of right foot; thats the beauty of MS single pedal operation.

Stevezzzz's example of I-70 [Eisenhower Tunnel West Portal > Silverthorne, CO] is one of the several Great Declines of the Western World. In an MS the driver *must* operate on the regen side of the coast point at all times, varying ever so slightly as needed to keep a steady speed. Yahoos in the leftmost lane will speed past you but give themselves away by periodically applying their brakes (FAIL!!). Their shiny cars no doubt have speed-scrubbing automatic transmissions that do downhill braking for them if they just understood that feature. My Forester does it just fine. MS does it even better since it is storing energy all the way down.
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