Regen vs. Coasting

[JohnQ]

Let's try this, solving for time rather than distance.

Again, let b = 1/2*(rho)*Cd*A

ma=-bv^2 **equivalency of forces
mdv/dt=-bv^2 **acceleration is the time derivative of velocity
-(m/b)/v^2 dv = dt **result of simple algebra
Integration across v1 to v2 leads to
(m/b)*(1/v2-1/v1)=t

I made some assumptions regarding certain constants (standard atmosphere, frontal area is 2.37m^2, mass is 2,108 kg). This leads to b=0.34989 kg/m.
Solving the equation with v1=75mph (33.5 m/s) and v2=55mph (24.6 m/s) gives
65 seconds

This would take about a mile or more. Even without rolling resistance this is a non-intuitive result.
Any idea where I dropped the ball here?

 

[arg]

Thought experiment:

1. You're at the top of a big steep hill going 55 mph. Coasting down the hill results in the car going 100 mph. At that point wind resistance is providing a LOT of drag and therefore you are wasting substantial amounts of energy. You arrive at the bottom of the hill and eventually coast down to 55 again; however, the pack has no more energy in it than when you started.

2. You're at the top of a big steep hill going 55 mph. You use regen to keep the speed at 55 mph. At that point wind resistance is modest, and you're pumping quite a bit of energy into the pack. When you get to the bottom of the hill you're still going 55, so you'll have to put the power back on a bit earlier than the previous example... but now you have a lot of extra energy stored in your pack.

I'm pretty sure in that extreme example using regen is better.

I had exactly the same thought as you describe here, but after a bit I realised that this was not a fair comparison.

The error here is that (in your example) the coasting has covered the distance in a shorter time. If you are happy to take longer over the journey, then you could have been driving more slowly in the first place.
So for a like-for-like comparison you should allow the coasting car to reduce to (say) 40mph for a while after it has reached the flat ground to give the same overall journey time as the regen car that did constant 55mph throughout.

With that adjustment, it is no longer clear cut that the regen car is doing better - it has more charge in the battery than the coasting car when it reaches the bottom of the hill, but it then needs to drive faster (or let the regen car drive slower) over the next couple of miles to catch up. I strongly suspect that there ought to be some relation between the delta between the two speeds (coasting case) and the regen efficiency that says regen is more efficient above a certain speed difference, but I haven't been able to work out what that relation is.


In the case of Todd's regen vs. coasting for a speed reduction on flat ground, the correct comparison isn't coasting vs. immediate regen down to the lower speed and then hold that speed: that causes the regen car to take longer, and again if you accept a longer journey time you could have been driving slower in the first place. The correct comparison has the regen car doing the regen somewhere around the middle of the zone where the other car was coasting - unfortunately it's not exactly the middle as the coasting is an exponential decay rather than linear. The right point balances the higher speed before doing the regen with the lower speed after the regen to give the same overall time as the coasting.

 

[Doug_G]

The error here is that (in your example) the coasting has covered the distance in a shorter time. If you are happy to take longer over the journey, then you could have been driving more slowly in the first place.
So for a like-for-like comparison you should allow the coasting car to reduce to (say) 40mph for a while after it has reached the flat ground to give the same overall journey time as the regen car that did constant 55mph throughout.

I did try to point that out in my thought experiment, by pointing out that the regen driver would have to get back on the throttle earlier. Maybe I wasn't clear enough.

I still think it's possible that the coasting driver might consume more energy with the same distance traveled. Aerodynamic losses go up quadratically with speed, whereas regen losses remain constant.

 

[hcsharp]

I still think it's possible that the coasting driver might consume more energy with the same distance traveled. Aerodynamic losses go up quadratically with speed, whereas regen losses remain constant.

Quadratically? Did you mean cubed? Wind resistance is relative to the square of the velocity, so wouldn't the energy loss go up with the cube of the velocity? What am I missing?

 

[JohnQ]

For those interested in the math and values. I adjusted the width of the vehicle to include the mirrors (I was incorrectly using track before) and the height is from the ground to the roofline and simply multiplied the two to get the area. This is not the strict orthographic projection of the frontal area of the car and the results can be quite sensitive to changes in area. It's likely overstated which would tend to reduce the time to slow the vehicle. In this case the change reduced the time to 49 seconds. Still significantly larger than expected ...

 

[Doug_G]

Quadratically? Did you mean cubed? Wind resistance is relative to the square of the velocity, so wouldn't the energy loss go up with the cube of the velocity? What am I missing?

Yes but because you're linearly moving faster, the effect on your range is only quadratic.

 

[Great Dane]

It is great that this question came up
because this is how I drive my ice Manual transmission car
I put it in nuetral all the time and coast
But I have never experienced any thing like the model S
It coast forever
I drive and look waaay ahead and i bump small click to N
and coast to red light,and then when I get close, I two click it down to D
and use the regen to do the final braking for me, Kind of fun experimenting to see
how low we can get the wh per mile also you want to try to keep the car moving all the time
because the biggest energy is used to get that 4700 lbs moving from a dead stop

 

[woof]

Wouldn't it be great if there were a foot peddle or paddle on the steering wheel that would disengage regen for as long as it's held down, then re-engage when released. One could then easily coast when desired without the trouble (and danger) of shifting out of Drive.

 

[hcsharp]

Yes but because you're linearly moving faster, the effect on your range is only quadratic.

I can see how your effect on range would be a quadratic function. Your original statement was about consuming energy and I assumed you meant consuming energy over a given time period vs overall trip energy at higher speed. With the word range, I realize you were referring to the latter which makes sense given the problem we're trying to solve. So your energy consumption (rate) increases cubically with speed, and your range decreases quadratically with speed.

- - - Updated - - -

Wouldn't it be great if there were a foot peddle or paddle on the steering wheel that would disengage regen for as long as it's held down, then re-engage when released. One could then easily coast when desired without the trouble (and danger) of shifting out of Drive.

A foot pedal would be more work than the existing method, and the paddle isn't much different from the column shifter we have now. Frankly, I wouldn't use either one. After a little practice you can coast pretty well by feathering your foot on the accelerator. And I find you almost never need to coast at exactly 0 watts anyway. Regardless of coasting method, it's fun solving for the speed at which it's more efficient to regen rather than coast.

 

[Great Dane]

Wouldn't it be great if there were a foot peddle or paddle on the steering wheel that would disengage regen for as long as it's held down, then re-engage when released. One could then easily coast when desired without the trouble (and danger) of shifting out of Drive.

Woof , I Like The idea, it would still put the car in nuetral , but so much easier then what i have been doing
By the way nuetral is done electronically, the motor is always connected via gears to the wheels
one reason for coasting so well is, there are way fewer components in the drive train.

 

[JohnQ]

So your energy consumption (rate) increases cubically with speed, and your range decreases quadratically with speed.

Yes. Power (energy per time unit) goes up as the cube of velocity due to air resistance.

 

[jerry33]

After a little practice you can coast pretty well by feathering your foot on the accelerator.

Right, and the car always behaves as expected.

And I find you almost never need to coast at exactly 0 watts anyway.

Yes. I've found that Wh/mi decreases more if the gliding is done with just a sliver of green showing. This is somewhat different than the Prius where keeping it at zero (display arrow unlit) works better.

 

[Sparrow]

From my observation I believe jerry33 is right about getting much better Wh/mi results by coasting with a small sliver of green showing.

 

[nexnbax]

Todd,
I have briefly followed this thread and when driving down a gradual hill, if I just gently move the shift lever down with a tap, the car goes to neutral. When I arrive at a point that I want to use the regen to come to a stop, I tap it again just a bit harder and it drops into drive.
Not sure if everyone does this... the car gives no lurching motion, just a smooth transition.
Please share your thoughts.
Thanks
Ivan
Sorry, but I do not follow the forum that much due to time constraints.

 

[Todd Burch]

For those interested in the math and values. I adjusted the width of the vehicle to include the mirrors (I was incorrectly using track before) and the height is from the ground to the roofline and simply multiplied the two to get the area. This is not the strict orthographic projection of the frontal area of the car and the results can be quite sensitive to changes in area. It's likely overstated which would tend to reduce the time to slow the vehicle. In this case the change reduced the time to 49 seconds. Still significantly larger than expected ...
View attachment 44736

JohnQ,
Using my Excel spreadsheet with dt = 0.1 sec and the same constants as you, and assuming zero rolling resistance or drivetrain losses, I calculate the time from 75 MPH to 55 MPH as 49.2 sec. So good agreement there.

Rolling resistance is F = Crr * N where Crr is the coefficient of rolling resistance and N is the normal force (essentially the weight) of the car. Since the car has a pretty good 50/50 weight distribution, this is a good estimate. A reasonable Crr value is 0.012. (I extracted this value from the Wikipedia article on rolling resistance, which refers to SAE tests on vehicle rolling resistance, which show Crr to be around this value).

If I then add rolling resistance into my calculation, I determine the time to go from 75 MPH to 55 MPH is 29.6 seconds.

Something that I found surprising is that even at 75 MPH, drag force is only about 2 times the rolling resistance. In other words, even at high speeds, rolling resistance is NOT negligible! This is in part due to the high mass of the Model S.

Determining losses from the drivetrain (reduction gearing, bearings, etc.) is much more difficult. I think I'll perform some tests, gather some numbers, and adjust my "other losses" until the calculated time matches what I really observe. If I assume 100 N of force due to drivetrain resistance, the time to coast from 75 MPH to 55 MPH is calculated as 25.5 seconds.

I've attached my Excel spreadsheet here:
http://www.sendspace.com/file/guht0f

Before I consider the regen case and go further, I'll gather some data and tweak my drivetrain resistance ("other losses") column. Anyone care to provide some data for "time to decelerate" on level roadway?

If anyone can provide a decent time to decelerate from one speed to another while coasting as best you can on a level roadway on a day with minimal winds, I'll use this to get a decent model going. I understand that numbers will be different for different configs (pano/non, 19 vs. 21s, twin vs. single chargers, child seats, etc.) but I just want to get relatively close here.

 

[hcsharp]

Before I consider the regen case and go further, I'll gather some data and tweak my drivetrain resistance ("other losses") column. Anyone care to provide some data for "time to decelerate" on level roadway?

If anyone can provide a decent time to decelerate from one speed to another while coasting as best you can on a level roadway on a day with minimal winds, I'll use this to get a decent model going. I understand that numbers will be different for different configs (pano/non, 19 vs. 21s, twin vs. single chargers, child seats, etc.) but I just want to get relatively close here.

I suspect you'll want to get ambient temperature with each data point that's sent to you. I calculated a full second difference from 75 -> 55 using an air density of 1.250 vs 1.225. The difference in temp between those two is about 5C. That will probably have a bigger impact than build config.

edit: Now that I think about it, you'll want their altitude as well. The difference between sea level and 5,000 ft is nearly 30 seconds using JohnQ's formula!

 

[jerry33]

I have briefly followed this thread and when driving down a gradual hill, if I just gently move the shift lever down with a tap, the car goes to neutral. When I arrive at a point that I want to use the regen to come to a stop, I tap it again just a bit harder and it drops into drive.
Not sure if everyone does this... the car gives no lurching motion, just a smooth transition.

I don't do this because if an emergency comes up, the time that it takes you to shift could mean the difference between an accident and accident avoidance. It's one of those things where the risk isn't worth the gain.

- - - Updated - - -

edit: Now that I think about it, you'll want their altitude as well. The difference between sea level and 5,000 ft is nearly 30 seconds using JohnQ's formula!

That sounds about right. Driving at 5,000 ft is far less energy intensive than driving at 600 ft.

 

[Jeff Miller]

JohnQ,
Using my Excel spreadsheet with dt = 0.1 sec and the same constants as you, and assuming zero rolling resistance or drivetrain losses, I calculate the time from 75 MPH to 55 MPH as 49.2 sec. So good agreement there.

Rolling resistance is F = Crr * N where Crr is the coefficient of rolling resistance and N is the normal force (essentially the weight) of the car. Since the car has a pretty good 50/50 weight distribution, this is a good estimate. A reasonable Crr value is 0.012. (I extracted this value from the Wikipedia article on rolling resistance, which refers to SAE tests on vehicle rolling resistance, which show Crr to be around this value).

If I then add rolling resistance into my calculation, I determine the time to go from 75 MPH to 55 MPH is 29.6 seconds.

Something that I found surprising is that even at 75 MPH, drag force is only about 2 times the rolling resistance. In other words, even at high speeds, rolling resistance is NOT negligible! This is in part due to the high mass of the Model S.

Determining losses from the drivetrain (reduction gearing, bearings, etc.) is much more difficult. I think I'll perform some tests, gather some numbers, and adjust my "other losses" until the calculated time matches what I really observe. If I assume 100 N of force due to drivetrain resistance, the time to coast from 75 MPH to 55 MPH is calculated as 25.5 seconds.

I've attached my Excel spreadsheet here:
http://www.sendspace.com/file/guht0f

Before I consider the regen case and go further, I'll gather some data and tweak my drivetrain resistance ("other losses") column. Anyone care to provide some data for "time to decelerate" on level roadway?

If anyone can provide a decent time to decelerate from one speed to another while coasting as best you can on a level roadway on a day with minimal winds, I'll use this to get a decent model going. I understand that numbers will be different for different configs (pano/non, 19 vs. 21s, twin vs. single chargers, child seats, etc.) but I just want to get relatively close here.

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)))

where

a = .5*rho*Cd*A
b = Crr*m*g

Using your numbers

rho <- 1.23
Cd <- .24
Crr <- .012
A <- 3.15
g <- 9.81
m <- 2108
vi <- 33.53
vf <- 24.59


I get Time = 29.6 seconds, in agreement with your numerical solution.

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...

 

[Discoducky]

I'm surprised to hear folks use neutral at high speeds. I wouldn't so that is out of my equation.

While I haven't done the calculations that folks are doing I'd bet that in just about every case that coasting is not only faster, but more efficient.

Now, what exactly is coasting? Like Jerry mentions, it might be just a bit of green or for some is neutral. For me it is no color at all. And it takes practice, but after driving over 4K miles 2X down to LA from Seattle in the 60kWh I find it 2nd nature now. With no color I find the car will accelerate much quicker on down hills at highway speed than if it has a sliver of orange or green (this seems to be the sweet spot). Allowing the car to accelerate to speeds of around 75MPH I think is the breakeven point between regen and no-color-coasting (NCC). Once the car hits 75MPH I let regen take the car down to 10+ over the speed limit and resume my normal driving.

With this method and other very polite, non-obtrusive, non-lasting drafting techniques I have been able to get my Wh/mile down into the low 200's even with hills in 70 degree weather. And I'm always with traffic or above the speed limit on highways.

 

[arg]

I did try to point that out in my thought experiment, by pointing out that the regen driver would have to get back on the throttle earlier. Maybe I wasn't clear enough.

Ah, OK. I'd taken that comment to mean that the regen car needs to go back on the throttle immediately at the bottom of the slope (to maintain 55mph), while the coasting car only needs to go back on the throttle when its speed has dropped to 55. That much is true, and still guarantees the regen car to have used less energy (so long as the hill is long enough - the extra due to going on the throttle sooner is constant, while the amount accumulated in the battery depends on the length of the hill).

However, to account for the time taken, the coasting car can stay off the throttle for a long time (or come back on the throttle at say 40mph for a while) to even out the journey time.

I still think it's possible that the coasting driver might consume more energy with the same distance traveled. Aerodynamic losses go up quadratically with speed, whereas regen losses remain constant.

I agree. Unfortunately, even in these simple cases there's no clear winner - it could go either way depending on the exact numbers involved.