Here's an interesting efficiency thought experiment for you. Those without an engineering/geeky mind may wish to move on . When driving at high speeds, drag consumes the most amount of energy. When driving at low speeds, other effects consume the most amount of energy. Imagine I'm going 75 MPH on the freeway and I take an off-ramp. Ahead, the ramp merges onto a road with a 55 MPH speed limit, and my goal is to enter that road having consumed the least amount of energy after leaving the highway. Obviously, braking to the target speed of 55 MPH is not the best thing to do. We all know that regen is more efficient than braking. So we have two remaining options to slow down efficiently: 1. Press the pedal enough to coast--no regen, no power delivery--until I hit my target speed of 55 MPH. 2. Allow the car to regen by some amount until I reach the target speed of 55 MPH. In option (1), I don't consume any power (or generate any), but I remain at a higher speed for a longer time. As a result, the car is exposed to higher drag forces for a longer time. In option (2), I regen at a maximum efficiency of about 85% (so I lose 15% due to regen alone), but I spend less time at the higher speeds where drag forces are higher. So the question: Which option is more efficient? In my college days, I (probably?) could have solved this. But with two young kids running around, I'm too lazy and/or time-constrained to make an effort. So maybe someone out there wants to impress us . Assumptions: 1. Drag force is 1/2 (rho) v^2*Cd*A. (Pick a reasonable density, area, and a Cd of 0.24). 2. Energy consumed by drag force is the drag force times the distance over which it applies (F * d). However, since F is not constant, you must use the integrated version: Energy = the integral of F over the distance d. 3. Speed is reducing from 75 MPH to 55 MPH. 4. Assume regen efficiency of 85%. Either assume a regen rate (like 30 kW), or--for bonus points--leave it as a variable to test the result at different values! 5. Ignore other energy consumption effects--(bearing friction, rolling friction, HVAC, etc.) as I believe they are relatively small compared to these two. Besides, most of those can be considered close enough to be identical in both situations. 6. The road is level. Anyone want to propose a solution? (This will require some paper and a writing utensil . So why do I ask? First, I'm curious. Not only is it not directly obvious to me which is better in this scenario, but I'm fairly certain it changes depending on the speeds involved. If going from 35 to 15 MPH, for instance, coasting is likely MORE efficient than regen, since drag forces are relatively small. At some speed, regen becomes more efficient. WHERE is that boundary? Second, you'll notice that when in cruise control and you drop the set speed, the car uses regen to reach that speed. Perhaps the engineers at Tesla already considered this--or perhaps it's required by law--but it would be cool if the car considered the current speed and the set speed, and used the most efficient method of getting there. Not a huge energy difference granted, but at the very least something to consider. OK...now solve, ye bright minds! Note: I solved this in Post #68 if you want to jump to the good stuff!

Coasting will always beat regen. Regen benefits the aggressive drivers by recapturing the energy that we would have lost by slamming our brakes on often. But the best efficiency is going to be using hypermiling techniques which means never using regen and never using your brakes and always coasting near 0kWh whenever possible.

Basically correct. Regen turns kinetic energy into electricity and heat. Gliding doesn't have any heat component so it's almost always more efficient. The rubs are 1) to gain the most distance you have to allow the glide to complete, if you have to slow down during the glide then distance is lost. 2) 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 When going for low kWh/mi using cruise control is a bad idea.

The cars lose an identical amount of energy during deceleration: 1/2m(v1-v2)^2. In the case of coasting, the car will go d1. In the case of regen, the car will go d2 (which is less than d1). So, the question is whether the energy stored through regen is more or less than the energy consumed to go (d1-d2) at 55mph. One could make a rough assumption that the regen is instantaneous to go from 75 to 55 mph, thus eliminating the loss to drag during deceleration in the regen example. Given that assumption, the energy loss as a result of decelerating from 75mph to 55mph through drag (we know that's 1/2m(v1-v2)^2) will be higher than the energy loss to travel that same distance at 55mph. Thus, it comes down to the efficiency of regen capture and whether that offsets those two energy differences. Simple enough to determine what the regen efficiency needs to be to break even.

Since it takes two conversions of energy when you regen AC>DC on regen, and then DC>AC when you use the power again, coasting would most likely be more efficient by quite a ways. I'm too lazy to do the math though.

I believe that assuming that coasting is always better is not correct. Consider a case where regen is 100% efficient. If in both cases the cars start at 70 MPH, you will consume less energy by regenning down to 17 MPH (as I recall the most efficient speed to drive in the Model S--that is, the speed that yields the highest range) and travelling the majority of the distance at the most efficient speed than to burn a lot of it through drag at higher, less efficient speeds. Naturally, as regen efficiency decreases below 100%, this benefit starts to decrease until at some point they both consume the same amount of energy. JohnQ is the closest so far, but has left off important details. Anyone willing to make detailed calculations? By the way, I'm a glider pilot. There are some things that are counterintuitive in glider flying. One is that if you are getting low and are a ways from an airport, it's usually better to fly faster (push the nose down) than to slow down. In that case, the speed at which the glider travels the most distance for a given elevation drop (called best L/D) is higher than the speed at which the glider loses the least elevation in a given amount of time (called min sink speed). Similar perhaps counterintuitive physics (to some) is at play here.

I would say in this example coasting definitely wins. The car slows down slower so you can let go off the 'gas' pedal earlier to start coasting which means you stop drawing power earlier. The average speed from the point when you let go of the accelerator to reaching 55 is lower than keeping the speed and regen later thus causing lower drag = less lost. Regen causes more friction in the tires = loss. Regen only captures a part of the energy back. The motor/inverter acting as a generator probably is only 80% efficient in that mode. The battery also doesn't 'absorb' 100% of what is fed into it during regen. Probably another 20% lost. Normal driving is 85-90% efficient, so yet another loss when reusing the energy that was captured during regen. According to the manufacturer of EV drive components the total real world efficiency of regen is round 30 percent. IOW if you got down a hill for 1 mile using regen, you will get only up 1/3 of that hill using the energy that you captured. So definitely try to coast as much as possible instead of regen when it comes to saving energy.

I haven't done any math, but I recall a similar thread where it posited that maintaining momentum always beat regen because you don't incur energy conversion losses (momentum to battery back to momentum). I also recall that same thread citing 60% as the conversion efficiency.

Coasting wins, or Newton was wrong. (Hint: he wasn't...) Every time you "push an electron" into or out of the battery system, there are losses in the energy transfer. As a corollary: the most efficient operation is when you're moving zero energy to or from the battery system. This means coasting... although we have to deal with the parasitic drag from the air we're pushing out of the way, and the fact that the tires have meaningful rolling resistance, and all the bushings and other random stuff inside the axles add a teeny bit of loss as well. To be even more precise, your best efficiency will be when you find exactly the right amount of "throttle" that takes you from your initial speed to your desired speed over the time period desired. If you spent 30 seconds at "20% throttle" and then 30 seconds at "0% throttle" that would also be less efficient than spending the whole minute at 10% throttle, because there are components in the loss function that are non-linear. But another interesting monkey wrench here is that by slowing down faster you also change the TOTAL net air resistance... meaning that if it takes you a minute to decelerate from 70mph to 50mph, your total drag loss is the integral of the air resistance vs. speed curve. If you slowed down faster (implying stronger regen) from 70->50, you would spend more time at lower speed, which means lower total drag, which means lower energy used. But I would further argue that this isn't an apples to apples comparison because we should all know that the dominant input for energy use is how fast we drive the car. Oh, crap. you said something about geeky types moving on... oops! ;-P

Ljwobker, I have a background in physics and engineering and am a degreed mechanical engineer. I am familiar with and have studied Newton, the laws of thermodynamics, etc. Saying that coasting always wins is just not correct. No offense to anyone (this is just a geeky question), but nobody has yet offered any numbers or equations that can support the statement that coasting is always better. I was speeking with a PhD physics colleague of mine today and he agreed that there is a speed threshold above which regen is more efficient. It's not a question with an obvious answer. Now it may be true that at speeds in the 75-55 MPH range, coasting is indeed more efficient--but there is a speed threshold above which regen is more efficient. I'm trying to find that speed. Yes, there are losses with regen and power production. Of course--all changes in energy form incur some sort of efficiency losses. But drag is also a loss (a very BIG one at high speeds, in fact). Yes, there is rolling resistance, there are bearing losses, etc...but those are fairly small compared to parasitic drag losses at higher speeds. And they are likely also not significantly different in the regen vs. coasting case. The second part of your statement--that slowing down faster changes the total air resistance, is exactly my point. In fact, you contradict your initial statement at the end of your post. As you said, if you slow down faster, you will use less energy to travel a given distance. (This is why the Model S's range at 17 MPH is more like 400-450 miles vs. 300 at 55 MPH. So coasting means you spend more time at higher speeds and incur higher drag losses, but you avoid energy conversion losses for regen. Using regen means you slow down faster and reduce drag losses, but incur energy conversion losses. I might have to get out some differential equations and actually solve this....

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.

2. you still need to drive longer to get to the distance of 1. and use the regen you recovered. but probably still better.

Doug, thanks for that extreme example. I think I might create an Excel spreadsheet and solve this through numeric methods. I'll post it here when (if) I figure it out.

In real world conditions, most hills end with a need to stop, or turn or something that alters the glide. On highways, maybe not. But coasting to the point of affecting cars behind you is probably not good. So the situations in which coasting can go on as long as momentum allows, are rare. Maybe in those situations, if the goal is to get as far as possible before stopping, it might be better. But where there's a defined endpoint, it seems that regen must win, by causing the car to arrive at that fixed point with more electrons in the battery. (I'm speaking as one with no actual knowledge of kinetics, electrophysics, or whatever else might supply some cred.) Along those lines, what I've wondered about is the relative benefit of low vs standard regen down a given long hill with a given end point. On a test drive (still don't have my car) I tried it both ways, and the needle was further in the green with lower regen, because with higher regen it required a bit of a touch to the accelerator to maintain speed. Assuming that's the case, and assuming one wants to maintain constant speed, I concluded that under those circumstances, the nod goes to lower regen. Wrong?

Given: a dale where the down side equals the up side. [You can vary H (height) to any value you want on either side, but we'll start with them being equal]. You already know the answer: If you begin at a very slow speed, like a roller coaster, the coasting car will never be able to reach the other top. Therefore reduce the other top down to where the car just makes it over. Now you have a true roller coaster hill for that particular car and you now know difference in height needed to achieve this balance. Now adjust the height (keeping the same differential) such that the car's max velocity is now 50 mph. If the car now crosses the top too fast you will have to decrease the differential. This, I suggest, will be ~the highest hill(s) that the car can make it over. Any higher and air losses will consume too much momentum. Any lower and there won't be enough momentum vs rolling friction. Air losses begin to get serious around 45 mph, so I chose 50 mph. Continuing with this model you can play around with these 2 variables: Vmax (velocity max) and Hd (Height differential) to find the max height that the car can roller coaster successfully. Obviously the car cannot use brakes as 100% of the energy absorbed (heat) is lost to atmosphere. At least with regen ~80% of energy absorbed is saved for later use. But regen can not improve roller coaster performance due to this ~20% net energy loss. --

Another thought experiment. There's a valley between you and your destination, and you're going to drive across it at a constant 55 mph. 1. The valley has steep slopes. You're having to use a lot of regen to maintain your speed, pumping that energy back into your pack. Now you climb up the other side. You use your regen energy to help climb the other side, and you get about 80% of that energy back after drive train losses. Clearly this took more power than driving across on a level bridge. 2. The valley has very gradual slopes. As you go down, it's such a gradual slope that you don't use any regen - in fact you have to keep the power on slightly to maintain speed. Then you climb up the other side. In this situation, I'd suggest that the energy consumed would probably be about the same if you had crossed the valley on a level bridge - just a slight increase because you covered more distance (the extra vertical up/down distance). Many situations could be between #1 and #2. Again, this shows that the details really matter.

Interesting thread; this is something I wonder about as I drive along hilly highways. At the margin, there are some additional energy draws that are dependent on time spent driving, e.g. HVAC. Adding these into the mix will shift (every so slightly) the optimal strategy. If you add the money value of time, your answer will differ still (but that's probably beyond the bounds of this thought exercise).

Update: I created an Excel spreadsheet last night that uses a controllable time differential (dt) to calculate drag force, rolling resistance, and drivetrain losses at every dt. I then run it out to determine the energy lost, amount of time to decelerate, and distance traveled over that deceleration. I'm still tweaking the sheet and need to verify with some more data, but even as a rough estimate--not shrinking down dt too much, and completely estimating drivetrain losses--my spreadsheet calculated a deceleration time from 55 MPH to 45 MPH of 14.3 seconds while coasting. My first test this morning in the car showed the actual time to be 16.1-ish seconds--so not bad for a first rough estimate. I'll keep refining (primarily on the drivetrain losses), then add in the regen case, and post results. The problem's actually not too hard to solve at all using numeric methods. But if you want to solve it using equations, it gets much more dicey... - - - Updated - - - Edit: Just realized I was using 0.25 for the drag coefficient and 4606 lbs for the weight. If I change it to 0.24 (closer to the actual published drag coefficient) and 4800 lbs (closer to my config with the pano roof), the spreadsheet estimates 16.35 sec! So actually I'm already within about 1.5% of the actual number--using a complete guess on drivetrain losses! Not bad--so this should hold out to be fairly accurate...

Todd-- As you're discovering, the numerical solution is straightforward. The analytic solution is much more challenging. I have a degree in Engineering Physics and was trying to dust off 25 year old mathematics and it's complex. I believe it has an analytic solution but I am not confident in the path I am going down. I've been attempting to solve for the distance it takes to decelerate the coasting vehicle via drag from 75mph to 55mph. Then one has the extra distance that the regen vehicle has to travel and it's trivial to solve for the extra energy used for the regen vehicle to travel that distance. Let's just say that it's been really messy. I'm coming up with (at an approximation) distance = (m/b)*log(v2/v1) where b is 1/2*(rho)*Cd*A. This gives a distance of approximately 810 meters or 2,657 ft, roughly 1/2 a mile. Your time approximates to 1/3 mile so I'm missing something, even given that I'm ignoring rolling resistance. I expect I've made a stupid error in trying to make the differential equation solvable. EDIT: Ah, your test was from 55-45 mph. Jury is still out on my solution I may have to do some test runs with GPS to see how my approximation holds up.

JohnQ, thanks for the input. Yeah, I decided my advanced math wasn't fresh enough to confidently try to solve it . I'll provide more input later, but yes...I could only do 55-45 this morning as there was a car in front of me.