I'm sure this question has been posed many times. In googling for an answer I found a thread on Reddit that provides some good info, but not exactly what I'm looking for. I am interested in getting an idea of the recovery rate in say the following scenario: You are driving up a 1 mile long hill with a 5% grade at 45mph. These are just example numbers. When you reach the crest you drive down the 5% grade hill for 1 mile at the same speed. So, if using enough just enough regen to not slow the card down to under 45mph (assume no actual use of brakes), how much of the power used to get up the hill is put back into the battery?

In my experience going up and down hills, you get about half your miles back. For instance, if you start at the bottom of hill with 250 miles and the hill is 10 miles to the top, by the time you get back to the starting point (20 miles total) you will be left with approximately 240 miles.

Wow. That is significant. I am in line for a M3 and had not thought about the benefits of regen until just recently. Your factoid gives me my first impression of how truly helpful regen is. Thanx.

This can't be right. It implies that 20 miles of driving (hilly driving at that) uses only 10 miles of range. In my experience, driving up and back down a ~3% grade (e.g. Sepulveda Pass in West LA) averages about 10% more expensive than driving the same distance on flat ground. So it may be practically "free" to drive down the hill (effectively coasting at 0kW), but will be more than double as expensive to drive up the hill. As hills get steeper and regen gets involved, this ratio gets worse. Driving up the CA Grapevine and back down (about a 6% grade) probably averages about 20% more expensive overall than driving the same distance on flat ground. It would be an interesting test.

I was typing roughly the same response as Ben. You definitely don't use less range going up and down than just driving flat. You use a bit more.

Can you explain again? If you drive 20 miles starting with 250 miles range, on a flat course, you should expect to have 230 miles remaining. Are you saying that going up and down a hill for that same 20 miles only uses 10 miles of range? That isn't getting 50% back, that's getting 150% back! **** Lifting 1.5 tons up 800 feet is about 1 kWh, IIHDMC. Converting that height back to kinetic energy (speed) by coasting is basically 100% efficient. If regenerative braking is done (either by you or the car), you could probably expect 80% back into the battery. You still need to pay for moving the car through air though. Thank you kindly. ETA: Beaten to the punch, cause I stopped to do math...

I've always wondered about using this method for setting the range record. For example, Beartooth Pass, in MT. WAG numbers. 30 miles uphill at 600wh/m (18kw/h's) 30 miles downhill at negative 300wh/m (Note: Teslas do better than "free" going downhill, it actually puts "fuel" back in the "tank") (-9kwh's) 60miles miles traveled, net used 9kwh, implied range in an 85kwh with 80kwh accessible, is 533 miles. As I said, I am guessing at numbers, but they seem plausible to me.

One immediate thing I noticed when I test drove a Model S is that the regen braking is much more aggressive than my Volt...even in L which, in the Volt is just a way to get more regen.

Sadly, no. 30 miles at 333 Wh/m plus 10kWh for 8000 feet = 20 kWh uphill then 30 miles at 333 Wh/m minus 10 kWh for -8000 feet = 0 kwh downhill net used 20 kWh versus 20 kWh for 60 miles flat at 333 Wh/m. Now if you use regen brakes instead of freewheeling, you might put that 10 kWh back in the battery. the 30 miles at 333 Wh/m = 10 kWh downhill (and 8.5 kWh in the battery (after efficiency losses) net use 30 - 8.5 = 21.5 kWh. [assuming all sorts of simplifying things, of course] Thank you kindly. p.s. Energy is measured in kWh (kiloWatts * hours) not kW/h (kiloWatts / hour).

Happily, yes. The downhill portion, through regenerative braking, gains 10kwh. The uphill costs 20kwh, the net is 10kwh, in this oversimplified scenario.

OK, so now you are approaching the bottom line. Just to be clear, because this is far more delicious than I realized, in some cases driving, say, 100 miles on a mountainous road could possible use about the same amount of energy as driving the same 100 miles on a flat road? (Again, in our simplified example, ignoring variables like wind, freewheeling, stopping to smell the roses, etc.)

The only way this is possible is if the downhill section is much longer than the uphill section (shallower grade). Another way it happens is if the hills are short enough so that the speed you gain going down one hill is enough to get you almost to the top of the next (too much speed just causes more aerodynamic losses). Both cases are rolling 19 using a d20. Generally you can never get back all the kinetic energy there is at the top of the hill because some will always be lost to heat.

Where I live is pretty flat, so I have less experience with this. However, compared to a gas car, EV's are going to actually gain charge when going downhill. Compare that to a gas car, which uses less fuel when going down a hill, but gains none. In an EV the uphills will take a big bite, but the downhills give a meaningful portion of it back.

Right. My reasoning for posing this question is that I will likely be the first in my (remote, mountainous) community to own a pure electric vehicle. I am posting a couple of essays in the local paper sort of evangelizing the coming of the long(er) range electric car. Our round trip commute to town is about 150 miles, so the Bolt and the M3 are really the first "affordable" solutions for making that run without having to worry about charging. So I'm sort of preemptively attempting to extinguish the concerns the locals might have about relying on an electric car. The idea that our local hills here can be used to advantage over an ICE car (making lemon aid from lemons) is pretty exciting, but I need to have my [theoretical] facts lined up before shooting off my mouth. As long as I don't oversell the concept of regenerative braking I think I can make the case now. Thanx to those who are chiming in to make this possible.

ps - I just changed my avatar. When I was a kid the amusement parks had these electric cars (Bumper cars or Dodger cars). You stepped on the pedal and they went. You let up on the peddle and they stopped. Who would have known that that's what the future held for the big boys and girls?

Here are my actual numbers driving between the Corning SC (276') and the Mt. Shasta City SC (~3550'). The distance is roughly 108 miles. I have driven uphill twice (331 wh/mi and 346 wh/mile) and driven downhill once (216 wh/mile). For all intents and purposes the interstate highway grades are identical between the northbound and southbound lanes. It is only the leg from Redding to Mt. Shasta City where the elevation changes dramatically. Redding is around 560' elevation, and is 49 miles from Corning. In comparison, from our home to the Manteca SC is 105 miles, with an elevation loss of about 250'. My average wh/mile in either direction is approximately 270. (I am not a hard or fast driver--leisurely is a more apt description.) I realize that this gross information does not indicate how much the regeneration actually replaces into the battery going downhill over a given stretch. But it might give someone with better mathematical skills than I an insight.

So Cpa's numbers would put about 337wh/mi up, and 216wh/m going down. Therefore averaging 276.5wh/mi for an up/down compared to nearly flat 105 miles route at 270wh/mi. Very small sample size, but points in the very general direction of being close to a wash. I think an important distinction to make is the elevation profile. Imagine you started at the base of a 10 mile uphill, then came down the same 10 mile downhill, with the uphill being all uphill and the downhill being all downhill. In this situation you would would put charge back into the battery the entire way down and consume the entire way up. In Cpa's situation, probably more real-world, the downhill portion is a net downhill but, I presume, includes uphills, and flats. This also means the uphill portion includes downhills and flats. The numbers start to get blurred together. Coming down a large hill in MN, with my energy graph set for the last 5 miles, I had something like -350wh/mi and projected range of 999 miles. I would really like to get a range charge at the bottom of a long, all uphill uphill road, and just go up and down and up and down and see how it all shakes out.

Wow. Much ado about something simple. The question the OP wants to ask is: how efficient is regenerative braking? Forget about hills and up and down and all that. The answer is in general, normal driving conditions, including hills (not the most extreme ones) and regular braking (not race track level) regen is in my experience somewhere around 50% efficient.