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Discussion Starter #1
Runners, bikers, and skiers know that a flat course is more efficient than a hilly one; but they don't regenerate energy. I always felt that my Prius benefited from hills but have no data. The BOLT allows some data gathering, and I think it supports my view that big hills increase fuel economy. Here is my reasoning.

I am confident I can get up a big hill (6000 ft in 50 miles, going 60-65 mph) getting 2.5 mi/kwh. Going down such a hill uses no energy. So that would be 100 miles using 20 kWh, for a rate of 5. But in fact the descent would store at least 4 kWh. So the total used is 16, giving an overall average of 100/16 or 6.25 mi/kWh. This is much more than I can get on flat ground at same speeds (60-65 mph).

Data to support: I can get from W Denver to the Eisenhower Tunnel (50 miles) using 21 kwh for a 2.38 mi/kwH average. But this includes several downhills. If it was pure up, I think my 2.5 estimate would apply, and I will try this sometime from Georgetown (8700 ft) to the tunnel (11100 ft), a section with no downs at all.

I can get from the tunnel to W Denver (11100 to 5200; 50 mi) using 2 kwH. The reason I have to use the 2 is that there a long flat stretch, and several good-sized uphills about halfway down.

One possible reason for some of the gain is that we know low temps. affect battery performance. But when going downhill at 0 energy, that becomes irrelevant. But the general principle here seems to transcend this. It just appears that the loss when going uphill is not that great, and the gain when going downhill is huge.

On my own small hill: 1.8 miles. 750 feet. Down: Gain 0.7 kwH. Up: Use 1.5 kwH. So total used is 0.8. That is a rate of 4.5 mi/kWh for the 3.6 miles. I know many of you have gotten 4.5. But my 4.5 is in cold temps. My first 2500 miles here have averaged 4.0. And my typical daily rides here (15-50 miles at 30-45 mph) average about 4.1 - 4.2 unless it is very cold.

I realize that this is not fully definitive and, naturally, my next time down to Denver I will record energy used on the pure downhill 20 miles from the Ike Tunnel to Georgetown and vice versa to see if the stats hold up.
 

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Combustion engine vehicles get better fuel economy in hilly terrain because climbing up the hill acts like a long "pulse" and descending down the back side is like a "glide". Combustion engines are extremely inefficient at partial throttle, and greatly improve efficiency near wide open throttle. At steady cruising speed and partial throttle, the engine might be 20% efficient at burning fuel and converting the heat into useful power. Climbing a hill loads the engine much more and the throttle will be closer to fully open. Towards the top range, the engine may be 30% efficient at converting fuel into useful power. This isn't a 10% increase in efficiency, but a whopping 50% boost in efficiency. This is why the "pulse and glide" method can greatly boost MPGs in a combustion engine vehicle.

Since the Bolt doesn't have a combustion engine, something else must be at play. My best guess is that your average speed going up the hill is slower than what your average speed would be on flat ground. That means less energy is being wasted overcoming aerodynamic drag. Essentially you're just driving slower to cover the given distance compared to an open and flat highway. The other consideration is that higher elevations have lower air density. Again, this reduces aerodynamic drag.

Aerodynamic drag is by far the #1 consumer of energy. Any reduction in drag will result in noticeably better efficiency.

 

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Discussion Starter #3
Yes, I was not thinking about drag and that reduction up here does help -- but that applies equally to all my driving.

And yes, I had heard that pulse and glide was optimal for ICEs. Also, I know that the experimenter -- me -- can bias things so I will carry on trying to get good data when occasion allows it. Eg, we go to Denver on nice days, when the moderate temps help the mileage.
 

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make sure you understand the prevailing winds. with airplanes, one direction is more efficient because of a tailwind and the return trip is terrible with the headwind.
 

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Discussion Starter #5
Almost always on this drive the tailwind would be eastbound (west wind) down the hill, which seems like it would have less effect than the corresponding headwind which would be on the expensive climb up the hill. That said, it was a mild day with very light winds.
 

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I can get from W Denver to the Eisenhower Tunnel (50 miles) using 21 kwh for a 2.38 mi/kwH average...
The real test is to measure kWh used, not the efficiency rating. So note the kWh used at the start and end of your segments and subtract to get the actual amount consumed.

My own, very informal test seemed to suggest that regen is somewhere in the range of 66% efficient - so going up and then down a hill will always use more power than going the equivalent distance on a flat road (other factors being equal).
 

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Combustion engines are extremely inefficient at partial throttle, and greatly improve efficiency near wide open throttle. At steady cruising speed and partial throttle, the engine might be 20% efficient at burning fuel and converting the heat into useful power. Climbing a hill loads the engine much more and the throttle will be closer to fully open. Towards the top range, the engine may be 30% efficient at converting fuel into useful power. This isn't a 10% increase in efficiency, but a whopping 50% boost in efficiency. This is why the "pulse and glide" method can greatly boost MPGs in a combustion engine vehicle.
A bit off-topic, but I believe this is the main operating principle for my Outlander PHEV when not in EV mode: it implements a hysteresis cycle. The engine is on for about two thirds of the cycle, propelling the car and recharging the battery a bit. This allows the engine to be off during the last third of the cycle. MPG during the first two thirds of the cycle is very poor. But after being compensated by the infinitive MPG during the last third of the cycle, the overall result is very good. I think it is easily explained: the total number of revolutions for the engine is two-thirds of what it would have been if the engine had run all the time. So, the amount of internal resistance to overcome is less.

For EV's like the Bolt, there is no such thing of 'turning off the engine'. The E-motor always spins along. Also, I have always assumed that losses in the system would be higher when the load was higher, even relatively. For example, when cables that lead from the battery to the E-motor transport more power they get hotter. So, more energy is lost. Of course, down hill they do not get hot. If my assumption is correct, and the load is for example 30 kW uphill / 0 kW down hill, the overall loss would be higher compared to driving on a flat road with a load of 15 kW.
 

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Discussion Starter #8
Sean: I did count kWh. My orig. post said: "I can get from W Denver to the Eisenhower Tunnel (50 miles) using 21 kwh." Yes, that is the proper way to do it and that is how I do it on all my experiments like this. As for 66%, that seems like only part of the story. Critical to this is how much the slowdown is for going up hill. But I agree that the data is not yet definitive and since I cannot drive a totally flat road in the same conditions it might not be possible to completely resolve. But even if they are not better, it seems clear that big hills are not a big liability for overall efficiency on a round-trip trip.
 

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A bit off-topic, but I believe this is the main operating principle for my Outlander PHEV when not in EV mode: it implements a hysteresis cycle. The engine is on for about two thirds of the cycle, propelling the car and recharging the battery a bit. This allows the engine to be off during the last third of the cycle. MPG during the first two thirds of the cycle is very poor. But after being compensated by the infinitive MPG during the last third of the cycle, the overall result is very good. I think it is easily explained: the total number of revolutions for the engine is two-thirds of what it would have been if the engine had run all the time. So, the amount of internal resistance to overcome is less.
I had a Ford C-Max Energi before getting my Bolt, and it followed a similar pattern. But I think the algorithm was driven by battery state and load demand (gas pedal position). At low demand, the car tries to use the battery first, since this is where the engine is least efficient. At very high demand, it uses both battery and engine combined. The interesting bit was in the middle range of demand. If the battery was low, it runs the engine at slightly higher output than is being asked for, and stores the excess in the battery. The engine is running closer to it's efficiency peak, so that's a good way to run it. That also builds up the battery, making it more likely that it can run on battery alone some point in the near future. If the battery is high, the threshold for starting the engine goes up, so it pulls the battery back down towards the middle of it's range by driving purely on battery more often. The results is what you see, the car alternates between running on engine and one battery, but the alternating is controlled by traffic patterns and hills more than anything else.
 

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But I think the algorithm was driven by battery state and load demand (gas pedal position). At low demand, the car tries to use the battery first, since this is where the engine is least efficient.
Exactly. This is why I wrote "when not in EV mode" ;-)
 

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Almost always on this drive the tailwind would be eastbound (west wind) down the hill, which seems like it would have less effect than the corresponding headwind which would be on the expensive climb up the hill. That said, it was a mild day with very light winds.
Are you coasting on the downhill faster than your average speed? In that case, a tailwind would reduce the aerodynamic penalty for going faster, more than it would penalize you climbing at a slower speed.
 

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Too bad the Outlander PHEV and Ford Energi have spark plugs in them. Otherwise they’d represent a comparator to the Bolt EV.

My own, very informal test seemed to suggest that regen is somewhere in the range of 66% efficient - so going up and then down a hill will always use more power than going the equivalent distance on a flat road (other factors being equal).
Up to now I’ve had that understanding as well. Based on prior posts like this one of 09-07-2017: ( I didn’t think that Teslatifosi-bias took away anything from the pure logic ):

…. regen physically only captures 80% of the kinetic energy….think about it this way - there are two factors going uphill

X kwh to go the "distance"
Y kwh to go "up"

you will always spend "X" - you can recovery some of "Y" on the trip back down - but will spend another "X" for the distance

if you spend 2 kWh to go 8 miles distance + 2 kWh because it's uphill (4 kWh total) - you can recover some significant percentage of the 2 uphill kWh going back down hill…

in both my bolt/Tesla you can basically go downhill for 'free' and stay either at the same battery %'age or gain a little bit - basically making the downhill portion free even though you still covered some distance - I've rarely seen a significant "gain" in battery percentage as a result of downhill recovery - but I often get the downhill portion for "free" - i.e. I have the same percentage of battery at the bottom of the hill as that top - which is free because I still drove the car some distance.
But now there are statistics that challenge the assumption that regeneration efficiency is perhaps in the range of 66%-80% and surely must always be less than 100%:

I am confident I can get up a big hill (6000 ft in 50 miles, going 60-65 mph) getting 2.5 mi/kwh. Going down such a hill uses no energy. So that would be 100 miles using 20 kWh, for a rate of 5. But in fact the descent would store at least 4 kWh. So the total used is 16, giving an overall average of 100/16 or 6.25 mi/kWh. This is much more than I can get on flat ground at same speeds (60-65 mph).
I don’t think my Bolt can attain that kind of efficiency. Gotta try an experiment on my own. Need a suitable hill.

According to Mr.Pythagorean, stanwagon’s theoretical hill must have a “run” of 49.987 miles. Because it has a 50-mile hypotenuse and a 1.1364-mile (6000 ft) vertical. Based on those characteristics the slope in percentage is 1.1364 / 49.987 (rise over run)= 2.27. And the slope in degrees is arctan (.0227)= 1.3°.

Hwy #401 east of Toronto has some hills. Perhaps I can identify a stretch of highway on a topo map, with 1.3° slope - - anyone know of one ?
 

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Cehjun, IIRC (it’s been more than 20 yrs), there’s a pretty long hill (up) going W on hwy 402 at London. Not rocky mtns length, but good enough for some instantaneous data?
 

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Hello StanWagon,

The distance from Lakewood, Colorado (near your West Denver(~?)) to the Eisenhower Tunnel is 52.5 miles according to Google Earth and is close to your stated 51 miles. You said you used 21 kWh. The elevation rise is from 5527 at Lakewood to 11030 at the tunnel entrance for an overall rise of 5506 feet. I see you used an elevation rise from 5200 to 11100 feet or a rise of 5900 feet.

If I understand ft-lbs correctly, it is the energy required to raise a pound of weight one foot. The conversion factor from foot-pounds to kwh is 3.76616 e-7.
The curb weight of the Bolt is 3563 pounds. Adding 2 adults and luggage, the Bolt weight would be around 4,000 pounds. Multiplying 5900 ft x 4000 lbs x 3.76616 x e-7 produces 8.888kWh that would have been used to raise your Bolt up about a mile in elevation. [Comments on this usage of ft-lbs to Kwhs would be appreciated.]

In other words 8.9 kwh of your 21 kwh would have been used just for the change in elevation leaving 12.1 kwh to go the horizontal distance of 51 miles. That works out to about 4.2 miles per kwh which is a reasonable usage rate without heat, A/C, and other electrical drains at lower end highway speeds. Not too shabby!!!

Going from the Eisenhower Tunnel to home, the Bolt would gain back those 8.9 kwh minus any inefficiencies due to Regen. If the pack fills, Regen can't be used. So, if there were no hills in between the Tunnel and home, I would expect your pack to be regen- erized by several kwh's until full. However, if you used 2 kwh, I assume that your pack was charged up along the way with the last miles being driven "on the flat". You'll know better of your situation.

I can see future onboard calculators, "Garmin GPSs", and Onstar apps taking elevation into an EVs use of kWhs to predict range. It will be a welcomed feature and maybe a way of sensing head and tail winds such as angle of the car from horizontal, speed, and anticipated energy usage versus actual.
 

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...it seems clear that big hills are not a big liability for overall efficiency on a round-trip trip.
Yes, coming from a Prius C which has comparatively limited regen and electric propulsion capabilities, I've been mightily impressed with just how efficient the Bolt is on the hills around town here in Vancouver. With the heater turned off and just running around our local streets I'm getting better than EPA rated efficiency on trips that go up and down multiple hills of a few hundred feet. That's way better than I saw in the Prius C, which was amazing on level ground but did substantially worse on a hilly route.

The only thing you have to watch out for is if you're planning a trip and your ending elevation is substantially higher than your starting elevation, in which case you can't gain back the power used to go up the hill. You'll gain it back eventually when you return home, but if you need to charge at the top of the hill you have to make sure you have the range to get there.

I made a trip up to the ski resort of Whistler, about 2000 feet above my home in Vancouver, and it took around 32 kWh to get up there, more than half my battery. I knew it would take less to get home again, but I charged up anyway "just in case". Sure enough, it only took around 22 kWh to drive back.
 

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For EV's like the Bolt, there is no such thing of 'turning off the engine'. The E-motor always spins along. Also, I have always assumed that losses in the system would be higher when the load was higher, even relatively. For example, when cables that lead from the battery to the E-motor transport more power they get hotter. So, more energy is lost. Of course, down hill they do not get hot. If my assumption is correct, and the load is for example 30 kW uphill / 0 kW down hill, the overall loss would be higher compared to driving on a flat road with a load of 15 kW.
If you are going downhill and not coasting in Neutral, then those same cables will be carrying current, possibly more current at times, upwards of 200A.
 

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Discussion Starter #17
peterE: Home is Silverthorne on the W side of the tunnel, not in Denver area like I think you thought. To repeat the basic numbers:
Home to Lakewood: 60 miles. Uses under 8 kwH. Then back up to Home, 60 more miles, uses about 21 kwH.
The complication is that on the section from Idaho Springs (about 8000 ft) down to Lakewood (5500 feet) there are several large hills: one very large one over a 2-mile stretch.

So the experiment is really best done on the section from Georgetown Exit to the W end of the tunnel, which is pure up all the way (8500 to 11100).

But it is certain that ambient temp. plays a big role. Today I went from home to Winter Park. 90 miles. Net loss of 1000 feet. I took the Kremmling route, so no big hills. 90 mi / 24 kwH = 3.75 mi/kwh. After a ski race I came home via Berthoud Pass (giant climb to 12000) and the Tunnel (another very big climb to 11100). 56 miles over 14 kWh. Rate of 4. Net gain of a thou. The drive home was at faster average speeds. Also warmer temps (40 vs 20 degs). But it had two very big climbs. So, not to belabor, but the big hills are just not a big problem.

Like many of you I am looking forward to spring and summer when ambient temps will improve the rate a lot. And, apropos of discussions elsewhere, I am not using N, just driving along at whatever seems reasonable on the downs.
 

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I can get from W Denver to the Eisenhower Tunnel (50 miles) using 21 kwh for a 2.38 mi/kwH average.

...

I can get from the tunnel to W Denver (11100 to 5200; 50 mi) using 2 kwH. The reason I have to use the 2 is that there a long flat stretch, and several good-sized uphills about halfway down.
Now *that* is why I read posts here. Thank you for the useful real-world information!

FWIW, another benchmark: Boulder to Ridge Road in Nederland is 13 miles and 3200 feet of ascent, in twisties at 30-45 mph. It takes about 8 kWh or a little under. The descent takes about -2 kWh [sic].
 

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If you are going downhill and not coasting in Neutral, then those same cables will be carrying current, possibly more current at times, upwards of 200A.
I did say 30K kW uphill, 0 kW down hill, didn't I? ;) Of course you are right, but you should read my post in the context of what was written by the TS. He suggested that going uphill using a lot of power and then downhill with 0 power might result in an average better than when driving on a flat road.

I guess my 'question' was: when is more heat dissipated in the cables (or in the battery, E-motor, ...): when consuming 30 kW for 10 minutes and then 0 kW for 10 minutes? Or when consuming 15 kW for 20 minutes? (30 and 15 just being random examples)

I think an ICE working quite hard for a bit and then not working at all for another bit is more efficient than an ICE working not so hard for two bits. But, depending on the answer to above question, with an electrical drivetrain it might be the other way around (especially as the E-motor cannot be unlinked front the wheels).

Not saying it is. Just wondering.

About going down hill: we always say that with an EV we should try to maintain constant speed to get the best efficiency. But I think we should let nature take it's cause. On a hilly road, we want to allow speed to increase when going down and allow it to go down again when we go up. And not try to fight gravity. That way we have less peaks in power consumption (or production). Only downside is that, although you average speed will still be roughly the same, the lower speed going up hill does not compensate for the higher speed down hill, from a drag perspective (speed squared and such).
 

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I think an ICE working quite hard for a bit and then not working at all for another bit is more efficient than an ICE working not so hard for two bits.
I am not so sure that pulse and glide actually works for ICE. The only engineer I ever heard comment on it said it doesn't work. He was a retired Ford engineer, and built efficiency event winning ICE vehicles.

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