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Hills and Energy Use

35210 Views 213 Replies 25 Participants Last post by  ARob
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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Here are some numbers: Butte, Montana, elevation at my house 1680 M, 5530 ft, Helena, Montana 1240 M, 4060 ft...

140 RT to Helena and then errands around Butte. Started the day fully charged at 205, used only 40 miles range on the trip to Helena...generically downhill, and then 100 on the uphill trip back to Butte...total range at that point would have been 203 and then I gained a few miles of range around Butte up to 215 total. Pretty much winter driving, temps from mid-teens to 30, intermittent snowpack on I-15. Pretty pleased overall with its performance.
Interesting. Shows that Bolt is very sensitive to grade. Don’t know exactly however, what algorithm GM is using to update the GOM numbers as the miles tick by. Would be better when possible to measure in kWh’s used.

From Helena it’s 61.5 miles on I15 (once onto I15 west off Prospect Ave) to Butte still on I15 (heading into the loop with I90). Not too steep, average slope is 0.6% or 0.35 degrees.

To simplify based on info earlier in this thread: uphill on that I15 stretch at 60mph may use (61.5 mi @ 3.5 mi /kWH) 17.5 kWh’s, add for temperature (say 32F versus 72F) 25% vs EPA test = 3.8 kWh, add for in-town 8.5 mile balance of trip at both ends (@ 4 mi /kWH) 2.1 kWh. Guesstimate total kWh for leg from Helena to Butte is 17.5+3.8+2.1= 23.4 kWh. Could look like 40% of a “tank”. Might correlate with the 100 the GOM showed, or not. Dependent on many other variables such as intra-hills, wind direction & speed, dry versus snow covered road surface.
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Answer: No* & No & No.
*- except the statement about air resistance in post #114 .

Roper buries the resistance due to friction into coefficient “b”. As indicated on the jpg in post #113 , his paper is alive and well here: http://www.roperld.com/science/ChevyBoltRange.htm

More specifically refer to picture below for illustration of increasing force necessary to overcome increasing friction going up a hill.


If there is anyone on this forum who does not deny the laws of physics. And who has some real life experience and / or (better yet) data or pictures to contribute about Bolt EV energy & hills. Please step forward.
One more time:
His equation for energy required to drive a distance d up a slope uses the linear coefficient in the second term.
The second term does NOT change based on slope

Frictional force is dependent on the normal force and that DECREASES with slope, as I have stated before. It is a maximum on a flat surface and ZERO at a 90 degree angle. Which is what another poster confirms as well.

If you need a diagram:
https://www.quora.com/Why-is-friction-less-on-an-inclined-plane
More specifically refer to picture below for illustration of increasing force necessary to overcome increasing friction going up a hill.
There are two terms in the equation in the chart. The first is the energy required to overcome gravity (notice no coefficient of friction). The second term is the energy required to overcome friction.

Note that the second term DECREASES with increased slope.

QED
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There are two terms in the equation in the chart. The first is the energy required to overcome gravity (notice no coefficient of friction). The second term is the energy required to overcome friction.

Note that the second term DECREASES with increased slope.

QED
Took me awhile to see where the confusion is occurring here. The example @Cehjun is showing is the push force required to get the car moving again on an incline. At rest, the friction is holding the car still on the incline. That force has increased, even though the friction is decreasing. Of course, at some point, if you continued to raise the incline, friction would fail to hold the car in place. So the force to move the car has to overcome the force holding it in place. Which really is the portion of the weight of the car due to the incline. Although the total force to move the car implies that the tires are not moving and your going to be dragging the car with the wheels locked. Not realistic. That's why man invented the wheel.
Frictional force is dependent on the normal force and that DECREASES with slope, as I have stated before. It is a maximum on a flat surface and ZERO at a 90 degree angle. Which is what another poster confirms as well.

If you need a diagram:
https://www.quora.com/Why-is-friction-less-on-an-inclined-plane
There are two terms in the equation in the chart. The first is the energy required to overcome gravity (notice no coefficient of friction). The second term is the energy required to overcome friction.

Note that the second term DECREASES with increased slope.

QED
No ͭ ⁴ ⁺ ͭ ⁵ .

It’s obvious that the post as referenced on quora.com emanated from a blogger who likes listening to himself while blithering too much garbage too long causing the thread to self-destruct. (which I’m concerned this thread is doing). Dummies.com is much more reliable.

It is accepted in the literature: in order to “slice” the energy required to hold a vehicle on a hill between normal force (the force perpendicular to the object) and friction force. The friction piece is denoted mgsinθ.

In picture in post #114 example: friction piece is 1,441 Newtons as shown bottom right.
Took me awhile to see where the confusion is occurring here. The example @Cehjun is showing is the push force required to get the car moving again on an incline. At rest, the friction is holding the car still on the incline. That force has increased, even though the friction is decreasing. Of course, at some point, if you continued to raise the incline, friction would fail to hold the car in place. So the force to move the car has to overcome the force holding it in place. Which really is the portion of the weight of the car due to the incline. Although the total force to move the car implies that the tires are not moving and your going to be dragging the car with the wheels locked. Not realistic. That's why man invented the wheel.
No. In assessing subject case (pic in post #117 ) one needs to contemplate static friction versus kinetic friction. And the assumed coefficient of friction.

The example represents a situation involving a Bolt on an incline. Whose tire to dry road coefficient of friction is estimated reasonably as ~0.8. And friction is static: the wheel & tire is capable of rolling down the incline but is held in place by way of the opposing Newtons as indicated.

The point of the exercise was to demonstrate that the friction piece increases as slope increases. That’s all.

For a more comprehensive analysis of vehicle propulsion energy, look at Van Sterkenburg’s formulas as referenced in post #107 .
I enjoy the theory discussed here and wish I had gone further in my math studies, but it seems we're overcomplicating the matter.

We know that more energy cannot be returned than spent going up a hill. The only way hills could be more efficient than flat steady cruising then comes down to factors such as greater powertrain efficiency at higher load (is there info on this?), slower average speeds, or reduced in air density.

It should be possible to get a relatively accurate test by several trials of using cruise control to cover some distance going up a hill and back, vs travel some flat distance and back. The trials would need to be be performed in the same general direction and timeframe to control for wind and temperature variables.

My biggest curiosity is how the powertrain efficiency varies with load. In conventional gasoline vehicles, efficiency goes up with higher load. I'm not so sure with EVs. There is a certain amount of power consumed to merely have an active system that isn't moving at all. This is needed to close the contactors, run the computers and sensors, etc.
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I enjoy the theory discussed here and wish I had gone further in my math studies, but it seems we're overcomplicating the matter.

We know that more energy cannot be returned than spent going up a hill. The only way hills could be more efficient than flat steady cruising then comes down to factors such as greater powertrain efficiency at higher load (is there info on this?), slower average speeds, or reduced in air density.

It should be possible to get a relatively accurate test by several trials of using cruise control to cover some distance going up a hill and back, vs travel some flat distance and back. The trials would need to be be performed in the same general direction and timeframe to control for wind and temperature variables.

My biggest curiosity is how the powertrain efficiency varies with load. In conventional gasoline vehicles, efficiency goes up with higher load. I'm not so sure with EVs. There is a certain amount of power consumed to merely have an active system that isn't moving at all. This is needed to close the contactors, run the computers and sensors, etc.

1. Overcomplicating: Yes, that’s a polite way of saying it. This was a great thread at the outset but has devolved into : 5% = good relevant information, 10% = reminiscing about one’s PHEV (whatever that is) or one’s former Nissan, 5% = MPGe eventually moved to another thread, and unfortunately 80% = sparring about the meanings and expressions of the “physics” involved.

2. Relatively accurate test…going up a hill & back vs flat distance:
There are (3) fairly good ones in this thread so far. With various slopes and distances: see posts #55 , #73 , #104 and discussion thereof, & stanwagon said he’d post more

3. Bolt EV powertrain efficiency as it varies with load: Great question. To me that means: how much power gets to the road compared to the power supplied by the battery? …(or compared to power delivered to the motor after it gets somewhat discombobulated by the inverter & the electronic controls?) … but also in a “dynamic” sense. That is, by “throttle opening” so to speak.
Google search is always one’s friend, there’s a chart showing a Leaf’s in here:
https://faculty.washington.edu/dwhm/2016/02/29/will-automation-benefit-electric-vehicle-efficiency/
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On March 13 I have to go to Boulder. My plan is to record miles and kWh used at

1. Top of hill (E end of Eisenhower Tunnel (instead of W).
2. Bottom of hill (Georgetown, where it flattens out.
3. Exit off I-70 to Rt 6 to Boulder after Idaho Springs (10 miles farther along on relatively flat, tho descending ground)

Then on return, repeat. So when I am done (and I will try to maintain constant speed). I will have a rate for both direx. of Idaho Springs <-> Georgetown (freeway, flattish) and GTown to Tunnel (very steep).
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Took me awhile to see where the confusion is occurring here. The example @Cehjun is showing is the push force required to get the car moving again on an incline. At rest, the friction is holding the car still on the incline. That force has increased, even though the friction is decreasing. Of course, at some point, if you continued to raise the incline, friction would fail to hold the car in place. So the force to move the car has to overcome the force holding it in place. Which really is the portion of the weight of the car due to the incline. Although the total force to move the car implies that the tires are not moving and your going to be dragging the car with the wheels locked. Not realistic. That's why man invented the wheel.
That's what I thought at first (static friction + dynamic friction). But that isn't what that equation is.

The first part of the equation is simply the energy required to lift the object (mass times gravity times the rise). There is no frictional component to that part of the equation.

The second part of the equation is the dynamic friction and decreases with slope
No ͭ ⁴ ⁺ ͭ ⁵ .

It’s obvious that the post as referenced on quora.com emanated from a blogger who likes listening to himself while blithering too much garbage too long causing the thread to self-destruct. (which I’m concerned this thread is doing). Dummies.com is much more reliable.

It is accepted in the literature: in order to “slice” the energy required to hold a vehicle on a hill between normal force (the force perpendicular to the object) and friction force. The friction piece is denoted mgsinθ.

In picture in post #114 example: friction piece is 1,441 Newtons as shown bottom right.

You are simply mistaken

Mass x Gravity x Height is the energy required to lift an object in a gravitational field, absolutely nothing to do with frriction. Sine of the angle = the height (because for purpose of calculating actual energy he is using 1 for the distance, ie the length of the hypotenuse)

Energy required to lift the object plus distance (1) times the dynamic frictional component = total energy

An increase in slope decreases dynamic frictional forces. At an angle of 90 degrees it is 0 and at a slope of 0 (flat) is is maximal.


There is no frictional component to that at all

Holding force involves a component of static friction, not dynamic. Static frictional forces are overcome once, not during the entire distance traveled up a slope. As such its insignificant and can be ignored when computing the energy required.
No. In assessing subject case (pic in post #117 ) one needs to contemplate static friction versus kinetic friction. And the assumed coefficient of friction.

The example represents a situation involving a Bolt on an incline. Whose tire to dry road coefficient of friction is estimated reasonably as ~0.8. And friction is static: the wheel & tire is capable of rolling down the incline but is held in place by way of the opposing Newtons as indicated.

The point of the exercise was to demonstrate that the friction piece increases as slope increases. That’s all.

For a more comprehensive analysis of vehicle propulsion energy, look at Van Sterkenburg’s formulas as referenced in post #107 .
In the dummies.com equation the friction component is multiplied by the cosine of the slope. It decreases in magnitude with slope.

Static frictional forces have nothing to do with the energy required to drive a Bolt down the road, dynamic frictional forces do. The coefficient of dynamic friction is multiplied by distance, you would not do that with a static frictional force.

In other words, ignore static frictional forces because they simply don't matter when computing energy required to drive up a hill.

Now a simple question for you when you pull an objectstright up, is the frictional force >0 or =0? Static or dynamic, it doesn't matter. Since there is zero force between the object and what it is being pulled up the answer is zero. Which is why when you increase slope the frictional force decreases. If you disagree then tell me how much MORE friction there is when you pull an object straight up. We already know what the force is to overcome gravity, it is m*g so tell me how much MORE force is required to overcome friction.

We already know that all the equations you have provided state that the energy required = m*g when the slope is 90 degrees. But that's just the force of gravity acting on the body, no friction involved.
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On March 13 I have to go to Boulder. My plan is to record miles and kWh used at

1. Top of hill (E end of Eisenhower Tunnel (instead of W).
2. Bottom of hill (Georgetown, where it flattens out.
3. Exit off I-70 to Rt 6 to Boulder after Idaho Springs (10 miles farther along on relatively flat, tho descending ground)

Then on return, repeat. So when I am done (and I will try to maintain constant speed). I will have a rate for both direx. of Idaho Springs <-> Georgetown (freeway, flattish) and GTown to Tunnel (very steep).
Good. Much appreciated. I've marked it down on my calendar to check back to this thread around March 14th.

Until then it's not worth reading this thread. As two posters are intent on continuously flicking turds into the punchbowl.
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Here is a detailed Word document from the University of Nebraska that explains friction on inclines in great detail:

https://www.google.com/url?sa=t&rct..._Incline.doc&usg=AOvVaw1ABN3Rc0_soR6Bfqk89xp9

It clearly states:

" Putting “Friction on an Incline” in Process terms: Friction on an incline is a generalization of friction on a horizontal surface. As the angle of the incline is increased, the force of friction will decrease due to the change in normal force"

Which is exactly what I have told you from the start. Doesn't matter if it is static or dynamic friction either, Less force pressing two surfaces together = less friction, doesn't that seem intuitive?
As two posters are intent on continuously flicking turds into the punchbowl.
Am I among them? 0:)
Today's experiment:

East side Eisenhower Tunnel (11000') to Georgetown exit (about 8500'). Very steep (3.8% grade). 12.6 miles. -1.2 kWh used on descent, 6.8 on ascent. Rate = 4.50 mi/kwh for the 25.2 miles.

G'town to Idaho Spgs. (this is not flat, but is the best I could do to get a comparison on roughly flat freeway, trying to drive roughly the same speed). 28.6 miles round trip. 6.3 kWh used (1.1 on descent, rest on ascent). Rate = 4.54 mi/kWh

So these rates are close enough to be considered the same, which is still interesting because the steep leg here is far far steeper than the nearly-flat leg. But I suspect that on a truly flat leg the rate would rise from 4.54 to 4.6 or more.

(I have now drives 4000 miles on the Bolt, with the average being 4.0, but maybe 1-2% of that is heat and batt. conditioning, so the true rate is under 4.1 for those varied miles at varied speeds on varied terrain.

Temps were in the 40s; road conditions good.
Today's experiment: East side Eisenhower Tunnel (11000') to Georgetown exit (about 8500'). Very steep (3.8% grade). 12.6 miles. -1.2 kWh used on descent, 6.8 on ascent. Rate = 4.50 mi/kwh for the 25.2 miles.

G'town to Idaho Spgs. (this is not flat, but is the best I could do to get a comparison on roughly flat freeway, trying to drive roughly the same speed). 28.6 miles round trip. 6.3 kWh used (1.1 on descent, rest on ascent). Rate = 4.54 mi/kWh

So these rates are close enough to be considered the same, which is still interesting because the steep leg here is far far steeper than the nearly-flat leg. But I suspect that on a truly flat leg the rate would rise from 4.54 to 4.6 or more.
Two things continue to surprise about your numbers stanwagon: 1.) how close to 100% regeneration “efficiency” you’re getting on those mountainous hills, and 2.) in general how relatively high your mi / kWh numbers are, as compared to what I get.

So I plugged your latest above numbers into my crude little worksheet “model” Bolt energy estimator. I had first tried out this model on paulgipe’s trip from Bakersfield CA to Pismo Beach CA and posted a pic of results in that thread of this picture here.

Making necessary adjustments for air density (altitude), temperature to 40F versus 72F and I used a one-Colorado person plus carryon GVW of 3765lbs,1708kg. And constant speed of 60mph.

Downhill: From Ike tunnel eastbound exit to Georgetown at I6 exit #228 . I split this into two pieces. From tunnel to the steep grade warning sign at mile 9.7 (average grade 3.55%), versus from mile 9.7 to exit #228 Georgetown (average grade 4.73%). But I can’t do regeneration yet. Have to work into the model, regeneration as a braking force at the wheel. And at the battery. But I did learn that in general at 60mph with no wind at 72F and at ~10000 feet, regen kicks-in going downhill at slope of 2.46% or steeper.

Uphill: Observed was 6.8 kWh. Model said: nominal= 2.39, with slope= 6.1, with slope & 40F= 6.34, with slope & 40F & headwind of 10mph*= 7.03. The numbers make sense.
* - nearest NWS weather station at Berthoud Pass indicates there was quite a strong west wind on Feb.27th

Flat: going from Georgetown exit #228 to Idaho Springs exit #240 elevation change is ~1000 feet or 1.6%. Two way observed was 6.3 kWh and 4.54 mi / kWh. Model two way said: 5.1 (using a total of 23.6 miles between those highway markers) and 4.63 mi / kWh. Again quite close on the mi / kWh number.

Your relatively high mi / kWh numbers in general ? Answer: unlike an ICE vehicle which gasps for air at higher elevations, altitude is your friend in Bolt. Density of air goes from 1.2 kg/cu.metre at Toronto to about 0.9 kg/cu.metre at 10000 feet. Big impact on Bolt’s energy efficiency ! Example: per above under Flat, model said 4.63 mi / kWh. Change the air density to 1.2 and it becomes 3.77 mi / kWh - - welcome to a typical Toronto number for 60mph at 40F.

Efficiency of regen ? Not sure. Might the west wind of February 27th have helped the downhill leg from tunnel to Georgetown ? And disadvantaged the uphill leg…

PS, I tried the model on a 95km (60 mi) trip I made to a lake north of Toronto on Feb 27. No hills steep enough to trigger regen. Broke it down into 9 segments each determined by change in elevation. Model said: one way 15.94 kWh and 3.68 mi / kWh. (I haven’t yet analyzed the return trip). Actual measured was 15.4 kWh and 3.83 mi / kWh. Variance in most of the 9 segments is explainable. I’m having more success with the little spreadsheet (but with much more labour involved) than any online trip planner/usage estimator that I’ve found so far. Although I expect that someone will eventually come up with a good online estimator specific to Bolt EV.
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And I think one explanation of the fact that these occasional trips to Denver yield numbers significantly better than my overall 4000 mile average is temperature related. I guess I will find out more on that score in spring and summer. But since it is clear (EPA ratings) that slower driving is better than faster for the Bolt, it is at least interesting that the opposite has been the case (on hilly terrain) in my experience. And note also that I am not driving in Neutral on the downhills, but I trying to keep it N like via the pedal. I don;t really want to use cruise control to get constant speed. That is the way to force constant speed, but it is not the most efficient way to drive.

Digression: China has just banned the letter "N" since people are using "N > 2" to refer to the their emperor's change in the law to allow life terms. Oh, "emperor" is also banned.

For the record, the last control point of my run was the W end of the small tunnel (known as the Twin Tunnels) about 2 miles E of Idaho Springs town.

And, yes, as has been mentioned air resistance could be the big thing re. the difference between Toronto and Silverthorne.

Cehjun: for the record, the trip back up had about 220 pounds more weight than the trip down.
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Ambient Temperature
Presently I’m using just a looky at the graph in L.David Roper’s paper here under ‘Range Versus Temperature’. Admittedly the original source is a study involving Nissan LeafI. Since I suspect the Bolt’s battery to be less volatile vs. temperature than the old Leaf, there obviously isn't enough reliable data to justify modeling temperature any further. And by using the old Leaf’s data I felt that any adjustment representing ambient temp below 72F would be overstated (conservative). Need experimental data specific to Bolt EV’s battery @ SOC & load versus ambient temperature.

220 lbs (100kg) more GVW on uphill leg from Gtown to tunnel
Moves the model from 6.34 kWh (no wind) to 6.56 kWh (no wind). Closer to the 6.8 observed.

About the current use of the letter N. Amongst all of it’s meanings Isaac Newton must be rolling in his grave.:eek:
Housekeeping re: the 3.76616E-07 conversion factor.

PeterE in post#14 suggested using the conversion formula from foot-pounds to kilowatt hours, in order to estimate the work required to "lift" a Bolt EV the amount of the vertical distance of the hill. Where, one (1) foot-pound (as a unit of work, not torque) equals 0.000000376616 kWh. XJ12 in post #63 laid out how foot-pounds translate through standard international (metric) units in order to get to kilowatt hours.

Some debate ensued as to the useability of this ft-lb to kWh conversion. Whether or not it needed further adjustment for energy losses between the road wheel and the Bolt's battery. However, use of this 3.76616E-07 conversion gave surprisingly close results to test data (posts #104 and #105 ) from a Bolt 's readouts at the battery level. Now I know why.

After more investigation. CONCLUSION: the 3.76616E-07 conversion is accurate for estimating slope impact when measuring force at the road wheel . But not as measured at the battery. The "proof" for this conclusion is detailed below. By way of matching the result of the 3.76616E-07 conversion, to the general formula for vehicle propulsion energy available from many authors. See bottom of this post for details. (caution: may cause headache).

Then the question becomes: what are the efficiencies (losses) from a Bolt's "battery to wheel" ? Absent someone’s live experiment, one way to estimate battery to wheel losses: compare the output of the "model" (see post #137 ) which reasonably estimates kWh's used at the battery level for each mini-trip within a trip, both with and without the effect of slope. Compare that to: kWh's used at the road wheel as calculated from the general formula for vehicle propulsion energy, which is known to give the same result as applying the 3.76616E-07 conversion method.

Surprise: Bolt's battery to wheel loss as estimated this way, is only 5.2%* on average. Much much lower loss than was shown in post #93 picture from the DoE’s webpage. More in line with single digit loss that XJ12 referenced with some source information in post #108 . (which I questioned, sorry:eek:). Certainly this 5.2% battery to wheel loss estimate is wrong, but it might be a good indicator -- given available information it's quite likely that the % battery to wheel loss is a single digit one.
*- calculation performed using all Bolt parameters and quantified variables as based on uphill leg from Georgetown CO to Eisenhower tunnel entrance. Details are available.

What this means: the 3.76616E-07 conversion provides a quick way of determining incremental slope impact as measured at the road wheels. One can envision an elevation map digitized. Such as one sees when elevations pop out on Google Earth when manually running the cursor over each lat&long second on the map. The calculation of energy use due to hills becomes a piece of cake.

And given that the further adjustment required to derive kWh's at the Bolt’s battery, from kWh's at the wheel, is likely no more than 1.11 . It all forms a nice way of handling elevation changes. Which represent a very significant variable in predicting trip usage.

Great thread stanwagon. I learned a lot and had fun doing it.
--------------------------------------------------------------------------------------------------
Proof: Conversion of foot-pounds directly to kWh’s yields energy as measured at the road wheels

Calculate the vehicle propulsion energy equation. (see post# 107 Van Sterkenburg, or other authors). Below example makes reference to trip leg assumptions using the grade above Georgetown CO to the Eisenhower tunnel. The assumed values for variables are not critical.

Energy required at the wheels for one second of time [Es] is:
Es = v*(½*Cd*A*ρ*v2 + m*g*(rw+sinθ) + m'*a)
where,
v – planned speed in metres per second. assume constant speed going into the start marker, same constant speed over the leg and same constant speed departing the finish marker. assume 60 mph or 28.62 metres/sec.
Cd – drag coefficient. 0.308 is latest published for Bolt EV.
A – frontal area of Bolt in square metres. assume 2.816. (conservative, assume no air gap below frt spoiler)
ρ – density (weight) of air in kg. per cubic metre. assume 0.9 high altitude Colorado mountains.
m – Bolt’s mass (weight GVW) in kilograms. assume 1707.8 or GVW of 3765 lbs one person + carryon.
g – force of gravity in metres per second2 . assume standard 9.8.
rw – rolling resistance coefficient. assume 0.0097881 for Bolt EV. estimate sourced here.
sinθ – slope in radians. this hill is 3.54% or 2.03 degrees or .0354 radians. this hill begins at 9,211 feet and ends at 11,026 feet over a distance of 9.7 miles.
a – acceleration a = ∆v / t . in this example I've made a zero by making ∆v zero.
Other assumptions: ambient temperature 72F as embedded in parameters, assumption of no accessories usage - - no heat, no aircon, etc, (and probably some more assumptions that I haven’t thought of).

Result for 582 seconds and with slope = 0, is 6.943E+106 joules or 1.929 kWh.
Result for 582 seconds and with slope = 3.55% is 16.2112E+106 joules or 4.503 kWh (going uphill)

Delta due to slope is 4.503 minus 1.929 equals 2.574 kWh. (A)

Now calculate the 3.76616E-07 conversion method.
Incremental work due to hill is ∆hlb * mlb * 3.76616E-07.
Put in the values, (11026-9211)*3765*0.000000376616,
Result is 2.574 kWh. (B)

(A) and (B ) are equal.
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