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Discussion Starter #1 (Edited)
Hi! Anyone here has data on the effect of elevation on either range or the Wh/mile energy consumption?

For example, assuming same temperature, driving speed (55 mph), road smoothness, driving profile, except changes in elevation

Range of 30 mile level ground, how many kWh used?
Range of 30 mile but a gain in elevation of 3,000 feet, how many kWh used?
Range of 30 mile but a drop in elevation of 3,000 feet, how many kWh used?

Or some similar data along those lines.


Assuming that the Chevy Volt is 3543 lbs and the driver is 150 lbs, the elevation change of 3,000 feet should equal to 4.17 kWh in terms of potential energy difference (mass * gravity * elevation).

I know that climbing up 3,000 ft, you will get penalized the full 4.17 kWh in addition to the typical consumption of 30 miles drive on level ground (about 7.98 kWh for the 30 mile run based on a typical 53 mile/full charge range) for a total of 12.15 kWh.

Going down wouldn't be the same though. Assuming you put it to cruise control at 55 mph using the L-mode. You still get that 7.98 kWh for the 30 mile run minus a fraction of the 4.17 kWh from regeneration of the downhill ride, perhaps we get 50% recapture efficiency, so that at the end of the downhill run, we only have used 5.90 kWh.

Would be interested to know if someone here have real data measured or read from the DIC as compared to theoretical assumed calculations. Thanks!
 

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Best you'll probably get is with a logging app on an smart phone like Torque or MyChevyVolt along with an ODBII reader.

I'm sure someone on the forums has done it.
 

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I can tell you it makes a difference. My wife's drive has about 1400 ft change. Going to work it takes 5 kW, coming home is about 11. The first mile from out house has a slight gain and the car gets about 2 m/kW. Then the next 10 miles which is down hill it gets 5+m/kW. Other factors make so much difference, temp, wind, speed that it's hard to put hard numbers on actual efficiency.
I am really amazed that electric efficiency varies so much as my tdi didn't vary that much. Usually the milage is 43-48 for a whole tank over 100k miles. There were a few outliers from road trips, higher and lower. I used to swing from 60 mpg going to work and 40 mpg coming home, but that still isn't the roughly half that we see with electric for uphill vs down hill. All I can assume is the efficiency of an ic engine changes depending on load, i.e. It gets better with higher load. Maybe for electrics it gets worse.
 

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Depending on the actual terrain, you could put it in neutral going downhill an use no energy at all ;-)

Towards the end of the run you could regen and end with more kWh than you started with :)
 

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On 18 miles for my work commute with 500 vertical feet at 9,100 feet to 9,600 feet, I usually burn about 2 KwH down to work and 4 KwH back uphill, 2 miles city and 7 miles freeway driving.
 

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Not easy to visualize how steep a road would be that drops 3,000 ft in 30 miles. If I’m driving down that hill in D with the cruise control set at 55 mph, is it steep enough so that gravity causes the car’s speed to increase? If so, I’m not really using any grid power at all, let alone at the rate of 7.98 kWh/30 miles used by the OP as level terrain consumption. If I shift into L, which then increases the regen and allows cruise control to maintain the 55 mph set speed, am I using any more battery power than when I was in D?

On the other hand, if it’s but a slight slope, and driving down the hill at 55 mph requires me to keep my foot on the accelerator to maintain speed, MGB is providing propulsion torque, and can’t also create regen at the same time. Turning on cruise control in D or L would require the same application of propulsion torque to maintain speed, and so no regen. Downhill ev mileage, of course, would be better than level terrain mileage. Perhaps under these conditions would the difference between uphill and downhill driving power consumption be related to the weight of the vehicle and passengers.

Don’t conflate braking regeneration, which recaptures some of the power that was originally used to increase the car’s momentum, with downhill regeneration, which uses gravity to create new energy.
 

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Not easy to visualize how steep a road would be that drops 3,000 ft in 30 miles. If I’m driving down that hill in D with the cruise control set at 55 mph, is it steep enough so that gravity causes the car’s speed to increase? If so, I’m not really using any grid power at all, let alone at the rate of 7.98 kWh/30 miles used by the OP as level terrain consumption. If I shift into L, which then increases the regen and allows cruise control to maintain the 55 mph set speed, am I using any more battery power than when I was in D?

On the other hand, if it’s but a slight slope, and driving down the hill at 55 mph requires me to keep my foot on the accelerator to maintain speed, MGB is providing propulsion torque, and can’t also create regen at the same time. Turning on cruise control in D or L would require the same application of propulsion torque to maintain speed, and so no regen. Downhill ev mileage, of course, would be better than level terrain mileage. Perhaps under these conditions would the difference between uphill and downhill driving power consumption be related to the weight of the vehicle and passengers.

Don’t conflate braking regeneration, which recaptures some of the power that was originally used to increase the car’s momentum, with downhill regeneration, which uses gravity to create new energy.
There may not be literally a regen if the slope is very gradual 3,000 ft over 158,000 ft run, which is about 1.89% grade, but it still helps a lot by reducing the force, hence the energy, required to keep the car moving because of the downhill, and maybe at a higher efficiency compared to regen itself.

You know, even if the car can't roll down by itself downhill, it is still way easier to push the car on the slight downhill slope than pushing it to a slight uphill road.
 

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For a 2% grade you basically have .02g's of downward force. You've also got the negative force related to the drag coefficient of the nose of the car (.27?), a little bit of rolling resistance, and your initial downward velocity (2% down) when you start the downhill. Here's a site to help you with the math https://www.grc.nasa.gov/www/k-12/airplane/falling.html . I have not done the math, but my guess is that you are going to pick up speed and will ultimately use negative energy as the regen keeps the car at a safe speed. If anyone has the time, I would be curious to see the calculation.
 
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