RGEV wrote:Redpoint5, that's interesting about the air density. Speeds were slower, seldom over 50 mph, but I don't think that's the major effect here.
If I calculate the potential energy (m*g*h) difference of raising the 3580# car 3,950 feet, I get
( (3580 + 200 lb) / 2.2lb/kg * 9.8m/s^2 * 3950/3.28 ft/m ) = 20.28x10^6 Joules
= 5.63 kWh
So climbing used 12.4 kWh for movement and 5.6kWh to raise the car and me.
Descending used 11.6 kWh for movement and put 5.6kWh back into the battery.
The change in potential energy due to elevation seems like the major effect.
I have to admit I skimmed your post and went off the thread title. Yes, elevation change
will affect range due to the electrical energy required to bring a mass (car) to a higher potential energy (elevation).
In petrol powered vehicles, medium grade hills are more efficient than flat driving because it acts like a prolonged "pulse and glide". Petrol engines are much more fuel efficient (like 2 to 3 times) operating near full throttle than partial throttle, so the hill has the engine operating near peak efficiency, while allowing it to idle down the hill. Obviously, hitting the brakes would be wasted energy, so the hill can't be too steep, or a lot of energy would be wasted.
I'm not so sure what the efficiency curve of an EV looks like. A flat surface is likely most efficient for an EV. Since an EV isn't anywhere near 100% efficient at recapturing kinetic energy (I've heard 1/3 efficient but I don't know for sure), it's still better to have a grade that doesn't require use of regen. That said, being able to recapture some of that energy on steep hills is certainly a strength of EVs over conventional vehicles.
Once I almost got a full charge in my Prius plug-in when I started at the top of Crater Lake with no EV range, and regened down to the highway nearly full. Only 13 miles of EV range, but it's better than wasting the energy as heat and brake dust.