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Can't you get that info from the logs?
There's elevation data as well, right?
The log's aren't detailed enough. But! the data on the CAN bus is...Can't you get that info from the logs?
I can't speak to what's theoretically possible, but my model is that when driving between 55 and 60 mph in good weather, I lose about 7 ideal miles for every 1,000 feet of elevation gain and get about half of that back for each 1,000 feet of drop. This model fits the data I've collected going over Snoqualmie Pass several times (3,00 feet) and Mt. Rainier once (6,500 feet).
I also drove back and forth over Stephen's pass (4,000 feet) in 23-degree F; that trip fits my model if I reduce range by about 10% to 15% due to the cold weather and wet roads.
It's going to be a challenge to do those same drives in a Leaf as there isn't currently any helpful charging along those routes.
Do you not have a thousand foot climb in your area?...And someone in SoCal will drive the route. ...
If I drive a loop that goes downhill for 5 miles (and drops 100 metres[1]) then up for 5 miles, the air resistance is largely the same (there could be a small effect from changing air density) and the rolling resistance is the same. So my consumption will be 3kWh plus whatever it costs for the electric motor to lift the mass of the car 100 metres less whatever was regained dropping the car 100 metres. If the regen, battery and electric motor drive were perfectly efficient (they aren't) then this would be net zero and my consumption would be the same on the down-up loop as on the flat loop.
We must also remember that, even though the battery-to-wheel conversion efficiency is pretty good (up to 80% or so), the energy makes a full circle back into the battery and it gets converted twice for a net efficiency of at most 80% * 80% = 64%.
According to this blog regen is 64% efficient, max.
According to this blog regen is 64% efficient, max.
The Magic of Tesla Roadster Regenerative Braking | Blog | Tesla Motors