Model S

Ideal Regenerative Braking Efficiency

edited November -1 in Model S
Under ideal conditions, what percentage of energy expended climbing a hill will be recovered on the downslope? By "ideal conditions" I mean that the angle of the descending slope is perfect for regeneration - the brakes never have to be used, and an adequate speed is maintained until the same lower elevation is reached at which the ascent began.

If this has been dealt with in previous posts I apologize, but these forums are difficult to search.


  • edited November -1
    It isn't a fixed. It depends on what version of firmware you are on. The Standard Regen isn't as robust as it used to be.

    The amount of regen and accelerator response has slowly decreased. Hopefully someday we will be able to select what level we are comfortable with. I think they are trying to find a balance that will be more acceptable to those that like to coast and for those that like a more responsive and aggressive deceleration and regen. More deceleration and regen seemed to cause complaints about transitioning in and out of cruise control and it was hard for former luxury ICE drivers to make the transition. I prefer to not use cruise control, so I'm ready for a sport mode. (These are my own intermediate conclusions based on observations and experience.)
  • edited November -1
    Haven't seen official data, but my WAG is 80-90% based on driving. It seems more like 90%, but I'm guessing 80% is more realistic from a physics standpoint.
  • edited November -1
    I just did the California Grape Vine climb and descent in both directions and watched the energy numbers. I initially expected that the total energy usage for the ascent and descent would be much more than if the same distance was flat. It turned out to be hardly any more, like 90%. It's hard to be precise because I stuck with only 55-60 mph on the uphill.
  • edited November -1
    JB Straubel put it at about 60% overall, due to the number stages of 5-10% efficiency losses.

    Battery to inverter to motor to inverter to battery (minus chemical conversion efficiency of battery itself).

    Stack this up, and you've lost about 40% on the round trip.

    If you just ask how efficient the harvest of momentum is (and ignore what was lost to create it), the loss is about half that.

    So if you measure from the top of a long hill to the bottom, and use no thrust or braking, just feather the regen to keep 20 mph, best case you might harvest and store 75 - 80% of the gravitational potential energy that you started with.
  • edited November -1
    Set cruise control for the desired descent speed and the car will use regen as necessary to maintain it.
  • edited November -1
    Just realized I phrased the question incorrectly. Since one doesn't travel straight up and down at a 90 degree angle, there is, of course, also some horizontal progress being made on both the ascent and descent, unless it is a corkscrew road leading to a single peak with the descent made on the same road.<P>So the question really is, how much energy is being lost as opposed to traversing the same horizontal distance on a perfectly flat road?<P>If one recoups 70% on a descent, this is really better than it sounds because some horizontal progress has also been accomplished.
  • edited November -1
    Here's an extreme example for which I have incomplete data, but the trip up Mt Evans, which is an annual Tesla event west of Denver, takes you to the top on the highest paved road in America. From the town of Idaho Springs, it's a 30 mile trip to the top 6700 ft above. The figure I don't have is how many rated miles I used going up, but the descent added 22 miles to my rated range, so I traveled 30 miles down for "net gain" of 52 rated miles.

    When I go into the mountains around here and see some inclines where my efficiency is as poor as 1000Wh/m, the return trip usually brings my overall average back to near 300Wh/m.
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