A few years back Drag Times reported these results on a Tesla S in the
quarter mile:

"As you can see in the video below the Tesla’s total energy consumption was
1.1 kWh at 114 MPH and after the car slowed down using the regenerative
braking .6 kWh was put back into the battery. This results in a net energy
usage of .5 kWh for the 1/4 mile pass. Pretty amazing that the car can
recover just over half of energy used and put it back into the battery using
it’s regenerative braking."

http://www.dragtimes.com/blog/tesla-...usage-and-cost

This agrees with the range of vehicle ke recouped by regen that I have
measured on my car by data logging battery current and voltage, taking their
product to estimate power, and dividing by the 1 sec time between samples to
estimate energy into/out of the pack during deceleration and acceleration.

Also a few years ago I estimated the percentage of energy expended driving
on level ground recouped by regen with the simplifications that deceleration
rate = acceleration rate, and the vehicle travels at constant speed between
stops.  Note that this assumes the vehicle is brought to a stop using only
regen, no mechanical braking, which may not be the case with some
manufactured EVs. I put this in a spreadsheet and did "what if" scenarios
varying things like vehicle mass and speed between start/stops each by a
factor of 3 to estimate relative effects.  See the summary of these below
the "examples".  The equation used is given for those who have an interest:

The percentage of the energy expended driving the car on level ground that
is regained through regen =
energy gained by regen/(energy used at constant speed + energy used to
accelerate), or:

n*(K.E. – L)/ [(d – 2*n*x) * (Fd + Fr) + n*(K.E. + L)], for 2*n*x < d

Where:
n = number of acceleration/deceleration cycles, L = sum of average losses
(need to average if travel at different constant speeds between stops, i.e.
loss at 35 mph, loss at 50 mph… just assume one constant speed between stops
for simplicity)
d is total distance traveled
x is the distance required to accelerate or decelerate to/from speed, so
total distance driven at constant speed is d – 2nx assuming the same rate
for acceleration and deceleration, and same constant speed between each
acceleration and deceleration for simplicity.

Examples:
(1) My car accelerates at 6 mph/sec to 60 mph, drives 10 miles, then
decelerates at the same rate to a stop, estimated percent energy regained
is: 2.7%

(2) Acceleration at 6 mph/sec to 35 mph and deceleration at same rate 10
times in 10 miles:12.7%

(3) Acceleration at 6 mph/sec to 35 mph and deceleration at same rate 20
times in 10 miles:20.7%

A more typical acceleration/deceleration rate might be 3 mph/sec, or 1.34
m/s (0 to 60 mph in 20 sec). Then the same three scenarios give 2.4%, 12%,
and 20%, so a factor of 2 slower rate doesn’t change the result that much.
At 2 mph/sec acceleration/deceleration rate the three scenarios give 2.2%,
11%, and 19%.

Increasing total miles traveled, d, to 30 in scenario (1) gives 0.9%.
Increasing stops in this scenario to 3, with 30 miles total, gives 2.7%.

Increasing vehicle mass increases the percentage of energy recovered, but
it’s a small effect for larger number of stop/starts. For example the first
scenario goes from 2.7% to 3.8%, second goes from 12.7% to 14.7%, third goes
from 20.7% to 23.2% if vehicle mass is doubled.  *So I would expect the
Tesla S to recover a bit higher proportion of total energy expended than my
much lighter vehicle.*

Decreasing losses in the motor/controller and drive train of course
increases the energy recovered. For example, decreasing motor/controller
loss to 15% in scenario (2) increases the energy gained from 12.7% to 16.5%.

Increasing drag coefficient or area, A, reduces percentage of energy
recovered, but it too is a small effect.
*

The largest effect is the constant speed between stop/starts.*  Varying this
from 10 to 35 mph results in a change in proportion of energy regained by a
factor of about 2.5.  So in heavy traffic that is creeping along at 10 mph
or less, proportion of expended energy regained will be very small, and
likely less than zero with climate control on.

--
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