Energy conservation strategies for electric vehicles on hills

Schiettekatte, François

doi:10.1119/10.0005854

Because the charge/discharge process of a battery is not perfectly efficient, there is an optimal point before the end of a downward slope at which a battery-based electric vehicle can save energy by engaging neutral (i.e., coasting). As an exercise in classical mechanics involving air and rolling resistance, we find an analytical expression for this distance on constant slopes, together with an expression for the energy saved. We compare the analytical solution to the numerical solutions for constant and sinusoidal hills. However, considering the characteristics of current electrical vehicles, the energy saved is only a few Wh per typical slope, resulting in negligible gains on any realistic trip.

I. INTRODUCTION

Several strategies have been developed to improve the range of battery-based electric vehicles, including reduced mass of the vehicle, better aerodynamics, lower tire resistance, and regenerative braking (hereafter called “regen”).¹ On highway roads, regen can be applied to maintain the vehicle at the driver-selected cruising speed $v_{0}$ on downhill slopes in order to charge the battery. This energy is then spent later when the vehicle reaches flatter portions of the road. Maintaining constant speed $v_{0}$ would indeed be the optimal energy-saving strategy if the charge-discharge process were 100% efficient (if we choose as a constraint that the speed must not go below $v_{0}$ ⁠). With a charge-discharge cycle efficiency $ε = 100 %$ ⁠, energy losses to resistance would only increase if the vehicle reached higher speeds.

However, this strategy is not necessarily optimal when $ε < 100 %$ ⁠. Indeed, if $ε = 0 %$ as with a vehicle equipped with an internal combustion engine, the optimal energy-saving strategy is to switch into neutral as soon as the slope suffices to maintain a minimum speed of $v_{0}$ ⁠. The vehicle can store any excess potential energy as kinetic energy and then coast until it returns to speed $v_{0}$ ⁠.

Here, as an exercise in classical mechanics, we solve the intermediate case where $0 % < ε < 100 % .$ We disregard the trivial energy-saving strategy of driving at slower speed, but rather consider the possibility of saving energy by going faster than the desired cruising speed during downhill portions of a trip, storing energy as kinetic energy. Speed limits and safe driving practices may impose an upper limit on the speed, but that limit is not considered. We also disregard any limitation due to regen power or stored energy.

A large body of literature exists about the numerical optimization of the speed profile (see, for example, Ref. 2). However, we could not find any analytical solution for the distance before the end of a slope where neutral should be engaged to achieve optimal energy savings, given a representative charge-discharge cycle efficiency $ε$ and a realistic speed-dependent resistance force. In Sec. II of this paper, using undergraduate-level classical mechanics, we derive such an analytical expression for a uniform slope. In Sec. III, we show through numerical integration of the equation of motion that the analytical expression also applies to the case of a sinusoidal elevation profile if inclination is moderate.

II. OPTIMAL POINT TO ENGAGE NEUTRAL

Consider a battery-equipped electric vehicle of mass $m$ cruising at speed $v_{0}$ and arriving at the top of a downward slope of angle $θ$ ⁠, height $h$ ⁠, and length $L_{1}$ ⁠, as illustrated in Fig. 1. Our strategy is first to consider the regen case where the motor is engaged all along and find how far the vehicle can go (distance $L_{2})$ using the energy stored over the distance $L \leq L_{1}$ from the end of the hill. Then, we consider the coasting case where the driver puts the vehicle in neutral at distance $L$ and find how far the vehicle can go, still in neutral, before slowing to its initial speed $v_{0}$ (distance $L_{2}^{'})$ ⁠. If $L_{2}^{'} > L_{2}$ ⁠, it means the vehicle saves energy by switching to neutral. We also determine the distance $L_{opt}$ over which the vehicle should coast in neutral to optimize energy savings.

Fig. 1.

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Slope of height h and total length $L_{1}$ ⁠, forming an angle $θ$ ⁠, followed by a flat section of road.

We consider a total (air and rolling) resistance force of the form

F_{r} (v) = {c v}^{2} + k,

(1)

that is, the usual quadratic air drag with coefficient

c

to which we add a constant

k

⁠. As we will see in Sec. III, the total resistance force of an actual vehicle (including rolling resistance, drive unit consumption, and ancillaries) is better represented by a third order polynomial, but solving the equation of motion with such a force would result in complicated mathematical expressions, if achievable analytically. We will see that Eq. accurately describes the total resistance force as a function of

v

⁠, provided that

v

is typical of highway speeds. It is also possible to fall back on the usual quadratic air resistance case by considering

k = 0

⁠.

Assuming that only gravity (⁠

g = 9.8 m / s^{2}

⁠) and

c v^{2} + k

resistance apply, Newton's second law indicates that

m \frac{d v}{d t} = m g sin θ - c v^{2} - k + F_{M},

(2)

where

F_{M}

is the force that the motor (or regen brakes) causes the road to apply to the vehicle.

F_{M}

is usually positive to maintain the speed but is negative in the case of regen. We can use Eq. to define the terminal speed

v_{ter} = \sqrt{(m g sin θ - k) / c},

(3)

that is, the speed the vehicle would reach in neutral (⁠

F_{M} = 0)

on a long slope of angle

θ

⁠, at which the resistance forces equilibrate the gravitational force (⁠

d v / d t = 0

⁠). In the case where the vehicle travels on a flat road (⁠

θ = 0

⁠) at constant speed (⁠

d v / d t = 0

⁠), Eq. simplifies to

F_{M} = c v^{2} + k,

(4)

that is, the force developed by the motor equals the resistance force.

A. Going downhill at constant speed with regen

Consider the situation where the vehicle is going downhill at constant speed

v_{0}

(i.e.,

d v / d t = 0

⁠), so that Eq. becomes

F_{M} = - m g sin θ + c v_{0}^{2} + k .

(5)

When the slope is steep enough that regen is engaged, it is useful to define the constant

A = \frac{m g sin θ}{c v_{0}^{2} + k} = \frac{c v_{ter}^{2} + k}{c v_{0}^{2} + k} > 1,

(6)

which is the ratio of the gravitational force to the drag force at the base speed

v_{0}

⁠. In this case,

F_{M}

is negative and

v_{ter} > v_{0}

⁠.

In the coasting phase of length

L

⁠, the battery stores a net energy

Δ E_{batt}^{slope} = - ε F_{M} L = (mgL sin θ - (c v_{0}^{2} + k) L) ε .

(7)

Here, the constant

ε

includes the efficiencies of both the charging and discharging processes, so that this energy can be used to maintain a constant speed once the vehicle reaches the flat road and experiences a drag force of

c v_{0}^{2} + k

⁠. The distance

L_{2}

that the vehicle travels before using all the stored energy is

L_{2} = ε (\frac{mgL sin θ}{c v_{0}^{2} + k} - L) = ε (A - 1) L .

(8)

B. Going downhill in neutral

Let us compare this expression to the one we obtain when, instead, the driver coasts in neutral, accelerating over this distance

L

⁠, and then continues, still on neutral, on the flat road section that follows until its speed comes back to

v_{0}

⁠. (The mathematical derivation is rather complicated. Here, we present only the main steps. Section 1 of the Appendix shows details of the solution.) For the downhill part of the journey, Eq. now becomes

m \frac{d v}{d t} = m g sin θ - c v^{2} - k .

(9)

Setting

t = 0

when neutral is engaged and integrating Eq. considering the boundary condition

v (t = 0) = v_{0}

⁠, we find

v (t) = v_{ter} \tanh (\frac{t + t_{0}}{τ}), t_{0} = τ arctanh \frac{v_{0}}{v_{ter}},

(10)

where

τ = m / c v_{ter}

and

t_{0}

corresponds to the time it would take for the vehicle, starting at

v = 0

⁠, to reach

v_{0}

on this slope, while

τ

is the characteristic time to reach

v_{ter}

⁠. Integrating the speed, we get the distance completed since

t = 0

x (t) = \frac{m}{c} \ln (\frac{\cosh (\frac{t + t_{0}}{τ})}{\cosh \frac{t_{0}}{τ}}),

(11)

because

v_{ter} τ = m / c

⁠. We use this expression to find the time

t_{L}

where

x (t_{L}) = L

⁠, that is, the time at which the vehicle arrives at the bottom of the slope. Then,

t_{L}

is substituted for

t

into Eq. to obtain the speed at the end of the slope

v_{L} = v_{ter} \tanh arccosh (\exp (\frac{c L}{m}) \cosh \frac{t_{0}}{τ}) .

(12)

The next step is to find the distance

L_{2}^{'}

the vehicle will coast on flat road after the slope before slowing down from

v_{L}

to

v_{0}

⁠. With

θ = 0

and

F_{M} = 0

⁠, the equation of motion (Eq. ) reduces to

m \frac{d v}{d t} = - c v^{2} - k .

(13)

Setting again

t = 0

at the beginning of the flat road section so the boundary condition is now

v (t = 0) = v_{L}

⁠, this equation is integrated to find

v (t)

⁠, and from that, the time

t_{r}

needed to slow back down to

v_{0}

⁠.

v (t)

is then integrated to find

x (t)

⁠, and

t_{r}

is substituted into

x (t)

to find

L_{2}^{'} = \frac{m}{2 c} \ln (\underset{A}{\underset{︸}{\frac{m g sin θ}{c v_{0}^{2} + k}}} (1 - e^{- 2 c L / m}) + e^{- 2 c L / m}) .

(14)

In Subsection II C, we compare

L_{2}

to

L_{2}^{'}

to determine whether regen or coasting yields a longer travel distance for zero net energy use. However, first we note that the constant

k

we added to the usual

c v^{2}

air drag force appears in the leading constant

A

in the logarithm, but not in the three other factors containing

c

⁠. Therefore, it would not have been obvious to guess Eq. from the solution of the problem considering only quadratic air drag.

Let us now consider the limits of Eq. . When

L ≫ m / 2 c

the characteristic slowing distance,

e^{- 2 c L / m} \to 0

and

L_{2}^{'} \approx \frac{m}{2 c} \ln A,

(15)

which is the maximum distance the car can go in neutral after a slope sufficiently long so that

v \to v_{ter}

⁠. Conversely, for

L ≪ m / 2 c

and

A

not much larger than unity, to first order,

L_{2}^{'} \approx \frac{m}{2 c} \ln (1 + \frac{2 c L}{m} (A - 1)) \approx (A - 1) L .

(16)

This is Eq. in the special case where

ε = 100 %

⁠. This is because for very short

L

⁠, the stored kinetic energy is fully recoverable because the speed remains close to

v_{0}

so the extra energy lost to air resistance is small. However, the second order term in the Taylor expansion is negative, so when

ε = 100 %

⁠, regen is always more efficient than coasting because some additional amount of energy is lost due to air resistance as the speed increases.

C. Extra distance

When the efficiency

ε < 100 %

⁠, there is a range of distance

L

for which Eq. gives larger values than Eq. . The energy that could be saved by coasting is found by multiplying the extra distance

Δ L_{2} = L_{2}^{'} - L_{2}

by the resistance force for traveling at speed

v_{0}

Δ E_{saved} = (c v_{0}^{2} + k) Δ L_{2} .

(17)

Subtracting Eq. for

L_{2}

from Eq. for

L_{2}^{'}

⁠, we obtain

Δ L_{2} (L) = \frac{m}{2 c} \ln (A (1 - e^{- 2 c L / m}) + e^{- 2 c L / m}) - (A - 1) ε L .

(18)

For large

L

⁠, the first term on the right-hand side tends towards a constant, Eq. , while the second term decreases linearly, so

Δ L_{2}

can become negative: more energy is lost to resistance than to charge-discharge inefficiency.

However, there is an optimal distance

L_{opt}

from the bottom of a hill (if

L_{opt} \leq L_{1}

⁠) where a maximum amount of energy is saved when coasting compared to regen. This is when

d (Δ L_{2}) / d L = 0

and it occurs at

L_{opt} = \frac{m}{2 c} \ln (1 + \underset{1 / A}{\underset{︸}{\frac{c v_{0}^{2} + k}{m g sin θ}}} (\frac{1}{ε} - 1)),

(19)

a relatively simple expression and the main result of this paper.

In order to get physical sense of this expression, consider the case where

ε \to 100 %

or

A ≫ 1

⁠. To first order approximation,

L_{opt} \approx \frac{v_{0}^{2} + k / c}{2 g sin θ} (\frac{1}{ε} - 1) .

(20)

Consequently, the steeper the slope, the closer to the end of the slope the vehicle should engage neutral to optimize energy savings. This is an expected result since, on steep hills, the acceleration and

v_{ter}

are larger, so engaging neutral too soon before the end of the slope would result in reaching high speeds and losing more energy to resistance than to the charge/discharge cycle. The leading constant of Eq. is also the ratio of the dissipative to the gravitation forces (⁠

1 / A

⁠) times the characteristic slowdown length on flat road,

m / 2 c

⁠, two of the three main influencing factors of the problem. For the third main factor, the efficiency

ε

⁠, we observe that

L_{opt}

vanishes as

ε \to 100 %

⁠, since it is optimal to apply regen all along in that case. Conversely,

L_{opt}

tends towards infinity for

ε \to 0 %

as mentioned for a car unable to apply regen. It does so logarithmically in the exact Eq. .

D. Energy saved

To conclude this section, we compute the optimal amount of energy saved by replacing

L

by

L_{opt}

⁠, Eq. , into Eq. , and use that to compute Eq. . We get

\begin{matrix} {Δ E}_{saved}^{opt} = (c v_{0}^{2} + k) \\ \times \{\frac{m}{2 c} \ln (A (1 - Λ) + Λ) + (A - 1) ε \frac{m}{2 c} \ln Λ\}, Λ = {[1 + \frac{1}{A} (\frac{1}{ε} - 1)]}^{- 1} . \end{matrix}

(21)

As detailed in Sec. 2 of the Appendix, this expression reduces to

\begin{matrix} Δ E_{saved}^{opt} = \frac{m}{2} (v_{0}^{2} + \frac{k}{c}) ([1 + κ] \ln \frac{A}{κ + 1} + κ \ln ε), κ = (A - 1) ε . \end{matrix}

(22)

This is the exact solution. In order to get a physical sense about the influence of the parameters, it is tempting to use the approximations we already developed, such as Eqs. and . It turns out that they only apply over a very limited range of

ε

and

A

values. It is better to start from Eq. and directly compute a second order expansion around

ε = 100 %

⁠. This tells us that

Δ E_{saved}^{opt} \approx \frac{m}{2} (v_{0}^{2} + k / c) (1 - \frac{1}{A}) \frac{{(1 - ε)}^{2}}{2} .

(23)

We see that savings increase with kinetic energy,

m v_{0}^{2} / 2

⁠, since they are a proportion of it. They also vanish as

A \to 1

since the car barely accelerate in such case, and

L_{2}^{'} \to 0

⁠. We further see that

Δ E_{saved}^{opt}

increases non-linearly as

ε

departs from 100%. In Sec. 3 of the Appendix, we show that about half of the losses to the regen inefficiency can be saved by coasting from the optimal distance.

We note that we obtained Eqs. (22) and (23) considering $L = L_{opt}$ ⁠. If $L_{opt} > L_{1}$ the length of the slope, neutral should be engaged from the beginning of the slope, and the energy saved can be computed by calculating $Δ L_{2} (L_{1})$ using Eq. (18) and putting the result in Eq. (17). We will come back to that point in Sec. III.

In conclusion to this section, we draw a few observations:

We found the exact expressions for the optimal distance from the end of a constant slope to engage neutral, Eq. (19), and the resulting energy savings, Eq. (22).
About half of the energy that would be lost to the charge/discharge process in the regen mode over the distance $L_{opt} + L_{2}^{'}$ can be saved by coasting over that distance from the end of a slope.
The steeper the slope, the later neutral should be engaged (if the slope is long enough) to avoid gaining too much speed that would make the vehicle lose more energy to speed-dependent resistance than in the charge/discharge process.
Energy savings resulting from engaging neutral increase quadratically with base cruising speed $v_{0}$ ⁠.
Savings depend non-linearly on the charge/discharge efficiency $ε$ and become appreciable only if this efficiency is low, which is of course undesirable.

However, the strategy of engaging neutral results in small gains in real-life conditions, as we will see in Sec. III.

III. MORE REALISTIC ELEVATION PROFILES

In this section, we compare the expressions we derived for a straight slope followed by a flat road stretch in Sec. II to more realistic cases. We do so using numerical integration of the equation of motion and optimization of the trajectory.

For these computations, we rely on the resistance force advertised by electric vehicle manufacturer Tesla for its model S.³ We extracted the data from the graph showing the force as a function of speed, to which it is necessary to fit a third order polynomial to reproduce the curve. This gives the following expression for the resistance forces in Newtons:

F_{f} (v) = \sum_{n = 0}^{3} a_{n} v^{n},

(24)

with

v

in m/s. The coefficients

a_{n}

are listed in Table I. The function is plotted as dots in Fig. 2. This force includes not only air resistance but also rolling resistance, drive unit consumption and ancillaries. The latter explains the curvature: this energy consumption is constant in time, so as speed decreases, the energy per unit distance increases. We could have considered only the air and rolling resistance, but we want to take full account of the energy consumption of a typical electric car. The negative third order term of Eq. makes the resistance force become negative at some speed above 38 m/s (maximum speed of Tesla data in Ref. 3). To avoid diverging and unrealistic numerical solutions, we artificially extend

F_{f} (v)

with an exponential. The parameters of this exponential are such that

F_{f} (v)

and its first derivative are continuous. The calculation results presented below rarely reach such a speed regime, but using this artifice helps the minimization algorithm to converge. In this section, we also set

m =

2300 kg and

ε =

60%,⁴ which are representative values for an early Tesla model S85, and

v_{0} =

26.8 m/s, which corresponds to 60 mph or 96 km/h. At that speed, Eq. predicts a force (or energy consumption) on flat road of 173 Wh/km, 278 Wh/mile, or 623 N.

Table I.

Coefficients $a_{n}$ in Eq. (24).

$n$	$a_{n}$
0	533.2 N
1	−24.61 kg/s
2	1.339 kg/m
3	−0.01099 kg s/m²

Fig. 2.

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(Color online) Force as a function of speed considering ${c v}^{2}$ (solid line), Eq. (24) (dashed), and an approximation of Eq. (24) by a relation $c v^{2} + k$ (dotted).

Figure 2 also presents a purely quadratic $c v^{2}$ air drag function (solid line) with $c$ = 0.824 kg/m adjusted so that the force at $v_{0} =$ 26.8 m/s is the same as Eq. (24). We see that such a curve features a slope that differs from that of Eq. (24) by a factor of about 2 at that speed. It turns out that this has a major effect: since many expressions in Sec. II featured a factor $m / 2 c$ ⁠, if $c$ is twice as large, it makes these relations off by the same factor. However, the form $c v^{2} + k$ ⁠, with $c =$ 0.424 kg/m and k = 319 N (dashed curve) fits the data very well over a realistic range of highway speeds (15 m/s< v < 38 m/s). We will use these values of $c$ and $k$ in the equations developed in Sec. II to compare these expressions to the results we obtain from the numerical optimizations below.

A. Straight slope

Let first consider straight slopes as we did in Sec. II, in order to have a direct comparison. In Fig. 3, we plot the energy saved according to a numerical integration of the equation of motion

m \frac{d v}{d t} = m g sin θ - F_{f} (v) + F_{M} (x) .

(25)

This is the same as Eq. except that we consider

F_{f} (v)

as in Eq. . Let us place the origin

x = 0

at the beginning of the slope and define

x_{0} = L_{1} - L

the position where neutral is engaged, so

F_{M} (x) = \{\begin{array}{l} F_{f} (v) - m g sin θ, & x \leq x_{0} \\ 0, & x > x_{0} . \end{array}

(26)

Fig. 3.

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(Color online) Optimized amount of energy saved (color scale) as a function of slope length $L_{1}$ and slope inclination $sin θ = h / L_{1}$ ⁠. Also plotted are the exact (dashed, Eq. (19)) and approximate [dotted, Eq. (20)] expression for $L_{opt}$ ⁠. The minimum of the vertical axis corresponds to the criterion of Eq. (6).

For all pairs of points (⁠ $L_{1}$ ⁠, $sin θ$ ⁠) covered in Fig. 3, we find the values of $x_{0}$ that maximizes $L_{2}$ ⁠. To do so, we scan $x_{0}$ between 0 and $L_{1}$ ⁠, and, at each step of this scan, we integrate numerically Eq. (25) (with boundary condition $v (x = x_{0}) = v_{0}$ ⁠) to find the distance after the slope where the vehicle slows down back to $v_{0}$ ⁠. The value of $x_{0}$ that maximizes $L_{2}$ for a given (⁠ $L_{1}$ ⁠, $sin θ$ ⁠) pair is used to compute the energy saved. This is the energy consumption between the $x_{0}$ and $L_{2}$ when the motor is engaged, computed as $\int F_{M} (x) d x$ when $F_{M} (x) \geq 0,$ and $\int ε F_{M} (x) d x$ when $F_{M} (x) < 0$ ⁠. Energy saved is reported as negative color-coded values.

In Fig. 3, Eq. (19) for $L_{opt}$ is plotted as a dashed line. It correctly predicts the inflection point in the contour lines: for larger distances, the contour lines are horizontal since no additional energy is saved when $L_{opt} < L_{1}$ ⁠. Neutral has to be engaged at $x_{0} = L_{1} - L_{opt}$ from the end of the slope in order to optimize savings as discussed in Sec. II. For shorter distances, neutral must be engaged from the beginning of the slope (⁠ $x_{0} = 0$ or $L = L_{1} < L_{opt}$ ⁠), but smaller energy savings are achieved. The approximate Eq. (20) (dotted line) gives reasonably good results for steep slopes but overestimate $d_{\max}$ by several percent for small inclinations.

We see from this graph that if slopes are moderate (below 5%), neutral should be engaged at almost 1 km from the end of the slope. Considering that slope inclinations are usually only a few percent and slope lengths are rarely more than 1 km long, neutral should, therefore, be engaged in most situations. Another striking feature of Fig. 3 is that the energy saved is a few Wh for any reasonable slope, a relatively small fraction of the energy consumption of the vehicle on a flat road at $v = v_{0}$ ⁠, about 173 Wh/km. This is an indication that such strategy does not save so much energy in any realistic setting.

B. Sinusoidal hills

We now study the more complex but perhaps more realistic situation of sinusoidal hills and valleys of different heights and lengths. As for the straight slope, we start the integration of the equation of motion at the top of the hill. At point

x_{0}

⁠, the motor is disengaged if

v_{0} < v_{ter}

(i.e., if the slope is sufficient for the vehicle to accelerate) and until the speed drops below

v_{0}

⁠. At that point, the motor then re-engages to maintain

v_{0}

⁠. The motor force profile is, therefore, given by

F_{M} (x) = \{\begin{array}{l} F_{f} (v) - m g sin θ, & v = v_{0,} \\ 0, & v > v_{0} and x > x_{0,} \end{array}

(27)

i.e., in comparison to Eq. , the condition where the motor force is applied now depends on the speed to avoid slowing down below

v_{0}

⁠. We scan the point

x_{0}

from the top to the bottom of the hill, integrating Eq. for each value and retain the value of

x_{0}

⁠, which optimizes energy-savings. We present two examples of the resulting optimizations in Fig. 4. On the left, we considered a path with a “wavelength” of

200 π = 628

m, that is, downward and upward slopes of

314

m each. The maximum slope along the path is 7%, and the maximum altitude reached is

200 \times 7 % = 14

m. The top graph shows the altitude as a function of the distance. In the center and bottom graphs, the blue dotted curves are the speed and force applied by the motor when in the regen mode in order to maintain the speed at

v_{0}

⁠. (Here, a negative force means that the vehicle is in regen, recharging the battery with efficiency

ε

⁠.) The force is shifted by a phase of

π / 2

with respect to the altitude, since the maximum force occurs in the middle of upward slopes, and conversely, maximum regen happens in the middle of downward slopes. In regen mode (dotted curve), the absolute value of the minimum force is smaller than the maximum force, the difference being due to the resistance force

F_{f} (v)

that the motor has to compensate.

Fig. 4.

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(Color online) Altitude, speed, and force as a function of distance for two different paths varying sinusoidally in altitude. Left: a downward slope of 314 m from top to bottom, maximum inclination of 7%. Right: a downward slope of 1257 m from top to bottom, maximum inclination of 11% (note the difference of horizontal axes span). Black solid curves correspond to the optimal speed and force profile to minimize energy consumption while maintaining a speed $v_{0} \geq$ 26.8 m/s. Blue dotted curves correspond to regen in order to maintain constant speed. Negative force corresponds to regen and battery charging. $ε = 60 %$ is applied to the regen part of the force.

If we optimize the speed profile to minimize energy consumption (solid black curves), it is seen that when the vehicle reaches a point where $v_{ter} > v_{0}$ ⁠, neutral is engaged (⁠ $F_{M} = 0$ ⁠), which results in a speed increase. The force reapplies when the vehicle slows down to $v_{0}$ ⁠, the force falling back to the value it would have had if constant speed $v_{0}$ was maintained. The regen mode is never used in this optimized speed profile.

On the right side of Fig. 4, we now considered a path with a “wavelength” of $800 π = 2.51$ km, the downward and upward slopes having a length of 1.26 km. This time the maximum slope is 11%, and the maximum elevation is 88 m. The force profile for the case of the regen strategy (bottom graph, blue dotted curve) has the same characteristics as in the example on the left. However, the optimal force profile (black curve) is characteristically different. The (imperfectly efficient) regen mode is used at the beginning of the downslope in order to prevent the speed at the bottom from being so great as to cause larger losses due to air resistance. (It is worth noting that while the force profiles on the left- and right-hand side of the figure appear similar, the horizontal axis features a markedly different length. Thus, neutral is engaged on a much longer distance on the right-hand side which, in combination with the steeper slope, results in a much higher maximum speed.)

We now want to compare this optimal distance to $L_{opt}$ obtained for straight slopes, Eq. (19). We perform the same calculations as in these two examples for a range of downward slope lengths and maximum slopes. The energy saved is plotted in Fig. 5. This is the difference of area under the solid black and dotted blue curves in the force profiles of Fig. 4, subtracting area when forces are negative. Hence, this graph is similar to Fig. 3, keeping in mind that the slope is not constant. We see that Fig. 5 indeed shares many characteristics with Fig. 3 such as lower savings for shorter slopes or smaller inclination. However, there is no horizontal part to the contour lines. The dashed-dotted curve in Fig. 5 passes through the minimum of the contour lines and is very close to the optimal point where neutral has to be engaged for a given slope value. We see that this curve is rather vertical for slopes shorter than about 1 km. For slopes steeper than 7%, it shows that neutral should always be engaged 1±0.1 km from the end, whatever is the slope length. This is somewhat surprising, but it appears to be the result of the slope shape. The inset of Fig. 5 gives an element of explanation. It shows two downward slopes with the maximum inclination and length indicated by the triangles of the corresponding color in the main figure. It is seen that in the last part, both curves overlap very well, hence resulting in a similar speed profile and energy savings despite being markedly different. This would not be the case of two such slopes if they were straight.

Fig. 5.

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(Color online) Optimized amount of energy saved (color scale) as a function of downward slope length and maximum slope inclination for sinusoidal paths as illustrated in Fig. 4. Also plotted are the exact expression for $L_{opt}$ ⁠, Eq. (19), considering as $sin θ$ the maximum slope (red dashed curve) or the average slope (green dashed curve). The blue dashed-dotted curve passes through the minimum of each contour line. The left and right red circles correspond to the left and right examples of Fig. 4. The cyan and purple triangles show the parameters of the cyan and dashed purple elevation profile in the inset.

The dashed-dotted curve in Fig. 5 is also very different than the result of Eq. (19) for $L_{opt}$ considering $sin θ$ as the maximum slope (red dashed curve). If we rather consider the average slope (which is a factor $2 / π$ smaller), we see it is not much better for maximum slopes above 7%, but it corresponds relatively well to the lower part of the dash-dotted curve: For low-inclination hills, where the magnitude of the gravitational force remains of the order of the resistance force and the speed reached remains close to $v_{0}$ ⁠, we can consider that the behavior of a vehicle is similar to the one on a straight slope of similar inclination, and that Eq. (19) for $L_{opt}$ considering the average slope applies. For steeper slopes, it appears that the shape of the slope influences the results in a way that shortens the distance from the end of the slope where neutral should be engaged for sinusoidal slopes.

Therefore, while an exact expression is not determined for sinusoidal hills and valleys, a practical approximation for the optimal distance from the end of a sinusoidal slope where a vehicle riding at $v_{0} =$ 26.8 m/s should go in neutral to save an optimal amount of energy as $\max (1 km, L_{opt})$ with $L_{opt}$ computed using Eq. (19) with the average slope.

And again, we find that for most typical slopes, which have an inclination of a few percent and length of less than 1 km, neutral should be engaged from the top. Savings achieved in these conditions are only a few Wh and are small compared to the usual consumption of an electric car on a flat road. Indeed, for the optimization of a trip of a few hundred km on the Interstate of a hilly region (not shown), we find savings of the order of 0.5 Wh/km for the whole trip. This is a small fraction of the energy consumption of an electric car on the highway. Similar savings can be achieved by reducing the cruse speed $v_{0}$ by a fraction of a percent.

IV. CONCLUSION

In principle, because the efficiency of the charge/discharge cycle of a battery is not 100%, a battery-equipped electric vehicle can save energy by coasting from a certain distance from the end of a slope and up to the point where the vehicle slows back to its original speed. As an application of undergraduate-level classical mechanics, we derived an exact expression for the optimal value of such distance in the case of constant slopes. We find that it also applies to more realistic slope profiles if they are not too steep. We show that about half of the energy lost to the charge/discharge process when using regen on the same distance can be avoided. However, the energy savings are only a few Wh for typical slopes and are therefore negligible on any realistic trip in comparison to the normal energy consumption of an electric car. Also, we have considered the optimization under the assumption that the vehicle speed could take any value greater than $v_{0}$ ⁠. Under a different set of assumptions, such as a traffic-friendly range that is capped at a speed limit and can be lowered no more than 75% of the speed limit, a different optimization strategy would result.

APPENDIX: MATHEMATICAL DETAILS

Section 1: Coasting distance

In this section, we detail the steps to obtain Eq. for

L_{2}^{'}

⁠. Equation can be rearranged as follows:

\frac{d v}{d t} = (g sin θ - \frac{k}{m}) (1 - \frac{v^{2}}{v_{ter}^{2}}) .

(A1)

Considering the boundary condition

v (t = 0) = v_{0}

⁠, we find

arc tanh v / v_{ter} - arc tanh (v_{0} / v_{ter}) = t / τ,

(A2)

where we defined

τ = \frac{v_{ter}}{g sin θ - k / m} = \frac{m}{\sqrt{c (m g sin θ - k)}} = \frac{m / c}{v_{ter}} .

(A3)

Isolating

v

in Eq. yields Eq. which, integrated, gives Eq. . To find the speed

v_{L}

reached at the end of the slope, we need the time

t_{L}

where

x (t_{L}) = L

⁠. From Eq. , we get

\frac{t_{L} + t_{0}}{τ} = arccosh (\exp (\frac{c L}{m}) \cosh \frac{t_{0}}{τ}) .

(A4)

Substituting this expression into Eq. , we obtain Eq. .

We now find the distance

L_{2}^{'}

the vehicle will travel on a flat road after the slope before slowing down from

v_{L}

to

v_{0}

⁠. Rearranging Eq. , we get

\frac{d v}{d t} = - \frac{k}{m} (1 + \frac{v^{2}}{v_{r}^{2}}), v_{r}^{2} = \frac{k}{c} .

(A5)

Taking again

t = 0

at the beginning of the flat road section, integrating this expression gives

\arctan v / v_{r} - \arctan (\frac{v_{L}}{v_{r}}) = - \frac{t}{τ_{r}}, τ_{r} = \frac{v_{r}}{k / m} = m / \sqrt{c k} .

(A6)

Solving for

v

⁠, we get

v = v_{r} tan \frac{- t + t_{2}}{τ_{r}}, t_{2} = τ_{r} \arctan (v_{L} / v_{r}) .

(A7)

This allows us to obtain the time

t = t_{r}

at which the vehicle has slowed down back to

v_{0}

⁠, i.e.,

v_{0} = v_{r} tan \frac{- t_{r} + t_{2}}{τ_{r}},

(A8)

which is

\frac{t_{r} - t_{2}}{τ_{r}} = - \arctan (\frac{v_{0}}{v_{r}}) .

(A9)

Integrating Eq. , we find the distance traveled during time

t_{r}

L_{2}^{'} = - v_{r} τ_{r} \ln (\frac{cos \frac{t_{r} - t_{2}}{τ_{r}}}{cos \frac{t_{2}}{τ_{r}}})

(A10)

or

L_{2}^{'} = \frac{m}{c} \ln {(\frac{cos \arctan (\frac{v_{0}}{v_{r}})}{cos \arctan (\frac{v_{L}}{v_{r}})})}^{- 1}

(A11)

since

v_{r} τ_{r} = m / c

⁠. Substituting Eq. for

v_{L}

into this equation, and in it,

t_{0} / τ

from Eq. , we get

L_{2}^{'} = \frac{m}{c} \ln (\frac{cos \arctan (\frac{v_{ter}}{v_{r}} \tanh arccosh (\exp (\frac{c L}{m}) \cosh arctanh \frac{v_{0}}{v_{ter}}))}{cos \arctan (\frac{v_{0}}{v_{r}})})

(A12)

or

L_{2}^{'} = \frac{m}{c} \ln (\frac{cos \arctan [β \tanh arccosh (\exp (z) \cosh arctanh α)]}{cos \arctan (α β)}),

(A13)

where we defined

z = \frac{c d}{m}, α = \frac{v_{0}}{v_{\lim}}, β = \frac{v_{\lim}}{v_{r}} .

(A14)

Fortunately, thanks to the following identities:

\cosh arctanh α = \frac{1}{\sqrt{1 - α^{2}}},

\tanh arccosh f = \sqrt{1 - 1 / f^{2}},

and

cos \arctan g = \frac{1}{\sqrt{1 + g^{2}}},

(A15)

Eq. reduces to

L_{2}^{'} = \frac{m}{c} \ln (\frac{1 + β^{2} (1 - e^{- 2 z} (1 - α^{2}))}{1 + α^{2} β^{2}}) .

(A16)

Since

β^{2} = mgh / L k - 1

and

α^{2} β^{2} = c v_{0}^{2} / k

⁠, this expression simplifies to Eq. .

Section 2: Energy saved

Here, we describe the steps to obtain Eq. from Eq. . First we note that

Λ = \frac{1}{1 + \frac{1}{A} (\frac{1}{ε} - 1)} = \frac{A ε}{A ε + (1 - ε)} = \frac{A ε}{(A - 1) ε + 1} .

(A17)

Then,

A (1 - Λ) + Λ = \frac{A (1 - ε)}{(A - 1) ε + 1} + \frac{A ε}{(A - 1) ε + 1} = \frac{A}{(A - 1) ε + 1},

(A18)

which differs from Eq. only by a factor

ε

⁠. Equation can, therefore, be rewritten as

\begin{matrix} Δ E_{saved}^{opt} = \frac{m}{2} (v_{0}^{2} + \frac{k}{c}) \\ \times ((1 + (A - 1) ε) \ln [\frac{A}{(A - 1) ε + 1}] + (A - 1) ε \ln ε) . \end{matrix}

(A19)

Section 3: Fraction of energy saved

Here, we compute the fraction of the energy, which would be lost to the charge/discharge process during regen that can be avoided by coasting over the optimal distance $L_{opt} + L_{2}^{'}$ ⁠.

When

ε < 100 %

⁠, regen losses at

v = v_{0}

are given by

E_{lost} (L) = (m g sin θ - (c v_{0}^{2} + k)) (1 - ε) L .

(A20)

They are proportional to the inefficiency

(1 - ε)

of the charging/discharging process. Evaluating that expression for

L = L_{opt}

⁠, Eq. , to consider the same distances as for Eqs. and , we find

\begin{matrix} E_{lost} (L = L_{opt}) = \frac{m}{2} (v_{0}^{2} + k / c) (A - 1) (1 - ε) \ln (1 + \frac{1}{A} (\frac{1}{ε} - 1)) . \end{matrix}

(A21)

A second order expansion around

ε = 1

00% gives

E_{lost} (L = L_{opt}) \approx \frac{m}{2} (v_{0}^{2} + \frac{k}{c}) (1 - \frac{1}{A}) {(1 - ε)}^{2} .

(A22)

This is twice Eq. . Therefore, about half of the losses a vehicle would suffer to the charge-discharge process over the distance

L_{opt} + L_{2}^{'}

can be avoided by going to neutral at

L_{opt}

⁠. A numerical examination of the ratio of exact Eqs. over shows that it becomes larger than 2 as

ε

departs from 100% reaches about 2.3 when

ε = 60 %

with a dependence on

A

⁠.

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Electric vehicles charge forward

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Energy conservation strategies for electric vehicles on hills

I. INTRODUCTION

II. OPTIMAL POINT TO ENGAGE NEUTRAL

A. Going downhill at constant speed with regen

B. Going downhill in neutral

C. Extra distance

D. Energy saved

III. MORE REALISTIC ELEVATION PROFILES

A. Straight slope

B. Sinusoidal hills

IV. CONCLUSION

APPENDIX: MATHEMATICAL DETAILS

Section 1: Coasting distance

Section 2: Energy saved

Section 3: Fraction of energy saved

REFERENCES

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Energy conservation strategies for electric vehicles on hills

I. INTRODUCTION

II. OPTIMAL POINT TO ENGAGE NEUTRAL

A. Going downhill at constant speed with regen

B. Going downhill in neutral

C. Extra distance

D. Energy saved

III. MORE REALISTIC ELEVATION PROFILES

A. Straight slope

B. Sinusoidal hills

IV. CONCLUSION

APPENDIX: MATHEMATICAL DETAILS

Section 1: Coasting distance

Section 2: Energy saved

Section 3: Fraction of energy saved

REFERENCES

Citing articles via

Submit your article

Sign up for alerts

Related Content

Resources

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