1. Introduction

2. Physical Value of Mechanical Motion

Figure 1. (a) turning money value into trajectories; (b) deriving physical transport value from inputs in engine power.

Figure 1. (a) turning money value into trajectories; (b) deriving physical transport value from inputs in engine power.

3. Some Simple One-Dimensional Transport Scenarii

3.2. Start-Stop Scenarii

The “fly-through” scenario discussed above is a fairly artificial one, only invented for motivating the entity of physical action to be a useful figure of merit for the hitherto undefined property of transport value. Realistic transport scenarii distinguish from this highly idealized case in that realistic scenarii start out at point $A$ from rest, speed up to a finite speed $v_{ab}>v_{AB}=L_{AB}/{\tau }_{AB}$ and slow down upon approach to endpoint $B$. In the following we will call such transport processes “start-stop” scenarii.

Within a “start-stop” scenario, mass $M$ starts out from rest at point $A$ and is accelerated by engine forces $F_{en}=Ma$ in time $\delta$ to attain a speed $v_{ab}$ after which it travels without any further acceleration until it approaches its endpoint $B$ at time $t={\tau }_{AB}-\delta$. Thereafter it becomes decelerated by brake forces $-a$ to come to rest at point $B$ again. Assuming that this “start-stop” transport occurs over the same length $L_{AB}$ and the same time ${\tau }_{AB}$ as in the “fly-through” scenario above, the accelerations and decelerations $a(L_{AB,}{\tau }_{AB},\delta )$ acting during these three phases need to be adjusted to the lengths $\delta$ of the acceleration and deceleration phases. Similarly, the velocity $v_{ab}(L_{AB,}{\tau }_{AB},\delta$) in the intermediate drift phase needs to be adjusted as well. These adjustments are shown in Figure 3.

In Figure 4a we show acceleration (red) and speed profiles (green) as well as the fractional distance $s\left(t\right)/L_{AB}$ covered up to time $t$ (blue), all as functions of the fractional transport time $t/{\tau }_{AB}$. In this first example we have assumed $\delta =0.2{\tau }_{AB}$. Figure 4b, on the other hand, shows the engine and brake power profiles (red) delivered to the mass $M$ close to points $A$ and $B$, as well as the resulting kinetic energy profiles $T\left(t\right)$ (green), both plotted once again against the fractional transport time $t/{\tau }_{AB}$. Also shown for comparison is the physical action $W(t,\delta )$ (blue) attained after time ${\tau }_{AB}$ and as normalized to the physical action $W_{AB}(M,L_{AB},{\tau }_{AB})$ that had been attained at the end of the “fly-through” scenario in Section 3.1 above.

Figure 5a,b, on the other hand, compare start-stop scenarii with increasingly longer acceleration and deceleration lengths $\delta$. While Figure 5a shows that, independent of $\delta$, all trajectories converge at endpoint $B$ after the time ${\tau }_{AB}$, Figure 5b shows that trajectories with larger $\delta ´s$ converge to increasingly larger values of transport value $W_{AB}$ as these approach endpoint $B$. With the principle of least action in mind [6,7], such an increase is expected as in a “start-stop”-scenario, mass $M$ had no longer been moving in a force-free field, but rather needed to be accelerated by engine forces close to point $A$ and decelerated by brake forces close to point $B$. Such acceleration and braking procedures are obviously associated with pieces of extra physical action and extra energy expense.

In order to arrive at more quantitative results concerning transport value generation and energy consumption, we note that the increase in physical action with increasing $\delta$ as displayed in Figure 5b is given by:

$$W_{start\_stop}\left(M,L_{AB},{\tau }_{AB,}\delta \right)=W_{AB}\left(M,L_{AB},{\tau }_{AB}\right)\left\{{\displaystyle \frac{1-\frac{4}{3}\left(\frac{\delta }{{\tau }_{AB}}\right)}{{\left[1-\left(\frac{\delta }{{\tau }_{AB}}\right)\right]}^2}}\right\}.$$

(13)

Here, $W_{AB}\left(M,L_{AB},{\tau }_{AB}\right)$ once again is the intrinsic value of a transport service that is obtained upon moving a mass $M$ within a force-free field along a distance $L_{AB}$ in a time ${\tau }_{AB}$. The other item of interest is the kinetic energy $T$ that needs to be generated to initiate the “start-stop” scenario close to point A, and the energy which needs to be withdrawn from mass $M$ to bring it to rest at point $B$:

$$T\left(M,L_{AB},{\tau }_{AB,}\delta \right)={\displaystyle \frac{1}{2}}M{v\left(\delta \right)}^2={\displaystyle \frac{1}{2}}M{\displaystyle \frac{{\left[\frac{L_{AB}}{{\tau }_{AB}}\right]}^2}{{\left[1-\left(\frac{\delta }{{\tau }_{AB}}\right)\right]}^2}}.$$

(14)

In order to address the issue of energy consumption, we compare in Figure 6 the traditional approach of generating transports by the use of internal combustion engines and dissipative brakes (Figure 6a), and in Figure 6b an all-electric scenario in which electrical energy is converted into mechanical energy close to point $A$ and in which the generated mechanical energy $T\left(M,L_{AB},{\tau }_{AB,}\delta \right)$ is recuperated into electrical energy by reverse-operating the electrical engine as a generator upon approach to point $B$.

Turning to the internal combustion scenario first, we note that in order to initiate the transport, the kinetic energy $T\left(M,L_{AB},{\tau }_{AB,}\delta \right)$ first needs to be generated by an internal combustion engine in which the chemical energy $E_{chem}$ in the fuel is converted into mechanical energy $E_{mech}$ with an efficiency of ${\eta }_{en}=E_{mech}/E_{chem}\cong 0.3-0.4$ [8]. Due to the limited engine efficiency, a relatively large amount of chemical energy needs to be used to generate the kinetic energy that is needed to initiate transport:

$$E_{acc}\left(M,L_{AB},{\tau }_{AB,}\delta \right)={\displaystyle \frac{1}{{\eta }_{en}}}T\left(M,L_{AB},{\tau }_{AB,}\delta \right).$$

(15)

In the deceleration phase the kinetic energy $T\left(M,L_{AB},{\tau }_{AB,}\delta \right)$ is irreversibly dissipated into heat by means of frictional braking [9] with an efficiency ${\eta }_{brake}=1$ [9]:

$$E_{dec}\left(M,L_{AB},{\tau }_{AB,}\delta \right)=T\left(M,L_{AB},{\tau }_{AB,}\delta \right).$$

(16)

Summing up both contributions, the total amount of chemical energy is obtained that has ultimately been converted into low-temperature heat:

$$E_{diss}\left(M,L_{AB},{\tau }_{AB,}\delta \right)=E_{acc}\left(M,L_{AB},{\tau }_{AB,}\delta \right)+E_{dec}\left(M,L_{AB},{\tau }_{AB,}\delta \right).$$

(17)

Dividing the transport value that had been generated through the energy $E_{diss}$, the benefit-to-cost ratio $BCR$ of mechanical motion is obtained:

$$BC{R}_{start\_stop}\left(M,L_{AB},{\tau }_{AB,}\delta \right)={\displaystyle \frac{W_{start\_stop}\left(M,L_{AB},{\tau }_{AB,}\delta \right)}{E_{diss}\left(M,L_{AB},{\tau }_{AB,}\delta \right)}}$$

(18)

Rearranging Equations 13-18, one obtains

$${BCR}_{start\_stop}\left(M,L_{AB},{\tau }_{AB,}\delta \right)=\left[{\displaystyle \frac{{\eta }_{en}}{{1+\eta }_{en}}}\right]\left[{\tau }_{AB}\right]\left[1-{\displaystyle \frac{4}{3}}\left({\displaystyle \frac{\delta }{{\tau }_{AB}}}\right)\right]$$

(19)

In the limit of transport paths $L_{AB}$ long compared to the acceleration and deceleration phases close to points $A$ and $B$ and in the limit of ideal engine efficiency of ${\eta }_{en}=1$, the benefit-cost ratio reduces to :

$$B{CR}_{start\_stop}\left(M,L_{AB},{\tau }_{AB,}\delta =0\right)={\displaystyle \frac{1}{2}}{\tau }_{AB}={\displaystyle \frac{1}{2}}{\displaystyle \frac{L_{AB}}{v_{AB}}}.$$

(20)

In this latter form, the function ${BCR}_{start\_stop}$ highlights the effect of decreasing benefit-to-cost relationships when the speed of transport is increased and its time-demand being limited.

Figure 6. Converting money value contained in stored energy into money value associated with mechanical motion; (a) energy flow during mechanical transport with the help of an internal combustion engine and dissipative brakes [9]; (b) initiation of transport with the help of an electrical motor and recuperation of energy by generating electrical energy during braking.

Figure 6. Converting money value contained in stored energy into money value associated with mechanical motion; (a) energy flow during mechanical transport with the help of an internal combustion engine and dissipative brakes [9]; (b) initiation of transport with the help of an electrical motor and recuperation of energy by generating electrical energy during braking.

3.3. Frictional Transport Scenarios

The transport scenarii discussed above still represent highly idealized versions in that these completely neglect the effects of friction. In a friction-dominated transport scenario, acceleration and deceleration processes are the result of an interplay of forward-driving engine ($F_{en}$) and retarding frictional forces ($F_{fr}$), giving rise to a stationary velocity $v_{AB}$ (Figure 7a):

$$M{\displaystyle \frac{dv\left(t\right)}{dt}}=F_{en}-F_{fr}=0$$

(21)

In road-traffic situations, cars usually experience frictional forces which scale quadratically with the momentary speed $v\left(t\right)$ [10]:

$$F_{fr}\left(t\right)={{\displaystyle \frac{1}{2}}\varrho }_{air}A{C}_d{v\left(t\right)}^2.$$

(22)

In this equation, $v\left(t\right)$ its momentary velocity, ${\varrho }_{air}$the air mass density, $C_d$ the dimensionless air drag resistance, and $A$the cross-sectional area of the vehicle.

Considering long transport distances $L_{AB}$ in which acceleration and deceleration phases close to start-points $A$ and endpoints $B$ do not play significant roles, engine power $P_{en}=F_{en}{v}_{AB}$ needs to be constantly applied to overcome the effect of air friction and to maintain a constant speed $v_{AB}$:

$$P_{en}\left(v_{AB}\right)={{\displaystyle \frac{1}{2}}\varrho }_{air}A{C}_d{v}_{AB}^3.$$

(23)

Double integration over the transport time ${\tau }_{AB}$ first yields the input in energy that is required for maintaining the speed $v_{AB}$ over the entire transport length $L_{AB}$ and then the physical action that is associated with the frictional transport over this same distance. With ${\tau }_{AB}=L_{AB}/v_{AB}$ the total energy input along the transport length $L_{AB}$ emerges as

$$E_{fr}\left(L_{AB},v_{AB}\right)={{\displaystyle \frac{1}{2}}\varrho }_{air}A{C}_d{v}_{AB}^2{L}_{AB},$$

(24)

and the generated transport value as:

$$W_{fr}\left(L_{AB},v_{AB}\right)={{\displaystyle \frac{1}{4}}\varrho }_{air}A{C}_d{v}_{AB}{L}_{AB}^2.$$

(25)

The benefit-to-cost ratio, resulting from Equations 24 and 25 once again leads back to Equation 19 with $\delta =0$. Depending on the engine power involved to overcome the effect of air friction, very different values of ${BCR}_{fr}$ are obtained. In the internal combustion scenario (ic), with ${\eta }_{en}\cong 0.3-0.4$,

$${BCR}_{ic}\left(M,L_{AB},{\tau }_{AB,}\delta =0\right)\cong {\displaystyle \frac{1}{6}}{\displaystyle \frac{L_{AB}}{v_{AB}}}={\displaystyle \frac{1}{6}}{\tau }_{AB}$$

(26)

is obtained. In the all-electric scenario (el.), in which ${\eta }_{en}\cong 1$, this same quantity transforms into

$${BCR}_{el}\left(M,L_{AB},{\tau }_{AB,}\delta =0\right)\cong {\displaystyle \frac{1}{2}}{\displaystyle \frac{v_{AB}}{L_{AB}}}={\displaystyle \frac{1}{2}}{\tau }_{AB}$$

(27)

With the values of $BCR$ having the physical dimension of time, $BCR$ values can be interpreted in terms of energy relaxation times ${\tau }_E$, i.e. as those fractions of the total transport time ${\tau }_{AB}$ in which the transport-enabling energy had stayed with the transported mass $M$.

Figure 7a: Start-stop scenario with $\delta =0$ and frictional forces acting along the transport path $L=\overrightarrow{AB}$. Arrows denote the direction of the transport (red), the inputs in engine energy (blue) and the withdrawal of energy through air friction and braking (brown). Energy relaxation times ${\tau }_f$ in friction-limited transport scenarios are normally in the order of several minutes [10,11].

Figure 7b: Left-hand ordinate: Physical action versus distance attained for frictional transport ($W_{fr})$, and frictionless “fly-through” transport ($W_{AB}$), all plotted versus transport length. Right-hand ordinate: transport value of frictional transport relative to frictionless “fly-through” transport versus transport length. In the evaluation a constant speed of $v_{AB}=100km/h$ had been assumed. The crossover point at $L_{AB}\cong 4km$ is only weakly dependent on $v_{AB}.$

The most interesting aspect in the context with frictional motion arises when the frictional transport scenario is compared to a frictionless start-stop scenario with $\delta =0$, in which the same mass $M$ is transported over the same length $L_{AB}$ and with the same average speed $v_{AB}$. While in this frictionless case the transport value increases linearly with the transport length $L_{AB}$ (see Equations 10 and 13),

$$W_{A{B}_{start}-stop}\left(M,L_{AB},{\tau }_{AB}\right)={\displaystyle \frac{1}{2}}M{v}_{AB}{L}_{AB},$$

(28)

The transport value in the frictional case in creases proportionately to $L_{AB}^2$ (Equation 25).

These different length dependencies are compared in Figure 7b, showing that there is a cross-over point between the frictionless and the friction-dominated “start-stop” transport scenarii at a transport lengths of around 4$km$. This cross-over indicates that at shorter distances inertial forces play a dominant role in requiring engine power inputs to initiate motion, while at larger distances engine power requirements and associated fuel consumption are increasingly dominated by frictional forces. With regard to traffic optimization this means that in the inertia-dominated domain the generation of transport value is in the hands of drivers and their decisions of applying acceleration and deceleration forces, while in the friction-limited transport domain the avoidance of excessively long transport distances and compliance with speed limitations are the parameters of choice which help to optimize transport values towards economic and environmental constraints.

4. Going Across Hills and Valleys

5. Onboard Measurement of Transport Value

6. Summary and Conclusions

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