# Electric-Vehicle Routing Planning Based on the Law of Electric Energy Consumption
## 3 Driving-Cycle-Based Energy Consumption Model
### Energy Consumption Models of EVs Based on Different Driving Cycles
Based on the law of uniform acceleration:  
\[s_{b1}=\frac{1}{2}a_{b1}t_{b1}^{2} \tag{9}\]  
the intensity of air resistance is changed by speed; then, based on the infinitesimal method:  
\[F_{w}\cdot s_{w,b1}=\int_{0}^{t_{b1}}\frac{1}{2}\rho_{a}C_{d}A_{f}v^{2}\cdot vdt \tag{10}\]  
This converts into:  
\[F_{w}\cdot s_{w,b1}=\frac{1}{8}\rho_{a}C_{d}A_{f}a_{b1}^{3}t_{b1}^{4} \tag{11}\]  
Therefore:  
\[E_{R}=fmg\cdot\frac{1}{2}a_{b1}t_{b1}^{2}+\delta ma_{b1}\cdot\frac{1}{2}a_{b1 }t_{b1}^{2}+mgsin(\alpha)\cdot\frac{1}{2}a_{b1}t_{b1}^{2}+\frac{1}{8}\rho_{a} C_{d}A_{f}a_{b1}^{3}t_{b1}^{4} \tag{12}\]  
The battery energy consumed during the acceleration phase of cycle B, \(E_{B1}\), is then calculated as below:  
\[E_{B1}=\frac{\frac{1}{2}mv_{b1}^{2}+\frac{1}{2}m(gf+\delta a_{b1}+gsin\, \alpha)a_{b1}t_{b1}^{2}+\frac{1}{8}\rho_{a}C_{d}A_{f}a_{b1}^{3}t_{b1}^{4}} \tag{13}\]  
In the second uniform motion phase, the acceleration resistance \(F_{j}\) is 0 due to 0 acceleration and constant speed, while air resistance \(F_{w}\) is constant. The battery energy consumed during this phase of cycle B, \(E_{B2}\), is determined as:  
\[ \begin{array}{l}E_{B2}=\frac{F_{f}+F_{g}+F_{w}\cdot s_{b2}}{\eta}\\ =\frac{(mgf+mgsin\alpha+\frac{1}{2}\rho_{a}C_{d}A_{f}v_{b1}^{2})\cdot v_{b1}(t _{b2}-t_{b1})}{\eta}\end{array}  \tag{14}\]  
In the last phase of deceleration, regardless of energy recovery, all the kinetic energy is converted into work to overcome the resistance, so the battery output power is 0, and the battery does not consume energy; that is, \(E_{B3}=0\). Based on the law of uniform acceleration and uniform deceleration:  
\[t_{b1}=\frac{v_{b1}}{a_{b1}} \tag{15}\]  
\[s_{b1}=\frac{v_{b1}^{2}}{2a_{b1}} \tag{16}\]  
\[s_{b3}=\frac{v_{b1}^{2}}{2a_{b2}} \tag{17}\]  
\(t_{b1}\) and \(s_{b1}\) are the driving time and distance in the acceleration phase, respectively, and \(s_{b3}\) is the driving distance in the deceleration phase. With known distance \(d\) between any two nodes, the distance traveled in the uniform motion phase is calculated as:  
\[s_{b2}=d-\frac{v_{b1}^{2}}{2a_{b1}}-\frac{v_{b1}^{2}}{2a_{b2}} \tag{18}\]  
Inserting Equations (15)-(18) into Equations (13) and (14), we have:  
\[E_{B1}=\frac{\frac{1}{2}mv_{b1}^{2}+\frac{v_{b1}^{2}}{2a_{b1}}m(gf+\delta a_{b1 }+gsin\;\alpha)+\frac{v_{b1}^{2}}{8a_{b1}}p_{a}C_{d}A_{f}}{\eta} \tag{19}\]  
\[E_{B2}=\frac{\Big{(}mgf+mgsin\;\alpha+\frac{1}{2}\rho_{a}C_{d}A_{f}v_{b1}^{2} \Big{)}\cdot\Big{(}d-\frac{v_{b1}^{2}}{2a_{b1}}-\frac{v_{b1}^{2}}{2a_{b2}} \Big{)}}{\eta} \tag{20}\]  
Therefore, the total battery energy consumed during the entire driving cycle B is determined as:  
\[E^{B}=E_{B1}+E_{B2} \tag{21}\]  
**Driving cycle A**, as in Figure 2, has five sequential phases: first acceleration, first deceleration, second acceleration, uniform motion, and second deceleration phase. Similar to the case of cycle B, in acceleration phases, the electric energy is converted into kinetic energy and the energy to overcome external resistance; in the uniform motion phase, the energy from the battery is used to overcome resistance; and in deceleration phases, the battery does not consume any energy.