﻿ 快慢车模式下轨道交通市郊线路通过能力计算
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 同济大学学报(自然科学版)  2019, Vol. 47 Issue (7): 1022-1030.  DOI: 10.11908/j.issn.0253-374x.2019.07.014 0

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TANG Lianhua, XU Xingfang. Calculation and Expression of Carrying Capacity of Suburban Line Under Fast-Slow Mode[J]. Journal of Tongji University (Natural Science), 2019, 47(7): 1022-1030. DOI: 10.11908/j.issn.0253-374x.2019.07.014

### 文章历史

Calculation and Expression of Carrying Capacity of Suburban Line Under Fast-Slow Mode
TANG Lianhua , XU Xingfang
Key Laboratory of Road and Traffic Engineering of Ministry of Education, Tongji University, Shanghai 201804, China
Abstract: In suburban rail transit line, the operation of fast-slow trains can be well adapted to uneven spatial and temporal distribution of the passenger flow. However, due to the use of non-parallel operation diagram, the proportion of fast and slow vehicles, the distribution and quantity of the overtaking station, the interval of train departure and so on, all have different influence on the carrying capacity of the line. In order to study the influence mechanism and internal relationship of carrying capacity with different parameters of suburban line under fast-slow mode, a calculation idea was put forward, and then the calculation and expression of the carrying capacity was established by considering the quantity of overtaking and the proportion of fast and slow vehicles. Shanghai Metro Line 16 being taken as an empirical study, the proposed methods were proved effective. This study results contribute to a theoretical reference for the calculation of the carrying capacity of suburban line under fast-slow mode.
Key words: suburban lines    carrying capacity    fast-slow mode    the quantity of overtaking    the proportion of fast and slow vehicles

1 快慢车线路通过能力计算方法

 $N_{\max }=3600 / \mathrm{h}$ (1)

 图 1 快慢车模式下通过能力计算思路 Fig.1 Calculating idea of carrying capacity under fast-slow mode

 $N = K \times 3\;600/{T_{周}} = \left( {m + n} \right) \times 3\;600/{T_{周}}$ (2)

2 快慢车组合的通过能力分析

2.1 无越行情况 2.1.1 快慢比m:n=1:1

 图 2 无越行、快慢比1:1时列车运行周期 Fig.2 Train diagram without overtaking and with the proportion of fast and slow 1:1

 ${T_{周}} = T_{{\rm{e}} - 1}^0 + T_{{\rm{l}} - {\rm{e}}}^0$ (3)
2.1.2 快慢比=m:n(m>1，n>1)

 图 3 无越行、快慢比m:n的列车运行周期 Fig.3 Train diagram without overtaking and with the proportion of fast and slow vehicles m:n

 ${T_{周}} = \left\{ {\begin{array}{*{20}{c}} \begin{array}{l} m \times \left( {T_{{\rm{e}} - 1}^0 + T_{1 - {\rm{e}}}^0} \right) + \left( {n - m} \right) \times T_{{\rm{l}} - {\rm{l}}}^0\\ n \times \left( {T_{{\rm{e}} - 1}^0 + T_{1 - {\rm{e}}}^0} \right) + \left( {m - n} \right) \times T_{{\rm{e}} - {\rm{e}}}^0 \end{array}&\begin{array}{l} m < n\\ m > n \end{array} \end{array}} \right.$ (4)

 $\begin{array}{*{20}{c}} {{T_{周}} = \min \left( {m,n} \right) \times \left( {T_{{\rm{e - l}}}^0 + T_{{\rm{l - e}}}^0} \right) + }\\ {\left( {{\rm{abs}}\left( {m - n} \right)} \right) * \left\{ {\begin{array}{*{20}{c}} {T_{{\rm{l - l}}}^0}&{m \le n}\\ {T_{{\rm{e - e}}}^0}&{m \ge n} \end{array}} \right.} \end{array}$ (5)

2.2 越行1次通过能力计算 2.2.1 快慢比m:n=1:1

 图 4 越行1次、快慢比1:1的列车运行周期 Fig.4 Train diagram with one overtaking and the proportion of fast and slow vehicles 1:1

 ${T_{周}} = {T_{{\rm{AC}}}} + {T_{{\rm{JA}}}} = T_{{\rm{l - e}}}^{\rm{l}} + T_{{\rm{e - l}}}^{\rm{l}}$ (6)

2.2.2 快慢比=m:n(m>1，n>1)

 图 5 越行1次、快慢比m:n(m < n)时列车运行周期 Fig.5 Train diagram with one overtaking and the proportion of fast and slow vehicles m :n(m < n)

 $\begin{array}{*{20}{c}} {{T_{周}} = T_{{\rm{e - l}}}^{1**} + \left( {n - m - 1} \right) \times T_{{\rm{1 - 1}}}^0 + T_{{\rm{1 - 1}}}^1 + }\\ {\left( {m - 1} \right) \times {T_{周\;i}} + T_{{\rm{1 - e}}}^1\;\;\;\;m < n} \end{array}$ (7)

 图 6 越行1次、快慢比m:n(m>n)时列车运行周期 Fig.6 Train diagram with one overtaking and the proportion of fast and slow vehicles m:n(m>n)

 $\begin{array}{*{20}{c}} {{T_{周}} = T_{{\rm{e - e}}}^1 + \left( {m - n - 1} \right) \times T_{{\rm{e - e}}}^0 + T_{{\rm{e - l}}}^{1 * } + }\\ {\left( {n - 1} \right) \times {T_{周\;i}} + T_{{\rm{l - e}}}^1\;\;\;m > n} \end{array}$ (8)

 ${T_{周}} = \left\{ \begin{array}{l} T_{{\rm{e - l}}}^{1 * * } + \left( {n - m - 1} \right) \times T_{{\rm{l - l}}}^0 + T_{{\rm{l - l}}}^1 + \\ \;\;\;\;\;\;\left( {m - 1} \right) \times {T_{周\;i}} + T_{{\rm{l - e}}}^1\;\;\;\;\;\;\;\;\;\;\;\;m < n\\ m \times {T_{周\;i}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;m = n\\ T_{{\rm{e - e}}}^1 + \left( {m - n - 1} \right) \times T_{{\rm{e - e}}}^0 + T_{{\rm{e - l}}}^{1 * } + \\ \;\;\;\;\;\;\left( {n - 1} \right) \times {T_{周\;i}} + T_{{\rm{l - e}}}^1\;\;\;\;\;\;\;\;\;\;\;\;\;m > n \end{array} \right.$ (9)

 图 7 Te-l1、Te-l1*和Te-l1**的关系 Fig.7 Relationship of Te-l1, Te-l1* and Te-l1**
 图 8 Te-e0与Te-e1和Tl-l0与Tl-l1的关系 Fig.8 Relationship of Te-e0 and Te-e1, Tl-l0 and Tl-l1

2.3 越行2次通过能力计算 2.3.1 快慢比=1:1(m=n)

 图 9 越行2次、快慢比1:1的列车运行周期 Fig.9 Train diagram with two overtaking and the proportion of fast and slow vehicles 1:1

 ${T_{周}} = T_{{\rm{l - e}}}^1 + T_{{\rm{e - l}}}^2$ (10)

2.3.2 快慢比=m:n(1≤m < n)

 图 10 越行2次，快慢比1:n(n≥2)的列车运行周期 Fig.10 Train diagram with two overtaking and the proportion of fast and slow vehicles 1:n(n≥2)

 ${T_{周}} = \left\{ \begin{array}{l} T_{{\rm{l - l}}}^2 + T_{{\rm{l - e}}}^1 + T_{{\rm{e - l}}}^{2 * }\;\;\;\;\;\;\;\;\;\;\;\;\;\;m = 1,n = 2\\ T_{{\rm{l - l}}}^{2 * } + T_{{\rm{l - e}}}^1 + T_{{\rm{e - l}}}^{1 * * } + \\ \;\;\;\left( {n - m - 2} \right) \times T_{{\rm{l - l}}}^0 + T_{{\rm{l - l}}}^{1 * }\;m = 1,n > 2 \end{array} \right.$ (11)

 图 11 越行2次、快慢比m:n(m < n)的列车运行周期图 Fig.11 Train diagram with two overtaking and the proportion of fast and slow vehicles m:n(m < n)

 $\begin{array}{l} {T_{周}} = \\ \left\{ \begin{array}{l} T_{{\rm{l - l}}}^2 + \left( {m - 1} \right) \times {T_{周{\rm{i}}}} + T_{{\rm{l - e}}}^1 + T_{{\rm{e - l}}}^{2 * }\;\;\;\;\;\;n = m + 1\\ T_{{\rm{l - l}}}^{2 * } + \left( {m - 1} \right) \times {T_{周{\rm{i}}}} + T_{{\rm{l - e}}}^1 + T_{{\rm{e - l}}}^{1 * * } + \\ \;\;\;\;\left( {n - m - 2} \right) \times T_{{\rm{l - l}}}^0 + T_{{\rm{l - l}}}^{1 * }\;\;\;\;\;\;\;\;\;\;\;\;\;n > m + 1 \end{array} \right. \end{array}$ (12)

 图 12 Te-l2、Te-l2*和Te-l1**的关系 Fig.12 Relationship of Te-l2, Te-l2* and Te-l1**
 图 13 Tl-l2, Tl-l2*和Tl-l1*的关系 Fig.13 Relationship of Tl-l2, Tl-l2* and Tl-l1*

2.3.3 快慢比=m:n(1≤n < m)

 图 14 越行2次，快慢比m:1(m≥2)的列车运行周期 Fig.14 Train diagram with two overtaking and the proportion of fast and slow vehicles m:1(m≥2)

 ${T_{周}} = \left\{ \begin{array}{l} T_{{\rm{l - e}}}^{\rm{1}} + T_{{\rm{e - e}}}^{\rm{2}} + T_{{\rm{e - l}}}^{{\rm{2}} * * }\;\;\;\;\;\;\;\;\;\;\;\;\;\;m = 2,n = 1\\ T_{{\rm{l - e}}}^{\rm{1}} + T_{{\rm{e - e}}}^{\rm{2}} + T_{{\rm{e - l}}}^{{\rm{1}} * } + \\ \;\;\;\;\left( {n - m - 2} \right) \times T_{{\rm{e - e}}}^{\rm{0}} + T_{{\rm{e - e}}}^{{\rm{2}} * }\;m > 2,n = 1 \end{array} \right.$ (13)

 图 15 越行2次、快慢比m:n(m>n)的列车运行周期 Fig.15 Train diagram with two overtaking and the proportion of fast and slow vehicles m:n(m>n)

 ${T_{周}} = \left\{ \begin{array}{l} \left( {n - 1} \right) \times {T_{周\;i}} + T_{{\rm{l - e}}}^1 + T_{{\rm{e - e}}}^2 + T_{{\rm{e - l}}}^{2 * * }\;\;\;m = n + 1\\ \left( {n - 1} \right) \times {T_{周\;i}} + T_{{\rm{l - e}}}^1 + T_{{\rm{e - e}}}^2 + T_{{\rm{e - l}}}^{1 * } + \\ \;\;\;\;\left( {n - m - 2} \right) \times T_{{\rm{e - e}}}^0 + T_{{\rm{e - e}}}^{2 * }\;\;\;\;\;\;\;\;\;\;m > n + 1 \end{array} \right.$ (14)

 图 16 Te-l1*与Te-l2**的关系 Fig.16 Relationship of Te-l1* and Te-l2**
 图 17 Te-e2和Te-e2*的关系 Fig.17 Relationship of Te-e2 and Te-e2*

 $\begin{array}{l} {T_{周}} = \\ \left\{ \begin{array}{l} \left\{ \begin{array}{l} T_{{\rm{l - l}}}^2 + \left( {m - 1} \right) \times {T_{周i}} + T_{{\rm{l - e}}}^1 + T_{{\rm{e - l}}}^{2 * }\;\;\;\;\;\;\;n = m + 1\\ T_{{\rm{l - l}}}^{2 * } + \left( {m - 1} \right) \times {T_{周i}} + T_{{\rm{l - e}}}^1 + T_{{\rm{e - l}}}^{1 * * } + \;\;\;\;\;\;\;\;\;m < n\\ \;\;\;\left( {n - m - 2} \right) \times T_{{\rm{l - l}}}^0 + T_{{\rm{l - l}}}^{1 * }\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;n > m + 1 \end{array} \right.\\ m \times \left( {T_{{\rm{l - e}}}^1 + T_{{\rm{e - l}}}^2} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;m = n\\ \left\{ \begin{array}{l} \left( {n - 1} \right) \times {T_{周i}} + T_{{\rm{l - e}}}^1 + T_{{\rm{e - e}}}^2 + T_{{\rm{e - l}}}^{2 * * }\;\;\;m = n + 1\\ \left( {n - 1} \right) \times {T_{周i}} + T_{{\rm{l - e}}}^1 + T_{{\rm{e - e}}}^2 + T_{{\rm{e - l}}}^{1 * } + \;\;\;\;\;\;\;\;\;m > n\\ \;\;\;\left( {n - m - 2} \right) \times T_{{\rm{e - e}}}^0 + T_{{\rm{e - e}}}^{2 * }\;\;\;\;\;\;\;\;\;\;\;m > n + 1 \end{array} \right. \end{array} \right. \end{array}$ (15)

2.4 寻优求解算法总结

(1) 结合线路数据及客流情况，预先设定快慢车开行比例、越行次数及越行站，确定一个快慢车组合内，快慢车的开行方式(如图 1所示)；

(2) 列出不同快慢车开行比例、越行次数下，T的总结式(5)、式(9)和式(15)；

(3) 利用计算机语言实现T表达式中关键变量的计算，最后应用式(2)计算得到不同快慢车开行比例、越行次数下线路的通过能力.

3 实例验证

(1) 无越行情况下，Te-l0=120、Tl-e0=857、Te-e0=120、Tl-l0=120，代入式(2)、式(5)，可得通过能力为

 $N = \frac{{3\;600\left( {m + n} \right)}}{{\left( {857 \times \min \left( {m,n} \right) + 120 \times \max \left( {m,n} \right)} \right)}}$ (16)

(2) 越行1次情况下，Tl-e1=316，Te-l1=371、Te-l1*=120、Te-l1**=120，Te-e1=687，Tl-l1=541，代入式(2)、式(9)，可得通过能力为

 $N = \left\{ \begin{array}{l} \frac{{3\;600\left( {m + n} \right)}}{{\left( {120n + 567m + 170} \right)}}\;\;\;\;\;m < n\\ \frac{{3\;600\left( {m + n} \right)}}{{687m}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;m = n\\ \frac{{3\;600\left( {m + n} \right)}}{{\left( {120m + 567n + 316} \right)}}\;\;\;\;\;m > n \end{array} \right.$ (17)

(3) 越行2次情况下，Tl-e1=316、Te-l2=146、Te-l2*=120、Te-l2**=120、Te-e2=462、Te-e2*=379、Tl-l2=316、Tl-l2*=421、Tl-l1*=120，代入式(2)、式(15)，可得通过能力为

 $N = \left\{ \begin{array}{l} \left\{ \begin{array}{l} \frac{{3\;600\left( {m + n} \right)}}{{\left( {462m + 290} \right)}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;n = m + 1\\ \frac{{3\;600\left( {m + n} \right)}}{{\left( {342m + 120n + 275} \right)}}\;\;\;\;\;\;\;\;\;n > m + 1 \end{array} \right.\\ \frac{{3\;600\left( {m + n} \right)}}{{462m}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;m = n\\ \left\{ \begin{array}{l} \frac{{3\;600\left( {m + n} \right)}}{{\left( {462n + 436} \right)}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;m = n + 1\\ \frac{{3\;600\left( {m + n} \right)}}{{\left( {342n + 120m + 575} \right)}}\;\;\;\;\;\;\;\;\;m > n + 1 \end{array} \right. \end{array} \right.$ (18)

 图 18 不同越行次数快慢车开行比例下线路通过能力 Fig.18 Carrying capacity under different quantity of overtaking and different proportion of fast and slow vehicles

(1) 对于相同越行次数，随着同种类列车(快车或慢车)开行列车的增加，线路通过能力不断上升，使得图形呈现“V”字型；

(2) 对于不同越行次数:当慢车开行列数相等，快车多于慢车时，对于相同的快慢比(如m:n=8:1)，快车越行慢车2次时线路的通过能力最小，无越行时线路通过能力最大，说明慢车由于被越行增加的停站时间对于线路通过能力有一定的损失；当快车数开行列车相等，且快车少于慢车时时，对于相同的快慢比(如m:n=1:8)，无越行时线路通过能力最小，快车越行慢车2次时线路通过能力最大，说明随着快车越行慢车次数的增加，前行慢车和后行快车之间的发车间隔逐渐缩短，从而带来线路通过能力的增加.

4 结语

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