Operation method of traffic vehicle in all-flight operation mode

Abstract

In order to solve the problem that the travel speed is too low due to the fact that public transport vehicles stop at stations along the way, the invention provides a multi-shift system stop-at-interval all-flight operation mode method, all-line running vehicles almost stop at the same time and stop at intervals, the risk of rear-end collision can be effectively avoided when driving is dense, and the vehicle safety is improved. Therefore, the passengers can arrive at the target station, the riding time of the passengers can be saved, the vehicle traveling speed can be increased, and the operation benefit is improved.

Description

Technical Field

[0001] The invention relates to a method for a public transport vehicle to stop at alternate stations, in particular to a method in one-way single-line operation (when in operation, there is only one lane in one direction where overtaking is not allowed), and belongs to the technical field of public transport vehicle stop at alternate stations. Background Art

[0002] In order to solve the traffic problem, people invented cars, trains, subways, maglev trains, and imagined the realization of vacuum tube trains in the future. The maximum speed that these transportation tools can reach is very fast, but the travel speed is not fast. For example, the city subway has a dedicated channel underground, without red and green lights, and is not affected by pedestrians and other vehicles. As a result, the travel speed (about 30Km / h) is only one-tenth of the high-speed rail speed (300Km / h to 350Km / h). The main reason is that there are many and dense stations along the way, making it difficult to increase the travel speed. In the one-way and single-line operation of the city subway, it is based on the conventions of buses and trains, and every station must be stopped. Due to the short distance between stations, generally only 2 to 3 minutes of travel, in order to take care of the comfort of passengers, the acceleration cannot be too large, generally not more than 1m / s^2, so that the subway has to slow down as soon as it accelerates, so the speed cannot be increased. When stopping at a station, the subway must go through the necessary steps of deceleration, stopping, opening doors, boarding and alighting passengers, closing doors, confirming door closing, starting, and accelerating. The whole process takes about two minutes. The time for boarding and alighting passengers varies depending on the number of people getting on and off. Generally, it takes 20 to 30 seconds when there are few people, and 8 to 12 seconds for confirming door closing. The subway does not move when stopping at a station. The travel time of most passengers who are not getting on or off at the current station is delayed by the few passengers who are getting on or off. As a result, each time a station is passed, the time of the stop is delayed. The more stations the subway stops at, the longer it takes to complete the journey, and the more passengers are stranded in the carriage due to travel. In order to solve the drawbacks caused by stopping at every station, a small number of subway lines abroad have been reformed into a two-shift "frog jump" method, which saves time by spacing out stations. Its rule is to divide the operating vehicles into two shifts, and the two shifts take turns to set off. The interval between stations for one shift is 1, 4, 7... and the interval between stations for the other shift is 2, 5, 8... (or 3, 6, 9...), that is, each shift stops at two stations and then takes one station (commonly known as "frog jump"). However, the disadvantage of this method is that some stations must be reached by transferring to the next train because the shifts do not stop. In addition, the distance between most stations is short, and the rear vehicle "flies" to the front vehicle's rear when it passes the interval station (the interval station is also called flying station), increasing the risk of rear-end collision between the front and rear vehicles. If the risk of rear-end collision is avoided, the time saved is not obvious. For vehicles operated in the "frog jump" style, the overall parking and moving cannot be synchronized ("flying" to the station at the same time), and this feature cannot cater to the operation of future vacuum tube trains. The fact that public transportation vehicles stop at every station along the way has seriously affected the travel speed. The "leapfrog" mode not only increases the risk of rear-end collisions, but also has no obvious effect on improving travel speed. A synchronous "flying" station operation mode is emerging and has come into being. Summary of the invention

[0003] The purpose of the present invention is to overcome the shortcomings of the previous technology and provide a method of stopping at alternate stations that can avoid traffic risks and significantly increase travel speed. The solution is to adopt a synchronous stop-and-go method when the vehicle stops at alternate stations. The multi-shift synchronous stop-and-go rule of the present invention is an operation plan formulated based on this principle. The following first introduces the concepts involved in this specification: Flying station: When public transport vehicles are in operation, if a vehicle passes a certain station without stopping, it is called an interval station, also called a flying station. The process of the vehicle flying through the station is called "flying".

[0004] Synchronous flying: When the vehicles are flying, for any adjacent front and rear vehicles, when one of the vehicles flies, the other vehicle always flies at a station with a fixed station distance (denoted as P, P ≥ 1 station distance) from the station when the other vehicle flies. This method of two adjacent vehicles flying is called synchronous flying. The vehicles that undergo synchronous flying do not necessarily fly at the same time. When the distance between the front and rear vehicles is greater than P, the front vehicle flies first and the rear vehicle flies later; when the distance between the front and rear vehicles is exactly equal to P, the front and rear vehicles fly at the same time; when the distance between the front and rear vehicles is less than P, the front vehicle flies later and the rear vehicle flies first. Using synchronous flying can avoid the above-mentioned "the rear vehicle flies while the front vehicle stops", which causes the rear vehicle to be forced to slow down and easily rear-end collision.

[0005] Synchronous flight operation: The operation mode in which the vehicles running on the entire line make synchronous flying stops as a whole, in sections or in part is called synchronous flight operation, which are respectively called overall synchronous flight (whole flight), section synchronous flight (separate flight) and partial synchronous flight (local flight).

[0006] Vehicle teams: In Qifei operation, we need all vehicles to stop and have different intervals. Therefore, we divide all vehicles into multiple teams. Vehicles in the same team stop and have the same intervals, while vehicles in different teams stop and have different intervals.

[0007] Vehicle shift system: The vehicle shift system is a dispatching system based on vehicle shifts. When there are Q vehicle shifts, the vehicle shift system is referred to as the Q shift system. According to the Q shift system, when dispatching, from shift 1 to shift Q in order, each shift dispatches and only dispatches one vehicle until the Qth vehicle is dispatched. The dispatched Q vehicles are called the first round. After each round is dispatched, the dispatch continues according to the dispatching rules of the first round.

[0008] M stops and N intervals: When a public transport vehicle is in operation, if the vehicle stops at several stations continuously and then "flies" to several stations continuously, it is called a number of stops and intervals, which is recorded as M stops and N intervals (M and N are positive integers). For example, two stops and three intervals means that the vehicle stops at two stations continuously and then "flies" to three stations continuously.

[0009] In the synchronous operation, according to the different train schedules Q, the number of consecutive stop sites M (M is a positive integer) and the number of consecutive interval sites N (N is a positive integer) are determined as follows: for all m and n that satisfy the equation m+n=Q (m, n, are all positive integers, Q is the number of schedules), M=km, N=kn (k is a positive integer). The Q corresponding to such values ​​can enable the vehicles running on the entire line to reach synchronous flying stations.

[0010] In synchronous flying station operation, the stops and intervals of each team are distributed periodically, and the period is recorded as T, T=kQ; the station distance between the two adjacent vehicles making the same stop (and flying station) is P=k. The arrangement method of vehicles for synchronous flying operation on the entire line is: when starting, from the first vehicle to the last vehicle, the first vehicle starts to operate in the M-stop N-space mode from the first station, and each subsequent vehicle lags behind (in the driving direction) P stations behind the station where the previous vehicle stops and has an interval, and makes the same stop and interval operation, so as to achieve the synchronous flying operation mode of the entire line. The different situations of Q values ​​are described as follows: 1: Two-shift system (Q=2) with M stops and N breaks: Because the rule is: m + n = Q (m∈N+, n∈N+), Q = 2, ∴m + n = 2. Solving this equation, we get m = 1, n = 1, because M = km, n = kn, T = kQ (k∈N+), P = k. So: (1) When k = 1, M = N = 1, T = 2, P = 1; (2) When k = 2, M = N = 2, T = 4, p = 2; (3) When k = 3, M = N = 3, T = 6, P = 3; … (k): M=N=k, T=2k, P=k. That is, there is one stop and one room, two stops and two rooms, three stops and three rooms... k stops and k rooms. One stop and one room means P=1, one shift stops at odd-numbered stations, and the other shift stops at even-numbered stations. The stops and intervals are distributed in a cycle of two stations; two stops and two rooms means P=2, one shift stops at two stations in a row and then stops at two stations, that is, stops at 1, 2, 5, 6, 9, 10... The stops and intervals of the other shift are exactly the opposite of the first shift, that is, stops at 3, 4, 7, 8, 11, 12... The stops and intervals of the stations are distributed in a cycle of four stations; three stops and three rooms means P=3, one shift stops at 1, 2, 3, 7, 8, 9... The stops of the other shift and the first shift are exactly the opposite. On the contrary, the stops at 4, 5, 6, 10, 11, 12... are distributed in a cycle with six stations in between; the k-stop k-interval is P=k, one shift stops at 1, 2, 3...k, 2k+1, 2k+2, 2k+3...3k... The other shift stops at the opposite of the first shift, stopping at k+1, k+2, k+3...2k, 3k+1, 3k+2, 3k+3...4k... The stops at 2k stations are distributed in a cycle with 2k stations in between.

[0011] 2: Three-shift system (Q=3) with M stops and N breaks: Since the rule is m+n=Q(m∈N+,n∈N+), Q=3, solving this equation yields: (1) m = 1, n = 2, (2) m=2, n=1.

[0012] ∵M=km,N=kn,T=kQ,there are also two groups of values ​​of M and N. The first group is when m=1,n=2: (1): When k = 1: M = 1, N = 2, T = 3, P = 1; (2): When k = 2: M = 2, N = 4, T = 6, P = 2; (3): When k = 3: M = 3, N = 6, T = 9, P = 3; … (k): M=k, N=2k, T=3k, P=k; The second group, when m=2, n=1, has: (1): When k = 1: M = 2, N = 1, T = 3, P = 1; (2): When k = 2: M = 4, N = 2, T = 6, P = 2; (3): When k = 3: M = 6, N = 3, T = 9, P = 3; … (k): M=2k, N=k, T=3k, P=k.

[0013] That is to say, the three-shift operation has one stop and two rooms, two stop and one room, two stop and four rooms, four stop and two rooms, three stop and six rooms, six stop and three rooms... k stop and 2k rooms, 2k stop and k rooms. Now we will focus on two stop and one room and one stop and two rooms.

[0014] The specific specification of "two stops and one room" is that the first shift: stops at 1 and 2 (i.e. stops at both stations 1 and 2), stops at 3 (i.e. stops at 3), stops at 4 and 5, stops at 6... stops at 3, 6, 9... that is, stops at 3n (n is a positive integer); the second shift: stops at 1, stops at 2, stops at 3 and 4, stops at 5, stops at 6 and 7, stops at 8... stops at 2, 5, 8... that is, stops at 3n-1 (n is a positive integer); the third shift: stops at 1, stops at 2 and 3, stops at 4, stops at 5, 6, stops at 7... stops at 1, 4, 7... that is, stops at 3n-2 (n is a positive integer). The specific specification of "one stop and two rooms" is exactly the opposite of the stops and intervals of "two stops and one room", that is, the third shift stops at 3, 6, 9... the second shift stops at 2, 5, 8... the first shift stops at 1, 4, 7... the stops and intervals are distributed in a cycle of three stations.

[0015] To summarize, the distribution pattern of the stops or interval sites of each shift in Qifei's three-shift operation is as follows: k stops for 2k periods and 2k stops for k periods are both based on the pattern of one stop for two periods or two stops for one period, with the number of original stops or interval sites expanded by k times.

[0016] 3: Four-shift system (Q=4) with M stops and N intervals: Because the rule is: m + n = Q (m ∈ N +, n ∈ N +), Q = 4, solving this equation gives three sets of solutions: (1): m = 1, n = 3; (2): m = 2, n = 2; (3): m=3, n=1.

[0017] There are also three sets of solutions for M=km, N=kn.

[0018] (1): M = k, N = 3k; (2): M = 2k, N = 2k; (3): m=3k, N=k.

[0019] Since M = km, N = kn, T = kQ, there are also three groups of values ​​for M and N: The first group: When m=1, n=3, we have: (1): When k = 1: M = 1, N = 3, T = 4, P = 1; (2): When k = 2: M = 2, N = 6, T = 8, P = 2; (3): When k = 3: M = 3, N = 9, T = 12, P = 3; … (k): M=k, N=3k, T=4k, P=k; The second group: When m=2, n=2, we have: (1): When k = 1: M = 2, N = 2, T = 4, P = 1; (2): When k = 2: M = 4, N = 4, T = 8, P = 2; (3): When k = 3: M = 6, N = 6, T = 12, P = 3; … (k): M=2k, N=2k, T=4k, P=k; The third group: When m=3, n=1, we have: (1): When k = 1: M = 3, N = 1, T = 4, P = 1; (2): When k = 2: M = 6, N = 2, T = 8, P = 2; (3): When k = 3: M = 9, N = 3, T = 12, p = 3; … (k): M=k, N=3k, T=4k, p=k; That is, the four-shift operation has one stop for three rooms, two stops for two rooms, three stops for one room, two stops for six rooms, four stops for four rooms, six stops for two rooms... k stops for 3k rooms, 2k stops for 2k rooms, 3k stops for k rooms. Now we will focus on the case of k=1 when one stop for three rooms, two stops for two rooms, and three stops for one room.

[0020] One stop three means the first shift stops at 1, 5, 9...4n-3 (n is a positive integer, recorded as n∈N+), the second shift stops at 4, 8, 12...4n (n∈N+), the third shift stops at 3, 7, 11...4n-1 (n∈N+), and the fourth shift stops at 2, 6, 10...4n-2 (n∈N+); two stop two means the first shift stops at 1, 2, 5, 6...4n-3, 4n-2 (n∈N+), the second shift stops at 1, 4, 5, 8, 9...4n-3, 4n (n∈N+), and the third shift stops at 1, 2, 5, 6...4n-3, 4n-2 (n∈N+), Stop at 3, 4, 7, 8...4n-1, 4n (n∈N+), the fourth shift stops at 2, 3, 6, 7...4n-2, 4n-1, the four-shift system of two stops and two rooms is different from the two-shift system of two stops and two rooms; three stops and one room means that the first shift has interval stations 4, 8, 12...4n (n∈N+), the second shift has interval stations 3, 7, 11...4n-1 (n∈N+), the third shift has interval stations 2, 6, 10...4n-2 (n∈N+), and the fourth shift has interval stations 1, 5, 9...4n-3 (n∈N+). When k takes different values, the method of stopping and spacing stations for each shift refers to the mode of k=1, except that the number of stops and interval stations is expanded by k times.

[0021] 4: Q shift system with M stops and N intervals: Because the rule is: m + n = Q (m ∈ N +, n ∈ N +), first solve this equation to get how many groups of solutions there are for all m and n, and then there are how many groups of solutions for M = km, N = kn. The arrangement method of the stops and intervals of the corresponding vehicle stations for each group of solutions is: each shift makes M stops and N intervals. When the train departs, the first train starts to make M stops and N intervals from the starting station, and the second train follows the first train to stop and interval at the station with a difference of k stations behind it (in the direction of travel), and the following trains all stop and interval at the station with a difference of k stations with the previous train. In this way, after the Q trains are dispatched, the second Q trains behind have the same stops and intervals as the previous trains. The distribution of the stops and intervals of the stations is based on the kQ cycle. In the operation of the simultaneous operation with M stops and N intervals, it is only when the distance between the front and rear trains is exactly P stations apart, that is, when they are separated by stations at the same time, when they fly stations synchronously, that is, when they fly stations synchronously. Even if the distance between the front and rear vehicles is not strictly P stations, and they are not at the same time, but have a time difference, the simultaneous flying operation can still operate normally. In other words, the simultaneous flying operation does not strictly require that the vehicles that have simultaneous flying stations are at the same distance, but has a large operating and fault tolerance space.

[0022] The first and / or last stop in Qifei's operation can be chosen to stop at every shift or according to the rules of each shift.

[0023] In simultaneous flying operation, the smaller the P value is, the denser the density of running vehicles will be in order for the front and rear vehicles of the synchronous flying station to fly at the same time, that is, the density of departures will be denser, but the departure density is dense enough that the rear vehicle "flies" first and the front vehicle "flies" later, resulting in the risk of rear-end collision. The maximum value is when the P value is larger, the sparser the density of running vehicles will be in order for the front and rear vehicles of the synchronous flying station to fly at the same time, that is, the density of departures will be sparser. When the departure density increases, the rear vehicle "flies" first and flies continuously to catch up with the front vehicle. In order to avoid the risk of rear-end collision, the departure density is appropriately reduced.

[0024] In the operation of Qifei, the number of stops and intervals is different, that is, the size of M and N in M-stop-N interval is different, and the effect and actual use are different. When M>N, there are few intervals, but there are vehicles that can go directly between any two stations, which is more suitable for buses with dense stations and slow speed, such as urban subways; when M=N, there are many intervals, but it is not guaranteed that there are vehicles that can go directly between any two stations. Except for the two-shift system M=N, all others can transfer to the current train. When the two-shift system M=N, if you cannot go directly to the target station, you can only choose to walk or other means of transportation after reaching the nearest station. It is more suitable for high-speed trains; when M<N, there are more intervals, but some of any two stations can be reached directly, some need to be transferred, and some cannot be reached and need to be changed to other means of transportation. It is more suitable for high-speed rail, maglev trains, and future vacuum tube trains; when M=1, N≥2, there are the most intervals, but there are very few direct stations, and you can only choose other means of transportation after reaching the nearest station. It is more suitable for maglev and future vacuum tube trains.

[0025] In Qifei operation, the smaller the Q value, the shorter the period of station stop or interval, and it is more suitable for low-speed traffic vehicles, such as urban subways; the larger the Q value, the longer the period of station stop or interval, and it is more suitable for high-speed trains, such as maglev trains and future vacuum tube trains.

[0026] The synchronous operation can be used in one direction or two directions in both directions. The advantage of the one-way mode is that when there are many stations between, the stations that cannot be reached directly can be reached by taking a return bus to the target station. In the two-way synchronous operation, since the departure is in the order of 1, 2, 3, when returning, if there is no change, it is still in the order of 1, 2, 3, but the order of the stations is reversed. This will either cause the corresponding stops and intervals of each team to be inconsistent in the two directions, or the synchronous flying station operation mode cannot be realized, so the sequence number of the team must be reversed, and the team identification must also be changed accordingly. If the running vehicles are exactly an integer multiple of the shift system, the shift sequence number is reversed when going, and the shift sequence number is changed back to the original when returning, and the vehicle identification is also changed back, that is, shift A and shift B are interchanged back and forth; if the running vehicles are not an integer multiple of the shift system, shift A and shift B cannot be interchanged when turning back and forth, but a large cycle of shift A to shift B, shift B to shift C, shift C to shift D, and then shift D to shift A. Whether it is an integer multiple or not has its own advantages and disadvantages. It is easy to interchange identifications back and forth for integer multiples, but some shifts are interchanged with each other (Q is an even number greater than 2, or a non-middle shift when Q is an odd number), and some are not interchanged (Q is equal to 2, or a middle shift when Q is an odd number), which causes the vehicles that stop at large stations (or small stations) to stop at large stations (or small stations) and cannot be changed; if it is not an integer multiple, the identification needs to be changed constantly, although it is easier to confuse, but the large and small stations where each vehicle stops are interchanged back and forth. These two methods can be selected by adjusting the number of added or deleted vehicles to an integer multiple.

[0027] The advantages of Qifei operation are: 1: The method of stopping at alternate stations is used, and the vehicle density is maximized (the reason is that when the rear vehicle flies to catch up with the front vehicle, the front vehicle also happens to be flying to the station); 2: It saves passengers’ time (including waiting time), shortens the total journey time by spacing out stops, increases vehicle turnover, and thus improves operational efficiency.

[0028] 3: It not only avoids the problem of rear-end collisions between vehicles when the trains stop at the interval stations of a single shift (including the "frog skipping" system of two shifts) during asynchronous flying, but also can increase the speed at the interval stations while appropriately adjusting the distance between vehicles.

[0029] 4: Shortening the total journey time will also bring about another effect. The number of passengers stranded in the carriage will decrease (this is mainly related to the interval between trains, but increasing the vehicle turnover rate can make the interval between trains shorter), reducing the load on the vehicle. The reduced load will make it easier to accelerate and brake, reduce wear on vehicles and tracks, reduce the impact on the surrounding areas (mainly vibration and noise), and allow for faster cornering speeds, among other benefits.

[0030] 5: The distance between two consecutive stops of the vehicle is artificially increased, and there is a relaxed time for speeding up and maintaining a relatively high speed between stations, which also saves the energy consumption of multiple acceleration and deceleration.

[0031] There are three main disadvantages of the mixed-flying operation: the first is that the waiting time increases; the second is that different mixed-flying modes may require transfers; and the third is that passengers are easily confused about the schedule. However, the two-lane fast and slow trains on the same line (one lane for fast trains and one lane for slow trains) also have these disadvantages. Although the third disadvantage is less (because people have the concept of fast and slow trains and are not easy to confuse the schedule), the first and second disadvantages are more serious. The first, more serious reason is that more express trains are needed to save waiting time. Even if there are so many express trains, they still need to stop at double platforms, otherwise the slow train will not move and the express train cannot stop. If it is a double-platform double-channel (operation mode), the simultaneous operation only changes the single-channel slow train and the single-channel express respectively, and the operation mode of the express and slow trains stopping at the double platform can be changed to simultaneous operation; the second, more serious reason is that different simultaneous operation modes, some (any two stations) do not require transfers, such as three shifts with two stops and one station, and some have a low transfer probability. Which mode to use can be selected, and for double-channel express trains, the transfer probability is higher and the mode cannot be changed.

[0032] Now, taking the urban subway as an example, we will introduce the three-shift, two-stop, one-room system operated by Qifei.

[0033] In a three-shift system with two stops and one room, the three shifts are given different names and different logos (two logos, vehicle logo and station sign logo), such as the front-line shift, the middle-line shift and the back-line shift. The logos can be simple and easy to distinguish, such as text, graphics, letters and colors. The name and (or) logo are marked in a conspicuous position inside and outside the vehicle, and the logo is marked above or below (or on the side) of each station on the station sign. The most typical approach is to divide the three shifts into yellow, green and red shifts. The external signs of the vehicles can be the conspicuous light colors corresponding to the vehicles of each shift, such as yellow light for yellow shift, green light for green shift, and red light for red shift. The internal signs of the vehicles can be lights or displayed on a display screen. The sign of each shift is to divide a small circle into three sectors evenly. The sector directly above is painted yellow as the yellow shift sign, the sector on the lower left is painted green as the green shift sign, and the sector on the lower right is painted red as the red shift sign. Then add signs in conspicuous positions inside and outside the vehicles of each shift. When marking the side of the station on each stop sign, pay attention to the fact that since each station has two or three shifts, there are two or three small circles next to the name of a station, and the entire stop sign looks dazzling with small circles. We make the team logo overlap into a small circle after transparent processing, that is, three sectors on a small circle. If there is a corresponding color in the corresponding position, it means one corresponding team, and if there are two or three corresponding positions, it means two or three corresponding teams. Put such a sign above or below (or next to) the station name on the bus stop sign. Each station name has only one small circle, which is simple and clear. When passengers check each shift, it is easy to understand which shift stops at which station and which station does not stop, so it is not easy to confuse the shift. Because passengers have different comprehension abilities, they can consult the staff. The platform staff can first train and then help answer questions. If the Qifei operation method can be recognized and popularized, it is hoped that the signs of different shift intervals and station sign reading will be written into textbooks.

[0034] With the overall understanding of Qifei's operation, let's calculate the time saved by passengers and the efficiency of vehicles. In the "two stops and one room" operation of the entire line, since the yellow, green and red teams correspond to intervals of 1, 4, 7... (3n-2), 2, 5, 8... (3n-1), 3, 6, 9... (3n), it can be seen that the team that stops is a cycle of every three stations, two teams stop at each station, and one team does not stop. When the passenger journey (number of stations) is the same, no matter which station the passenger takes the bus from, the result of calculating the saved time is the same, so only one of the stations is taken to calculate the average time saved by the passenger, which is the average time saved by all passengers. Since the shuttle bus stops at three stations in a cycle, the stations selected during the calculation can be arbitrary, but the stations that the passengers want to arrive at must be three consecutive stations to be representative. Before the calculation, make assumptions based on probability theory, use the assumed conditions to calculate the results and then compare them with the actual situation, and try to reduce the error with authenticity. Before calculation, assume that all stations are equally spaced, the stop times are equal, and the number of people getting on and off at all stations is equal. Assume that the total number of stations is M, and the average distance traveled by each passenger is 1 / 3 of the total number of stations. Take station 4 as an example: the stations that passengers need to arrive at are (M-1) / 3+3, (M-1) / 3+4, and (M-1) / 3+5. When calculating the specific route, these three stations take integer values ​​(i.e. (M-1) / 3 takes an integer value). Let the yellow shift be at the (M-1) / 3+3rd station, then the green shift is at (M-1) / 3+4, and the red shift is at (M-1) / 3+5. Similarly, let the time of stopping at each station be t1, and the time interval between buses at the time of departure be consistent as t2. There are three stations 1, 2, and 3 behind station 4 (referring to the driving direction). Since the stations where the three shifts stop are distributed in a cycle of three stations, the buses of the three shifts arriving at station 4 are t1 earlier than stopping at every station, but the bus interval time t2 remains unchanged.

[0035] First calculate the time saved by passengers arriving at the (M-1) / 3+3 station: Because the yellow bus does not stop at station 4 or (M-1) / 3+3, passengers can directly reach the destination by taking any of the following two buses. When waiting for the bus, there is a 2 / 3 probability that the bus interval is half of the bus interval time t2, that is, t2 / 2, and a 1 / 3 probability that the bus interval is half of the bus interval 2t2, that is, t2. The average waiting time is (t2 / 2)×2 / 3+t2 / 3=2t2 / 3. The average waiting time for "one stop at a time" is t2 / 2, so the waiting time is delayed by 2t2 / 3-t2 / 2=t2 / 6. The number of stations between reaching the destination station is: (M-1) / (3×3)=(M-1) / 9, and the time saved by stopping at every other station is (M-1)t1 / 9. The total time saved by reaching (M-1) / 3+3 stations is (M-1)t1 / 9-t2 / 6.

[0036] Now let’s calculate the time saved by passengers arriving at (M-1) / 3+4 stations: The yellow bus does not stop at station 4, and the green bus does not stop at station (M-1) / 3+4. There are two ways to reach station (M-1) / 3+4: one is to take the red bus directly, and the other is to take the green bus and transfer to the yellow bus. In order to calculate the average waiting time for passengers, the time for passengers to arrive at the platform and wait for the bus is divided into three periods, and the time in each period is equal.

[0037] The first period: The yellow class has just arrived and the red class has just arrived and the gate is opened and closed. Since the red class can go directly to the destination station, they can get on the bus and ride. The average waiting time is t2 / 2, which is the same as before the reform. That is, there is no delay in waiting for the bus compared with before the reform. The total time saved is: (M-1)t1 / 9; The second period: the red bus just reaches the point where the green bus arrives and the gate opens and closes. Since the green bus does not stop at the destination station, the red bus can take the green bus to transfer to the yellow bus (because the yellow bus does not stop at Station 4). The average waiting time is still the same as before the reform, which is t2 / 2. In other words, there is no delay compared with before the reform. However, waiting for the next bus (yellow bus) during the transfer delays t2. The total time saved is: (M-1)t1 / 9-t2. The third period: The green class has just arrived when the yellow class arrives (the yellow class does not stop). Since the yellow class does not stop, you can only take the red class behind it. The average waiting time is t2 more than the waiting time in the first period. The total time saved is: (M-1)t1 / 9-t2.

[0038] The average time saved by taking the bus at stop 4 is the sum of the three periods divided by 3: [(M-1)t1 / 9+(M-1)t1 / 9-t2+(M-1)t1 / 9-t2] / 3=(M-1)t1 / 9-2t2 / 3 And calculate the time saved by passengers arriving at (M-1) / 3+5 stations: The red bus does not stop at (M-1) / 3+5 stations. After taking the red bus, you still need to transfer to the next green bus, which saves the same amount of time as taking the green bus directly. The green bus can reach the destination directly. The interval between the green buses is 3t2, and the average waiting time is: 3t2 / 2. The average waiting time is 3t2 / 2-t2 / 2=t2 of the time before the reform. The time saved by the interval station is (M-1)t1 / 9. The total time saved by arriving at (M-1) / 3+5 stations is: (M-1)t1 / 9-t2.

[0039] Compared with the passenger travel time of "one stop per bus", the average time saved is: T=【(M-1)t1 / 9-t2 / 6+(M-1)t1 / 9-2t2 / 3+(M-1)t1 / 9-t2」 / 3=(M-1)t1 / 9-11t2 / 18 Analyze the equation T = (M-1) t1 / 9-11t2 / 18. There are two terms on the right side of the equation. The first term (M-1) t1 / 9 is the time saved by the 1 / 3 distance interval station. It is related to the distance of the journey. The longer the journey, the more time is saved. The time saved by the 1 / 2 distance interval station is: (M-1) t1 / 6, and the time saved by the 1 / 3 distance interval station is: (M-1) t1 / 3; the second term 11t2 / 18 is the extra time for random boarding compared to the conventional mode of stopping at every station. It can be reduced by letting passengers get on the bus on time through the shuttle bus schedule. If the timing is just right, this term can be negative, that is, -t2 / 2. -t2 / 2 is the time to stop at each station to wait for the bus. The actual value of the second term is between 11t2 / 18 and -t2 / 2. This time is controlled by the passengers' familiarity with the arrival time of the shuttle bus.

[0040] The total time saved by the interval between stations is not only the stop time t1, but also the time saved by the vehicle maintaining high-speed driving, which is t3. Let t3 = t1 / 2 (which is equivalent to the actual situation), and then let the latter term 11t2 / 18 equal to 0 (because most passengers are familiar with the arrival time of the bus due to work habits and repeatedly take the same bus). Then add these two instructions to T and recalibrate T, then: T=(M-1)(t1+t3) / 9-0=(M-1)t1 / 6 This result is the average time saved by all passengers, and also the time saved by a single passenger if the distance traveled is 1 / 3 of the full journey. If the distance traveled is 1 / 2 of the full journey, the time saved is (M-1)t1 / 4, and if the distance traveled is the full journey, the time saved is (M-1)t1 / 2 The benefits of improved subway operations are: (M-1) (t1 + t3) / (3Mt4) × 100% ≥ (M-1) (t1 + t3) / (3Mt2) × 100% ≈ t1 / (2t2) × 100% Where: t4 is the time it takes for the subway to complete one station (because the departure density does not exceed one train per station, so t4≤t2) In order to have a clearer understanding of the calculation results, we will give an example and list facts in the example to demonstrate the deviation between probability calculation and reality. In the example calculation, we do not use the existing formula to directly calculate the result, but re-examine the calculation process of probability theory with the actual situation to see whether the result deviates in a favorable direction (the favorable direction means saving more time and having higher benefits) or an unfavorable direction.

[0041] Actual route: Wuhan Metro Line 2, a total of 38 stations, one train every 3 minutes, and stops for about one minute at a station, so we calculate it as one minute. Since it is a practical consideration, we have to make precise calculations for each sampling number, and analyze them one by one below: 1:38 sites: The data is real.

[0042] 2: One train every 3 minutes: This is a probability. During off-peak hours, there is a train every 3.4 minutes, and during peak hours, there is a train every 2.9 minutes.

[0043] 3: The parking time is 1 minute: it includes stopping (stationary), opening the door, waiting, closing the door, confirming the door closing, and starting. The door opening and closing time is about 25 seconds during low peak hours, 40 to 60 seconds during peak hours, and the door closing confirmation time is 10 to 12 seconds. The total parking time during peak hours (excluding the time for deceleration and acceleration) will exceed 1 minute; 4: Assumptions of equal station spacing and equal stop times: The spacing is basically equal in station arrangement, and the stop times are not much different. These two data have little impact on the calculation results and can be ignored.

[0044] 5: The assumption that the number of passengers getting on and off at each station is equal: First, if the number is not equal, it will be inconvenient to simplify the calculation. Second, the calculation result is only related to the distance of the passengers' journey, and has nothing to do with whether the number of passengers getting on and off at each station is equal.

[0045] 6: Assumption of average journey: First, passengers who transfer between different lines are counted as two rides, not one ride, because they have to wait for the bus again when transferring between different lines, and then the average journey is evaluated. The average journey must be less than 1 / 2, because when the average journey is equal to 1 / 2, basically every passenger arrives at the terminal, which is impossible in actual situations (assuming that the terminal has the same number of passengers and the journey is 0 stops, then, starting from the starting station, every passenger arrives at the terminal, and the average journey of the passenger is exactly 1 / 2 of the whole journey).

[0046] 7: Let’s look at the calculation process again: According to the probability distribution, the number of passengers getting on and off at each station is equal, but this is not the case in reality. When passengers take any subway (operating with two stops and one station), theoretically 2 / 3 of the passengers can get off directly at the station, and 1 / 3 of the passengers cannot get off at the station (they cannot get off and have to transfer). In reality, these two data are somewhat different compared with the probability, because people always choose the most time-saving way to travel, and they will automatically adjust the delay time to the minimum. Some of the passengers who want to transfer arrive at their destination between two stations; some passengers transfer to buses, and subway stations and bus stations on many lines often have public stops, so these two groups of passengers do not need to transfer; and some passengers originally have other transportation methods that they prefer not to take, so they simply do not take the subway if it is time-consuming, and take the subway again if it saves time (this alone, the reform plan is also a positive side for operational economic benefits). Due to these factors, the passengers who originally transferred will flow to get off directly at the station. The proportion of passengers who get off directly at the station will be larger than 2 / 3, and the proportion of passengers who need to transfer will be smaller than 1 / 3, and the average time saved by passengers will be greater. The original calculation (without instruction correction) T = (M-1) t1 / 9-11t2 / 18 = 13 / 6 minutes, so the actual T> 13 / 6 minutes, the theoretical result is T = 13 / 6 minutes, which shows that the subway will change the stop at each station to a three-shift system with stops at every other station, and the passengers will save an average of 13 / 6 minutes. When the journey of a single passenger is less than 1 / 3 of the whole journey, the time saved is less than 13 / 6 minutes, and vice versa; after adding instruction correction, T = (M-1) t1 / 6 = 6 minutes. If the journey is 1 / 2 of the whole journey, the time saved is 9 minutes, and the total time saved is 18 minutes.

[0047] 8: Improved efficiency of subway operations: U=(M-1)(t1+t3) / (3Mt4)×100% ≥(M-1)(t1+t3) / (3Mt2)×100% ≈t1 / (2t2)×100%=13.3% U: The benefits of subway improvement; M: total number of sites; t1: time of stop; t2: shuttle bus interval time; t3: The time saved by stopping at every other station to maintain high speed (excluding the time of stopping); t4: The time required for the shuttle bus to complete one station distance; This data is only related to the number of interval stations, the proportion of stops and driving time. Because the stopping time is longer during peak hours, the theoretical calculation will be lower than the actual situation, which means that the actual situation during peak hours will be more favorable.

[0048] 9: Compared with the two-stop-one-station operation and the one-stop operation, the average time saved by passengers is: T = (M-1)t1 / 6-11t2 / 18. Among the two terms on the right side of the equation, the first term is the time saved by the interval between stations and the speed increase, which is calculated based on 1 / 3 of the average journey. If the journey is long, the first term will be larger and the time saved will be more. People who originally chose to drive may choose to take the subway. If the journey is short, it may waste time. Passengers may not take the subway and choose other modes of transportation (such as shared bicycles) instead. Therefore, the actual time is If the journey is long and time-saving, more passengers will be attracted. If the journey is short and time-consuming, fewer passengers will be attracted. In fact, the time saved is better than the data shown in the first item (time saving). The second item is the time that passengers wait for the train when it stops at each station. In fact, if the subway strengthens management and arranges the arrival time of the yellow, green and red groups at the station, most passengers can take the train at the time. A few passengers, although they don’t know the time, will reserve time for the ride and will not delay their business. Therefore, this item can basically be regarded as zero. Therefore, from the perspective of probability calculation based on actual effects, the actual effect of Qifei operation will not only be much greater than the effect value calculated in theory, but also the improved benefits will make the vehicles redundant. On the other hand, the reform of the subway’s stop at intervals shortens the journey time, and the subway speeds up and saves more time on the ride. The reform of the urban subway’s stop at intervals has brought a feasible solution to the subway speed-up, saved time for passengers, and brought benefits to the operation of the subway.

[0049] By referring to the theoretical calculation of the three-shift two-stop and one-room system and the actual value correction method, it can be concluded that in actual operation, the other modes of synchronous flying stations with M stops and N rooms can save riding time, shorten the time for vehicles to complete the entire journey and improve operating efficiency.

Claims

1. A public transportation vehicle operates in a one-way and one-line mode in a synchronous flight operation mode, wherein the vehicle reaches the synchronous flight station through the operation methods of whole flight, separate flight and partial flight during operation.

2. According to claim 1, the synchronous station flying is characterized in that when the public transportation vehicles are in operation, for the adjacent front and rear vehicles, when one of the vehicles flies at the station, the other vehicle always maintains a station flying distance with the station when the first vehicle flies at the station, which includes simultaneous station flying and non-simultaneous station flying.

3. According to claim 1, the synchronous flying station is characterized in that according to the different public transportation vehicle shifts Q, the number of consecutive stop sites M and the number of consecutive interval sites N are determined as follows: for all m and n that satisfy the equation m+n=Q (m, n, Q are all positive integers, Q is the number of shifts), take M=km, N=kn (k is a positive integer).

4. According to claim 1, the public transportation vehicle refers to a vehicle used for public transportation and having multiple stops along the way, including buses, trains, high-speed trains (EMUs), subways, maglev trains, and future vacuum tube trains used as public transportation vehicles.