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With scheduled transportation modes the operating characteris t ics (q k and V) can be decided by the central controller , of course, subject totheir sat isfying q = k . VFor scheduled modes the three terms in this formula mean: q(frequency);

k (number of vehicles/length of route); V (average speed including stops) .In public transportation systems. for instance. one can select th e density,

k by fixing the number of buses on a l ine; one can also (to a certain extent)fix their average round-trip speed by providing different numbers of stops.Once this is done, the frequency of service, q. is a u t o m a t i ~ a l l y fixed.

The values of k and V , however, cannot be arbitrari ly selected becauseof safety reasons. For a certain value of k, there is a value V(k) whichcannot be exceeded; this relationship is analogous to the one in the previoussection and, thus, defines a fundamental d i a ~ r a m ; i . e . , for any given k we canfind V(k) and q = kV(k) , the maximum safe speed (including stops) and flow .However, since speeds smal l er than V(k) are also safe, (q, k. u) combinationsin the shaded area (see figure below) are also feasible.

kDENSITY

The density-volume diagram for railroads operating on a single track with a"Block Signal Control System" (a system of t raff ic l ights that always ensures aminimum distance, b , between the rear end of a train and th e beginning of thenext one) is obtained as an example. We assume that there are no stops.

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37We know that the spacing between t rains must be larger than th e length ofa t ra in , i , and the block distance, b ; thus:and

spacing ~ b + ik < I-b+i

The maximum safe operating speed of the t rains is selected to ensure thata t ra in ca n brake to a stop in less than b distance units . Fo r this example,we wil l assume that th e t ra in must be able to stop in b/2 distance units .

I t is well known that the braking distance of a moving object a t a constantdeceleration ra te , f . g , (f is the coefficient of f r ic t ion and g the acceletion of gravity) is :

v2braking distance = ~ 2 - : - ( f ~ . - g ' : " ' " ) This formula can be obtained with the methods in Section 3.1. Fo r safety we hav

and

is :

v2 b--


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38

One can now explore what happens to the maximum operating speed, density, andflow (point P above) when . the block length, b, and/or the length of thet rains is changed.

The same type of analysis can be carried out when there are stops on th eway; however, in that case i t is more convenient to obtain th e maximum flowdirectly from th e time-space diagram. The diagram below depicts the time-spacet ra jector ies of ~ o consecutive t rains as they stop at ~ o continguous sta t ions:

xb + 2.V'

STATION 2~ average speed excluding stop

= b + 2.STATION 1

t

I t can be seen that th e trajectory of the following train is ident ical to th etraj ec tory of the f i rs t t rain but shif ted to the r ight by an amount: d + (b + l ) /V I .

This is the. minimum safe headway and, thus, the maximum frequency i s :1 V'~ x d + b + 2. b + 1 + V! dV'

The time-space diagram is very useful to analyze si tuations where th e t ra-jectory of one vehicle is affected by the t rajectory of other vehicles. Theuses of the time-space diagram are best explained by means of examples. Thenext section provides some solved problems (some of which have appeared inprevious exams) covering the material in th is chapter and i l lustrat ing differentuses of th e time-space diagram. 87


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39

3.4. ExamplesProblem 1

Rapid t rans i t service is offered on a 40 km. long (round tr ip) route witha to tal of 10 stops (one stop every 4 kms.) Trains stop a t th e stations for 30sees. I f the average speed of the trains excluding stops is 80 km/hr., calculatethe maximum possible service frequency and the required number of trains i f trainmust be separated at a l l times by a t least 1 km. and they are 200 meters long.

Solution:From a time space diagram such as the last one, we ' find that

v'~ x = b + i + V'd ____ ; : ; . . ; O ~ _ _ _ _ : ~ D 42.86 veh/hr.1 + 0.2 + 80 x ~ 3600The average speed of the t rains is

v 40 40- - ~ ' - - - - = ~ - ~ : . . . . - ~ - = 68.57 km./hr.Round Trip Time 40 x 3080 + 10 3600and th e density

~ 42.86k :z -V- = 68.57 :z 0.625

The number of trains is N k x 40 .. 25.

At this point we note that instead of q " V k we could have used (thisis always possible for public t ransportat ion) q K N/(Round Trip Time) sincev = LengthTrip Time

Problem 2

and Number of Vehiclesk = ~ ~ ~ ~ ~ ~ ~ ~ ~Length

We now study a one-way railroad l ine without stops. Assuming that the topspeed of a train is V* (a function of the length of the train and the number andpower of the engines) find the block length that allows th e largest possiblefrequency. gg


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SolutionWe f i r s t note that V < V* and V 5.. /b . f g' . Thus:

and k < 1- b+"IAs was d i s c u s ~ e d before. q (the ordinate of point P in th e figuremax

before the las t one) is given by:

_m_in--.::{_V*---!-....I.r'b:,....:-::--_f_. g : : . . : . . . ' = { ~ . (lJ . f ' g'}~ x = b + i min b + i b + iWe shal l find the value of b that maximizesof b that makes V* = (b , f ' g; i . e,

Ib' f g'~ x - --b + i

Let bO be the valueThe function

has a maximum (you should check this by taking derivatives) forb i (which is also the sought maximum i f i < bO " , Case l I in the figurebelow)

If the plot of vs. i ha s th e maximum a t b bO (seealso th e figure below).

Mi'- b+.tI V*,rjb+II

~ - - - - - - ~ - - - - - - - - ~ - - - - - - b ~ - - - - ~ - L - - - - - - - - - - - - b

89ase 1: t> b. I) Case II: t < bOThe solution can be written more concisely as:

b - min (h n ; t }


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Problem 3This problem i l lus tra tes us e of th e time-space diagram to analyze th e interaction of vehicles in a narrow way.

\'--________1 \_________EAST WEST\ ~ ___ J ,

+,L___ _m_i_l_e_s_____J 1: ::::5 J____ _mi_l_e_s__jThe above waterway is wide enough for one ship only, except in the cent ral sidinwhich is wide enough fo r two ships. Ships can t ravel at an average speed of sixmiles/hour; they must be spaced at leas t one-half mile apart while moving in thewaterway and 0.25 miles apart while stopped in the siding. Westbound ships trav fu l l of cargo and are thus given high priori ty by the canal authori t: over th eeastbound ships which travel empty.

Westbound ships t ravel in four ship convoys which are regularly scheduledevery 3-1/2 hours and do not stop at the siding.1. Find the maximum daily t ra f f ic of eastbound ships.2. Find the maximum daily t ra f f ic of eastbound ships i f the siding is expanded

to one mile in length on both sides to a tota l of three miles.Note: We assume that eastbound ships wait exactly six minutes to enter

either one of the straightaways af te r a w e s t b o u ~ d convoy ha s clearedi t . We do this to take into account that ships do not accelerateinstantaneously.

SolutionWe f i r s t draw the time-space diagram for the problem a t an adequate scale.

Next, we plot the t rajectories of the high-priority (westbound) convoys. Thesehave been plotted in the f igure. The dashed band (width - 1 mile) represents the

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siding, where eastbound and westbound trajectories may cross. These dashed lineswill help us draw th e eastbound t ra jector ies .

Part 1:We s ta r t by drawing th e t rajectory of a ship entering th e western end of

the canal at 3:30 p.m. (the earl ies t possible time for that part icular gapin between convoys). Note haw i t must stop a t the eastern end of the sidingto yield the r ight of way to th e las t ship of the westbound convoy; note alsohow i t makes i t within the 5 min. allowance to the eastern end of th e canal.The same process is followed successfully with the second t rajectory. In thatcase we must also watch for th e safe spacings while moving and s t o p p e ~ . :Thethird ship, however, would not be able to arrive to the western end of the sidingwithin the 5 min. allowance and i t cannot be dispatched.

Capacity 2 2(ships per ~ hours) x 2i(hours) - 13.71 ships/day.32(hours)

Part 2:Enlarging the siding diminishes the length of the straightaways and, thus,

more ships can make i t in time (see Figure).

Capacity - 4 x ~ = 27.42 ships /day.~ 2Test your scheduling sk i l l s :

1See i f you can find a way of scheduling six ships per 32 hour period insteadof the 4 depicted in th e figure. (Hint: you may have to stop some of th e eastbounships in th e siding) .

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