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Unformatted text preview: Time Space Diagrams
Transportation System Capacity
Dr. Antonio A. Trani
Professor of Civil and Environmental Engineering
Virginia Polytechnic Institute and State University Blacksburg
Spring 2011 Virginia Tech 1 Why Time Space Diagrams?
To estimate the following:
• Headway between operations at various transportation
facilities • Spacing between operations • Capacity of transportation systems • Determine basic level of service Virginia Tech 2 Sample Time Space Diagrams
We examine time space diagram applications for the
following systems:
a) Rail (done in class)
b) Automobile (done in class)
c) Airport (air transportation)
The rest of the handout applies timespace diagram
principles to estimate the capacity of a single runway at
an airport. Virginia Tech 3 Factors Affecting Runway Capacity
There are numerous factors that affect runway capacity.
Here are some of the most relevant:
• Runway conﬁguration (number of runways in use,
location of runway exits, etc.) • Aircraft mix (percent of aircraft in various wake vortex
categories) • Weather conditions (visibility, ceiling, wind direction
and speed) • Airport equipage (type of navaids, ATC equipment) • Operating procedures (noise considerations, special
approach and departure procedures)
Virginia Tech 4 Sample Use of Technology to Use Multiple Runways
• Radar surveillance is required at large airports to allow
simultaneous use of parallel runways (shown in the Figure)
Independent arrival streams
Runway 1 Airport Terminal 4,300 ft. or more Runway 2 Virginia Tech 5 Independent Triple and Quadruple Approaches
To Parallel Runways (IFR)
• The idea behind this concept is to allow triple and quadruple
parallel approaches to runways separated by 5,000 feet using
standard radar systems (scan update rate of 4.8 seconds) at
airports having ﬁeld elevations of less than 1,000 feet • Increase to 5,300 ft. spacing between runways for elevations
above 5,000 ft.
p
Runway 1
Runway 2 Runway 3
R 5,000 ft. or more 2
Virginia Tech 6 Independent Departures and Standard Radar
• Simultaneous departures can be conducted if two parallel
runways are located 2,500 ft.
p Runway 1 2,500 ft. Runway 2 Virginia Tech 7 Independent Departures and Arrivals with Standard
Airport Radar
• Simultaneous departures and arrivals can be conducted if two
parallel runways are located 2,500 ft. Departure
Stream Runway 1 2,500 ft. Runway 2 Virginia Tech Arrival
Stream 8 TimeSpace Analysis
• A simple technique to assess runway and airspace
capacity if the headway between aircraft is known • The basic idea is to estimate an expected headway,
E(h), and then estimate capacity as the inverse of the
expected headway 1
Capacity = E(h)
E ( h ) is expressed in time units (e.g., seconds) Virginia Tech 9 TimeSpace Analysis Nomenclature δ ij is the minimum separation matrix (miles). For this
class we assume includes air trafﬁc control buffers
times. T i is the arrival time (to the runway) of the lead aircraft
T j is the arrival time (to the runway) of the following
aircraft T ij is the headway between two successive aircraft (s)
δ is the minimum arrivaldeparture separation (miles)
Virginia Tech 10 ROT i is the runway occupancy time for aircraft i
(s) V i is the speed of aircraft i (lead aircraft) in miles per
hour Virginia Tech 11 TimeSpace Analysis Nomenclature V j is the trailing aircraft speed (miles per hour)
γ is the common approach length (miles). This is the
distance outside the runway where aircraft ﬂy a
common path aligned with the runway. Virginia Tech 12 Final Approach and Landing Processes
Space Runway ROTi TDi ROTj Tj Ti
V
i γ Time V
j Entry Gate Virginia Tech 13 Possible Outcomes of a Single Runway TimeSpace Diagram
Since aircraft approaching a runway arrive in a random
pattern we distinguish between two possible scenarios:
• Closing case  Instance when the approach of the lead
aircraft is less than that of the trailing aircraft ( V i ≤ V j ) • Opening Case  Instance when the approach speed of
lead aircraft is higher than trailing aircraft ( V i > V j ) Virginia Tech 14 Closing Case (Equations)
Headway ( T ij = T j – T i ) assuming control is exercised
as the lead aircraft passes the entry gate (at a distance
γ ) from the runway is, δ ij
T ij = Vj
NOTE: the distance γ does not inﬂuence the outcome
of this analysis because the following aircraft (fast) is
“closing on” the lead vehicle (slow). Virginia Tech 15 Closing Case Diagram (Arrivals Only)
Space Runway ROTi ROTj Ti Tj V
i γ Time δij 1 V <V
i
j V
j
1 Entry Gate Virginia Tech 16 Opening Case (Equations)
Headway ( T ij = T j – T i ) is, δ
1 1T ij = ij + γ  –  V j V i
Vj
assuming control is exercised as the lead aircraft passes
the entry gate.
NOTE: The second term in the previous equation
measures the time aircraft (i) and (j) space themselves
further over a distance γ . This term is important
because
Virginia Tech 17 Opening Case Diagram (Arrivals Only)
Space Runway ROTi ROTj Ti Tj
V
j V
i γ 1 1 Time
V >V
i
j δij
Entry Gate Virginia Tech 18 Mixed Operations (Arrivals/Departures)
Space Runway TDi ROTi Ti T1 ROTj T2 Tj Time
δ V
i γ G V
j T1 = Ti + RΟΤι
T2 = T j  δ / V j
Gap (G) exist if T2  T1 > 0 Entry Gate TDi is the departure
runway occupancy time Virginia Tech 19 Air Trafﬁc Control ArrivalArrival Wake Vortex
Separations Near Runways
Table 2. Typical Intrail Separations Near Runways in miles for Large
Hub Airports (includes buffers added by air traffic control). Trailing Aircraft
Lead
Aircraft
Heavy
Large
Small Heavy Large Small 5.00
3.00
3.00 6.00
3.00
3.00 7.00
5.00
3.00 Virginia Tech 20 DepartureDeparture Aircraft Separations
Table 2. Typical Intrail Departure Separations Near Runways in
seconds. Trailing Aircraft LEAD
ACFT.
Heavy
Large
Small Heavy Large Small 90
60
60 120
60
60 120
60
60 Virginia Tech 21 Example Problem (1)
Determine the saturation capacity of an airport serving
two groups of aircraft: a) heavy (70% of the population)
and b) small (30% of the population). Assume the
common approach length γ to be 7 miles.
The aircraft performance characteristics are given in the
following table.
Table 3. Aircraft Characterictics.
Aircraft Group Runway Occupancy
Time (seconds) Approach Speed
(m.p.h.) Heavy 60 150 Small 40 70 Virginia Tech 22 Determine Aircraft Mix and Probabilities
The following is a probability matrix establishing the
chance that an aircraft of type (i) follows aircraft of type
(j). We assume random arrivals.
Table 4. Probability Matrix (Pij). Aircraft (i) follows aircraft (j).
Trailing Aircraft
Lead Aircraft Heavy Small Heavy =(0.7) x (0.7) = 0.49 = (0.7) x (0.3) = 0.21 Small = (0.3) x (0.7) = 0.21 = (0.3) x (0.3) = 0.09 NOTE: verify that ∑P ij = 1.0 i, j Virginia Tech 23 Compute Headways Between Successive
Arrivals
Closing case:
Lead = small, Following = heavy aircraft TS – H δS – H
3  = 0.02 hours
=  = VH
150 Usually is convenient to express headway in seconds. 3δS – H
T S – H =  =  3600 = 72 seconds 150
VH
Virginia Tech 24 Compute Headways Between Successive
Arrivals
Closing case (apply this case when speeds are equal):
Lead = small, Following = small aircraft TS – S 3δS – S  3600 = 154 seconds
=  = 70
VS Lead = heavy, Following = heavy aircraft 5δH – H
T H – H =  =  3600 = 120 seconds 150
VH
Virginia Tech 25 Compute Headways Between Successive
Arrivals
Opening case:
Lead = heavy, Following = small aircraft TH – S δH – S
1 1  –  seconds
=  + γ V S V H
VS 711T H – S =  3600 + 7  –  3600 = 552 70 70 150
seconds Virginia Tech 26 Arrival Aircraft Headway Table
The following table summarizes the computed
headways for all cases when an aircraft of type (i)
follows aircraft of type (j). We assume random arrivals.
Table 5. Headways (seconds) when aircraft (i) follows aircraft (j).
Trailing Aircraft
Lead Aircraft Heavy Small Heavy 120 552 Small 72 154 Virginia Tech 27 Compute Expected Value of Headway
The expected value of the headway is: E ( T ij ) = ∑P T
ij ij for all i,j pairs i, j E ( T ij ) = P H – H × T H – H + P S – H × T S – H + P H – S × T H – S
+ PS – S × TS – S
E ( T ij ) = 0.49 ( 120 ) + 0.21 ( 72 ) + 0.21 ( 552 ) + 0.09 ( 154 ) E ( T ij ) = 203.7 seconds
Virginia Tech 28 Compute ArrivalsOnly Capacity
• The capacity as the inverse of the expected headway 1Capacity =  vehicles per second
E ( T ij )
E ( T ij ) is expressed in time units (e.g., seconds)
Using more standard units of capacity (aircraft per
hour), 3600Capacity =  vehicles per hour
E ( T ij )
Virginia Tech 29 ArrivalsOnly Capacity
For the single runway example the arrivalsonly
capacity is, 3600
C arrivals =  = 17.7 aircraft arrivals per hour
203.7
NOTE: this value is low for a busy airport. At busy
airports small aircraft are generally handled at a
different runway if possible to improve the capacity of a
runway operated by heavy aircraft. Virginia Tech 30 Analysis of Runway Gaps
Gaps can be studied for all four possible instances
studied so far. For example, if a heavy aircraft is
followed by a small one, there is a headway of 552
seconds between two successive arrivals. This leaves a
large gap that be exploited by air trafﬁc controllers to
handle a few departures on the same runway.
The gap for a heavysmall case is, δ
G H – S = T 2 – T 1 = T S –  – ( T H + ROT H ) V S Virginia Tech 31 Gap Analysis
Assume the arrival of the heavy aircraft occurs at time
t=0 seconds. GH – S 2 552 –  3600 – ( 0 + 60 )
= 70 G H – S = 389 seconds
The expected time between successive departures at this
airport is 83 seconds (see Table 2 adjusted by the
probability values computed). A gap of 389 seconds is
sufﬁcient to “launch” four departures. You can do the Virginia Tech 32 same analysis for all other instances and estimate
the departure capacity of the runway per hour.
Gap: Lead aircraft = small, following aircraft = small GS – S 2 154 –  3600 – ( 0 + 40 )
= 70 G S – S = 11 seconds
One departure can be injected when a small aircraft
follows another small aircraft. While 11.1 seconds is
small gap, the fact is any gap > 0 will in theory result in
one departure as long as the pilot responds quickly to
ATC commands.
Virginia Tech 33 Gap: Lead aircraft = small, following aircraft = heavy GS – H 2 72 –  3600 – ( 0 + 40 )
= 150 G S – H = – 16 seconds
No departures can be scheduled when a small aircraft
follows a heavy aircraft. Virginia Tech 34 Gap: Lead aircraft = heavy, following aircraft =
heavy GH – H 2 120 –  3600 – ( 0 + 60 )
= 150 G H – H = 12 seconds
One departure (on the average) can be scheduled
between a heavy aircraft followed by another heavy
aircraft. Virginia Tech 35 The analysis of gaps for four arrival instances is
presented in Table 6. The number of departures per gap
is also presented in Table 6.
Table 6. Gaps (seconds) when aircraft (i) follows aircraft (j).
Successive departures per gap are shown in parenthesis.
Expected value of departure occupancy time is E(TDi) = 83
seconds).
Trailing Aircraft
Lead Aircraft Heavy Small Heavy 12 (1) 389 (4) Small 16 (0) 11 (1) Virginia Tech 36 Analysis of Arrival Gaps
The ﬁnal question that needs to be answered is: how
many times each gap happens during the period of
interest?
From our analysis of arrivals only, we determined that
on the average hour 17.7 arrivals could be processed at
the runway. Since two successive arrivals are needed to
form a gap, we can infer that around 16.7 gaps are
present in one hour.
The probabilty of each one of the four arrival instances
is known and has been calculated in Table 4. Thus using
these two pieces of information we estimate the number
Virginia Tech 37 of times gaps will occur during one hour. Consider a
heavy aircraft leading another heavy aircraft. Forty nine
percent of the time this instance occurs at the airport.
Thus for 16.7 gaps per hour this represents an equivalent
number of hourly departures per arrival instance
( E D H – H ), ED H – H = TG ( P H – H ) ( DG H – H )
where: TG is the total number of gaps per hour, P H – H is
the probablity that a heavy aircraft follows another
heavy, and DG H – H is the number of departures per gap
for each instance (numbers in parentheis in Table 6). Virginia Tech 38 ED H – H = 16.7 ( 0.49 ) ( 1 ) = 8.18
equivalent departures per hour
Similarly, ED H – S = 16.7 ( 0.21 ) ( 4 ) = 14.03
ED S – H = 16.7 ( 0.21 ) ( 0 ) = 0
ED S – S = 16.7 ( 0.09 ) ( 1 ) = 1.50
equivalent departures per hour Virginia Tech 39 Departures with Arrival Priority
Table 7 summarizes the number of departures per hour
per instance.
Table 7. Equivalent departures per hour per arrival instance
when aircraft (i) follows aircraft (j).
Trailing Aircraft
Lead Aircraft Heavy Small Heavy 8.18 14.03 Small 0.00 1.50 Total departures per hour = 23.7 departures per hour Virginia Tech 40 Recapitulation of Results so Far C arrivals 3600 = 17.7
= arrivals per hour
203.7 C departures = 23.7 departures per hour
These results indicate that a single runway can process
17.7 arivals per hour and during the same period process
23.7 departures per hour using the gaps formed by the
arrivals.
Total operations = 41.4 aircraft per hour Virginia Tech 41 Final Note
If only departures are processed at this runway (no
arrivals), the departures only capacity is the reciprocal
of the departure headway (83 seconds), 3600
C dep – NA =  = 43.3 departures per hour
83
Airport engineers use a capacity diagram illusrated in
the ﬁgure to display all three hourly capacity results in a
single diagram. These diagrams represent a Pareto
frontier of arrivals and departures. The airport can be
operated inside the Pareto boundary. Virginia Tech 42 Arrivals per Hour ArrivalDeparture Capacity Diagram
20 B (23.7,17.7) A (0,17.7) 10 0 C (43.3,0) 10 20 30 40 Departures per Hour
Virginia Tech 43 Interpretation of ArrivalDeparture Diagram
• Line segement AB represents a region where arrivals
are given priority over departures. 17.7 arrivals per hour
are processed and up to 23.7 departures per hour. • Line segment BC represents a tradeoff region. Here we
increase the separation between successive arrivals to
allow more departures. In the limiting case (no
arrivals), only departures and processed at a rate of 43.3
per hour. • Any operating point inside the Pareto frontier is
feasible. Points outside the boundary encompassed by
line segments AB and BC cannot be sustained for
long periods of time.
Virginia Tech 44 ...
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