runway capacity
Post on 06-Feb-2016
80 Views
Preview:
DESCRIPTION
TRANSCRIPT
Runway Capacity
Ability to accommodate Departures Arrivals
Minimize delays Computational models
Minimum aircraft separation FAA Handbook
Basic Concepts
γ
δij
vi
vj
δd
tij
δjivj
vi
A-A δij or δji (mi)
D-D tij (sec)
D-A δd (mi)
A-D Clear runway
Entry Gate
Time
Example 1 (1/3)
Entry gate 7 miles; D-D 120 sec; D-A 2 miles; A-A: J-K 3 miles, J-J K-K 4 miles, K-J 5miles Arrival times: J 280 sec, K 245 sec; Runway occupancy and lift off roll 40 sec
Runway capacity for pattern K K J K J?Note: j is slower, but also smaller aircraft than k (5 miles for wake vortex)
Example 1 (2/3)
7 mi
K JK K J
K 35 sec/mi; J 40 sec/mi
K-K Same speed
K-J Opening
K-J Opening
J-K Closing
Note: ignore slopes of lines, first two K’s should be steeper
Example 1 (2/3)
7 mi
K JK
4
5
3
K J
5
K 35 sec/mi; J 40 sec/mi
K-K Same speed
K-J Opening
K-J Opening
J-K Closing
Example 1 (2/3)
7 mi
K JK
4
5
3
K J
5
285 425 635 740
910245
K 35 sec/mi; J 40 sec/mi
K-K Same speed
K-J Opening
K-J Opening
J-K Closing
425-40-(7-5)*35
315
285+4*35
245+40 315+(7*40)+40
635-40-(7-3)*35
455
740-40-(7-5)*35
630
630+(7*40)+40
950
455+(7*35)+40
1400Note: pattern could repeat
starting at 770s… why?
Example 1 (3/3)
7 mi
K JK K J
285 425 635 740
9102452 mi
175
K 35 sec/mi; J 40 sec/mi
315
385 595 700
515 640 830
950
Note: need 120 s between successive departures… can not have two in a row with this repeating pattern of arrivals
Example 1 (3/3)
7 mi
K JK K J
285 425 635 740
910245
2 mi
Capacities
Avg time of arrivals 770/5 = 154 sec CA = 3600/154 = 23.4 A/hr
Three departures for 5 arrivals (0.60) CM = (3600/154)(1+.60) = 37.4 Ops/hr
Note: if next K arrives at gate at 770 … then have 5 arrivals in 770s (different than book which would recommend 910). This assumes exact repeat pattern kkjkj. Book allows for varying pattern but same proportions.
Error Free Operations
Arrival & departure matrices Same rules Inter-arrival time
vi≤ vj Tij = δij/vj
vi>vj Tij = (δij/vi) +γ [(1/vj) –(1/vi)] control in airspace (separation inside gate)
Tij = (δij/vj) +γ [(1/vj) –(1/vi)] control out of airspace (separation outside of gate)
D-A min time δd/vj
Closing case
Opening case
Example 2 (1/3)
Entry gate 7 miles; D-D 120 sec; D-A 2 miles; A-A: J-K 3 miles, J-J K-K 4 miles, K-J 5miles; Arrival times: J 280 sec, K 245 sec; Runway occupancy and lift off roll 40 sec; Control in airspace. Speeds: K 103 mph; J 90 mph
Runway capacity for error free operations for K 60% and J 40%? (note: proportion same as previous problem, but order not specified here so may have different pattern, e.g., kkjkj or kkkjj or kjkjk.)
Example 2 (2/3)
Speeds K 103 mph; J 90 mph
TijK-K δij/vj = (4/103) 3600 = 140 secJ-J δij/vj = (4/90) 3600 = 160 secJ-K δij/vj = (3/103) 3600 = 105 secK-J (δij/vi) +γ [(1/vj) –(1/vi)] =(5/103 +7(1/90 -1/103))3600 = 210 sec
Pij J K
J .16
.24
K .24
.36
Trail
LeadE(Tij) = ΣPijTij = 16(160)+.24(210)+.24(105)+.36(140)
= 151.6 secCA = 3600/151.6 = 23.7 Arr/hr (note slight difference from example 1)
J K
J 160
210
K 105
140
Trail
LeadTij
0.4*0.6 = expected proportion of Ks following Js
Faster, bigger plane
Example 2 (3/3)
E(δd/vj) = 0.6 [2(3600)/103] + 0.4 [2(3600)/90] = 74 sec = average time available until plane touches down from 2 miles out
E(Ri) = 40 sec = time to clear RW E(td) = 120 sec = time between departures
For departures between arrivals, how much time does it take?E(Tij) = E(δd/vj) +E(Ri) + (n-1) E(td)
For 1 departure E(Tij) = 74 + 40 + (1-1) 120 = 114For 2 departures E(Tij) = 74 + 40 + (2-1) 120 = 234
Pij J K
J .16
.24
K .24
.36
Trail
Lead
Tij J K
J 160
210
K 105
140
Trail
Lead
Total Pij 0.76
CM = (3600/151.6)(1.76) = 41.8 Ops/hr
Note: highlighted area provides long enough times to release one departure. Never time to release two.
Example 2 (3/3)
What if want at least 2 departures 20% of the time?
Increase some Tij to 234 sec
For 2 departures required E(Tij) = 74 + 40 + (2-1) 120 = 234 sec
Pij J K
J .16
.24
K .24
.36
Trail
Lead
Tij J K
J 160
234
K 105
140
Trail
Lead
CM = (3600/157.4)(1 + 1 (.16+.36) + 2 (.24)) = 45.7 Ops/hr
E(Tij) = ΣPijTij = .16(160)+.24(234)+.24(105)+.36(140) = 157.4 sec
Position Error Operations
Aircraft can be ahead or behind schedule
Need for buffer to avoid rule violation Aircraft position is normally
distributed Buffer (Bij)
vj > vi zσ vj<vi zσ – δ[(1/vj)-(1/vi)]
where σ standard deviation; z standard score for 1-Pv; Pv probability of violation
Closing case
Opening case (use zero if negative)
See p. 318
Example 3 (1/2)
For same operations, assume a Pv 10% and σ= 10 sec and estimate new capacity.K-K σ z = 10 (1.28) = 12.8 secJ-J σ z = 10 (1.28) = 12.8 secJ-K σ z = 10 (1.28) = 12.8 secK-J σ z -δij [(1/vj) –(1/vi)] =(12.8 -5(3600/90 -3600/103) = -12.44
… use 0 sec
Bij
T’ij J K
J 172.8
210
K 117.8
152.8
Trail
Lead
CA = 3600/161.3 = 22.3 Arr/hr
E(Tij) = ΣPijTij = .16(172.8)+.24(210)+.24(117.8)+.36(152.8)
= 161.3 sec
Tij J K
J 160
210
K 105
140
Trail
LeadK 103 mph; J 90 mph
Example 3 (2/2)
E(δd/vj) = 0.6 [2(3600)/103] + 0.4 [2(3600)/90] = 74 sec
E(Ri) = 40 sec E(td) = 120 sec
For departures between arrivalsE(Tij) = E(δd/vj) +E(Ri) + (n-1) E(td) + E(Bij)
For 1 departure E(Tij) = 74 + 40 + (1-1) 120 + 9.7 = 123.7For 2 departures E(Tij) = 74 + 40 + (2-1) 120 +9.7 = 243.7
Pij J K
J .16
.24
K .24
.36
Trail
Lead
Tij J K
J 172.8
210
K 117.8
152.8Tr
ail
Lead
Total Pij 0.76
CM = (3600/161.3)(1.76) = 39.3 Ops/hr
E(Bij) = 12.8(0.76)=9.7 sec
Runway Configuration
Approach works for single runway Adequate for small airports Charts and software is used for more
than one runways
Delay & Runways
0
1
2
3
4
5
6
7
8
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1
Relationship between average aircraft delay in minutes and ratio of annual demand to annual service volume
Example 4
For a demand of 310,000 operations, maximum delay of 5 minutes,and MI 90 VFR, 100 IFR determine possible runway configurations
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.10
1
2
3
4
5
6
7
8
Demand/Service ratio
Dela
y (
min
/op)
Possible Options
C ASV 315000D ASV 315000L ASV 315000
Demand/Service 310000/315000 = .98
Delays 1-3.5 min All OK
Factors for Capacity (see p. 303)
Aircraft mix Class A (single engine, <12,500 lbs) Class B (multi-engine, <12,500 lbs) Class C (multi-engine, 12,500-300,000 lbs) Class D (multi-engine, > 300,000 lbs)
Operations Arrivals Departures Mixed
Weather IFR VFR
Runway exits
Nomographs, see AC 150/5060-5
Example 5 (1/3)
• Two parallel runways; • Aircraft classes: A 26%; B 20%; C 50%;
D 4%; • Touch and go 8%; • 2 exits at 4,700 ft and 6,500 ft from
arrival threshold; • 60% arrivals in peak hour.
Capacity?
Annual Service Volume
Runway use schemes Weighted hourly capacity (Cw) Annual service volume
ASV = Cw D H
where D daily ratio; H hourly ratio
Mix Index
H D
0-20 7-11 280-310
21-50 10-13
300-320
51-180 11-15
310-350
Weighted Capacity
Cw = Σ Ci Wi Pi/ Σ Wi Pi … where Pi percent of time for Ci; Wi weight
Percent of
Dominant
Capacity
VFRAll
IFR Mix Index
0-20 21-50 51-180
>91 1 1 1 1
81-90 5 1 3 5
66-80 15 2 8 15
51-65 20 3 12 20
0-50 25 4 16 25Dominant Capacity: Greatest percent time use
weights
Example 6 (1/3)
VFR IFR
70% - 110 ops 80% - 88 ops
20% - 88 ops 0% - 0 ops
10% - 40 ops 20% - 55 ops
VFR 85%, MI 60; IFR 15% MI 95
A
B
C
capacity
Example 6 (2/3)
WeatherRunwa
y Percent Capacity
VFR A 60 110
B 17 88
C 8 40
IFR A 12 88
B 0 0
C 3 55
% of Dominant Capacity
100
80
36
80
0
50
Weight
1
15
25
15
-
25
Cw = Σ Ci Wi Pi/ Σ Wi Pi = 770/5.70= 74.0 ops/hr
WP CWP
.60 66.0
2.55 224.4
2.00 80.0
1.80 158.4
0 0.0
.75 41.25
85% x 70% = 59.5%88/110
top related