1. Runway Capacity

2. Runway Capacity Ability to accommodate Minimize delays
Departures
Arrivals
Minimize delays
Computational models
Minimum aircraft separation
FAA Handbook

3. Basic Concepts
Time
δij
A-A δij (mi) vi γ
δij vj vj vi
Entry Gate

4. Basic Concepts γ Time tij δd δij A-A δij or δji (mi) vi D-D tij (sec)
D-A δd (mi) vj
δji
A-D Clear runway vj vi
Entry Gate

5. 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)

6. Note: ignore slopes of lines, first two K’s should be steeper
Example (2/3)
K 35 sec/mi; J 40 sec/mi
K-K Same speed
7 mi
K-J Opening
J-K Closing
K-J Opening K K J K J
Note: ignore slopes of lines, first two K’s should be steeper

7. Example 1 (2/3) 7 mi K 35 sec/mi; J 40 sec/mi K-K Same speed
K-J Opening
J-K Closing
K-J Opening K K J K J

8. Note: pattern could repeat starting at 770s
Example (2/3)
455+(7*35)+40
285+4*35
245+40
315+(7*40)+40
630+(7*40)+40
285
425
635
740
950
245
910 3
K 35 sec/mi; J 40 sec/mi 4
K-K Same speed
7 mi 5 5
K-J Opening
J-K Closing
K-J Opening K K J K J
Note: pattern could repeat starting at 770s
… why?
140
315
455
630
(7-5)*35
(7-5)*35
(7-3)*35

9. Example 1 (3/3) 7 mi K 35 sec/mi; J 40 sec/mi
Note: need 120 s between successive departures… can not have two in a row with this repeating pattern of arrivals
Example (3/3)
285
425
635
740
950
910
245
385
595
700
2 mi
175
315
515
640
830
7 mi
K 35 sec/mi; J 40 sec/mi K K J K J

10. 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.
Example (3/3)
285
425
635
740
245
910
2 mi
Capacities
Avg time of arrivals
770/5 = 154 sec
CA = 3600/154 = 23.4 A/hr
7 mi
Three departures for 5 arrivals (0.60)
CM = (3600/154)(1+.60) = 37.4 Ops/hr K K J K J

11. 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

12. 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.)

13. Example 2 (2/3) Tij Speeds K 103 mph; J 90 mph Tij Pij
Faster, bigger plane
Tij
Lead J K
160
210
105
140
Trail
Speeds K 103 mph; J 90 mph
Tij
K-K δij/vj = (4/103) 3600 = 140 sec
J-J δij/vj = (4/90) 3600 = 160 sec
J-K δij/vj = (3/103) 3600 = 105 sec
K-J (δij/vi) +γ [(1/vj) –(1/vi)] =(5/103 +7(1/90 -1/103))3600 = 210 sec
Lead
Pij
E(Tij) = ΣPijTij = 16(160)+.24(210)+.24(105)+.36(140) = sec J K
.16
.24
.36
Trail
CA = 3600/151.6 = 23.7 Arr/hr (note slight difference from example 1)
0.4*0.6 = expected proportion of Ks following Js

14. Example (3/3)
E(δd/vj) = 0.6 [2(3600)/103] [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)
Note: highlighted area provides long enough times to release one departure. Never time to release two.
For 1 departure E(Tij) = (1-1) 120 = 114
For 2 departures E(Tij) = (2-1) 120 = 234
Lead
Lead
Pij J K
.16
.24
.36
Tij J K
160
210
105
140
Total Pij 0.76
Trail
Trail
CM = (3600/151.6)(1.76) = 41.8 Ops/hr

15. Example 2 (3/3) What if want at least 2 departures 20% of the time?
For 2 departures required E(Tij) = (2-1) 120 = 234 sec
Increase some Tij to 234 sec
E(Tij) = ΣPijTij = .16(160)+.24(234)+.24(105)+.36(140) = sec
Lead
Lead
Tij J K
160
234
105
140
Pij J K
.16
.24
.36
Trail
Trail
CM = (3600/157.4)(1 + 1 ( ) + 2 (.24)) = 45.7 Ops/hr

16. 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

17. Aircraft Position
δij
σ P
δij
Error

18. Lead
Example (1/2)
K 103 mph; J 90 mph
Tij J K
160
210
105
140
Trail
For same operations, assume a Pv 10% and σ= 10 sec and estimate new capacity.
Bij
K-K σ z = 10 (1.28) = 12.8 sec
J-J σ z = 10 (1.28) = 12.8 sec
J-K σ z = 10 (1.28) = 12.8 sec
K-J σ z -δij [(1/vj) –(1/vi)] =( (3600/ /103) = … use 0 sec
Lead
T’ij J K
172.8
210
117.8
152.8
E(Tij) = ΣPijTij = .16(172.8)+.24(210)+.24(117.8)+.36(152.8) = sec
Trail
CA = 3600/161.3 = 22.3 Arr/hr

19. Example (2/2)
E(δd/vj) = 0.6 [2(3600)/103] [2(3600)/90] = 74 sec
E(Ri) = 40 sec
E(td) = 120 sec
E(Bij) = 12.8(0.76)=9.7 sec
For departures between arrivals
E(Tij) = E(δd/vj) +E(Ri) + (n-1) E(td) + E(Bij)
For 1 departure E(Tij) = (1-1) = 123.7
For 2 departures E(Tij) = (2-1) = 243.7
Lead
Lead
Pij J K
.16
.24
.36
Tij J K
172.8
210
117.8
152.8
Total Pij 0.76
Trail
Trail
CM = (3600/161.3)(1.76) = 39.3 Ops/hr

20. Runway Configuration Approach works for single runway
Adequate for small airports
Charts and software is used for more than one runways

21. Runway Configurations

22. Runway Configuration Selection
Annual demand
Acceptable delays
Mix Index
C+3D percentages

23. Delay & Runways
Relationship between average aircraft delay in minutes and ratio of annual demand to annual service volume

24. Example 4
For a demand of 310,000 operations, maximum delay of 5 minutes,
and MI 90 VFR, 100 IFR determine possible runway configurations
Possible Options
C ASV
D ASV
L ASV
Demand/Service
310000/ = .98
Delays min All OK

25. 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, ,000 lbs)
Class D (multi-engine, > 300,000 lbs)
Operations
Arrivals
Departures
Mixed
Weather
IFR
VFR
Runway exits

26. Nomographs, see AC 150/5060-5

27. 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?

28. Example (2/3)
C= 92* 1* 1 = 92 ops/hr

29. Example (3/3)
C= 113* 1.04* 0.90 = 106 ops/hr

30. Annual Service Volume Runway use schemes Weighted hourly capacity (Cw)
ASV = Cw D H
where D daily ratio; H hourly ratio
Mix Index H D
0-20
7-11
21-50
10-13
51-180
11-15

31. Percent of Dominant Capacity
Weighted Capacity
Cw = Σ Ci Wi Pi/ Σ Wi Pi … where Pi percent of time for Ci; Wi weight
Percent of Dominant Capacity
VFR
All
IFR Mix Index
0-20
21-50
51-180
>91 1
81-90 5 3
66-80 15 2 8
51-65 20 12
0-50 25 4 16
weights
Dominant Capacity: Greatest percent time use

32. Example 6 (1/3) capacity VFR IFR 70% - 110 ops 80% - 88 ops AB
10% - 40 ops
20% - 55 ops C
VFR 85%, MI 60; IFR 15% MI 95

33. Example 6 (2/3) Cw = Σ Ci Wi Pi/ Σ Wi Pi = 770/5.70= 74.0 ops/hr
88/110
Weather
Runway
Percent
Capacity
VFR A 60
110 B 17 88 C 8 40
IFR 12 3 55
% of Dominant Capacity
100 80 36 50
Weight 1 15 25 - WP
CWP
.60
66.0
2.55
224.4
2.00
80.0
1.80
158.4
0.0
.75
41.25
Cw = Σ Ci Wi Pi/ Σ Wi Pi = 770/5.70= 74.0 ops/hr

34. Example (3/3)
Annual demand: 294,000 ops; average daily traffic 877 ops; peak hour 62, MI 90 VFR/ 100 IFR
What will be the Annual Service Volume that could be accommodated for the runway system shown?
ASV = Cw D H = 74 (294000/877) (877/62) = 350,900 ops/year