Runway Capacity

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

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

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

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)

Note: ignore slopes of lines, first two K’s should be steeper Example 1 (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

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

Note: pattern could repeat starting at 770s Example 1 (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? 140 315 455 630 425-40-(7-5)*35 740-40-(7-5)*35 635-40-(7-3)*35

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 1 (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

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 1 (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

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) 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) = 151.6 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

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) Note: highlighted area provides long enough times to release one departure. Never time to release two. For 1 departure E(Tij) = 74 + 40 + (1-1) 120 = 114 For 2 departures E(Tij) = 74 + 40 + (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

Example 2 (3/3) What if want at least 2 departures 20% of the time? For 2 departures required E(Tij) = 74 + 40 + (2-1) 120 = 234 sec Increase some Tij to 234 sec E(Tij) = ΣPijTij = .16(160)+.24(234)+.24(105)+.36(140) = 157.4 sec Lead Lead Tij J K 160 234 105 140 Pij J K .16 .24 .36 Trail Trail CM = (3600/157.4)(1 + 1 (.16+.36) + 2 (.24)) = 45.7 Ops/hr

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

Aircraft Position δij σ P δij Error

Lead Example 3 (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)] =(12.8 -5(3600/90 -3600/103) = -12.44 … 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) = 161.3 sec Trail CA = 3600/161.3 = 22.3 Arr/hr

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 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) = 74 + 40 + (1-1) 120 + 9.7 = 123.7 For 2 departures E(Tij) = 74 + 40 + (2-1) 120 +9.7 = 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

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

Runway Configurations

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

Delay & Runways 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 Possible Options C ASV 315000 D ASV 315000 L 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?

Example 5 (2/3) C= 92* 1* 1 = 92 ops/hr

Example 5 (3/3) C= 113* 1.04* 0.90 = 106 ops/hr

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 7-11 280-310 21-50 10-13 300-320 51-180 11-15 310-350

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

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

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

Example 6 (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