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Runway Capacity

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**Runway Capacity Ability to accommodate Minimize delays**

Departures Arrivals Minimize delays Computational models Minimum aircraft separation FAA Handbook

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

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**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

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**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)

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**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

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**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

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**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

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**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

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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

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**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

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**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.)

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**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

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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

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**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

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**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

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Aircraft Position δij σ P δij Error

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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

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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

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**Runway Configuration Approach works for single runway**

Adequate for small airports Charts and software is used for more than one runways

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**Runway Configurations**

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**Runway Configuration Selection**

Annual demand Acceptable delays Mix Index C+3D percentages

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

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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

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**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

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Nomographs, see AC 150/5060-5

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**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?

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

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

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**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

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**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

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**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

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**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

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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

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