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

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Presentation on theme: "Runway Capacity.  Ability to accommodate Departures Arrivals  Minimize delays  Computational models Minimum aircraft separation FAA Handbook."— Presentation transcript:

1 Runway Capacity

2  Ability to accommodate Departures Arrivals  Minimize delays  Computational models Minimum aircraft separation FAA Handbook

3 Basic Concepts γ δ ij vivi vjvj A-A δ ij (mi) Entry Gate Time vjvj vivi

4 Basic Concepts γ δ ij vivi vjvj δdδd t ij δ ji vjvj vivi A-A δ ij or δ ji (mi) D-D t ij (sec) D-A δ d (mi) A-D Clear runway Entry Gate Time

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 Example 1 (2/3) 7 mi K J K K J K 35 sec/mi; J 40 sec/mi K-K Same speed K-J Opening J-K Closing Note: ignore slopes of lines, first two K’s should be steeper

7 Example 1 (2/3) 7 mi K J K K J 5 K 35 sec/mi; J 40 sec/mi K-K Same speed K-J Opening J-K Closing

8 Example 1 (2/3) 7 mi K J K K J K 35 sec/mi; J 40 sec/mi K-K Same speed K-J Opening J-K Closing (7-5)* * (7*40) (7-3)* (7-5)* (7*40) (7*35) Note: pattern could repeat starting at 770s … why?

9 Example 1 (3/3) 7 mi K J K K J mi 175 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

10 Example 1 (3/3) 7 mi K J K K J mi Capacities Avg time of arrivals 770/5 = 154 sec C A = 3600/154 = 23.4 A/hr Three departures for 5 arrivals (0.60) C M = (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.

11 Error Free Operations  Arrival & departure matrices  Same rules  Inter-arrival time v i ≤ v j T ij = δ ij /v j v i >v j T ij = (δ ij /v i ) +γ [(1/v j ) –(1/v i )] control in airspace (separation inside gate) T ij = (δ ij /v j ) +γ [(1/v j ) –(1/v i )] control out of airspace (separation outside of gate)  D-A min time δ d /v j 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) SpeedsK 103 mph; J 90 mph T ij K-Kδ ij /v j = (4/103) 3600 = 140 sec J-J δ ij /v j = (4/90) 3600 = 160 sec J-Kδ ij /v j = (3/103) 3600 = 105 sec K-J(δ ij /v i ) +γ [(1/v j ) –(1/v i )] =(5/103 +7(1/90 -1/103))3600 = 210 sec P ij JK J K.36 Trail Lead E(T ij ) = ΣP ij T ij = 16(160)+.24(210)+.24(105)+.36(140) = sec C A = 3600/151.6 = 23.7 Arr/hr (note slight difference from example 1) JK J K Trail Lead T ij 0.4*0.6 = expected proportion of Ks following Js Faster, bigger plane

14 Example 2 (3/3) E(δ d /v j ) = 0.6 [2(3600)/103] [2(3600)/90] = 74 sec = average time available until plane touches down from 2 miles out E(R i ) = 40 sec = time to clear RW E(t d ) = 120 sec = time between departures For departures between arrivals, how much time does it take? E(T ij ) = E(δ d /v j ) +E(R i ) + (n-1) E(t d ) For 1 departure E(T ij ) = (1-1) 120 = 114 For 2 departures E(Tij) = (2-1) 120 = 234 P ij JK J K.36 Trail Lead T ij JK J K Trail Lead Total P ij 0.76 C M = (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.

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

16 Position Error Operations  Aircraft can be ahead or behind schedule  Need for buffer to avoid rule violation  Aircraft position is normally distributed  Buffer (B ij ) v j > v i zσ v j

17 Aircraft Position Error δ ij σ P

18 Example 3 (1/2) For same operations, assume a P v 10% and σ= 10 sec and estimate new capacity. K-Kσ z = 10 (1.28) = 12.8 sec J-J σ z = 10 (1.28) = 12.8 sec J-Kσ z = 10 (1.28) = 12.8 sec K-Jσ z -δ ij [(1/v j ) –(1/v i )] =( (3600/ /103) = … use 0 sec B ij T’ ij JK J K Trail Lead C A = 3600/161.3 = 22.3 Arr/hr E(T ij ) = ΣP ij T ij =.16(172.8)+.24(210)+.24(117.8)+.36(152.8) = sec T ij JK J K Trail Lead K 103 mph; J 90 mph

19 Example 3 (2/2) E(δ d /v j ) = 0.6 [2(3600)/103] [2(3600)/90] = 74 sec E(R i ) = 40 sec E(t d ) = 120 sec For departures between arrivals E(T ij ) = E(δ d /v j ) +E(R i ) + (n-1) E(t d ) + E(B ij ) For 1 departure E(T ij ) = (1-1) = For 2 departures E(Tij) = (2-1) = P ij JK J K.36 Trail Lead T ij JK J K Trail Lead Total P ij 0.76 C M = (3600/161.3)(1.76) = 39.3 Ops/hr E(B ij ) = 12.8(0.76)=9.7 sec

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 / =.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 150/5060-5see 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 5 (2/3) C= 92* 1* 1 = 92 ops/hr

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

30 Annual Service Volume  Runway use schemes  Weighted hourly capacity (C w )  Annual service volume ASV = C w D H where D daily ratio; H hourly ratio Mix Index HD

31 Weighted Capacity C w = Σ C i W i P i / Σ W i P i … where P i percent of time for C i ; W i weight Percent of Dominant Capacity VFR All IFR Mix Index > Dominant Capacity: Greatest percent time use weights

32 Example 6 (1/3) VFRIFR 70% ops80% - 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

33 Example 6 (2/3) WeatherRunwayPercentCapacity VFRA60110 B1788 C840 IFRA1288 B00 C355 % of Dominant Capacity Weight C w = Σ C i W i P i / Σ W i P i = 770/5.70= 74.0 ops/hr WPCWP % x 70% = 59.5% 88/110

34 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 = C w D H = 74 (294000/877) (877/62) = 350,900 ops/year


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