Runway Capacity Runway Capacity w Ability to accommodate
Runway Capacity
Runway Capacity w Ability to accommodate l l Departures Arrivals w Minimize delays w Computational models l l Minimum aircraft separation FAA Handbook
A/C Separation Video
Basic Concepts Arri vals rule ! Time δij A-A δij (mi) vi γ vj vj δij vi Entry Gate
Dep Basic Concepts artu res d rool Time tij δd δij vi A-A δij or δji (mi) D-D tij (sec) γ vj vj D-A δd (mi) δji A-D Clear runway vi Entry Gate
Error Free Example 1 – pattern known w Entry gate 7 miles; w D-D 120 sec; D-A 2 miles; w A-A (closest point): J-K 3 miles, J-J K-K 4 miles, K-J 5 miles* w Arrival times: J 280 sec, K 245 sec; w Runway occupancy and lift off roll 40 sec Runway capacity for pattern K K J? * j is slower, but also smaller aircraft than k (5 miles for wake vortex)
Example 1 (2/3) Where to put them? ? ? K 35 sec/mi; J 40 sec/mi K-K Same speed 7 mi K-J Opening J-K Closing K-J Opening “gate” K K J Note: ignore slopes of lines, first two K’s should be steeper
Example 1 (2/3) 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 “gate” K K J
Example 1 245+40 (2/3) 455+(7*35)+40 285+4*35 315+(7*40)+40 285 425 635 740 630+(7*40)+40 950 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 “gate” K 0 K J 140 K 315 455 425 -40 -(7 -5)*35 J 630 635 -40 -(7 -3)*35 740 -40 -(7 -5)*35 Note: pattern could repeat starting at 770 s … why?
Note: need 120 s between successive departures… can not have two in a row with this repeating pattern of arrivals Example 1 285 245 175 (3/3) 385 315 635 740 425 910 595 700 515 640 7 mi 950 D-A 2 miles 830 K 35 sec/mi; J 40 sec/mi “gate” K K J
Example 1 285 Note: if next K arrives at gate at 770 … then have 5 arrivals in 770 s (different than book which would recommend 910). This assumes exact repeat pattern kkjkj. Book allows for varying pattern but same proportions. (3/3) 780 635 425 465 950 1055 NO! 740 245 175 385 315 595 700 515 640 860 910 830 7 mi D-A 2 miles Capacities: Avg time of arrivals 770/5 = 154 sec CA = 3600/154 = 23. 4 A/hr Two departures for 5 arrivals (0. 40) CM = (3600/154)(1+. 40) = 32. 7 Ops/hr “gate” K 0 K 140 J K 315 J 455 630 K 770
Error Free (If pattern is unknown …) w Arrival & departure matrices w Same rules w Inter-arrival time l l Closing case Opening case 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) w D-A min time δd/vj γ = distance to gate
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 5 miles; 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. )
Faster, bigger plane K 103 mph; J 90 mph Trail Example 2 Speeds Lead Tij J K J 160 210 K 105 140 (2/3) δij/vj = (4/103) 3600 = 140 sec δij/vj = (4/90) 3600 = 160 sec δij/vj = (3/103) 3600 = 105 sec (δij/vi) +γ [(1/vj) –(1/vi)] =(5/103 +7(1/90 -1/103))3600 = 210 sec Tij K-K J-J J-K K-J Lead Trail Pij J K J . 16 . 24 K . 24 . 36 E(Tij) = ΣPij. Tij =. 16(160)+. 24(210)+. 24(105)+. 36(140) = 151. 6 sec 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(td) = 120 sec = time between departures E(Ri) = 40 sec = time to clear RW 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 Tij J K Pij J K J 160 210 J . 16 . 24 K 105 140 K . 24 . 36 Trail Lead Total Pij 0. 76 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) = ΣPij. Tij =. 16(160)+. 24(234)+. 24(105)+. 36(140) = 157. 4 sec Lead Tij J K Pij J K J 160 234 J . 16 . 24 K 105 140 K . 24 . 36 Trail Lead CM = (3600/157. 4)(1 + 1 (. 16+. 36) + 2 (. 24)) = 45. 7 Ops/hr Bonus: this seems fishy … how can we get more capacity by delaying JK arrivals?
Position Error Operations w Aircraft can be ahead or behind schedule w Need for buffer to avoid rule violation w Aircraft position is normally distributed Closing case w Buffer (Bij) l l Opening case (use zero if negative) 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 See p. 318
Aircraft Position δij σP Error δij
Example 3 Trail Lead Tij J K J 160 210 K 105 140 K 103 mph; J 90 mph (1/2) For same operations, assume a Pv 10% and σ= 10 sec and estimate new capacity. Bij Bonus: why do you think σ z = 10 (1. 28) = 12. 8 sec this is negative? σ z = 10 (1. 28) = 12. 8 sec σ z -δij [(1/vj) –(1/vi)] =(12. 8 -5(3600/90 -3600/103) = -12. 44 … use 0 sec K-K J-J J-K K-J Trail Lead T’ij J K J 172. 8 210 K 117. 8 152. 8 E(Tij) = ΣPij. Tij =. 16(172. 8)+. 24(210)+. 24(117. 8)+. 36(152. 8) = 161. 3 sec 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(td) = 120 sec E(Ri) = 40 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 Tij J K Pij J K J 172. 8 210 J . 16 . 24 K 117. 8 152. 8 K . 24 . 36 Trail Lead CM = (3600/161. 3)(1. 76) = 39. 3 Ops/hr Total Pij 0. 76
Runway Configuration w Approach works for single runway w Adequate for small airports w Charts and software is used for more than one runways
Runway Configurations
Runway Configuration Selection w Annual demand w Acceptable delays w Mix Index l C+3 D 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 8 Possible Options 7 Demand/Service 310000/315000 =. 98 Delays 1 -3. 5 min All OK 6 Delay (min/op) C ASV 315000 D ASV 315000 L ASV 315000 5 4 3 2 1 0 0 0, 1 0, 2 0, 3 0, 4 0, 5 0, 6 0, 7 0, 8 0, 9 Demand/Service ratio 1 1, 1
Factors for Capacity (see p. 303) w Aircraft mix l l 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) w Operations l l l Arrivals Departures Mixed w Weather l l IFR VFR w Runway exits
Nomographs, see AC 150/5060 -5
Example 5 Given: • 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 (VFR, IFR)? (1/3)
Example 5 IFR C= 92* 1* 1 = 92 ops/hr (2/3)
Example 5 VFR C= 113* 1. 04* 0. 90 = 106 ops/hr (3/3)
Annual Service Volume w Runway use schemes w Weighted hourly capacity (Cw) w Annual service volume l 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 VFR All >91 IFR Mix Index 0 -20 21 -50 51 -180 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 25 Dominant Capacity: Greatest percent time use weights
Example 6 capacity (1/3) VFR IFR 70% - 110 ops 80% - 88 ops 20% - 88 ops 0% - 0 ops 10% - 40 ops 20% - 55 ops A B C VFR 85%, MI 60; IFR 15% MI 95
Example 6 85% x 70% = 59. 5% 88/110 (2/3) Weather Runway Prop. P Capacity C % of Dominant Capacity VFR A . 60 110 100 1 . 60 66. 0 B . 17 88 80 15 2. 55 224. 4 C . 08 40 36 25 2. 00 80. 0 A . 12 88 80 15 1. 80 158. 4 B . 00 0 0 - 0 0. 0 C . 03 55 50 25 . 75 41. 25 IFR Weight W Wx. P Cx. Wx. P Cw = Σ Ci Wi Pi/ Σ Wi Pi = 570/7. 70= 74. 0 ops/hr
Example 6 Given: 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 (3/3)
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