Runway Capacity

Runway Capacity

Ability to accommodate Departures Arrivals

Minimize delays Computational models

Minimum aircraft separation FAA Handbook

Basic Concepts

γ

δij

vi

vj

δij

A-A δij (mi)

Entry Gate

Time

vj

vi

Basic Concepts

γ

δij

vi

vj

δd

tij

δjivj

vi

A-A δij or δji (mi)

D-D tij (sec)

D-A δd (mi)

A-D Clear runway

Entry Gate

Time

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)

Example 1 (2/3)

7 mi

K JK K J

K 35 sec/mi; J 40 sec/mi

K-K Same speed

K-J Opening

K-J Opening

J-K Closing

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

Example 1 (2/3)

7 mi

K JK

4

5

3

K J

5

K 35 sec/mi; J 40 sec/mi

K-K Same speed

K-J Opening

K-J Opening

J-K Closing

Example 1 (2/3)

7 mi

K JK

4

5

3

K J

5

285 425 635 740

910245

K 35 sec/mi; J 40 sec/mi

K-K Same speed

K-J Opening

K-J Opening

J-K Closing

425-40-(7-5)*35

315

285+4*35

245+40 315+(7*40)+40

635-40-(7-3)*35

455

740-40-(7-5)*35

630

630+(7*40)+40

950

455+(7*35)+40

1400Note: pattern could repeat

starting at 770s… why?

Example 1 (3/3)

7 mi

K JK K J

285 425 635 740

9102452 mi

175

K 35 sec/mi; J 40 sec/mi

315

385 595 700

515 640 830

950

Note: need 120 s between successive departures… can not have two in a row with this repeating pattern of arrivals

Example 1 (3/3)

7 mi

K JK K J

285 425 635 740

910245

2 mi

Capacities

Avg time of arrivals 770/5 = 154 sec CA = 3600/154 = 23.4 A/hr

Three departures for 5 arrivals (0.60) CM = (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.

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)

Speeds K 103 mph; J 90 mph

TijK-K δij/vj = (4/103) 3600 = 140 secJ-J δij/vj = (4/90) 3600 = 160 secJ-K δij/vj = (3/103) 3600 = 105 secK-J (δij/vi) +γ [(1/vj) –(1/vi)] =(5/103 +7(1/90 -1/103))3600 = 210 sec

Pij J K

J .16

.24

K .24

.36

Trail

LeadE(Tij) = ΣPijTij = 16(160)+.24(210)+.24(105)+.36(140)

= 151.6 secCA = 3600/151.6 = 23.7 Arr/hr (note slight difference from example 1)

J K

J 160

210

K 105

140

Trail

LeadTij

0.4*0.6 = expected proportion of Ks following Js

Faster, bigger plane

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)

For 1 departure E(Tij) = 74 + 40 + (1-1) 120 = 114For 2 departures E(Tij) = 74 + 40 + (2-1) 120 = 234

Pij J K

J .16

.24

K .24

.36

Trail

Lead

Tij J K

J 160

210

K 105

140

Trail

Lead

Total Pij 0.76

CM = (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.

Example 2 (3/3)

What if want at least 2 departures 20% of the time?

Increase some Tij to 234 sec

For 2 departures required E(Tij) = 74 + 40 + (2-1) 120 = 234 sec

Pij J K

J .16

.24

K .24

.36

Trail

Lead

Tij J K

J 160

234

K 105

140

Trail

Lead

CM = (3600/157.4)(1 + 1 (.16+.36) + 2 (.24)) = 45.7 Ops/hr

E(Tij) = ΣPijTij = .16(160)+.24(234)+.24(105)+.36(140) = 157.4 sec

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

Error

δij

δijσ P

Example 3 (1/2)

For same operations, assume a Pv 10% and σ= 10 sec and estimate new capacity.K-K σ z = 10 (1.28) = 12.8 secJ-J σ z = 10 (1.28) = 12.8 secJ-K σ z = 10 (1.28) = 12.8 secK-J σ z -δij [(1/vj) –(1/vi)] =(12.8 -5(3600/90 -3600/103) = -12.44

… use 0 sec

Bij

T’ij J K

J 172.8

210

K 117.8

152.8

Trail

Lead

CA = 3600/161.3 = 22.3 Arr/hr

E(Tij) = ΣPijTij = .16(172.8)+.24(210)+.24(117.8)+.36(152.8)

= 161.3 sec

Tij J K

J 160

210

K 105

140

Trail

LeadK 103 mph; J 90 mph

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

For departures between arrivalsE(Tij) = E(δd/vj) +E(Ri) + (n-1) E(td) + E(Bij)

For 1 departure E(Tij) = 74 + 40 + (1-1) 120 + 9.7 = 123.7For 2 departures E(Tij) = 74 + 40 + (2-1) 120 +9.7 = 243.7

Pij J K

J .16

.24

K .24

.36

Trail

Lead

Tij J K

J 172.8

210

K 117.8

152.8Tr

ail

Lead

Total Pij 0.76

CM = (3600/161.3)(1.76) = 39.3 Ops/hr

E(Bij) = 12.8(0.76)=9.7 sec

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

0

1

2

3

4

5

6

7

8

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1

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

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.10

1

2

3

4

5

6

7

8

Demand/Service ratio

Dela

y (

min

/op)

Possible Options

C ASV 315000D ASV 315000L 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) Annual service volume

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

VFRAll

IFR Mix Index

0-20 21-50 51-180

>91 1 1 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 25Dominant Capacity: Greatest percent time use

weights

Example 6 (1/3)

VFR IFR

70% - 110 ops 80% - 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

Example 6 (2/3)

WeatherRunwa

y Percent Capacity

VFR A 60 110

B 17 88

C 8 40

IFR A 12 88

B 0 0

C 3 55

% of Dominant Capacity

100

80

36

80

0

50

Weight

1

15

25

15

-

25

Cw = Σ Ci Wi Pi/ Σ Wi Pi = 770/5.70= 74.0 ops/hr

WP CWP

.60 66.0

2.55 224.4

2.00 80.0

1.80 158.4

0 0.0

.75 41.25

85% x 70% = 59.5%88/110

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