Solved Problems- Airport Planning and Layout

1. Using the wind data given here, construct a wind rose for an airport with Airport Reference Code BII and indicate what would be the best orientation for a runway based on these prevailing winds. If available, check your results with the latest version of the FAA's computer program, 'Airport Design for Microcomputers".

Wind Direction Percentage of Winds17-21 kt 22-27 kt >27 kt

123456789101112131415161718192021222324252627282930313233343536

0.20.40.10.50.20.30.10.90.80.80.91.31.11.00.90.90.70.80.50.20.10.10.20.70.60.50.40.60.50.80.41.10.50.30.40.2

0.1

0.1

0.10.10.10.10.20.2030.1

0.1

0.1

0.10.10.20.10.20.10.20.10.10.10.1

0.1

0.1

0.10.10.1

<0.1

<0.1

0.10.10.1

0.1

20.0 3.0 1.0Percent of Wind 0-10 kt =60.4Percent of Wind 11-16 kt =15.6

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Solution:

According to FAA – [AC: 150/5300-13], Allowable cross-wind component for the Airport Reference Code "BII" equals 13 Knots

The shown direction of the airport could guaranty about 92% of the wind directions and speeds to be within the acceptable ranges.

Since the Airport Reference Code BII, it is not considered as low speed approach for small aircrafts, and then the above result can be considered acceptable.

Note: for the purpose of this exercise, we have assumed that " Percent of Wind 11-16 kt =15.6" is within the direction that gives acceptable crosswind component.

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2. A utility airport is being planned to serve a city of 35,000 people. The visual runway is to be 3600 ft. long. Indicate whether or not the following objects will be considered obstructions to air navigation by the FAA. (HINT: 1 nautical mile = 6076 ft.) a) A 220-ft radio tower that is not in the landing approach, located 3.1 miles from

the airport reference point. The ground elevation at the tower is 25ft higher than the established airport elevation.

b) A planned 75-ft-high office building within the landing approach 1/2 mile from the end of the runway.

Solution:

Part (a)

Population = 35,000Runway length = 3,600 ft = 1,097 mThe tower height = 220 + 25 = 245 ftThe tower is at a distance of 3.1 miles = 16,270 ft from the airport reference point.

According to FAR_Part 77, Obstructions to Navigation:

An object constitutes an obstruction to navigation if: If 200 ft. above ground level or 200 ft. above the airport elevation (whichever

is greater) up to 3 miles (for runway lengths > 3200 ft.) from the airport.Increase 100 ft. every mile up to 500 ft. at 6 miles from the ARP (airport reference point)

Is 500 ft. or more above ground level at the object site If penetrates an imaginary surface (a function of the precision of the runway) If penetrates the terminal obstacle clearance area (includes initial approach

segment) If penetrates the enroute obstacle clearance area (includes turn and termination

areas of federal airways)

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Horizontal surface plus conical surface lengths = 5,000 ft + 4,000 ft = 9,000 ft

The tower is located at a distance = 16,270 ft from the airport reference point which is beyond the conical surface, and consequently the tower is not considered an obstruction to air navigation.

Part (b)

A planned 75-ft-high office building within the landing approach 1/2 mile = 2,624 ft from the end of the runway.

The approach surface length = 5,000 ft and the approach surface slope = 20:1

The permissible height at a distance of 2,624 ft = 2,624 ft / 20 = 131.2 ft, and consequently, the object is not considered an obstruction to air navigation.

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3. A new air carrier airport is being planned to serve a city of 300,000 people. The airport will have dual 8000-ft precision instrument runways. Indicate whether or not the following objects constitute an obstruction to air navigation. The established airport elevation is 1220 ft.

a) A railroad within the approach path located 1.2 miles from the end of the runway with an elevation of 1310 ft. Assume the elevation of the end of the runway is 1220 ft.

b) A 298-ft radio tower not in the approach path located 3.5 nm from the airport reference point. The ground elevation at the tower is 1275 ft.

Solution:

Part (a)

The railroad elevation is 1310 ft and the airport elevation is 1220 ft. which means that the railroad is 90 ft above the airport elevation, at a distance of 1.2 miles=6300 ft.

The runways are precision instrument. The object is located at a distance of 6300 ft which means that the approach surface slope at this distance is 50:1 , and consequently, the permissible distance is 6300/50 = 126 ft. above the airport level. And consequently, the railroad does not constitute an obstruction to air navigation.

Part (b)

The radio tower is at a distance of 3.5 nm = 21,266 ft = 4 miles from the ARP.

The maximum distance of the surfaces = 1000 + 1050 + 10,000 + 4,000 = 16,050 ft

This means that the tower is at a distance far than the distances of all the surfaces. An consequently, the tower does not constitute an obstruction to air navigation.

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4. What length of runway is required for a smaller airport that is 6000 ft above sea level and has a normal maximum temperature of 75°F? The effective gradient is 1.2 percent. The airport must accommodate small airplanes having more than 10 seats.

Solution:

According to the above charts, the required runway length is as follows:75% of fleet, runway length = 4900 ft95% of fleet, runway length = 7050 ft100% of fleet, runway length = 7150 ft

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5. What length of runway is required for a Boeing 757 Series 232 aircraft given the following design conditions?

Normal maximum temperature 75°F ≈ 24°CAirport elevation 4000 ft (1219 m)Take-off weight 220,000 lb (99,873 kg)Maximum landing weight 175,000 lbEffective runway gradient 0.85 percent

Solution:

According to the tables below,

Landing conditions (Table 18-1): runway length = 5570 ft = 1698 m

Take-off conditions (Table 18-2) – by interpolation:

Reference Factor = [(1220-1000) / (1500-1000)]*(67.6-61.4) + 61.4 = 64.172Runway length = [(64.172-60)/ (70-60)]* (2912-2462)] + 2462 = 2650 m

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6. Profile grade data for a proposed airport runway are given here. Does the proposed longitudinal grade design conform with the requirements of the FAA? The first vertical curve is 1600 ft long and the second is 1000 ft. long. Use the Airplane Design Group VI and Approach Category D.

Station Grade (Percent) Comment0+00

30 + 50

55 + 50

92 + 00

-0.85

+0.65

-0.40

Begin runway

PI station no. I

PI station no. 2

End runway

Solution:

According to the longitudinal grade limitations shown below, we can conclude the following:

Between stations 0+00 to 30+50, the grade should not exceed 0.8. Between stations 30+50 to 55+50, the grade (+0.65) is acceptable. And the

vertical curve length = 1600 ft > [(0.8+0.65)*1000 ft = 1450 ft], and consequently the length of the first vertical curve is acceptable.

Between stations 50+50 to 92+50, the grade (-0.40) is acceptable. And the vertical curve length = 1000 ft > [(0.65+0.40)*1000 ft = 1050 ft], and consequently the length of the first vertical curve is acceptable.

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7. Determine the following dimensions for an air carrier airport designed to accommodate a B-727-000 aircraft.

a) Runway centerline to taxiway centerline b) Taxiway width on tangent c) Taxiway width on curve d) Radius of taxiway centerline on curves

Solution:

According to FAA & ICAO classification shown below, B-727-000 is considered as Design group III

Assume Approach category = C. then, from table 18-4a shown below, and from tables 4-1&4-2 extracted from AC 150/5300-13, we can conclude the following values:

Runway centerline to taxiway centerline = 3oo ft Taxiway width on tangent = 50 ft Taxiway width on curve = 50/2 + (100-55) = 70 ft Radius of taxiway centerline on curves = 100 ft

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