Airport engg.

1. Module – 4
• Airport Planning

2. Air transport characteristics
•Unbroken Journey: Air transport provides unbroken
journey over land and sea. It is the fastest and quickest
means of transport.
•Rapidity: Air transport had the highest speed among
all the modes of transport.
•Expensive: Air transport is the most expensive means
of transport. There is huge investment in purchasing
aero planes and constructing of aerodromes.
•Special Preparations: Air transport requires special
preparations like wheelers links, meteorological
stations, flood lights, searchlights etc.

3. •High Speed: The supreme advantage of air transport is its high
speed. It is the fastest mode of transport and thus it is the most
suitable mean where time is an important factor.
•Comfortable and Quick Services: It provides a regular,
comfortable, efficient and quick service.
•No Investment in Construction of Track: It does not require
huge capital investment in the construction and maintenance of
surface track.
•No Physical Barriers: It follows the shortest and direct route as
seas, mountains or forests do not come in the way of air
transport.
•Easy Access: Air transport can be used to carry goods and
people to the areas which are not accessible by other means of
transport.
Advantages

4. •Emergency Services: It can operate even when all other
means of transport cannot be operated due to the floods or
other natural calamities. Thus, at that time, it is the only
mode of transport which can be employed to do the relief
work and provide the essential commodities of life.
•Quick Clearance: In air transport, custom formalities can be
very quickly complied with and thus it avoids delay in
obtaining clearance.
•Most Suitable for Carrying Light Goods of High Value: It is
most suitable for carrying goods of perishable nature which
require quick delivery and light goods of high value such as
diamonds, over long distances.
Advantages

5. •National Defence: Air transport plays a very important
role in the defence of a country. Modern wars have been
fought mainly by aeroplanes. It has upper hand in
destroying the enemy in a very short period of time. It also
supports over wings of defence of a country.
•Space Exploration: Air transport has helped the world in
the exploration of space.
Advantages

6. Disadvantages
In spite of many advantages, air transport has the following
limitations:
•Very Costly: It is the costliest means of transport. The fares of
air transport are so high that it is beyond the reach of the
common man.
•Small Carrying Capacity: Its carrying capacity is very small and
hence it is not suitable to carry cheap and bulky goods.
•Uncertain and Unreliable: Air transport is uncertain and
unreliable as it is controlled to a great extent by weather
conditions. Unfavorable weather such as fog, snow or heavy rain
etc. may cause cancellation of scheduled flights and suspension
of air service.
•Breakdowns and Accidents: The chances of breakdowns and
accidents are high as compared to other modes of transport.
Hence, it involves comparatively greater risk.

7. Disadvantages
•Large Investment: It requires a large amount of capital
investment in the construction and maintenance of
aeroplanes. Further, very trained and skilled persons are
required for operating air service.
•Specialized Skill: Air transport requires a specialized skill and
high degree of training for its operation.
•Unsuitable for Cheap and Bulky Goods: Air transport is
unsuitable for carrying cheap, bulky and heavy goods because
of its limited capacity and high cost.
•Legal Restrictions: There are many legal restrictions imposed
by various countries in the interest of their own national unity
and peace.

42. haul operations

50. land which is free from residential or industrial developments.

53. The layout of an airport mainly depends on the basic
configurations of the runways. The other airport elements are
then correlated in such a way that an integrated layout is
developed giving smooth flow of traffic, keeping in mind the
taxi distances to a minimum, providing shortest route for the
passengers. A proper airport layout provides full functional
efficiency with the minimum space utilization. An engineer
should attempt to provide the simplest design which yields
the optimum service to air passengers. A good airfield layout
should posses the following characteristic:
Typical airport layouts
•Landing, taxing and taking off as independent operations
without interference.
•Shortest taxiway distance from loading runway end.
•Safe runway length

54. •Safe approaches
•Excellent control tower visibility
•Adequate loading apron space
•Sufficient terminal building facilities
•Sufficient land area to permit subsequent expansion
•Lowest possible cost of construction.
Typical airport layouts

55. Typical airport layouts

56. Typical airport layouts

57. Module – 5
Airport Design

58. According to the International Civil Aviation Organization (ICAO)
a runway is a "defined rectangular area on a land aerodrome
prepared for the landing and takeoff of aircraft".

71. The following is the wind date at proposed airport site when
wind intensity is above 6.4 kmph. Determine the best direction
to orient the runway and hence calculate total wind coverage
and calm period.
Direction
Duration of wind, %
6.4 -25
kmph
25 -40
kmph
40 -60
kmph
N 4.2 1.6 0.2
NNE 7.1 3.6 0.3
NE 5.2 2.3 0.5
ENE 2.2 1.4 0.4
E 1.8 0.2 0.0
ESE 1.3 0.7 0.0
SE 1.6 0.4 0.0
SSE 3.1 0.9 0.0
S 6.2 1.3 0.5
SSW 10.3 4.2 0.5
SW 7.6 2.1 0.3
WSW 5.0 0.9 0.1
W 2.2 1.4 0.4
WNW 1.7 0.3 0.0
NW 0.8 0.2 0.0
NNW 4.0 1.0 0.0

72. Solution
Direction
Duration of wind, % Total in each
direction
%
6.4 -25
kmph
25 -40
kmph
40 -60
kmph
N 4.2 1.6 0.2 6
NNE 7.1 3.6 0.3 11
NE 5.2 2.3 0.5 8
ENE 2.2 1.4 0.4 4
E 1.8 0.2 0.0 2
ESE 1.3 0.7 0.0 2
SE 1.6 0.4 0.0 2
SSE 3.1 0.9 0.0 4
S 6.2 1.3 0.5 8
SSW 10.3 4.2 0.5 15
SW 7.6 2.1 0.3 10
WSW 5.0 0.9 0.1 6
W 2.2 1.4 0.4 4
WNW 1.7 0.3 0.0 2
NW 0.8 0.2 0.0 1
NNW 4.0 1.0 0.0 5
Total Coverage in all Directions 90

74. Total Coverage in all Directions = 90 %
Calm period = 100 – Total Coverage in all Directions
= 100 – 90
= 10 %
Best orientation of runway is NNE and SSW
Wind coverage = N+NNE+NE+S + SSW + SW + Calm period
= 6 + 11 + 8 + 8 + 15 + 10 + 10
= 68 %

75. Direction
Duration of wind, %
6.4 -25
kmph
25 -40
kmph
40 -60
kmph
N 7.4 2.7 0.2
NNE 5.7 2.1 0.3
NE 2.4 0.9 0.5
ENE 1.2 0.4 0.2
E 0.8 0.2 0.0
ESE 0.3 0.1 0.0
SE 4.3 2.8 0.0
SSE 5.5 3.2 0.0
S 9.7 4.6 0.0
SSW 6.3 3.2 0.5
SW 3.6 1.8 0.3
WSW 1.0 0.5 0.1
W 0.4 0.1 0.0
WNW 0.2 0.1 0.0
NW 5.3 1.9 0.0
NNW 4.0 1.3 0.3
The following is the wind date at proposed airport site when
wind intensity is above 6.4 kmph. Determine the best direction
to orient the runway and hence calculate total wind coverage
and calm period.

85. Problems on Basic Runway Length
1. The length of the runway under the Standard condition is
1620 m. The airport site has an Elevation of 270m. And the
reference temperature of the airport is 32.90 0 C . It is
decoded to construct the runway with can effective Gradient
of 0.20 %. Determine the Corrected length of the Runway
Solution
Given: L = 1620 m, Airport reference temperature = 32.90 0 C
Elevation = 270m and effective Gradient = 0.20 %

86. Problems on Basic Runway Length
Step 1 : Correction for Elevation
Correction = 7×L× Elevation /(100×300)
= 7×1620×270/(100×300)
= 102.06 m
Corrected Runway Length, L1 = L+ Correction
= 1620+102.06
=1722.06 m

87. Problems on Basic Runway Length
Step 2: Correction for Temperature
Standard Atmospheric temperature = 15 0 C -0.0065× Elevation
= 15 0 C -0.0065× 270
=13.25 0 C
Rise of the temp. = Airport reference temp. – Std. Atm. temp.
=32.90 0 C - 13.250 C
=19.65 0 C
Correction = L1 × Rise of the temperature /100
=1722.06×19.65/100
= 338.38m

88. Problems on Basic Runway Length
Corrected Length, L2 = L1 + Correction for temp.
= 1722.06+338.38
=2060.44 m
Step 3. Combined Correction
=27.18%
As per ICAO this combined correction Should not Exceed 35 %
Hence the correction is ok

89. Problems on Basic Runway Length
Step 4. Correction for Gradient
Applying Correction for the Effective Gradient at the rate of
20% for Each 1 % effective Gradient
Correction = 20 × L2 × Effective Gradient /100
= 20 × 2060.44 × 0.20 /100
=82.41m
Corrected Length = 2060.44 + 88.41
= 2060.44 + 88.41
=2142.85 m
above value may be rounded to the nearest 10 m, then the
corrected length
Correction Length = 2140 m

90. Problems on Basic Runway Length
1. The length of the runway under the Standard condition is
2100 m. The airport be provided at an Elevation of 410 m
above MSL. The airport reference temperature is 32 0 C . The
construction plans provides the following data. Determine the
length of Runway also apply correction as per ICAO and FAA
specication.
Given: L = 1620 m, Airport reference temperature = 32.90 0 C
End to end of
Runway (m)
0 -300 300-900 900-1500 1500- 1800 1800-2100 2100-2700 2700-3000
Grade ( %) + 1.00 - 0.50 + 0.50 + 1.00 - 0.50 - 0.40 - 0.10

91. Problems on Basic Runway Length
Solution
Given: L = 2100 m, Airport reference temperature = 32 0 C
Elevation = 410 m
Step 1 : Correction for Elevation
Correction = 7×L× Elevation /(100×300)
= 7×2100×410/(100×300)
= 200.90 m
Corrected Runway Length, L1 = L+ Correction
= 2100+ 200.90
=2300.90 m

92. Problems on Basic Runway Length
Step 2: Correction for Temperature
Standard Atmospheric temperature = 15 0 C -0.0065× Elevation
= 15 0 C -0.0065× 410
=12.335 0 C
Rise of the temp. = Airport reference temp. – Std. Atm. temp.
=32 0 C - 12.335 0 C
=19.665 0 C
Correction = L1 × Rise of the temperature /100
= 2300.90 ×19.665/100
= 452.47m

93. Problems on Basic Runway Length
Corrected Length, L2 = L1 + Correction for temp.
= 2300.90 +452.47
=2753.37 m
Step 3. Combined Correction
=31.112%
As per ICAO this combined correction is less than 35 % Hence the
correction is ok

94. Problems on Basic Runway Length
Step 4. Correction for Gradient
The elevations of different points as per proposed grading plan
Chainage (m) 0 300 900 1500 1800 2100 2700 3000
Elevation (m) 100 103 100 103 106 104.5 102.1 101.8

95. Problems on Basic Runway Length
Correction = 20 × L2 × Effective Gradient /100
= 20 × 2753.37 × 0.183 /100
= 100.77m
Corrected Length = L2 + Correction
= 2753.37 + 100.77
=2854.14 m
above value may be rounded to the nearest 10 m, then the
corrected length
Correction Length = 2860 m

96. Problems on Basic Runway Length
2. The data below refers to the daily temperature for the
hottest month of the year 1988 for given airport site.
Determine the airport reference temperature
Date
Temperature, ͦ C
Date
Temperature, ͦC
Maximum Average Maximum Average
1 42.5 25.5 16 43.7 26.2
2 42.5 25.5 17 43.8 25.9
3 42.7 25.7 18 44.0 26.3
4 43.0 25.9 19 44.8 26.3
5 43.0 25.9 20 44.1 26.3
6 43.0 25.9 21 44.3 26.5
7 42.8 25.8 22 44.3 26.9
8 43.0 25.9 23 44.5 26.5
9 43.0 25.9 24 44.6 26.5
10 43.1 25.0 25 44.6 26.9
11 43.3 26.3 26 44.7 27.0
12 43.5 26.4 27 44.6 27.0
13 43.3 26.3 28 44.7 27.0
14 43.5 26.4 29 44.8 26.2
15 43.6 26.3 30 45.0 27.2

97. Problems on Basic Runway Length
Solution:
Mean of the maximum daily temperature, tm = 1312.3/30
= 43.74 ͦ C
Mean of the average daily temperature, ta = 787.4 /30
= 26.25 ͦ C
= 32.08 ͦ C