Airport Engineering MCQ Quiz - Objective Question with Answer for Airport Engineering - Download Free PDF
Last updated on Mar 28, 2025
Latest Airport Engineering MCQ Objective Questions
Airport Engineering Question 1:
What are the maximum longitudinal and effective gradients, respectively, for runways of 'A' type airport recommended by ICAO?
Answer (Detailed Solution Below)
Airport Engineering Question 1 Detailed Solution
Explanation:
Maximum Longitudinal and Effective Gradients for 'A' Type Airport Runways
The International Civil Aviation Organization (ICAO) has set guidelines for the design and construction of airport runways. One of the important aspects to consider in runway design is the gradient, which affects the performance and safety of aircraft during landing and takeoff. The maximum allowable gradients help ensure that the aircraft can safely maneuver on the runway without excessive effort or risk.
The ICAO recommends that the maximum longitudinal gradient for 'A' type airport runways should be 1.50%. This ensures that the runway slope is not too steep for safe aircraft operations.
The maximum effective gradient, which considers the overall slope of the runway, should be 1.00%. This helps in maintaining a consistent and manageable slope for aircraft movement
Airport Engineering Question 2:
Determine the turning radius of the taxiway according to Horonjeff equation for a supersonic transport aircraft with a wheel base of 30 m and main loading gear of 6 m for a design turning speed of 50 kmph. Assume the coefficient of friction between the tyre and pavement surface is 0.13 and width of taxiway pavement is 22.5 m.
Answer (Detailed Solution Below)
Airport Engineering Question 2 Detailed Solution
Concept:
According to Horonjeff, the radius of a taxiway for a supersonic transport aircraft is given by:
\( R = \frac{0.388 \cdot W^2}{\left( \frac{T}{2} - S \right)} \)
Where:
W = Wheel base of aircraft = 30 m
T = Width of taxiway pavement = 22.5 m
S = Distance from midpoint of gear to pavement edge = \( 6 + \frac{\text{wheel tread}}{2} = 6 + \frac{6}{2} = 9 \, \text{m} \)
Step-by-step Calculation:
\( \frac{T}{2} = \frac{22.5}{2} = 11.25 \, \text{m} \)
\( \left( \frac{T}{2} - S \right) = 11.25 - 9 = 2.25 \, \text{m} \)
Now, \( R = \frac{0.388 \cdot 30^2}{2.25} = \frac{0.388 \cdot 900}{2.25} = \frac{349.2}{2.25} = 155.2 \, \text{m} \)
Airport Engineering Question 3:
The length of a runway under standard conditions is 1840 m. The airport site has an elevation of 250 m. Its reference temperature is 32°C. Assuming all the other conditions to be under standard level, determine the correction for elevation.
Answer (Detailed Solution Below)
Airport Engineering Question 3 Detailed Solution
Concept:
According to ICAO, the basic runway length should be increased by 7% for every 300 m rise in elevation above mean sea level.
\(Correction = \frac{7}{100} \times \text{Runway Length} \times \frac{\text{Airport Elevation}}{300}\)
Given:
Runway length = 1840 m
Airport elevation = 250 m
Calculation:
\(Correction = \frac{7}{100} \times 1840 \times \frac{250}{300}\)
\(= 0.07 \times 1840 \times 0.8333 = 107.52 \, \text{m}\)
Airport Engineering Question 4:
Where the aircraft is standing (parking), the facility for loading and unloading operation in front of terminal building is known as:
Answer (Detailed Solution Below)
Airport Engineering Question 4 Detailed Solution
Airport Engineering Question 5:
At an airport, the turning radius of a taxiway is given by
Answer (Detailed Solution Below)
Airport Engineering Question 5 Detailed Solution
Top Airport Engineering MCQ Objective Questions
At a certain station, the mean of the average temperature is 25° C and mean of the maximum daily temperature is 40° C. What is the airport reference temperature (ART) ?
Answer (Detailed Solution Below)
Airport Engineering Question 6 Detailed Solution
Download Solution PDFConcept:
Airport reference temp. (Tr) \(= {T_a} + \frac{1}{3}\left( {{T_m} - {T_a}} \right)\)
Ta = Avg. temp of the hottest month
Tm = Monthly mean of maximum daily temp. of same month.
Calculation:
TA = 25°C
Tm = 40°C
\({T_\pi } = {T_a} + \frac{1}{3}\left( {{T_m} - {T_a}} \right)\)
\(= 25 + \frac{1}{3}\left( {40 - 25} \right)\)
= 25 + 5 = 30°C
Match the designated component in List 1 with its purpose at airports in List 2.
List 1
A. Basic Runway length
B. Runway Capacity
C. Runway geometric design
D. Hanger
List 2
1. Width and length of safety area of airport
2. Housing, servicing of aircrafts
3. Location of exit taxiways
4. Engine failure class
Answer (Detailed Solution Below)
Airport Engineering Question 7 Detailed Solution
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Basic Runway length |
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Runway capacity |
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Runway geometric design |
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Hanger |
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The point of intersection of the obstruction clearance line and the extended plane of the runway surface and the other end of the runway is called as:
Answer (Detailed Solution Below)
Airport Engineering Question 8 Detailed Solution
Download Solution PDFExplanation:
Effective length of runway:
For takeoff, effective length of the runway means the distance from the end of the runway at which the takeoff is started to a point at which the obstruction clearance plane associated with the other end of the runway intersects the runway centerline.
Additional Information
Basic Runway Length:
- It refers to the length of an airport runway under the following assumptions:
- No wind is blowing on runway;
- Runway is levelled (No effective gradient);
- Airport is at sea level and Standard Temperature at the airport is 15°C;
- Aircraft is loaded to its capacity;
- No wind is blowing on the way to destination Standard temperature prevails along the way.
For a runway at an elevation of 2000 m above MSL and airport reference temperature of 12°C, the rise in temperature to be considered as per ICAO is
Answer (Detailed Solution Below)
Airport Engineering Question 9 Detailed Solution
Download Solution PDFConcept:
The standard temperature at the airport site can be determined by reducing the standard mean sea level temperature of 15°C at the rate of 6.5°C per thousand meter rises in elevation.
Standard temperature = 15 - 0.0065 × elevation
Rise in temperature = (Reference Temp. - Standard Temp.)
Calculation:
Standard temperature =15 - 0.0065 × 2000 = 2°C.
Rise in temperature = 12 - 2 = 10°C
As per ICAO, for A, and B type of airports, maximum effective grade is
Answer (Detailed Solution Below)
Airport Engineering Question 10 Detailed Solution
Download Solution PDFExplanation:
Effective gradient: As per ICAO Maximum effective gradient
(i) For A, B and C type = Max 1%
(ii) For D and E type = Max 2%
Transverse gradient: Provided for drainge purpose. As per ICAO
(i) For A, B type = Max. value 1.5% and Min value 0.5%
(ii) For C, D and E type = Max. value 2% and min value 0.5%
Rate of change longitudinal gradient: AS per ICAO
(i) For A, B and C type = Max. 0.10% per 30 m for vertical curve
(ii) For E and D type = 0.4% per 30 m for vertical curve.
Airport elevation is the reduced level above Mean Sea Level of
Answer (Detailed Solution Below)
Airport Engineering Question 11 Detailed Solution
Download Solution PDFConcepts:
Airport elevation is the reduced level above the Mean Sea Level of the highest point of the landing area.
Important PointsMean Sea Level
- Mean Sea Level (MSL) is the datum for measurement of elevation and altitude.
- Mean Sea Level is the equipotential surface of the Earth.
- Mean Sea Level (MSL) adopted by a survey of India for reference is located at Mumbai High.
- India's mean elevation is 160 m and is taken from Mumbai High.
Runway length required under standard conditions is 1500 m then actual elevation of the site is 1100 m above M. S. L Runway length corrected for altitude will be
Answer (Detailed Solution Below)
Airport Engineering Question 12 Detailed Solution
Download Solution PDFBasic runway length is based on some assumptions:-
a) Derived at Mean Sea level
b) Derived for a standard temperature of 15°C
c) The gradient must be zero.
The actual runway need not follow all these guidelines and hence we must apply some corrections.
The sequential correction is applied in the basic runway length.
The first correction is altitude correction, the second correction is temperature correction and the last correction is gradient correction.
Altitude Correction:
Basic runway length is increased by 7% for every 300 m rise above mean sea level.
This is done due to the decrease in drag force on the aircraft owing to the decrease in density of air as the altitude increases.
Given:
Height above MSL = 1100 m, and Basic Runway length = 1500 m
After applying elevation correction,
\({\rm{Correction}} = 1500 \times \frac{7}{{100}} \times \frac{{1100}}{{300}} = 385{\rm{\;m}}\)
Corrected Runway Length = 1500 + 385 = 1885 m
Tips and Tricks:-
Temperature correction and gradient corrections are applied on the subsequently corrected runway length.Which one of the following instances of performance of aircraft is not considered for determining basic runway length?
Answer (Detailed Solution Below)
Airport Engineering Question 13 Detailed Solution
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Factors affecting basic runway length |
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Aircraft characteristics |
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Safety requirements
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Airport environment |
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Important point:
Basic runway length is calculated with the following assumptions:
i) It is calculated at Mean sea level (MSL)
ii) It is calculated for standard temp 15°C at MSL.
iii) The gradient is assumed to be zero.
Due to the variance of these assumptions, some corrections are applied.
i) Elevation Correction: 7% increase in basic runway length for every 300 m rise above MSL
ii) Temperature correction: 1% increase for every 1°C rise in airport reference temperature with respect to standard temperature at elevation.
Where,
Standard temperature at elevation = Temperature at MSL – 0.0065 × Elevation
This increase is made on the already corrected runway length for elevation.
iii) Gradient correction: 20% increase for 1% of effective gradient.
The height of the pilot's eye above the runway surface is assumed as
Answer (Detailed Solution Below)
Airport Engineering Question 14 Detailed Solution
Download Solution PDFExplanation:
The height of the pilot's eye above the runway surface is assumed as 3 meters.
The height of the driver’s eye above the road surface is assumed as 1.2 meters.
Eye’s level of the driver/pilot is very essential factor in deciding the various geometric design factor for highway and runway respectively.
Identify the mis-matched pair about type of Airport and the maximum value of longitudinal gradient of a taxiway, as per ICAO.
Answer (Detailed Solution Below)
Airport Engineering Question 15 Detailed Solution
Download Solution PDFExplanation:
Taxiway:
- Taxiways as defined by ICAO are a defined path on a land aerodrome established for the taxiing of aircraft and intended to provide a link between one part of the aerodrome and another in an expeditious manner.
Design criteria for taxiway:
1) Longitudinal gradient as per ICAO
Type or Class of airport | Permissible gradient |
A/B | 1.5 % |
C/D/E | 3 % |
2) Transverse gradient as per ICAO
Type/Class of airport | Permissible gradient |
A/B/C | 1.5 % |
D/E | 2 % |
3) Rate of change of longitudinal gradient as per ICAO
Rate of change | Class of airport |
1% / 30 m of vertical curve | A/B/C |
1.2% / 30 m of vertical curve | D/E |