## Geometric Design of Runway Numerical Question of Airport Engineering

Geometric Design of Runway Numerical Question of Airport Engineering, In Part 1, we set the stage by dissecting the foundational elements of geometric runway design. From the critical dimensions that govern runway length to the graceful curves and strategic alignments that guide every take off and landing, this is where the blueprint for aviation excellence takes shape.

## Correction for basic runway length (l):

### Correction for elevation (ICAO):

As per the recommendation of ICAO, the basic runway length should be increased at the rate of 7% per 300 m rise in elevation of the airport above the mean sea level (MSL).

Corrected length of runway, (l_{1}) = (1+(x/300 * 7/100))

Here, x is the elevation above MSL, and l is basic runway length.

### Correction for temperature (ICAO)

After the basic length is corrected for the elevation of airport, it is further increased at the rate of 1% for every 1°C rise in airport reference temperature (ART) above the standard atmospheric temperature at that elevation (SATx).

Airport reference temperature, ART = T_{a} + (T_{m}- T_{a}/3)

Here, Ta is the monthly mean of the average daily temperature for the hottest month of the year, and T is the monthly mean of the maximum daily temperature for the same month.

The temperature gradient of the standard atmosphere from the mean sea level to the altitude at which the temperature becomes (SAT) = 15°C - 0.0065 × x, where x is the elevation above the mean sea level.

So, Rise in temperature ΔT = ART - (SAT)x

Corrected length (l_{2}) = l_{1} * (1 + ΔT/100)

**Note: **As per ICAO, cumulative correction for elevation and temperature together **≥** 35%

cumulative % correction = (l_{2}-l/l) 100 **≥** 35%

If the correction exceeds 35%, it should be further checked up by conducting specific studies at the site by model tests.

**Notes**

Effective gradient is defined as the maximum difference in elevation between the highest and lowest point of the runway divided by the total length of the runway.

### Correction for gradient (FAA)

According to FAA (Federal Aeration Administration), the runway length, after being corrected for elevation and temperature, should further be increased at the rate of 20% for every 1% of the effective gradient.

**Q 1)** Which of the following factors are considered for estimating the required runway length for an aircraft landing?

1) Normal maximum temperature

2) Airport elevation

3) Maximum landing weight

4) Effective runway gradient

a) 1, 2, 3, 4

b) 1, 3 and 4

c) 2 and 3

d) 2 and 4

**Sol:** Factors considered for calculation of required runway length:

1) Maximum temperature

2) Maximum landing weight

3) Airport elevation above MSL

4) Effective runway gradient

So, correct answer is **a)**.

**Q 2)** Effective gradient of a runway is 0.5 percent. If the corrected runway length after elevation and temperature correction is 2890 m, then the revised runway length for effective gradient will be:

a) 2880 m

b) 2990 m

c) 3079 m

d) 3179 m

**Sol.** l_{2} = 2890

effective gradient (G) = 0.5%

Corrected length = l_{2} x (1+20% x G)

Corrected length = 3179 m

So, Correct answer is **d)**

**Q 3)** If the airport reference temperature of a runway at an elevation of 1000 m above MSL is 19°C, then the rise in temperature to be considered as per ICAO is:

a) 27.5°C

c) 10.5°C

b) 15°C

d) 13.25°C

**Sol: **SAT_{x} = 15° - 0.0065 × MSL

SAT_{x} = 15° - 0.0065 × 1000

SAT_{x}= 8.5°C

Change in temperature (ΔT) = ART - SAT_{x}

ΔТ = 19- 8.5

ΔТ = 10.5

So, correct option is **c).**

**Q 4)** If the monthly mean of average daily temperature and monthly mean of the maximum daily temperature of the hottest month of the year are 38°C and 47°C, respectively, then the airport reference temperature is:

a) 38°C

b) 41 °C

c) 47°C

d) 50°C

**Sol.** Correct option is **b)**

**Q 5)** If the aeroplane reference field length of a runway located at 420 m above the MSL is 1900 m, then what will be the approximate corrected runway length for elevation?

a) 1900 m

b) 2087 m

c) 2187 m

d) 2320 m

**Sol.** Correct option is **c)**

**Q 6)** Consider the following data for the proposed longitudinal section of a runway:

### End to end of runway

- 0.0 to 5.0 chains
- 5.0 to 15.0 chains
- 15.0 to 30.0 chains
- 30.0 + 40.0 chains

### Gradient

- +1.2 %
- -1.4 %
- +0.7%
- +0.9%

If the length of one metric chain is 20 m, then the effective gradient of runway is

**Sol:** 100 m is considered as the initial R.L. (reduced level) of chainage at O.

Maximum difference in elevation = 102.3 - 98.4 = 3.9 m

Total runway length = 40 × 20 = 800 m

Therefore, effective gradient of runway = [(3.9 / 800) * 100] = = 0.4875%

Hence, the effective gradient of the runway is 0.49%.

**Q 7)** The basic length of the runway is 1700 m, and the elevation of the runway site is 2600 m above MSL. The reference temperature of the airport location is 32.90°C. If the runway is to be constructed with an effective gradient of 0.20 percent, then what will be the corrected length of the runway?

**Sol:** **i)** Corrected length for elevation = 1700 x (1 + 7/100 x 260/300) = 1803.133 m

**ii)** Correction for temperature:

Standard atmospheric temperature of airport site

= 15 - 0.0065 × 260 = 13.31 °C

Rise of temperature = 32.90 - 13.31 = 19.59 °C

Corrected length for temperature = 1803.133 x (1 + 19.59/100) = 2156.356

Hence, corrected length of the runway after elevation and temperature correction is 2157 m.

**iii)** Check for the total correction for elevation plus temperature:

Total correction in percentage = [(2157 - 1700)/1700] x 100 = 26.88

According to ICAO, elevation plus temperature correction should not exceed 35%.

**iv) **Correction for gradient (20/100 × 2157 × 0.2) = 86.28

Corrected length = 2157 + 86.28 = 2243.28 m

Hence, the corrected runway length is 2250 m.

**Q 8)** Required runway length for take-off at a level site at sea level in standard atmospheric condition is 2400 m, and the required runway length for landing at sea level in the standard atmospheric conditions is 2000 m. Reference temperature is 24°C, and the aerodrome elevation is 220 m. Temperature in the standard atmosphere for an elevation of 220 n is 15°C. If the runway slope is 0.5%, then determine the length of the runway after applying a correction to runway length.

**Sol:** Given

Basic runway take-off length, L_{t} = 2400 m

Basic runway landing length, L_{L} = 2000 m

Elevation = 220 m

Standard temperature, T_{s} = 15°C

Reference temperature, T_{R} = 24°C

Runway slope, g = 0.5%

**Correction to runway take-off length**

**i) Correction for elevation:**

Correction for elevation, as recommended by ICAO, is 7% per 300m elevation

Therefore, correction = 7/100 x elevation/300 x L_{T} = 123.2 m

Corrected length, L' = 2400 + 123.2 = 2523.2m

**ii) Correction for temperature:**

Rise in temperature = T_{R} - T_{S} = 24 - 15 = 9°C

As per ICAO, correction for temperature is 1% for every 1°C rise in temperature.

Correction = 1/100 x rise in temperature/1 x L'

Where L' = runway take-off length after correction for elevation is applied

Correction = 1/100 x 9/1 x 2523.2 = 227.088 m

Corrected length (L") = L' + 227/088 = 2750.288 m

Check:

As per ICAO, the total correction for elevation plus temperature should not exceed 35% of the basic runway length.

Total correction for elevation and temperature

= 123.2 + 227.088 = 350.288 m

Total correction (in percentage) = 350.288/2400 x 100 = 14.595% , 35 %

Hence, **OK**

**iii) Correction for gradient:**

FAA recommends that runway length after having been corrected for elevation and temperature should be further increased at the rate of 20% for every 1% of effective gradient.

Correction for gradient = 20/100 x g x L" = 275.0288 m

Final runway take-off length = L"+ 275.0288 = 2750.288 + 275.0288 = 3025.317 m = 3025 m

**2) Correction to runway landing length**

**i) Correction for elevation**

Correction = 7/100 x elevation/300 x L_{L}

= 7/100 x 200/300 x 2000 = 93.33 m

Corrected runway landing length = 2000 + 93.33 = 2093.33 m

No corrections are needed to landing length for temperature and gradient.

Actual runway length to be provided would be greater than (1) and (2)

Therefore, the corrected runway length is 3025 m

## previous year's questions

**Q 1)** The appropriate design length of a clearway is calculated on the basis of 'Normal Take-off condition. Which one of the following options correctly depicts the length of the clearway? (Note: None of the options is drawn to scale)

**[GATE-2020, SET-I]**

**Sol: c)**

**Q 2)** The longitudinal section of a runway provides the following data:

End-to-End runway (m) | Gradient (%) |

0 to 300 | + 1.2 |

300 - 600 | - 0.7 |

600 - 1100 | + 0.6 |

1100 - 1400 | - 0.8 |

1400 - 1700 | - 1.0 |

**[GATE-2021, SET-I]**

**Sol: (0.32)**

**Q 3)** A runway is being constructed in a new airport as per the International Civil Aviation Organization (ICAO) recommendations. The elevation and the airport reference temperature of this airport are 535 m above the mean sea level and 22.65°C, respectively. Consider the effective gradient of the runway as 1%. The length of runway required for a design-aircraft under the standard conditions is 2000 m. Within the framework of applying sequential corrections as per the ICAO recommendations, the length of runway corrected for the temperature is

a) 2223 m

b) 2250 m

c) 2500 m

d) 2750 m

**[GATE-2017, SET-I]**

**Sol: c)**

### Keywords

Geometric design of runway

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