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Question 1
ప్రశ్న 1
In the conventional preliminary runway-length correction method, what does basic runway length represent?
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Explanation:
Basic runway length is the starting point for the conventional preliminary correction method. It is determined for specified reference conditions rather than actual airport conditions. These reference conditions include sea-level standard atmosphere, still air and zero runway slope. Corrections may be needed when the proposed airport differs from these conditions. The final runway length must ultimately satisfy the relevant aircraft-performance and operating requirements.
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Question 2
ప్రశ్న 2
Which sequence correctly describes the conventional preliminary calculation of corrected runway length when elevation, temperature and effective-gradient corrections are all required?
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Explanation:
The conventional calculation begins with the selected basic runway length. An elevation correction is applied first to reflect conditions above mean sea level. The specified temperature correction is then applied to the elevation-corrected length. The gradient correction is applied to the length obtained after the preceding corrections. Because the stages are successive, the overall percentage increase generally differs from the sum of the individual percentages.
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Question 3
ప్రశ్న 3
Under the conventional general correction method described in ICAO aerodrome-design guidance, what is the preliminary runway-length correction for airport elevation?
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Explanation:
The conventional general method uses a 7% increase for each 300 m of elevation above mean sea level. Greater elevation generally means lower air density under comparable atmospheric conditions. Lower air density can reduce aircraft takeoff performance. In this preliminary calculation, the elevation correction is applied to the basic runway length. Aircraft-specific approved performance data should be preferred where available, rather than assuming this general factor is universally applicable.
వివరణ:
Question 4
ప్రశ్న 4
The basic runway length for a preliminary airport design is 2000 m and the airport elevation is 600 m above mean sea level. Using an elevation correction of 7% of basic length per 300 m, what is the elevation-corrected length?
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Explanation:
The given elevation is equivalent to two increments of 300 m. At 7% per increment, the total elevation correction is 14% of basic length. The additional length is 2000 times 0.14 = 280 m. Adding this to the basic length gives 2000 + 280 = 2280 m. This is the elevation-corrected length, not the final length if temperature and gradient corrections remain applicable.
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Question 5
ప్రశ్న 5
For a preliminary runway design, basic length is 2400 m and the airport elevation is 450 m above mean sea level. Using a linear elevation allowance of 7% per 300 m, what length results after elevation correction alone?
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Explanation:
The elevation factor is 0.07 times 450/300. This gives a fractional increase of 0.105, or 10.5%. The extra length is 2400 times 0.105 = 252 m. Hence the elevation-corrected length is 2400 + 252 = 2652 m. The calculation assumes the conventional linear elevation-correction method stated in the question.
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Question 6
ప్రశ్న 6
Under the standard-atmosphere approximation T_s = 15 - 0.0065H, where T_s is in degrees Celsius and H is elevation in metres, what is the standard atmospheric temperature at an airport elevation of 600 m?
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Explanation:
The standard atmospheric temperature is 15 degrees Celsius at mean sea level under the stated approximation. Temperature decreases at 0.0065 degrees Celsius per metre of elevation. At 600 m, the temperature decrease is 0.0065 times 600 = 3.9 degrees Celsius. Subtracting gives 15 - 3.9 = 11.1 degrees Celsius. This standard temperature is the reference against which airport reference temperature is compared.
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Question 7
ప్రశ్న 7
The monthly mean of average daily temperatures during the hottest month is 27 degrees Celsius, and the monthly mean of daily maximum temperatures for that month is 36 degrees Celsius. The older ICAO Aerodrome Design Manual describes a conventional runway-correction reference temperature T_old = T_1 + (T_2 - T_1)/3, whereas ICAO Annex 14 defines aerodrome reference temperature using the monthly mean of daily maximum temperatures for the hottest month. Which pair gives the older formula result and the Annex 14 reference temperature, respectively?
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Explanation:
The older ICAO Aerodrome Design Manual formula uses the monthly mean of average daily temperatures as T_1 and the monthly mean of daily maxima as T_2. Substitution gives T_old = 27 + (36 - 27)/3 = 30 degrees Celsius. The ICAO Annex 14 aerodrome reference-temperature definition instead uses the monthly mean of daily maximum temperatures for the hottest month, averaged over the specified period of years. Using the supplied data, that definition gives 36 degrees Celsius. These two definitions must be distinguished rather than treated as interchangeable in modern airport-design questions.
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Question 8
ప్రశ్న 8
An airport is at 900 m elevation, where standard temperature is calculated as 15 - 0.0065H degrees Celsius. Its airport reference temperature is 28.15 degrees Celsius. How many degrees does the reference temperature exceed the standard temperature at that elevation?
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Explanation:
At 900 m elevation, the standard temperature decrease is 0.0065 times 900 = 5.85 degrees Celsius. Therefore the standard temperature is 15 - 5.85 = 9.15 degrees Celsius. The airport reference temperature is 28.15 degrees Celsius. The excess is 28.15 - 9.15 = 19.0 degrees Celsius. Under the stated conventional temperature correction, this excess is the relevant temperature difference.
వివరణ:
Question 9
ప్రశ్న 9
A preliminary runway length has already been corrected for elevation to 2140 m. Airport reference temperature exceeds the standard temperature at that elevation by 8 degrees Celsius. At 1% additional length per degree Celsius of excess, what is the temperature-corrected length?
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Explanation:
Temperature correction is applied to the elevation-corrected length in the stated conventional procedure. The temperature excess is 8 degrees Celsius. At 1% per degree, the multiplier is 1.08. Multiplying 2140 m by 1.08 gives 2311.2 m. This answer includes elevation and temperature effects but not any additional gradient correction.
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Question 10
ప్రశ్న 10
A runway has a basic length of 2000 m and lies at an elevation of 300 m. Assume a 7% elevation increase per 300 m and an additional 5% temperature correction applied to the elevation-corrected length. Ignoring gradient, what is the final corrected length for this exercise?
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Explanation:
The elevation correction adds 7% to the basic 2000 m. The elevation-corrected length is therefore 2000 times 1.07 = 2140 m. The subsequent temperature correction is 5% of 2140 m, not 5% of the original length. Multiplying 2140 by 1.05 gives 2247 m. The total increase over basic length is 247 m, or 12.35%, because the corrections compound.
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Question 11
ప్రశ్న 11
In the conventional ICAO general-correction method, a 2000 m basic runway at 600 m elevation receives a 14% elevation correction and then a 20% temperature correction on the elevation-corrected length. Which conclusion is correct regarding the historical 35% combined-correction screening condition?
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Explanation:
The elevation-corrected length is 2000 times 1.14 = 2280 m. Applying the 20% temperature increase gives 2280 times 1.20 = 2736 m. The total increase is 736 m above the 2000 m basic length. This is 36.8%, which exceeds the 35% screening threshold described in the conventional guidance. A specific aeronautical study is therefore indicated rather than treating the extrapolated result as a final approved runway length.
వివరణ:
Question 12
ప్రశ్న 12
Along a proposed runway centreline of length 3000 m, the maximum elevation is RL 111 m and the minimum elevation is RL 93 m. What is the effective longitudinal gradient, calculated as the elevation range divided by runway length and expressed as a percentage?
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Explanation:
Effective longitudinal gradient uses the difference between maximum and minimum centreline elevations. The elevation range is 111 - 93 = 18 m. The total runway length is 3000 m. Therefore the effective gradient is 18/3000 times 100 = 0.60%. It need not be the same as the gradient between the two runway endpoints if an intermediate high or low point exists.
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Question 13
ప్రశ్న 13
In the conventional general-correction procedure given in the ICAO Aerodrome Design Manual, what runway-slope correction is specified where the basic runway length governed by takeoff is at least 900 m?
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Explanation:
The ICAO Aerodrome Design Manual's conventional general procedure specifies a 10% runway-length increase for each 1% runway slope under the stated takeoff-length condition. The applicable basic takeoff length must be at least 900 m for that general provision. Some conventional Indian textbook problems instead prescribe a 20% per 1% effective-gradient calculation. Those are different calculation conventions and should not be substituted without identification. Actual aircraft-specific performance and the governing aerodrome requirements take precedence over an illustrative factor.
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Question 14
ప్రశ్న 14
A conventional Indian civil-engineering textbook exercise explicitly prescribes a 20% runway-length increase for each 1% of effective gradient. If the length after elevation and temperature corrections is 2100 m and effective gradient is 0.5%, what is the corrected length?
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Explanation:
This question explicitly adopts the 20% per 1% gradient convention used in the stated textbook exercise. At an effective gradient of 0.5%, the corresponding increase is 10%. The length before gradient correction is 2100 m. Multiplying 2100 by 1.10 gives 2310 m. This answer follows the specified textbook rate and should not be confused with calculations that prescribe a different gradient-correction coefficient.
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Question 15
ప్రశ్న 15
Using an explicitly specified ICAO-style preliminary gradient allowance of 10% per 1% runway slope, the runway length after elevation and temperature corrections is 2200 m. For an effective gradient of 0.75%, what is the gradient-corrected length?
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Explanation:
The assumed gradient rate is 10% additional length for a 1% effective gradient. An effective gradient of 0.75% therefore produces a 7.5% increase. The relevant length before this correction is 2200 m. The calculation is 2200 times 1.075 = 2365 m. This result uses the rate explicitly stated in the question and does not imply that the alternative 20% textbook rate has been applied.
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Question 16
ప్రశ్న 16
Under a textbook exercise that explicitly uses a 20% length increase for each 1% effective gradient, the elevation-and-temperature-corrected length is 2200 m. If effective gradient is 0.75%, what final length is obtained?
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Explanation:
The exercise prescribes the 20% gradient-correction convention. At a gradient of 0.75%, the correction percentage becomes 20 times 0.75 = 15%. The length before gradient correction is 2200 m. Increasing it by 15% gives 2200 times 1.15 = 2530 m. The distinction from the 10% coefficient must be preserved because it produces a different answer from the same input gradient.
వివరణ:
Question 17
ప్రశ్న 17
The basic runway length is 2000 m and the airport elevation is 300 m. Adopt a 7% elevation increase per 300 m, a 10% temperature increase applied next, and a gradient allowance of 10% per 1% effective gradient. If effective gradient is 0.5%, what is the final corrected length?
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Explanation:
The elevation correction produces 2000 times 1.07 = 2140 m. The temperature correction is then applied to this value, giving 2140 times 1.10 = 2354 m. A gradient of 0.5% under the stated rate produces a 5% further increase. Hence the final length is 2354 times 1.05 = 2471.7 m. Each correction is applied successively, so using a direct sum of percentages would produce an incorrect result.
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Question 18
ప్రశ్న 18
A conventional Indian textbook runway-length exercise specifies the following: basic length 1800 m, elevation 600 m, elevation increase 7% per 300 m, temperature increase 5% on the elevation-corrected length and gradient increase 20% per 1% effective gradient. If effective gradient is 1%, what final corrected length is obtained?
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Explanation:
At 600 m elevation, the elevation increase is 14%, giving 1800 times 1.14 = 2052 m. Applying the specified 5% temperature increase gives 2052 times 1.05 = 2154.6 m. The stated textbook gradient convention adds 20% for the 1% effective gradient. The final result is 2154.6 times 1.20 = 2585.52 m. This is a compounded preliminary calculation using the explicitly supplied factors, not a substitute for verified aircraft-specific performance requirements.
వివరణ:
Question 19
ప్రశ్న 19
A basic runway length of 1000 m is increased by 7% for elevation and then by 10% for temperature, both corrections being applied successively. What is the overall percentage increase over the basic length?
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Explanation:
The elevation factor is 1.07. The temperature factor is 1.10 and is applied after the elevation correction. The combined multiplier is 1.07 times 1.10 = 1.177. Thus the final length is 1177 m for a 1000 m basic length. The increase is 177 m, equivalent to 17.7%, which is greater than the arithmetic sum of 17%.
వివరణ:
Question 20
ప్రశ్న 20
A corrected runway length of 2520 m has been obtained by multiplying the basic length successively by an explicitly specified temperature factor of 1.20 and a gradient factor of 1.05. No elevation correction applies. What was the basic runway length?
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Explanation:
The total correction multiplier is 1.20 times 1.05. This equals 1.26. The basic length is found by dividing the final corrected length by the combined multiplier. Therefore 2520/1.26 = 2000 m. Reversing compounded corrections requires division by their factors rather than subtraction of their nominal percentage increases from the final length.
వివరణ:
Question 21
ప్రశ్న 21
A runway starts and ends at the same elevation but has a pronounced intermediate crest. Which statement about its effective gradient is correct?
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Explanation:
Effective runway gradient is calculated from the centreline elevation range over the runway length. The relevant range is the difference between its maximum and minimum elevations. Equal elevations at the two ends do not rule out an intermediate crest or depression. Such internal elevation variation can therefore give a nonzero effective gradient. Local longitudinal slopes, changes of slope and visibility requirements must also be checked separately.
వివరణ:
Question 22
ప్రశ్న 22
A proposed airport has an airport reference temperature below the standard atmospheric temperature at its elevation. Which treatment is most defensible when using a conventional general temperature-correction method?
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Explanation:
The conventional positive temperature-correction rule addresses temperatures exceeding the standard value at the airport elevation. It does not justify unlimited shortening of the runway when the reference temperature is lower. Colder conditions can improve some aircraft-performance characteristics, but the actual outcome is aircraft- and condition-dependent. Other constraints may still determine the required takeoff or landing distance. The engineer should therefore rely on the applicable correction instructions and approved performance evidence rather than extrapolate a general factor beyond its intended use.
వివరణ:
Question 23
ప్రశ్న 23
In a conventional general-correction screening exercise, elevation is 600 m and elevation correction is 7% per 300 m. Temperature correction is 1% per degree Celsius of excess and is applied to the elevation-corrected length. Approximately what maximum temperature excess would keep the combined elevation-plus-temperature increase at 35% of basic length?
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Explanation:
At 600 m elevation, the elevation multiplier is 1.14. To produce a total correction of 35%, the final combined multiplier must be 1.35. Therefore 1.14 times (1 + 0.01 times the temperature excess) equals 1.35. Solving gives a temperature excess of approximately 18.42 degrees Celsius. Under the conventional guidance, exceeding the 35% combined-correction threshold indicates the need for a specific study rather than unrestricted use of the general factors.
వివరణ:
Question 24
ప్రశ్న 24
After applying the explicitly specified local-condition correction procedure, the required takeoff-based runway length is 2700 m and the independently determined landing-based requirement is 2600 m. Assume both must be satisfied by the same paved runway, with no stopway, clearway or displaced threshold. What is the minimum governing paved length?
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Explanation:
Takeoff and landing impose separate minimum length requirements. The runway must satisfy both under the stated assumptions. The required paved length is therefore the larger of 2700 m and 2600 m. This gives 2700 m, governed by the takeoff requirement. Adding the two lengths together would not represent the required length of a shared runway.
వివరణ:
Question 25
ప్రశ్న 25
A preliminary runway correction has been calculated using an assumed effective gradient from an earlier 1800 m alignment. The planned runway is subsequently extended and regraded to 2300 m. What is the correct engineering response before finalizing the runway length?
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Explanation:
Effective gradient depends on the elevation range and length of the actual runway profile. Extending or regrading the runway can change both quantities. An earlier gradient calculation may therefore no longer describe the revised alignment. The applicable aircraft-performance calculations and other runway geometric constraints must be checked for the updated proposal. Final design should be based on the actual approved runway configuration rather than an outdated preliminary estimate.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. The selected runway length needed for the intended aircraft under sea-level standard atmospheric conditions, still air and zero runway slope, before local-condition corrections. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: D. Correct the basic length for elevation, apply the temperature correction to the elevation-corrected length and then apply the stated gradient correction to the resulting length. సమాధానం: D.
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Question 3 ప్రశ్న 3Answer: A. An increase of 7% of the basic length for every 300 m of airport elevation above mean sea level. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: C. 2280 m. సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: B. 2652 m. సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: A. 11.1 degrees Celsius. సమాధానం: A.
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Question 7 ప్రశ్న 7Answer: D. 30 degrees Celsius and 36 degrees Celsius. సమాధానం: D.
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Question 8 ప్రశ్న 8Answer: C. 19.0 degrees Celsius. సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: B. 2311.2 m. సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: A. 2247 m. సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: D. The corrected length is 2736 m, corresponding to a 36.8% increase, and the general method indicates the need for a specific study. సమాధానం: D.
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Question 12 ప్రశ్న 12Answer: C. 0.60%. సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: B. 10% additional length for every 1% of the specified runway slope. సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: A. 2310 m. సమాధానం: A.
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Question 15 ప్రశ్న 15Answer: D. 2365 m. సమాధానం: D.
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Question 16 ప్రశ్న 16Answer: C. 2530 m. సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: B. 2471.7 m. సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: A. 2585.52 m. సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: D. 17.7%. సమాధానం: D.
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Question 20 ప్రశ్న 20Answer: C. 2000 m. సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: B. Effective gradient can be nonzero because it is based on the highest and lowest centreline elevations along the runway, not merely the difference between endpoints. సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: A. Do not automatically apply an unsupported negative general correction; assess the stated method and use approved aircraft-performance data for the actual conditions. సమాధానం: A.
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Question 23 ప్రశ్న 23Answer: D. 18.42 degrees Celsius. సమాధానం: D.
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Question 24 ప్రశ్న 24Answer: C. 2700 m. సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: A. Reassess the actual longitudinal profile, effective gradient and aircraft-performance requirements for the revised geometry before adopting the final runway length. సమాధానం: A.