Question 105-1 : A commercial air transport flight is to be conducted in a single engined piston aircraft at the most critical point during the flight the distance to be travelled to a safe landing place is 10 nm given the following information what is the lowest altitude the aircraft should fly over this critical ? [ Exam pilot ]

Question 105-2 : Given the following information determine the maximum defactored landing distance for a multi engine performance class b aeroplane operated for commercial air transportlanding distance available 4970 ft runway surface grass less than 20 cm long not very short factor 115 runway condition wet factor ?

2 628 ft.

For a multi engined performance class b aeroplane operated for commercial air transport the landing distance required ldr must not exceed 70% of the landing distance available lda which equates to a factor of 143for this question the lda needs to be factored down according to the field length requirements on cap 698 section 3 mep1 page 17 143 for the regulations 115 for the grass runway up to 20 cm long 115 for the wet condition of the runway no allowance is permitted for upslopetherefore the defactored landing distance will be 4970 ÷ 115 ÷ 115 ÷ 143 = 2628 ftnote all the factors are cumulative and can be applied in any order

Question 105-3 : You are the pilot of an airline whose fleet is composed of only single engine aircraft for flight planning purposes and considering the following data what distance should you use in the appropriate landing distance graph to get the maximum permissible landing mass landing distance available 3150 ft ?

1790 ft.

Refer to cap 698 section 2 data for single engine piston aeroplane sep1 52 landing requirements field length requirementsa the landing distance from a screen height of 50 ft must not exceed 70% of the landing distance available ie a factor of 143b if the landing surface is grass up to 20 cm long on firm soil the landing distance should be multiplied by a factor of 115c if the metar or taf or combination of both indicate that the runway may be wet at the estimated time of arrival the landing distance should be multiplied by a factor of 115d the landing distance should be increased by 5% for each 1% downslope no allowance is permitted for upslopee the despatch rules for scheduled planned landing calculations are in jar ops 1550 c in this exercise we should start with the given landing distance available lda and divide by each factor to obtain the defactored landing distance the factors we will be using are 143 for the 70% lda 115 for the 15% runway wet 107 for the 14% downslope 5 x 14 = 7% therefore the defactored landing distance will be 3150 ÷ 143 ÷ 115 ÷ 107 = 1790 ftnote all the factors are cumulative and can be applied in any order

Question 105-4 : With regard to the graph for the light twin aeroplane will the accelerate and stop distance be achieved in a take off where the brakes are released before take off power is set 2147 ?

No the performance will be worse than in the chart.

Admin on the associated condtions you can read 'full power before brake release'if you release the brake before the engine is in full power your acceleration will be slower a longer runway distance is required to accelerate until v1

Question 105-5 : The critical engine inoperative ?

Increases the power required and the total drag due to the additional drag of the windmilling engine and the compensation of the yaw moment.

When an engine failure occurs in a multi engine aircraft asymmetric thrust and drag produce the following effects on the aircraft's axes of rotation pitch down along the lateral axis loss of accelerated slipstream over the horizontal stabilizer produces less negative lift roll down toward the inop engine along the longitudinal axis wing produces less lift on side of failed engine due to loss of accelerated slipstream yaw toward the inop engine along the vertical axis loss of thrust and increased drag from the windmilling propellerto compensate for these effects a pilot must add additional back pressure deflect the ailerons into the operating engine and apply rudder pressure on the side of the operating enginea loss of one engine on a multi engine results in a loss of 50% of all available power and up to 80% of the aircraft's excess power and climb performance due to increased drag from the inoperative engine

Question 105-6 : Following a take off determined by the 50ft 15m screen height a light twin climbs on a 10% over the ground climb gradientit will clear a 900 m high obstacle in relation to the runway horizontally situated at 10 000 m from the 50 ft clearing point with an obstacle clearance of ?

115 m.

Climb gradient = change in height distance travelledchange in height = gradient x distance travelledchange in height = 10 x 10000 = 100000divide 100000 by 100 to express climb gradient as percentage 100000 100 = 1000 madd 15m screen height = 1015 mthe aircraft will clear a 900 m high obstacle with an obstacle clearance of 115 m

Question 105-7 : A runway is contaminated with 05 cm of wet snowthe flight manual of a light twin nevertheless authorises a landing in these conditionsthe landing distance will be in relation to that for a dry runway ?

Increased.

Question 105-8 : Following a take off limited by the 50 ft screen height a light twin climbs on a gradient of 5%it will clear a 160 m obstacle in relation to the runway horizontally situated at 5 000 m from the 50 ft point with an obstacle clearance margin of ?

105 m.

A 5% climb on 5000 m means a 250 m gain in height 5% > for 100 m horizontal distance you gain 5 m in height + 50 ft 15 m it gives 265 m265 160 = 105 m

Question 105-9 : The pilot of an aircraft has calculated a 4000 m service ceiling based on the forecast general conditions for the flight and a take off mass of 3250 kgif the take off mass is 3000 kg the service ceiling will be ?

Higher than 4000 m.

Question 105-10 : The flight manual of a light twin engine recommends two cruise power settings 65 and 75 %the 75% power setting in relation to the 65 % results in ?

Question 105-11 : At a given mass the reference stall speed of a twin engine turboprop aircraft is 100 kt in the landing configurationthe minimum speed a pilot must maintain in short final is ?

123 kt.

Question 105-12 : Given oat 25°cpressure altitude 3000 ftrwy 24lwind 310°20kttake off mass 4400 lbsheavy duty brakes installedother conditions as associated in the header of the graphwhat is the accelerate and stop distance under the conditions given 2129 ?

Question 105-13 : Use performance manual mep1 figure 32with regard to the graph for the light twin aeroplane if the brakes are released before take off power is achieved the acceleratestop distance will be err a 032 378 ?

Longer than the graphical distance.

On the associated condtions you can read 'full power before brake release'if you release the brake before the engine is in full power your acceleration will be slower a longer runway distance is required to accelerate until v1

Question 105-14 : For this question use annex ecqb 032 016 v2015 10 or performance manual mep 1 figure 37 given the following conditions what is the one engine inoperative rate of climb oat 20°cpressure altitude 14000 ftgross mass 4000 lbother conditions as associated in the header of the graph err a 032 387 ?

175 ftmin.

Img com encom032 387jpg

Question 105-15 : Use performance manual mep 1 figure 31given oat 15°cpressure altitude 4000 ftrwy 12rwind 080°12 ktstake off mass 4000 lbsother conditions as associated in the header of the graphwhat is the take off distance under the conditions given err a 032 388 ?

1550 ft.

Runway 12 120° wind from 080° it is a headwindheadwind component 12 kt x cos angle between the wind and the runway 12 x cos 40° = 92 kt com encom032 388pngwe are looking for the total take off distance over 50ft obstacle not for the ground roll distance

Question 105-16 : Use performance manual mep 1 figure 37given oat 20°cpressure altitude 18000 ftgross mass 4000 lbsmixture leaned to 25°f rich of peak egtother conditions as associated in the header of the graphwhat is the two engine rate of climb for the conditions given err a 032 389 ?

1050 ftmin.

Img com encom032 389jpg

Question 105-17 : Other conditions as associated in the header of the graph oat 10°cpressure altitude 2000 ftgross mass 3750 lbsmixture full richgiven the following conditions what is the two engine rate of climb 2124 ?

1770 ftmin.

1110you must use the longer lines for the full rich mixture setting

Question 105-18 : Use performance manual mep 1 figure 31given oat 15°cpressure altitude 4000 ftrwy 12rwind 080°12 kttake off mass 4000 lbsother conditions as associated in the header of the graphwhat is the ground roll distance under the conditions given err a 032 391 ?

1270 ft.

Runway 12 120° wind from 080° it is a headwindheadwind component 12 kt x cos angle between the wind and the runway 12 x cos 40° = 92 kt com encom032 391pngwe are only looking for the ground roll distance not for the total take off distance over 50ft obstacle

Question 105-19 : Given oat 24°cpressure altitude 3000 ftrwy 12lwind 080°12 kttake off mass 3800 lbsother conditions as associated in the header of the graphwhat is the take off distance under the conditions given 2122 ?

1700 ft.

Runway 12 120° wind from 080° it is a headwindheadwind component 12 kt x cos angle between the wind and the runway 12 x cos 40° = 92 kt 1112we are looking for the total take off distance over 50ft obstacle not for the ground roll distance

Question 105-20 : Given oat 24°cpressure altitude 3000 ftrwy 30rwind 060°4 kttake off mass 3800 lbsother conditions as associated in the header of the graphwhat is the take off distance under the conditions given 2120 ?

2000 ft.

1132 babar350 pay attention that 30r does not mean 030°m but 300°m absolutelyrunway 30r 300° wind from 060° it's a tailwind 1115tailwind component 4 kt x cos angle between the wind and the runway = 4 x cos 60° = 2 kt

Question 105-21 : Given oat 20°cpressure altitude 14000 ftgross mass 4000 lbsmixture full richother conditions as associated in the header of the graphwhat is the two engine rate of climb for the conditions given 2119 ?

1300 ftmin.

1113you must use the longer lines for the full rich mixture setting

Question 105-22 : Use performance manual mep 1 figure 31given oat 24°cpressure altitude 3000 ftrwy 12lwind 080°12 kttake off mass 3800 lbsother conditions as associated in the header of the graphwhat is the ground roll distance under the conditions given err a 032 395 ?

1350 ft.

Runway 12 120° wind from 080° it is a headwindheadwind component 12 kt x cos angle between the wind and the runway 12 x cos 40° = 92 kt com encom032 395pngwe are only looking for the ground roll distance not for the total take off distance over 50ft obstacle

Question 105-23 : Given oat 24°cpressure altitude 3000 ftrwy 30rwind 060°4 kttake off mass 3800 lbsother conditions as associated in the header of the graphwhat is the ground roll distance under the conditions given 2116 ?

1670 ft.

Runway 30r 300° wind from 060° it is a tailwind 1115tailwind component 4 kt x cos angle between the wind and the runway = 4 x cos 60° = 2 kt 1133we are only looking for the ground roll distance not for the total take off distance over 50ft obstacle

Question 105-24 : Given oat 20°cpressure altitude 2000 ftrwy 07rwind 120° 15 kttake off mass 4500 lbsheavy duty brakes installedother conditions as associated in the header of the graphwhat is the accelerate and stop distance under the conditions given 2117 ?

3450 ft.

Runway 07r 070° wind from 120° it is a headwindheadwind component 12 kt x cos angle between the wind and the runway 12 x cos 50° = 97 kt 1116accelerate and stop distance = 3700 ftbut heavy duty brakes are installed we must reduce distance by 7% 3700 x 093 = around 3450 ft

Question 105-25 : What is the accelerate and stop distance under the conditions given oat 25°cpressure altitude 3000 ftrwy 26lwind 310°20kttake off mass 4400 lbheavy duty brakes installedother conditions as associated in the header of the of the graph 2114 ?

3500 ft.

Runway 26l 260° wind from 310° it is a headwindheadwind component 20 kt x cos angle between the wind and the runway 20 x cos 50° = 13 kt 1117accelerate and stop distance = 3700 ftbut heavy duty brakes are installed we must reduce distance by 7% 3700 x 093 = around 3500 ft

Question 105-26 : What is the accelerate and stop distance under the conditions given oat 10°cpressure altitude 4000 ftrwy 12rwind 180°10 kttake off mass 4600 lbheavy duty brakes installedother conditions as associated in the header of the graph 2115 ?

3550 ft.

Head wind component 10 kt x cos angle between the wind and the runway 10 x cos 60° = 5 kt 1118accelerate and stop distance = 3800 ftbut heavy duty brakes are installed we must reduce distance by 7% 3800 x 093 = 3534 ft

Question 105-27 : Use performance manual mep 1 figure 32given oat 25°cpressure altitude 3000 ftrwy 24lwind 310°20 kttake off mass 4400 lbsheavy duty brakes installedother conditions as associated in the header of the graphwhat is the accelerate and stop distance under the conditions given err a 032 400 ?

3750 ft.

Head wind component 20 kt x cos angle between the wind and the runway 20 x cos 70° = 7 kt com encom032 400jpgaccelerate and stop distance = 4000 ftbut heavy duty brakes are installed we must reduce distance by 7% 4000 x 093 = 3720 ft

Question 105-28 : What is the accelerate and stop distance under the conditions given given oat 20°cpressure altitude 2000 ftrwy 24lwind 120° 8 kttake off mass 4500 lbheavy duty brakes installedother conditions as associated in the header of the graph 2112 ?

4200 ft.

Runway 24l 240° wind from 120° it is a tailwindtailwind component 8 kt x cos angle between the wind and the runway 8 x cos 60° = 4 kt 1120accelerate and stop distance = 4500 ftbut heavy duty brakes are installed we must reduce distance by 7% 4500 x 093 = around 4200 ft

Question 105-29 : What is the accelerate and stop distance under the conditions given oat 10°cpressure altitude 4000 ftrwy 30lwind 180°10 kttake off mass 4600 lbheavy duty brakes installedother conditions as associated in the header of the graph 2111 ?

4250 ft.

Runway 30l 300° wind from 180° it is a tailwindtailwind component 10 kt x cos angle between the wind and the runway 10 x cos 60° = 5 kt 1121accelerate and stop distance = 4580 ftbut heavy duty brakes are installed we must reduce distance by 7% 4580 x 093 = around 4250 ft

Question 105-30 : Given oat 10°cpressure altitude 2000 ftgross mass 3750 lbsother conditions as associated in the header of the graphwhat is the one engine inoperative rate of climb for the conditions given 2109 ?

430 ftmin.

1134

Question 105-31 : Given oat 0°cpressure altitude 18000 ftgross mass 3750 lbsmixture leaned to 25°f rich of peak egtother conditions as associated in the header of the graphwhat is the two engine rate of climb for the condions given 2110 ?

1050 ftmin.

1122short lines are for a lean mixture

Question 105-32 : Which engine is considered critical in the event of an engine failure during take off for an aeroplane propelled by two clockwise rotating piston engines ?

The left engine.

Ecqb03 july 2016on a jet engines aeroplane you don't have the additional 'critical engine' notion added to your engine failureexplanation clockwise rotation as viewed from the pilot's seat the critical engine will be the left engine com encom080 341jpgmulti engine aeroplanes are subject to p factor just as single engine aeroplanes are the descending propeller blade of each engine will produce greater thrust than the ascending blade when the aeroplane is operated under power and at positive angles of attack the descending propeller blade of the right engine is also a greater distance from the center of gravity and therefore has a longer moment arm than the descending propeller blade of the left engine as a result failure of the left engine will result in the most asymmetrical thrust adverse yaw as the right engine will be providing the remaining thrust

Question 105-33 : Unless otherwise specified in the afm for a performance class b aeroplane landing on a short grass runway what factor must be applied to the landing distance ?

115.

Ecqb03 july 2016

Question 105-34 : A pilot is flying a twin engine piston aeroplane with all engines operatinggiven the following information what will be the vertical clearance over the obstacle listed below mtom 4750 kgairport elevation 500 ftobstacle elevation 644 ftdistance of the obstacle from the end of the tod 2000 ?

150 ft.

Ecqb03 july 2016multi engine class b multi engine class b aircraft have a requirement to clear obstacles by 50 ft from the end of the tod up to 1500ft using net performance after which the aircraft is considered to be en routea an operator shall ensure that the take off flight path of aeroplanes with two or more engines determined in accordance with this sub paragraph clears all obstacles by a vertical margin of at least 50 ft but performance class b multi engined aircraft piston required a minimum of 4% climb gradient at take off all engine operating at 2000 m or 6560 ft we are already at a height of 35 ft4% of 6560 ft = 262 ft35 + 262 = 297 ft will be our minimum height 2000m after todobstacle height = 644 ft 500 ft = 144 ft297 ft 144 ft = 153 ft

Question 105-35 : What is the minimum obstacle clearance above obstacle given perf class bcloud base above reference 0 300 ftwind calmobstacle distance since end of todr 15000 ftobstacle height above reference 0 600 ftrate of climb all engines 1830 ftminrate of climb single engine 400 ftmintas 101 kt ?

215 ft.

Perf class b failure of the critical engine is assumed to occur where visual reference is lostthe gradient to engine failure height is the all engine gradient x 077 giving net gradient 112415000 ft x 03048 = 4572 m101 kt x 1852 = 187 kmhto climb 250 ft at 1830ftmin x 077 it takes 250 1830x077 0177 minuteat 187 kmh the distance to climb 250 ft during 0177 min will be 187 60 minutes = 312 kmmin312 x 0177 = 0552 km or 552 mit remains 4572 552 m before the obstacle 4020 mthe time taken for a distance of 4008 m is 4020 km 312 kmmin = 129 minduring this time you will climb of 129 min x 400 ftmin = 516 ft300 + 516 = 816 ft816 600 = 216 ft

Question 105-36 : During take off the third segment begins ?

When acceleration to flap retraction speed is started.

The first segment starts at 'reference zero' and ends when the gear comes upthe second segment lasts until levelling off for flap retractionthe third segment ends when ready for the enroute climbit is usually a level burst at 400 ft during which acceleration is made to climb speed flaps are retracted and power is reduced to max continuous

Question 105-37 : What is the maximum vertical speed of a three engine turbojet aeroplane with one engine inoperative n 1 and a mass of 75 000 kg using the following tas 202 ktdrag 553000nthrust per engine 300000ng = 10 ms²1 kt = 100 ftminsin angle of climb = thrust drag weight ?

+1267 ftmin.

Calculation for the climb gradient aircraft weiht in newton 750000 nthrust 2 engines = 300000 x 2 = 600000 nsin angle of climb = thrust drag weightsin angle of climb = 600000 553000 750000 = 00626in percent it is 626%rate of climb = climb gradient x tasrate of climb = 626 x 202 = 1265 ftmin

Question 105-38 : During the certification flight testing of a twin engine turbojet aeroplane the real take off distances are equal to 1547 m with all engines running 1720 m with failure of critical engine at v1 with all other things remaining unchangedthe take off distance adopted for the certification file is ?

1779 m.

Cs 25113 take off distance and takeoff run 2 115% of the horizontal distance along the take off path with all engines operating from the start of the take off to the point at which the aeroplane is 11 m 35 ft above the take off surfaceall engine take off distance is 1547 x 115 = 1779 m one engine take off distance is 1720 m1779 m is the greatest distance

Question 105-39 : For a turboprop powered aeroplane a 2200 m long runway at the destination aerodrome is expected to be 'wet' the 'dry runway' landing distance should not exceed ?

1339 m.

2200115 x 07 = 1339 mnotice 07 turboprop06 turbojetfactor 115 for a wet runway brudef isn't it asked the 'dry runway' landing distance in that case 115 shouldn't be taken into accountthe runway at destination is expected to be 'wet' you must be able to stop your aeroplane in the wet required landing distance even if at the time of arrival the runway is dry

Question 105-40 : Characteristics of a three engine turbojet aeroplane are as follows thrust = 50 000 newton engineg = 10 ms²drag = 72 569 nminimum gross gradient 2nd segment = 27%sin angle of climb = thrust drag weightthe maximum take off mass under 2nd segment conditions is ?

101 596 kg.

Sin angle of climb = thrust drag weightor weight = thrust drag sin angle of climb weight = 50000x2 72569 0027weight = 1015960 ndivide newtons by g = 101596 kgas we are calculating maximum take off mass for a path to avoid obstacles we must assume one engine out unless the question states otherwise our available thrust is only 2x50000 nnote 27% express as a decimal is 0027angle = arctan 0027 = 155°sin 155° = 0027