Maximum endurance for a piston engine aeroplane is achieved at ?
Attainment > AIM
The maximum indicated air speed of a piston engine aeroplane without turbo charger in level flight is reached ?
At reference or see Performance Manual SEP1 1 Figure 2 4 With regard to the graph for landing performance what is the minimum headwind component required in order to land at Helgoland airport Given ?
How does thrust of a propeller vary during take off r assuming unstalled flow conditions at propeller blades the thrust Decreases while aeroplane speed builds up. the angle of attack as in case of a aircraft wing defined as angle between chord line relative airflow with a propeller however relative airflow the resultant of airflow due to rotation forward speed change either of these values the angle of attack will change the angle of attack decreases similar to a decreases of angle of attack an aircraft wing less lift will be produce a propeller it results in less thrust. 
In a given configuration endurance of a piston engine aeroplane only depends on Altitude speed mass fuel on board. oszklarska the fuel on board weight contained in airplane mass it not ? therefore 'altitude speed mas answer equivalent to 'altitude speed mass fue no you need to know quantity of fuel on board to be able to calculate endurance (to stay airborne as longer as possible). 
At a higher gross mass on a piston engined aeroplane in order to maintain a given angle of attack configuration and altitude The airspeed must be increased the drag will also increase. at a higher gross mass we need to increase lift (if we consider to maintain same altitude) lift = cl x 1/2 rho v² x s cl = lift coefficient rho = density v = tas (in m/s) s = surface since density surface lift coefficient (e g change in angle of attack) do not change only speed can changed therefore airspeed must be increased the drag will also increase. 
On a reciprocating engine aeroplane to maintain a given angle of attack configuration and altitude at higher gross mass An increase in airspeed power required. at a higher gross mass we need to increase lift (if we consider to maintain same altitude) lift = cl x 1/2 rho v² x s cl = lift coefficient rho = density v = tas (in m/s) s = surface since density surface lift coefficient (e g change in angle of attack) do not change only speed can changed therefore power required increases airspeed will increased the drag will also increase notice whatever propulsion system (prop or jet) it doesn't play a role (as seen its absence in lift formula). Question 79-8
An aeroplane with reciprocating engines flying at a constant angle of attack mass and configuration with increasing altitude drag Remains unchanged but tas increases. with increasing altitude density (rho) decreases angle of attack mass configuration remain constant to maintain lift tas has to increase drag = cd x 1/2 x rho x v² (velocity = tas) v increases while rho decreases drag remains constant. Question 79-9
On a reciprocating engine aeroplane with increasing altitude at constant gross mass angle of attack and configuration power required Increases the tas increases the same percentage. this question compares an aircraft at a certain weight angle of attack a first altitude then in same configuration at a second altitude you have to understand that aircraft in straight level flight (it not climbing) our aircraft has to produce same lift lift (l) = 1/2 rho cl s v² for a higher altitude density reduces cl s remain unchanged thus only tas (v) can (and must) increase for that reason power required must increase. Question 79-10
At reference or see performance manual sep 1 figure 2 4 with regard to landing chart the single engine aeroplane determine landing distance from a height of 50 ft given o a t 27 °cpressure altitude 3000 ftaeroplane mass 2900 lbstailwind component 5 ktflaps landing position down runway tarred and dry 2146 Approximately 85 feet. . Question 79-11
performance manual sep 1 figure 2 4 with regard to landing chart the single engine aeroplane determine landing distance from a height of 50 ftgiven o a t isa +15°cpressure altitude 0 ftaeroplane mass 2940 lbtailwind component 10 ktflaps landing position down runway tarred and dry 2145 Approximately 93 feet. . Question 79-12
for this question use annex 032 005 or performance manual sep 1 figure 2 4 with regard to landing chart the single engine aeroplane determine landing distance from a height of 50 ft given oat isa +15°cpressure altitude 0 ftaeroplane mass 2940 lbheadwind component 10 ktflaps landing position down runway short and wet grass firm soil err _a_032 295 Approximately 8 feet. Img /com_en/com032 295 jpg oat isa +15°c so oat +30°c at 0 ft pressure altitude runway wet grass landing factor short grass 1 15 landing factor wet runway 1 15 1275 ft x 1 15 = 1466 ft 1466 ft x 1 15 = 1686 ft approximately 1800 feet our answer. Question 79-13
With regard to take off performance chart the single engine aeroplane determine maximum allowable take off mass given o a t isapressure altitude 4000 ftheadwind component 5 ktflaps uprunway tarred and dryfactored runway length 2000 ftobstacle height 50 ft 2143 Approximately 8 feet. complete graph with data you will find 3200 lbs. Question 79-14
At reference or see performance manual sep 1 figure 2 2 with regard to take off performance chart the single engine aeroplane determine take off distance to a height of 50 ft given oat 7°cpressure altitude 7000 ftaeroplane mass 2950 lbsheadwind component 5 ktflaps approach settingrunway tarred and dry err _a_032 297 Approximately 2 5 ft. Complete graph with data img /com_en/com032 297 jpg you find 1900 ft closest answer 2050 ft. Question 79-15
At reference or see performance manual sep 1 figure 2 1 with regard to take off performance chart the single engine aeroplane determine take off speed 1 rotation and 2 at a height of 50 ft given o a t isa+10°cpressure altitude 5000 ftaeroplane mass 3400 lbsheadwind component 5 ktflaps uprunway tarred and dry err _a_032 298 Approximately 2 5 ft. Img /com_en/com032 298 jpg . Question 79-16
At reference or see performance manual sep 1 figure 2 2 with regard to take off performance chart the single engine aeroplane determine take off distance to a height of 50 ft given o a t 38°cpressure altitude 4000 ftaeroplane mass 3400 lbstailwind component 5 ktflaps approach settingrunway dry grasscorrection factor 1 2 err _a_032 299 Approximately 396 ft. Img /com_en/com032 299 jpg do not forget to apply 'grass correction factor' 3300 ft x 1 2 = 3960 ft. Question 79-17
With regard to climb performance chart the single engine aeroplane determine climb speed ft/min o a t isa + 15°cpressure altitude 0 ftaeroplane mass 3400 lbsflaps upspeed 100 kias 2139 Approximately 396 ft. youssef92 for me i find that answer 1370ft/min if you start at 15°c instead of 30°c you will find around 1350 1370 ft but question states o a t isa + 15°c it means an outside temperature of 30°c at sea level. Question 79-18
At reference or see performance manual sep 1 figure 2 2 with regard to take off performance chart the single engine aeroplane determine take off distance over a 50 ft obstacle height given o a t 30°cpressure altitude 1000 ftaeroplane mass 2950 lbstailwind component 5 ktflaps approach settingrunway short wet grass firm subsoil if runway surface wet grass a 1 3 factor must beused err _a_032 301 Approximately 396 ft. Img /com_en/com032 3587 jpg 1750 ft x surface factor 1 3 = 2275 ft. Question 79-19
Using landing diagram single engine aeroplane determine landing distance from a screen height of 50 ft in following conditions given pressure altitude 4000 fto a t 5°caeroplane mass 3530 lbsheadwind component 15 ktflaps downrunway tarred and drylanding gear down 2135 Approximately 396 ft. 1320 ft (close to 1350 ft). Question 79-20
The pilot of a single engine aircraft has established climb performance the carriage of an additional passenger will cause climb performance to be Approximately 396 ft. 1320 ft (close to 1350 ft). Question 79-21
An extract of flight manual of a single engine propeller aircraft reproduced in annex airport characteristics hard dry and zero slope runwayactual conditions are pressure altitude 1 500 ftoutside temperature +18°cwind component 4 knots tailwindfor a take off mass of 1 270 kg take off distance will be 2137 Approximately 396 ft. 1270 kg = 2800 lbs convert 1520 ft to meters 1520 x 0 3048 = 463 m. Question 79-22
With regard to landing chart the single engine aeroplane determine landing distance from a height of 50 ft given o a t isapressure altitude 1000 ftaeroplane mass 3500 lbstailwind component 5 ktflaps landing position down runway tarred and dry 2136 Approximately 7 feet. you will find 1760 ft (close enough to 1700 feet). Question 79-23
With regard to take off performance chart the single engine aeroplane determine take off distance to a height of 50 ft given o a t 30°cpressure altitude 1000 ftaeroplane mass 3450 lbstailwind component 2 5 ktflaps uprunway tarred and dry 2134 Approximately 247 feet. . Question 79-24
For this question use reference or performance manual sep 1 figure 2 1with regard to take off performance chart the single engine aeroplane determine maximum allowable take off mass given oat isapressure altitude 4000 ftheadwind component 5 ktflaps uprunway tarred and dryfactored runway length 2000 ftobstacle height 50 ft err _a_032 359 Approximately 247 feet. Complete graph with data img /com_en/com032 359 jpg you will find 3240 lbs. Question 79-25
Consider graphic representation of power required versus true air speed tas a piston engined aeroplane with a given mass when drawing tangent from origin point of contact a determines speed of 2133 Maximum specific range. maximum specific range reached at minimum drag speed (vmd) a piston engined aeroplane vmd the speed maximum range in a prop aircraft vmd the speed maximum endurance in a jet aircraft. Question 79-26
For this question use reference sep 1 figure 2 3 using climb performance chart the single engine aeroplane determine ground distance to reach a height of 1500 ft above reference zero in following conditions given o a t at take off isaairport pressure altitude 5000 ftaeroplane mass 3300 lbsspeed 100 kiaswind component 5 kt tailwind err _a_032 362 Maximum specific range. At 5000 ft isa t° 5°c 1500/1080= 1 39 minute ground speed 100 + 5 = 105 kt (105/60) x 1 39 = 2 43 nm 2 43 = 4 5 km = 14722 ft img /com_en/com032 362 jpg closest answer 15640 ft normally we should use climb gradient since question gives a speed in ias but rate of climb can also lead closely to correct answer ebbr1000 tas= 108 kt height difference x 100/ climb gradient= (1500 x 100)/ 9 9 = 15151 ft (15151 x gs)/ tas= (15151 x 113)/ 108 = 15852 ft. Question 79-27
For this question use reference using climb performance chart the single engine aeroplane determine rate of climb and gradient of climb in following conditions given o a t at take off isaairport pressure altitude 3000 ftaeroplane mass 3450 lbsspeed 100 kias err _a_032 363 Maximum specific range. Isa at 3000 ft 15° (2° x 3) = +9°c for a given ias true airspeed about 2% higher than ias per 1000ft of altitude above sea level img /com_en/com032 363 jpg rate of climb 1140 ft/min gradient of climb 10 6% atplea figure cap698 figure 2 3. Question 79-28
At reference or use performance manual sep 1 figure 2 1airport characteristics hard dry and runway slope zeroactual conditions are pressure altitude 1500 ftoutside temperature +18°cwind component 4 kt tailwindfor a take off mass of 2800 lbs take off distance will be err _a_032 405 Maximum specific range. Img /com_en/com032 405 png . Question 79-29
Unless otherwise specified in afm a performance class b aeroplane landing on a downhill runway what factor must be applied each 1% of downslope Maximum specific range. Ecqb04 october 2017 the factor to be applied each 1% of downslope 5% (or 1 05). Question 79-30
Unless otherwise specified in afm a performance class b aeroplane taking off on a uphill runway what factor must be applied each 1% of uphill slope Maximum specific range. Ecqb04 october 2017 the factor to be applied each 1% of uphill slope 5% (or 1 05). Question 79-31
The following conditions are observed at an airport runway 13 wind 140° 30 kt a pilot can determine a crosswind component of Maximum specific range. vt = sin 10° x 30 = 5 kt. Question 79-32
Maximum crosswind demonstrated equal to 0 2 vs0 and following conditions are observed at an airport vs0 70kt landing runway 35 wind 300° 20 kt Maximum specific range. 350 300=50° crosswind sin50x20=15 3kt demonstrated crosswind 70x0 2=14kt. Question 79-33
Which margin above stall speed provided the landing speed reference vref Maximum specific range. 350 300=50° crosswind sin50x20=15 3kt demonstrated crosswind 70x0 2=14kt. Question 79-34
With regard to graph the light twin aeroplane will accelerate and stop distance be achieved in a take off where brakes are released before take off power set 2147 No performance will be worse than in chart. on associated condtions you can read 'full power before brake release' if you release brake before engine in full power your acceleration will be slower a longer runway distance required to accelerate until v1. Question 79-35
The critical engine inoperative Increases power required the total drag due to additional drag of windmilling engine the compensation of yaw moment. When an engine failure occurs in a multi engine aircraft asymmetric thrust drag produce following effects on aircraft's axes of rotation pitch down along lateral axis loss of accelerated slipstream over horizontal stabilizer produces less negative lift roll down toward inop engine along longitudinal axis wing produces less lift on side of failed engine due to loss of accelerated slipstream yaw toward inop engine along vertical axis loss of thrust increased drag from windmilling propeller to compensate these effects a pilot must add additional back pressure deflect ailerons into operating engine apply rudder pressure on side of operating engine a loss of one engine on a multi engine results in a loss of 50% of all available power up to 80% of aircraft's excess power climb performance due to increased drag from inoperative engine. Question 79-36
Following a take off determined the 50ft 15m screen height a light twin climbs on a 10% over ground climb gradient it will clear a 900 m high obstacle in relation to runway horizontally situated at 10 000 m from 50 ft clearing point with an obstacle clearance of Increases power required the total drag due to additional drag of windmilling engine the compensation of yaw moment. Climb gradient = change in height / distance travelled change in height = gradient x distance travelled change in height = 10 x 10000 = 100000 divide 100000 100 to express climb gradient as percentage 100000 / 100 = 1000 m add 15m screen height = 1015 m the aircraft will clear a 900 m high obstacle with an obstacle clearance of 115 m. Question 79-37
A runway contaminated with 0 5 cm of wet snow the flight manual of a light twin nevertheless authorises a landing in these conditions the landing distance will be in relation to that a dry runway Increases power required the total drag due to additional drag of windmilling engine the compensation of yaw moment. Climb gradient = change in height / distance travelled change in height = gradient x distance travelled change in height = 10 x 10000 = 100000 divide 100000 100 to express climb gradient as percentage 100000 / 100 = 1000 m add 15m screen height = 1015 m the aircraft will clear a 900 m high obstacle with an obstacle clearance of 115 m. Question 79-38
Following a take off limited the 50 ft screen height a light twin climbs on a gradient of 5% it will clear a 160 m obstacle in relation to runway horizontally situated at 5 000 m from 50 ft point with an obstacle clearance margin of Increases power required the total drag due to additional drag of windmilling engine the compensation of yaw moment. A 5% climb on 5000 m means a 250 m gain in height (5% > 100 m horizontal distance you gain 5 m in height) + 50 ft (15 m) it gives 265 m 265 160 = 105 m. Question 79-39
The pilot of an aircraft has calculated a 4000 m service ceiling based on forecast general conditions the flight and a take off mass of 3250 kgif take off mass 3000 kg service ceiling will be Increases power required the total drag due to additional drag of windmilling engine the compensation of yaw moment. A 5% climb on 5000 m means a 250 m gain in height (5% > 100 m horizontal distance you gain 5 m in height) + 50 ft (15 m) it gives 265 m 265 160 = 105 m. Question 79-40
The flight manual of a light twin engine recommends two cruise power settings 65 and 75 % the 75% power setting in relation to 65 % results in An increase in speed fuel consumption fuel burn/distance. A 5% climb on 5000 m means a 250 m gain in height (5% > 100 m horizontal distance you gain 5 m in height) + 50 ft (15 m) it gives 265 m 265 160 = 105 m.
Exclusive rights reserved. Reproduction prohibited under penalty of prosecution.
3119 Free Training Exam