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Attainment > AIM : Maximum endurance for a piston engine aeroplane is achieved at ?

Question 79-1 : The speed that approximately corresponds to the maximum rate of climb speed the speed for maximum lift coefficient the speed for minimum drag the speed that corresponds to the speed for maximum climb angle

.maximum endurance in a piston engine aeroplane will occur at vmp minimum power speed you want to use as little fuel as possible in order to stay airborne as longer as possible you are not concerned with distance..the best rate of climb will occur where you have excess power available that is vy best rate of climb speed and in a piston engine aeroplane that is definitely slower than speed for minimum drag vmd thus maximum endurance for a piston engine aeroplane is achieved at a speed that approximates the best rate of climb speed vmp.a common mistake is done between the trhust required curve showing drags and power required curve showing required power . 1135.for the propeller driven aircraft curve the lowest point of the curve is the tas at wich the least power is needed as opposed to producing the least drag and is therefore the best for endurance in level flight it is also the maximum rate of climb speed because the gap between power required and power available is greatest more power is needed above and below the minimum power speed exemple 179 the speed that approximately corresponds to the maximum rate of climb speed.the speed that approximately corresponds to the maximum rate of climb speed.

The maximum indicated air speed of a piston engine aeroplane without turbo ?

Question 79-2 : At the lowest possible altitude at the optimum cruise altitude at the service ceiling at the practical ceiling

.maximum ias will always be achieved at sea level because this is where density is highest the dynamic pressure is 1/2 rho v² which is sensed by the pitot system exemple 183 at the lowest possible altitude.at the lowest possible altitude.

At reference or see performance manual sep1 1 figure 2 4 .with regard to the ?

Question 79-3 : 10 kt no wind 5 kt 15 kt

.please find hereafter the full completed graph . 1072.be aware that we must assume a factored runway a piston engine airplane has to land in the 70% of the runway length .1300 ft x 0 7 = 910 ft with this distance none of the answers are correct exemple 187 10 kt.10 kt.

How does the thrust of a propeller vary during take off run assuming unstalled ?

Question 79-4 : Decreases while the aeroplane speed builds up increases while the aeroplane speed builds up varies with mass changes only has no change during take off and climb

.the angle of attack as in the case of a aircraft wing is defined as the angle between the chord line and relative airflow with a propeller however the relative airflow is the resultant of the airflow due to rotation and forward speed change either of these values and the angle of attack will change . 1079.the angle of attack decreases similar to a decreases of angle of attack for an aircraft wing less lift will be produce for a propeller it results in less thrust exemple 191 decreases while the aeroplane speed builds up.decreases while the aeroplane speed builds up.

In a given configuration the endurance of a piston engine aeroplane only ?

Question 79-5 : Altitude speed mass and fuel on board altitude speed and mass speed and mass speed mass and fuel on board

Oszklarska .the fuel on board weight is contained in airplane mass is it not .therefore the 'altitude speed and mass' answer is equivalent to 'altitude speed mass and fuel'. .no you need to know the quantity of fuel on board to be able to calculate endurance to stay airborne as longer as possible exemple 195 altitude, speed, mass and fuel on board.altitude, speed, mass and fuel on board.

At a higher gross mass on a piston engined aeroplane in order to maintain a ?

Question 79-6 : The airspeed must be increased and the drag will also increase the airspeed must be decreased and the drag will increase the lift/drag ratio must be increased the airspeed must be increased and the drag will remain unchanged

.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 and lift coefficient e g change in angle of attack do not change only speed can changed .therefore the airspeed must be increased and the drag will also increase exemple 199 the airspeed must be increased and the drag will also increase.the airspeed must be increased and the drag will also increase.

On a reciprocating engine aeroplane to maintain a given angle of attack ?

Question 79-7 : An increase in airspeed and power is required a higher coefficient of drag is required an increase in airspeed is required but power setting does not change requires an increase in power and decrease in the airspeed

.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 and lift coefficient e g change in angle of attack do not change only speed can changed .therefore power required increases airspeed will increased and the drag will also increase .notice whatever the propulsion system prop or jet it doesn't play a role as seen by its absence in the lift formula exemple 203 an increase in airspeed and power is required.an increase in airspeed and power is required.

An aeroplane with reciprocating engines is flying at a constant angle of attack ?

Question 79-8 : Remains unchanged but the tas increases increases at constant tas decreases and the cas decreases too because of the lower air density remains unchanged but the the cas increases

.with increasing altitude density rho decreases angle of attack mass and 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 exemple 207 remains unchanged but the tas increases.remains unchanged but the tas increases.

On a reciprocating engine aeroplane with increasing altitude at constant gross ?

Question 79-9 : Increases and the tas increases by the same percentage increases but tas remains constant decreases slightly because of the lower air density remains unchanged but the tas increases

.this question compares an aircraft at a certain weight and angle of attack for a first altitude and then in the same configuration at a second altitude .you have to understand that the aircraft is in straight and level flight it is not climbing.our aircraft has to produce the same lift .lift l = 1/2 rho cl s v²..for a higher altitude density reduces cl and s remain unchanged thus only tas v can and must increase .for that reason the power required must increase exemple 211 increases and the tas increases by the same percentage.increases and the tas increases by the same percentage.

At reference or see performance manual sep 1 figure 2 4 .with regard to the ?

Question 79-10 : Approximately 1850 feet approximately 1700 feet approximately 1370 feet approximately 1120 feet

. 1091 exemple 215 approximately : 1850 feet.approximately : 1850 feet.

Performance manual sep 1 figure 2 4 .with regard to the landing chart for the ?

Question 79-11 : Approximately 1930 feet approximately 950 feet approximately 1400 feet approximately 750 feet

. 1128 exemple 219 approximately : 1930 feet.approximately : 1930 feet.

For this question use annex 032 005 or performance manual sep 1 figure 2 4.with ?

Question 79-12 : Approximately 1800 feet approximately 1300 feet approximately 2000 feet approximately 1450 feet

/com en/com032 295 jpg.. oat isa +15°c so oat is +30°c at 0 ft pressure altitude.runway wet grass.landing factor for short grass is 1 15 and landing factor for wet runway is 1 15..1275 ft x 1 15 = 1466 ft..1466 ft x 1 15 = 1686 ft..approximately 1800 feet is our answer exemple 223 approximately: 1800 feet.approximately: 1800 feet.

With regard to the take off performance chart for the single engine aeroplane ?

Question 79-13 : 3200 lbs 3000 lbs 2900 lbs > 3650 lbs

.complete the graph with the data . 1093.you will find 3200 lbs exemple 227 3200 lbs.3200 lbs.

At reference or see performance manual sep 1 figure 2 2 .with regard to the ?

Question 79-14 : Approximately 2050 ft approximately 1150 ft approximately 2450 ft approximately 1260 ft

Complete the graph with the data. /com en/com032 297 jpg..you find 1900 ft the closest answer is 2050 ft exemple 231 approximately: 2050 ft.approximately: 2050 ft.

At reference or see performance manual sep 1 figure 2 1 .with regard to the ?

Question 79-15 : 71 and 82 kias 65 and 75 kias 73 and 84 kias 68 and 78 kias

/com en/com032 298 jpg. exemple 235 71 and 82 kias.71 and 82 kias.

At reference or see performance manual sep 1 figure 2 2 .with regard to the ?

Question 79-16 : Approximately 3960 ft approximately 3680 ft approximately 4150 ft approximately 5040 ft

/com en/com032 299 jpg..do not forget to apply 'grass.correction factor' .3300 ft x 1 2 = 3960 ft exemple 239 approximately: 3960 ft.approximately: 3960 ft.

With regard to the climb performance chart for the single engine aeroplane ?

Question 79-17 : 1290 ft/min 1370 ft/min 1210 ft/min 1150 ft/min

. 1129..youssef92 .for me i find that the answer is 1370ft/min. .if you start at 15°c instead of 30°c you will find around 1350 1370 ft but the question states o a t isa + 15°c .it means an outside temperature of 30°c at sea level

At reference or see performance manual sep 1 figure 2 2 .with regard to the ?

Question 79-18 : 2275 ft 1900 ft 1600 ft 2000 ft

/com en/com032 3587 jpg...1750 ft x surface factor 1 3 = 2275 ft exemple 247 2275 ft.2275 ft.

Using the landing diagram for single engine aeroplane determine the landing ?

Question 79-19 : 1350 ft 1550 ft 1020 ft 880 ft

. 1097.1320 ft close to 1350 ft exemple 251 1350 ft.1350 ft.

The pilot of a single engine aircraft has established the climb performance ?

Question 79-20 : Degraded improved unchanged unchanged if a short field take off is adopted

exemple 255 degraded.degraded.

An extract of the flight manual of a single engine propeller aircraft is ?

Question 79-21 : 465 m 615 m 540 m 395 m

. 1099.1270 kg = 2800 lbs.convert 1520 ft to meters .1520 x 0 3048 = 463 m exemple 259 465 m.465 m.

With regard to the landing chart for the single engine aeroplane determine the ?

Question 79-22 : Approximately 1700 feet approximately 1150 feet approximately 1500 feet approximately 920 feet

. 1102.you will find 1760 ft close enough to 1700 feet exemple 263 approximately : 1700 feet.approximately : 1700 feet.

With regard to the take off performance chart for the single engine aeroplane ?

Question 79-23 : Approximately 2470 feet approximately 1440 feet approximately 2800 feet approximately 2200 feet

. 1131 exemple 267 approximately : 2470 feet.approximately : 2470 feet.

For this question use reference or performance manual sep 1 figure 2 1.with ?

Question 79-24 : 3240 lbs 3000 lbs 2900 lbs greater than 3650 lbs

Complete the graph with the data. /com en/com032 359 jpg..you will find 3240 lbs exemple 271 3240 lbs.3240 lbs.

Consider the graphic representation of the power required versus true air speed ?

Question 79-25 : Maximum specific range maximum endurance maximum thrust critical angle of attack

. 1085.maximum specific range is reached at minimum drag speed vmd for a piston engined aeroplane. 1135..vmd is the speed for maximum range in a prop aircraft .vmd is the speed for maximum endurance in a jet aircraft exemple 275 maximum specific range.maximum specific range.

For this question use reference sep 1 figure 2 3 .using the climb performance ?

Question 79-26 : 15640 ft 18909 ft 20109 ft 16665 ft

At 5000 ft isa t° is 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. /com en/com032 362 jpg..closest answer is 15640 ft normally we should use the climb gradient since the question gives a speed in ias but rate of climb can also lead closely to the 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 exemple 279 15640 ft.15640 ft.

For this question use reference .using the climb performance chart for the ?

Question 79-27 : 1140 ft/min and 10 6% 1030 ft/min and 8 4% 1170 ft/min and 9 9% 1140 ft/min and 8%

Isa at 3000 ft .15° 2° x 3 = +9°c..for a given ias the true airspeed is about 2% higher than ias per 1000ft of altitude above sea level. /com en/com032 363 jpg..rate of climb 1140 ft/min .gradient of climb 10 6%. atplea .figure cap698 figure 2 3 exemple 283 1140 ft/min and 10.6%.1140 ft/min and 10.6%.

At reference or use performance manual sep 1 figure 2 1.airport characteristics ?

Question 79-28 : 1500 ft 2000 ft 1750 ft 1350 ft

/com en/com032 405 png. exemple 287 1500 ft.1500 ft.

Unless otherwise specified in the afm for a performance class b aeroplane ?

Question 79-29 : 1 05 1 03 1 15 1 1

Ecqb04 october 2017...the factor to be applied for each 1% of downslope is 5% or 1 05 exemple 291 1.051.05

Unless otherwise specified in the afm for a performance class b aeroplane ?

Question 79-30 : 1 05 1 03 1 01 1 15

Ecqb04 october 2017...the factor to be applied for each 1% of uphill slope is 5% or 1 05 exemple 295 1.051.05

The following conditions are observed at an airport runway 13 wind 140° for ?

Question 79-31 : 5 kt 18 kt 15 kt 10 kt

.vt = sin 10° x 30 = 5 kt exemple 299 5 kt.5 kt.

Maximum crosswind demonstrated is equal to 0 2 vs0 and the following conditions ?

Question 79-32 :

.350 300=50°..crosswind sin50x20=15 3kt..demonstrated crosswind 70x0 2=14kt exemple 303

Which margin above the stall speed is provided for the landing speed reference ?

Question 79-33 : 1 30 vso 1 05 vso 1 10 vso 1 2 vmca

exemple 307 1.30 vso1.30 vso

With regard to the graph for the light twin aeroplane will the accelerate and ?

Question 79-34 : No the performance will be worse than in the chart it does not matter which take off technique is being used the chart has been made for this situation performance will be better than in the chart

.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 exemple 311 no, the performance will be worse than in the chart.no, the performance will be worse than in the chart.

The critical engine inoperative ?

Question 79-35 : Increases the power required and the total drag due to the additional drag of the windmilling engine and the compensation of the yaw moment does not affect the aeroplane performance since it is independent of the power plant decreases the power required and increases the total drag due to the additional drag of the windmilling engine and the compensation of the yaw moment increases the power required and decreases the total drag due to the windmilling engine

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 propeller.to 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 engine.a 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 exemple 315 increases the power required and the total drag due to the additional drag of the windmilling engine and the compensation of the yaw moment.increases the power required and the total drag due to the additional drag of the windmilling engine and the compensation of the yaw moment.

Following a take off determined by the 50ft 15m screen height a light twin ?

Question 79-36 : 115 m 100 m 85 m it will not clear the obstacle

Climb gradient = change in height / distance travelled..change in height = gradient x distance travelled..change in height = 10 x 10000 = 100000..divide 100000 by 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 exemple 319 115 m.115 m.

A runway is contaminated with 0 5 cm of wet snow .the flight manual of a light ?

Question 79-37 : Increased unchanged reduced substantially decreased

exemple 323 increased.increased.

Following a take off limited by the 50 ft screen height a light twin climbs on ?

Question 79-38 : 105 m 90 m 75 m it will not clear the obstacle

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 m .265 160 = 105 m exemple 327 105 m.105 m.

The pilot of an aircraft has calculated a 4000 m service ceiling based on the ?

Question 79-39 : Higher than 4000 m less than 4000 m unchanged equal to 4000 m only a new performance analysis will determine if the service ceiling is higher or lower than 4000 m

exemple 331 higher than 4000 m.higher than 4000 m.

The flight manual of a light twin engine recommends two cruise power settings ?

Question 79-40 : An increase in speed fuel consumption and fuel burn/distance same speed and an increase of the fuel burn per hour and fuel burn/distance an increase in speed and fuel burn/distance but an unchanged fuel burn per hour same speed and fuel burn/distance but an increase in the fuel burn per hour

exemple 335 an increase in speed, fuel consumption and fuel-burn/distance.an increase in speed, fuel consumption and fuel-burn/distance.

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