Aeronautical knowledge Flashcards

(66 cards)

1
Q
  • What is the load factor with a bank of 60° ?
A

2 (Load Factor = 1 / Cos ∝°)

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2
Q
  • When the thrust reversers have their best efficiency ?
A

High speed

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3
Q
  • What is the freezing point of the Jet A1 ?
A

-47 °C

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4
Q

What is the vertical speed required on a 3° descent path (ILS) ?

A

(GS / 2) *10 (ex: 140kts => 700 ft/min)

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5
Q
  • What is the great circle distance on a meridian ?
    (or any points on the same meridian and also along the equator)
A

count the amount of degrees and minutes and transform into Nm.
1 minute = 1 Nm and 1 degree = 60 Nm

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6
Q
  • What is the great circle distance on a specific parallel ?
A

count the amount of degrees and minutes and transform into Nm.
1 minute = 1 Nm and 1 degree = 60 Nm
multiply by the Cos of the latitude (ex: Cos 60° = 0,5)

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7
Q
  • What is the most sensible type of wing to crosswind and why ?
A

High swepted wing
The swepted wing facing the crosswind will create more lift thus inducing a bank roll at liftoff.

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8
Q
  • What is the only parameter that determine Vmca AND Vmcg ?
A

Engine Thrust

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9
Q

On a Mercator chart, what is the main characteristic of the meridians ?

A

Bending towards the north pole

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10
Q
  • What is the phenomenon created when a cold front overtakes a warm front ?
A

An occlusion

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11
Q

What is the TAS at FLxxx if IAS is xxx ?

A

TAS = IAS + FL/2 or 2 kts increase per 1000 feet
ex : FL200 with IAS 250kts or 2% increase per 1000 feet
=> 250 + 100 = 350 kts

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12
Q
  • What is the CG position that is less favourable concerning the fuel consumption ?
A

Forward CG

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13
Q
  • What are the criteria of a stabilized approach ?
A

a. 1 dot on Loc & GS
b. Speed between Vref & Vref+20
c. Landing configuration
d. Normal Power setting
e. Briefing & Checklist complete

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14
Q

Whait is the temperature at 39000 ft if tropopause is only above that level?

A

2deg/1000ft! 39*2=78 Standard temp is 15, so 15-78=-63 Celsius.

If it were the tropopause the answer would be -56,50C

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15
Q

Calculate distance between 30N and 45 N at 50 W on a meridian.

A

45-30=15! 15*60=900NM (1degree is 60NM)

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16
Q

Vertical speed on a 3 degree glide?

A

5*GS=ROD

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17
Q

Calculate distance from 30Wand 20W at 60 N

A

30-20=10, 1060=600NM 600NMcos60=600*0,5=300NM

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18
Q

What is the CG position, which is less favourable concerning fuel consumption?

A

Forward CG

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19
Q

What is TAS at 9300 feel altitude, with IAS of 160?

A

TASD=IAS + ALT/2 160+93/2=206,5KTS

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20
Q

On a mercator Chart a great Circle between to points is bending to the pole or to the
equator?

A

Great circle is bending to the pole, because the meridians are straight at the mercator
projection.

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21
Q

What is the parameter that determines Vmca and Vmcg?

A

Engine thrust

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22
Q

Which type of wing is more sensitive on take off and why? high swwp or low sweep?!?

A

High swept wing, because the crosswind will create more lift on 1 side inducing a bank roll
and a yaw at lift off.

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23
Q

Which instruments are affected during climb if the pitot tube is blocked and how?

A

Airspeed indicator –reading 0, stuck or act like an altitude indicator
Vertical speed indicator- 0
Altimeter- will be stuck

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24
Q

Has X-wind more influence @ take off with swept or straight wing? Explain!

A

a. Swept wing
b. Only wind perpendicular to the chord line creates lift

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25
Great circle @ northern hemisphere bent towards Northpole or equator?
This can be interpreted both ways. The top side of the curve is bent towards the northpole, but the circle bents towards the equator. Not sure what the correct answer is. But if you explain this and make a small drawing of what you mean, it is probably okay. b. Equator
26
Freezing point
a. Jet A1 : -47C b. Jet A : -40C
27
Meridian Distance =
21600 NM (360x60)
28
Rate one turn
bank angle = 15% x TAS
29
ILS minimums
IIIa : DH 50’ à 100’ RVR 200m + Auto land IIIb : DH 0’ à 50’ RVR 75m à 200m + Roll out
30
ALT correction
1Mb = 30’ b. TEMP Corr = 4 x DISA x ALT/1000 c. High à Low: look below
31
SAT =
TAT - 100 x MACH – 50
32
TAS =
6 x MACH x 100 kts à TAS = 10 x MACH NM/Min
33
Approx KTAS from Mach, multiply Machnumber by 570
Example: Mach = .77 à.77x570 = 439 KTas
34
GS
10 x DIS DME in 36 sec
35
Turn diameter
TAS/100 The radius of a standard rate turn in NM is equal to 1% of 1⁄2 the airspeed i. Example: TAS =200kts à 200:2=100 à1% of 100 = 1.0 Nm radius b. The diameter of the turn (twice the radius) is equal to 1% of the airspeed
36
App category D:
Vref 141 – 165
37
22. Wake category heavy =
+ 136 tons
38
EPR
outlet press/inlet pressure
39
N1
Fan speed, better to measure, reliable to measure, no true indication of thrust, needs only 1 sensor
40
Contaminated RWY
When more than 25% of at least 1/3rd of the runway is covered by one or more of the following: 3mm standing water or slsuh, compacted snow, dry snow, frost…25% of rwy is covered by:
41
Aquaplaning:
a. Dynamic Vp = 9 x Sqrt Psi (tire pressure) b. New tires have lower speeds: Vp = 6 x Sqrt Psi i. Main gear 200 – 205 Psi ii. Old tires Vp = 9 x Vrw= 128kts iii. New tires Vp = 6 x Vrw= 85 kts
42
Swept wing:
a. increases critical MACH number b. Reduces drag at high speed (wave drag) c. Reduced liftà at low speeds a lot of high lift divices d. Wing tip stall à loss of aileron control e. Tortial stiffening of wing needed f. Aileron reversal g. Less vulnerable for turbulence (less lift)
43
L=
1/2 * r *S * CL a. r = air density b. Wingsurface c. Lift coefficient (determined by airfoil & Angle of attack) d. LSS = 20,05 x Sqrt T e. T air temp in K (00 C = 273,15 K)
44
Normal climb gradient SID
3,3%
45
Enroute min climb gradient 1,6% Gross (1,1 Net)
Net should clear all higher terrain by 2000’ and low terrain by 1000’
46
Approach climb gradient 2,7% min
Landing gear retracted, app flaps, N-1
47
Landing climb gradient
3,2% min a. LG extended, Landing flaps, All engines
48
Act landing distance
= 60% req landing distance a. (Required = 1,67 x act) Dry b. Wet required landing distance = Dry x 1,15
49
VMCg
VMCG is the lowest speed at which the takeoff may be safely continued following an engine failure during the takeoff run. When an engine fails during the takeoff run, the thrust yawing moment will force a displacement of the airplane on the runway. If the airspeed is not high enough and hence, the rudder generated side force is not powerful enough, the airplane deviates from the runway centerline and might even veer off the runway if the asymmetrical thrust setting is maintained.
50
Explain balanced field length
a. ASDR = TODR b. The balanced field length is the shortest field length at which a balanced field takeoff can be performed.
51
TORA
Take Off Run Available Distance de piste disponible pour le roulage au décollage De début de piste → fin de piste Sans stopway ni clearway
52
TODA
Take Off Distance Available Distance disponible pour décoller + monter TODA = TORA + Clearway La clearway = zone dégagée au-delà de la piste (sans obstacle)
53
ASDA
Accelerate Stop Distance Available 👉 Distance disponible pour accélérer puis s’arrêter ASDA = TORA + Stopway La stopway = zone renforcée après la piste pour freinage d’urgence
54
LDA
Landing Distance Available Distance disponible pour l’atterrissage
55
Pitot blocked on descent, what happen to ASI (airspeed indicator)
- Constant altitude: Read don't change - Descent: indicate lower speed than the actual speed - Climb: indicate a higher speed than the actual speed
56
Density altitude
DA= PA + 120ft/C° per C° above ISA DA= PA - 120ft/C° per C° below ISA
57
Pressure altitude
PA= elevation + 30 x (1013-QNH)
58
Effect of center of gravity on fuel consumption
Aft CG decreases the fuel consumption. The elevator has to create an upward force which creates lift and reduces fuel consumption but less stable FWD CG increase fuel consumption. The elevator has to create an downward force which reduces lift and increase fuel consumption but more stable
59
Thrust reverser
More effective at high speed. More air mass passes through the engine.
60
Rate on turn
AoB = (TAS/10) + 7
61
Turn diameter
TAS/100 a. The radius of a standard rate turn in NM is equal to 1% of ½ the airspeed Example:TAS =200kts à 200:2=100 à1% of 100 = 1.0 Nm radius b. The diameter of the turn (twice the radius) is equal to 1% of the airspeed
62
Turn radius
TAS/200
63
Landing distance requirement
Dry: Required filed length = 1,67 x landing distance Wet: 1,15 x required field length for dry runway
64
Angle for crosswind component on approach
30 degrés = 50% crosswind 45 degrés 70% crosswind 60 degrés = 80-90% crosswind 90 degrés = 100% crosswind
65
ISA
ISA = 15°C; 1013,25 HPa; 29,92 Hg; Sound speed 340m/s a. -2°/ 1000ft b. 30ft/HPa
66
Symbole vent carte
Petite barre :5 kts Grande barre: 10 kts Triangle noir: 50 kts Les barres sont du côtés d'où vient le vent.