Calculation Questions Flashcards

1
Q
  1. If a hold inbound is 300 and left-hand hold, if flying 030 what entry type?
A

a. Calculating entry, out bound +70 gives you sectors.
120+70
=190
HDG 030, entry is parallel

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2
Q
  1. Calculating distance on the arc. 56 degrees, 15 mile arc?
A

a. 56/(60/15)
=56/4
=14 nm on arc

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3
Q
  1. Inbound 270, standard hold, heading 030 what entry
A

a. Standard holds are right hand
Offset entry

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4
Q
  1. Inbound hold axis 270 degrees. You are tracking a heading of 180 to the hold, what is your entry procedure?
A

Parallel

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5
Q
  1. Draw a hold and work out 1,2,3 sector entry.
A

Standard hold is right hand.
For calculating entries, left holds, LADD, add 70 degrees to inbound. Right hand holds, -70 from inbound.

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6
Q
  1. ROC required with 6% and 140 kts?
A

a. Gradient x GS x 1.013
140kts x 6 x 1.013
850 fpm

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7
Q
  1. Your 50 nm from VOR at FL290, you need to descend to 8000 ft by VOR, at 271 kts GS, what is minimum ROD required?
A

a. 271 kts = 4.5 nm/min
Traveling for 50 nm will take you (50/4.5) 11 mins
Height to descend (29 000 – 8000 = 21 000) 21 000 in 11 minutes (21 000/11) = 1909 ft/min RoD (answers all 100ft of each other)

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8
Q
  1. 6% profile on VOR approach, at 140 kts, what is the ROD required?
A

a. Gradient x GS x 1.013
6 x 140 x 1.013
850 fpm

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9
Q
  1. While flying an ILS with a 3 degree path. Your ground speed decreases from 160kts to 130kts. By how much should you decrease your rate of descent?
A

a. ROD @ 160kts = 800ft/min
ROD @ 130kts = 650ft/min
800-650 = 150 ft
150fpm

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10
Q
  1. Flying an approach with a 6% gradient at 140kts groundspeed what ROD is required to maintain profile
A

a. Climb gradient x GS x 1.013
6 x 140 x 1.013
=850

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11
Q
  1. Your ground speed is 250 knots and you are at FL160. If you need to be overhead the VOR at a 1000’ and your TOD is 50nm, what’s the ROD required?
A

a. 16,000 – 1,000 = 15,000 to lose
Time = Distance / Speed
50 / 250 = 0.2
0.2 x 60 = 12 minutes
15,000 ft / 12 min = 1,250 fpm

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12
Q
  1. It takes 22 minutes to climb to altitude in still air conditions, you cover 66 nm in this time. If there was a 35kt head wind, what distance would you cover?
A

a. 66/22=3
3x60 = 180 kts (GS)
180-35 kts head wind = 145 kts head wind
145/60 = 2.4 nm/minute traveling for 22 minutes
Cover 53 nm

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13
Q
  1. Climb to FL190 22min 110nm covered with planned 79kt headwind, what is the distance covered with 30 kt headwind
A

a. 110/22=5
5x60=300 kts (GS)
(79-30)=49kts
300+49=349
349/60 = 5.8
5.8x22
128nm

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14
Q
  1. An aircraft flight planning chart states that the time to reach FL190 at a given weight is 22 mins with a still air distance of 66nm. What will be the distance travelled with an average head wind component of 35 kts?
A

a. 66/22=3 nm/min
3x60=180 kts G/S
180kts-35kts=145kts new G/S
145/60=2.41 nm/min
2.41x22=53nm

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15
Q
  1. Travelling at 290kts, for 58 nm, how long will it take you
A

a. 290/60=4.8nm/min
58/4.8=12mins

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16
Q
  1. Travelling at 178kts, how far will you go in 7 minutes?
A

a. 178/60=2.9nm/min
2.9x7=20nm

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17
Q
  1. 480nm distance +60 kts headwind on the way there, 380 TAS. How long does it take?
A

a. 380-60=320 kts GS
320/60=5.3nm/min
480/5.3=90min
2hours 30 min

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18
Q
  1. Planned groundspeed =240kts over a 480nm trip. Dep at 1000hrs. After 150nm, your 2 mins late. What is your revised ETA using a new G/S?
A

d. 480/150=3.2.
If you are 2 minutes late at 1/3 of the way, you will be six minutes late in total.
AETA
1206

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19
Q
  1. An aircraft travelling a distance of 450nm from A to B. After covering 150nm, is two mins behind ETA. At what time will it reach B? Departed A at 1000. GS- 240 Kts
A

a. 450/150=3
6 minutes behind in total
240/60=4nm minute
450/4=112minutes
ATD 1000 +112=ETA 1152
+6 minutes AETA= 1158

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20
Q
  1. Flying A – B (dept time = 10.40 am & total distance is 430 NM): following 55 nm you arrive 2 mins later than expected, what is your new ETA at B?
A

a. 430/55=7.8
7.8x2=15.6 minutes late in total

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21
Q
  1. Aircraft is flying from A-B, Distance is 480nm, TAS of 380kts. From A-B there is a 60kt headwind, wind remains constant. How long will it take to go from A-B and back to A
A

a. GS A-B = 380 kts – 60 kts = 320 kts GS
320/60=5.3
480/5.3=90 minutes A-B
GS B-A= 380+60=440kts
440/60=7.3
480/7.3=65minutes
Total elapsed time = 90+65=155minutes

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22
Q
  1. One aircraft is indicating 7000ft with QNH 1023 set and another aircraft is indicating 8000ft with QNH 1013 set. What is the height difference between the two aircraft
A

b. 1023-1013=10
10x30
300ft variant, so 700 ft in separation

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23
Q
  1. Climb to FL190 22min 110nm covered with planned 70kt headwind, what is the distance covered with 30kt headwind
A

a. 110/22=5nm/min
5x60=300kt GS
300+(70-30)=340kts new GS
340kts/60=5.6nm minute
5.6x22minutes=124nm

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24
Q
  1. Descent from FL270 -To FL160, G/S 256kts and ROD 800ft/min. You need to be 6nm from the VOR station by bottom of descent. How far away do you start your descent?
A

a. 27000-16000=11000ft
11000/800=13.75minutes
256/60=4.28nm/min
4.28x13.75=59nm
59+6=65nm TOD

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25
Q
  1. Calculating head and crosswind components
A

a. 30 degrees off of runway heading is 90% HW and 50% XW
40 degrees off of runway heading is 70% HW and 60% XW
50 degrees off of runway heading is 60% HW and 70% XW
60 degrees off of runway heading is 50% HW and 90% XW
If you use 110-degrees off you will get the head wind percentage
110-40 degrees off = 70% headwind component

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26
Q
  1. Taking off from runway 36, with a wind of 030°/30kts. What are the headwind and crosswind components?
A

a. XW = Angular difference = 30 degrees = 50% crosswind (clock code method, eg. 30 degrees = 30 minutes on a clock which = 50%)
XW=15 kts
HW = 110-30 = 80%
30kts x 80%
HW=24kts

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27
Q
  1. Runway 21, wind 255/40, approach TAS 120. Headwind/crosswind components and heading on final?
A

a. XW angular diff = 45 degrees = 70%
40x0.7=28kts
HW= 110-45= 65%
40x0.65=26kts

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28
Q
  1. 120 track, wind 240/40. Calculate crosswind and drift.
A

a. XW use track recipical
120=300
Angular difference = 60 degrees
Tailwind = 110-60 = 50%, 40x0.5 = 20 kts
X/Wind = 60 degrees = 90 % = 40 x 0.9 = 36 kts
= 90%

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29
Q
  1. An aircraft is conducting an ILS to runway 21. Surface wind 255/40 and the approach speed is 120kts. What is the XW component, and required heading for this approach?
A

a. XW angular diff = 45 degrees = 70%
40x0.7=28kts
28kts XW
heading 225 (how do you calculate a heading from crosswind component?)

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30
Q
  1. Elevation 371’, QNH 998 hPa. What is the pressure altitude?
A

a. 1013-998=15
15x30=450
450+371= 821PA

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31
Q
  1. 998hpa elevation 1171, what’s the pressure altitude?
A

a. 1013-998=15
15x30=450
1171+450=1621PA

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32
Q
  1. 4500’ QNH 1013 ISA +10 what is your true altitude
A

a. PA=4500ft
ISA @ 4500ft=15-(2x4.5=9)=6
Local temp = (6+10) = 16 degrees ambient
16-6=10 is temp diff between ambient and ISA
10x120
1200+4500ft
DA=5780ft

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33
Q
  1. Density alt Queenstown? 1171 ft elevation, 996 QNH, 23 degrees temp
A

a. 1013-996=17
17x30=510
510+1171=1681 PA
15-(2x1.5)=12
23-12=11
11x120=1320
1681+1320=3001 DA
b. Answer wasn’t exact, had to round to 3400ft which was closest

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34
Q
  1. How to write ISA-18 figures ect
A
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35
Q
  1. Aircraft travels from A-B, 480 NM, then back to A. Speed 320 kts, wind +60. How long does it take?
A

a. 3 hours
G/S A-B=380kts
380/60=6.3nm/min
480/6.3=76min
G/S B-A= 260kts
260/60=4.3nm/min
480/4.3=111mins
EET= 111+76=187 mins, 3 hours, 7 mins

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36
Q
  1. Time at 145E is 0600 what is the time in UTC
A

a. 145/15=9.6 = 9 hours, 36 minutes
East is ahead of time, so minus 9 hours 36 minutes from 0600
2024 UTC and back one day

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37
Q
  1. An aircraft departs IAH at xxxx local on Wednesday and lands at Auckland 0605 local Friday morning. IAH is UTC -5 and Auckland is UTC + 12. How long is the flight?
A

a. always reference UTC, Get both arrival and departures into UTC and work the difference between the UTC times.
0605-1200=1805
Time of arrival-time of departure = Elapsed time

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38
Q
  1. Take off out of Houston at 1800 local time (UTC - 5) and arrive in Auckland at xxx local time (UTC+12) – what is the total flight time?
A

a. Huston UTC=1800+0500=2300UTC
Auckland UTC= xxx-1200=yyyy UTC
Elapsed time = arrival UTC- departure UTC = 14 hours

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39
Q
  1. How far off track are you if 4° off after 45nm?
A

60/(Distance gone)=(Track Error)/(Distance off)
60/46=4/x
Finish rearranging this formula

40
Q
  1. Surface temperature is + 10 degrees. Dew point is + 5 degrees what altitude so you expect cumuliform cloud to form?
A

a. DALR 3 degrees/1000ft.
10-5=5
5/3=1.6
Cloud base = 1600ft

41
Q
  1. 1:1,000,000 scale how many nm is 3.8 cm?
A

a. 3,800,000cm
3,800,000cm/100=3800m
3800/100=38km
38/0.53 (0.53 nm per km) = 20nm

42
Q
  1. How many feet and km are in 1 nm?
A

a. 1nm = 6080ft
1nm = 1.82km

43
Q
  1. At 200 kts how long to fly 20nm or 10nm
A

a. 200kts = 3.3nm minute
20nm/3=6 minutes
10nm/3=3 minutes

44
Q
  1. Fuel calculation. It was 2500kg fuel, 500kg burn, 250kg app fuel needed how long do we have left?
A

a. 2500-250=2250
2250/500=4.5 hours

45
Q
  1. Fuel calculation in the hold I got asked by ‘controller’ what my hold endurance was. Ask second pilot fuel on board and fuel flow 2000kg @ 700kg/hr, 350kg reserve .
A

a. 1650/700
Close to two hours and 20 minutes

46
Q
  1. Entry plus outbound. How many laps of the hold can we do with 4 Tonne of fuel and burning 3 tonne an hour (1 tonne reserves)
A

a. 4-1=3, therefore 1 hour hold

47
Q
  1. How to calculate rate one.
A

a. 10% airspeed + 7

48
Q
  1. Hold entry: What distance have you travelled 40 degrees around 15 nm arc?
A

a. 15 nm arc = 60/15=4
40/4=10 nm

49
Q
  1. In hold: What distance have you travelled 30 degrees around 10 nm arc?
A

a. 60/10=6
30/6=5nm

50
Q
  1. How many miles from threshold do we need to start descending from 4000 ft on ILS?
A

a. ILS is a 3 degree profile, 13nm ish (because 3x13=39)

51
Q
  1. 60 Degrees of ARC on a 12 NM ARC How many miles?
A

a. 60/12
= 5 (every 5 degrees = 5 nm)
60/5
=12nm

52
Q
  1. 24 NM in a 6 mins, G/S?
A

a. V=d/t
24/6
=4nm/min
4x60
240 kts G/S

53
Q
  1. 18NM in 27 mins. G/S?
A

a. V=d/t
18/27
=0.7
60x0.7
42 kts G/S

54
Q
  1. On a SID 3.3% (200ft/min) what alt do you need to be at 10 NM?
A
55
Q
  1. Descent form FL160 to 6000FT at 1000FT/min. How long will it take?
A

a. 16 000 – 6000
= 10 000ft
@1000 ft/min
= 10 minutes

56
Q
  1. Distance around a 15nm ARC?
A

a. 60/15
= 4 degrees per nm

57
Q
  1. G/S 170 kts. How far in have you gone in 24 mins?
A

a. 170/60
=2.8
24x2.8
68nm

58
Q
  1. What is 832 divided by 16?
A

832/16=52

59
Q
  1. Descent from 9000FT to 2000FT. With a groundspeed of 210kts and a 1000ft/min ROD what is TOD?
A

a. 9000-2000
=7000ft descent
@1000ft/min
=7 mins
210 kts = 3.5 nm min
7x3.5
24.5 nm TOD

60
Q
  1. ROD Required on the ILS at 170KT G/S?
A
61
Q
  1. You are travelling at 240KTS how many miles would you have travelled in 12 mins?
A

a. 240 kts = 4 nm min
4x12
=48 nms

62
Q
  1. You travel 44nm in 24mins, what is your G/S?
A

a. 44/24
=2
2x60
120 kts groundspeed

63
Q
  1. 48 degrees on a 15nm ARC is how many miles?
A

a. 60/15
=4
48/4
=12 nm

64
Q
  1. At 210kts how many miles will you travel in 16mins?
A

a. 210 kts = 3.5 nm/min
16x3.5
56 nm

65
Q
  1. You need to descend from 25 000 ft to 10 000 ft by 13DMEusing a profile of 3nm/1000ft. what is your TOD?
A

a. 25 000 – 10 000
= 15 000
15 000 / 1000
= 15
15 x 3
45 nm
+ 13
58nm TOD

66
Q
  1. Your landing speed is 120kts, on final for RWY25 the wind is reported as 300/25. What’s the crosswind?
A

a. 300-250
50-degree difference
Using clock code methos. 50 degrees = 50 minutes on watch face, so 5/6ths so 80%
25 kts x 0.8
20 kts x/w

67
Q
  1. At 230kts how many miles will you travel in 18 mins?
A

a. 230 kts = 4 nm/min
4x18
=72nm

68
Q
  1. Your track is 080 and the crosswind is 22kts. At 180kts TAS, How much drift do you need to apply?
A
69
Q
  1. At 300kts, how long will it take for you to travel 75 degrees on a 12nm ARC?
A

a. 60/12
=5
75/5
=15 nm
300 kts = 5 nm/min
15/5
3 minutes

70
Q
  1. What’s the headwind component on the RWY 14 if the wind is 170/30?
A

a. 170-140
= 30 degrees
Clock code 30 minutes = half
30 kts/2
15 x/wind
30 degrees = 90% h/wind
27 kts head wind

71
Q
  1. G/S 290kts, time to cover 58nm?
A

a. 290 kts = 5 nm/min
58/5
11 mins

72
Q
  1. 22nm in 12mins, G/S?
A

a. 22/12
2 nm/min
2x60
120 kts groundspeed

73
Q
  1. RWY 24, wind 270/30, what is the xwind?
A

a. 270-240
30-degree angular difference
30 x 0.5
15 kts x/wind

74
Q
  1. 210kts, time to cover 35nm?
A

a. 210 kts = 3.5 nm/min
35/3.5
10 mins

75
Q
  1. TAS 240kts, track 340, wind 310/30, what is the drift angle?
A

a. 340-310
30 degrees

76
Q
  1. 310/20, RWY 34, what is the headwind component?
A

a. 340-310
30 degrees
x/wind (50%)= 10 kts
h/wind (90%) = 18 kts

77
Q
  1. At 258 kts how long will it take to cover 58nm?
A

a. 258/60
=4.3 nm/min
58/4.3
13 mins

78
Q
  1. RWY 34, wind 010/30, x/wind?
A

a. 010-340
=30 degrees
x/wind (50%) = 15 kts
h/wind (90%) = 27 kts

79
Q
  1. Distance around a 12nm and 15nm arc?
A

a. 12 nm = 5 degrees/nm
15 nm = 4 degrees/nm

80
Q
  1. After travelling 128nm in 58mins, what’s your G/S?
A

a. 128/58
2.2 nm/min
X 60
132 kts groundspeed

81
Q
  1. At 240kts, how far will you travel in 58mins?
A

a. 240/60
= 4 nm min
X 58 mins
232 nm

82
Q
  1. On a SID 3.3% (200ft/nm), what alt do you have to be at 10nm?
A
83
Q
  1. G/S 170kts, how far would you have travelled in 24mins?
A

a. 170/60
3nm min
X 24
72 nm

84
Q
  1. What does the VSI read to maintain at 120kts g/s on a SID?
A
85
Q
  1. How many NM between 270 and 300 radials on a 10nm arc?
A

a. 30 degrees
30/6
=5 nm

86
Q
  1. What speed will an aircraft with a single wheel pressure of 100psi hydroplane?
A

a. 9 x √100
=9x10
90 kts

87
Q
  1. What’s the pressure alt of an airfield if the elevation 700ft and QNH is 1003hpa?
A

a. 1013-1003
10
X30
300 +700
PA = 1000 ft

88
Q
  1. Add these: 1:24, 2:48, 2:27 =
A
89
Q
  1. Tire pressure is 64psi, at what speed will the a/c aquaplane?
A

a. 9 x √64
9 x 8
=72 kts

90
Q
  1. Cruising at 10000ft at 180kts, what’s you TAS?
A

a. ? 2% for CAS for evert 1000 ft
2x10
20% faster
180x0.2
=216 kts TAS

91
Q
  1. What’s the aquaplaning speed for a tyre pressure of 121psi?
A

a. 9 x √121
9 x 11
99 kts

92
Q
  1. If 2 aircraft were approaching each other head on and are 225nm apart at 0000utc. What time will they pass if aircraft 1 is doing 420kts and aircraft 2 is doing 480kts?
A

a. 420/60
=7nm min
480/60
= 8nm min
8+7=15 nm
225/15
=15 minutes

93
Q
  1. 8kg/nm burn rate, what’s your range with 1000kgs of gas on board?
A

a. 1000/8
125 nm

94
Q
  1. Hydroplaning speed of a rotating wheel at 100psi?
A

a. 9 x √100
9 x 10
90 kts

95
Q
  1. How many ft/nm is 2.5%, 3.3%, 4.9%?
A
96
Q
  1. At 400kts how many nm will you travel in 24mins?
A

a. 400/60
=7
24x7
168nm