Calculation Questions Flashcards
1
Q
- 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
2
Q
- Calculating distance on the arc. 56 degrees, 15 mile arc?
A
a. 56/(60/15)
=56/4
=14 nm on arc
3
Q
- Inbound 270, standard hold, heading 030 what entry
A
a. Standard holds are right hand
Offset entry
4
Q
- Inbound hold axis 270 degrees. You are tracking a heading of 180 to the hold, what is your entry procedure?
A
Parallel
5
Q
- 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.
6
Q
- ROC required with 6% and 140 kts?
A
a. Gradient x GS x 1.013
140kts x 6 x 1.013
850 fpm
7
Q
- 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)
8
Q
- 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
9
Q
- 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
10
Q
- 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
11
Q
- 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
12
Q
- 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
13
Q
- 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
14
Q
- 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
15
Q
- Travelling at 290kts, for 58 nm, how long will it take you
A
a. 290/60=4.8nm/min
58/4.8=12mins
16
Q
- Travelling at 178kts, how far will you go in 7 minutes?
A
a. 178/60=2.9nm/min
2.9x7=20nm
17
Q
- 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
18
Q
- 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
19
Q
- 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
20
Q
- 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
21
Q
- 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
22
Q
- 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
23
Q
- 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
24
Q
- 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
25
21. Calculating head and crosswind components
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
26
22. Taking off from runway 36, with a wind of 030°/30kts. What are the headwind and crosswind components?
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
27
23. Runway 21, wind 255/40, approach TAS 120. Headwind/crosswind components and heading on final?
a. XW angular diff = 45 degrees = 70%
40x0.7=28kts
HW= 110-45= 65%
40x0.65=26kts
28
24. 120 track, wind 240/40. Calculate crosswind and drift.
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%
29
25. 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. XW angular diff = 45 degrees = 70%
40x0.7=28kts
28kts XW
heading 225 (how do you calculate a heading from crosswind component?)
30
26. Elevation 371’, QNH 998 hPa. What is the pressure altitude?
a. 1013-998=15
15x30=450
450+371= 821PA
31
27. 998hpa elevation 1171, what’s the pressure altitude?
a. 1013-998=15
15x30=450
1171+450=1621PA
32
28. 4500’ QNH 1013 ISA +10 what is your true altitude
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
33
29. Density alt Queenstown? 1171 ft elevation, 996 QNH, 23 degrees temp
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
34
30. How to write ISA-18 figures ect
35
45. Aircraft travels from A-B, 480 NM, then back to A. Speed 320 kts, wind +60. How long does it take?
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
36
47. Time at 145E is 0600 what is the time in UTC
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
37
49. 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. 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
38
50. 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. Huston UTC=1800+0500=2300UTC
Auckland UTC= xxx-1200=yyyy UTC
Elapsed time = arrival UTC- departure UTC = 14 hours
39
52. How far off track are you if 4° off after 45nm?
60/(Distance gone)=(Track Error)/(Distance off)
60/46=4/x
Finish rearranging this formula
40
53. Surface temperature is + 10 degrees. Dew point is + 5 degrees what altitude so you expect cumuliform cloud to form?
a. DALR 3 degrees/1000ft.
10-5=5
5/3=1.6
Cloud base = 1600ft
41
2. 1:1,000,000 scale how many nm is 3.8 cm?
a. 3,800,000cm
3,800,000cm/100=3800m
3800/100=38km
38/0.53 (0.53 nm per km) = 20nm
42
3. How many feet and km are in 1 nm?
a. 1nm = 6080ft
1nm = 1.82km
43
8. At 200 kts how long to fly 20nm or 10nm
a. 200kts = 3.3nm minute
20nm/3=6 minutes
10nm/3=3 minutes
44
11. Fuel calculation. It was 2500kg fuel, 500kg burn, 250kg app fuel needed how long do we have left?
a. 2500-250=2250
2250/500=4.5 hours
45
13. 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. 1650/700
Close to two hours and 20 minutes
46
14. 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. 4-1=3, therefore 1 hour hold
47
15. How to calculate rate one.
a. 10% airspeed + 7
48
21. Hold entry: What distance have you travelled 40 degrees around 15 nm arc?
a. 15 nm arc = 60/15=4
40/4=10 nm
49
22. In hold: What distance have you travelled 30 degrees around 10 nm arc?
a. 60/10=6
30/6=5nm
50
23. How many miles from threshold do we need to start descending from 4000 ft on ILS?
a. ILS is a 3 degree profile, 13nm ish (because 3x13=39)
51
1. 60 Degrees of ARC on a 12 NM ARC How many miles?
a. 60/12
= 5 (every 5 degrees = 5 nm)
60/5
=12nm
52
2. 24 NM in a 6 mins, G/S?
a. V=d/t
24/6
=4nm/min
4x60
240 kts G/S
53
3. 18NM in 27 mins. G/S?
a. V=d/t
18/27
=0.7
60x0.7
42 kts G/S
54
4. On a SID 3.3% (200ft/min) what alt do you need to be at 10 NM?
55
5. Descent form FL160 to 6000FT at 1000FT/min. How long will it take?
a. 16 000 – 6000
= 10 000ft
@1000 ft/min
= 10 minutes
56
6. Distance around a 15nm ARC?
a. 60/15
= 4 degrees per nm
57
7. G/S 170 kts. How far in have you gone in 24 mins?
a. 170/60
=2.8
24x2.8
68nm
58
8. What is 832 divided by 16?
832/16=52
59
9. Descent from 9000FT to 2000FT. With a groundspeed of 210kts and a 1000ft/min ROD what is TOD?
a. 9000-2000
=7000ft descent
@1000ft/min
=7 mins
210 kts = 3.5 nm min
7x3.5
24.5 nm TOD
60
10. ROD Required on the ILS at 170KT G/S?
61
11. You are travelling at 240KTS how many miles would you have travelled in 12 mins?
a. 240 kts = 4 nm min
4x12
=48 nms
62
12. You travel 44nm in 24mins, what is your G/S?
a. 44/24
=2
2x60
120 kts groundspeed
63
13. 48 degrees on a 15nm ARC is how many miles?
a. 60/15
=4
48/4
=12 nm
64
14. At 210kts how many miles will you travel in 16mins?
a. 210 kts = 3.5 nm/min
16x3.5
56 nm
65
15. You need to descend from 25 000 ft to 10 000 ft by 13DMEusing a profile of 3nm/1000ft. what is your TOD?
a. 25 000 – 10 000
= 15 000
15 000 / 1000
= 15
15 x 3
45 nm
+ 13
58nm TOD
66
16. Your landing speed is 120kts, on final for RWY25 the wind is reported as 300/25. What’s the crosswind?
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
17. At 230kts how many miles will you travel in 18 mins?
a. 230 kts = 4 nm/min
4x18
=72nm
68
18. Your track is 080 and the crosswind is 22kts. At 180kts TAS, How much drift do you need to apply?
69
19. At 300kts, how long will it take for you to travel 75 degrees on a 12nm ARC?
a. 60/12
=5
75/5
=15 nm
300 kts = 5 nm/min
15/5
3 minutes
70
20. What’s the headwind component on the RWY 14 if the wind is 170/30?
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
21. G/S 290kts, time to cover 58nm?
a. 290 kts = 5 nm/min
58/5
11 mins
72
22. 22nm in 12mins, G/S?
a. 22/12
2 nm/min
2x60
120 kts groundspeed
73
23. RWY 24, wind 270/30, what is the xwind?
a. 270-240
30-degree angular difference
30 x 0.5
15 kts x/wind
74
24. 210kts, time to cover 35nm?
a. 210 kts = 3.5 nm/min
35/3.5
10 mins
75
25. TAS 240kts, track 340, wind 310/30, what is the drift angle?
a. 340-310
30 degrees
76
26. 310/20, RWY 34, what is the headwind component?
a. 340-310
30 degrees
x/wind (50%)= 10 kts
h/wind (90%) = 18 kts
77
27. At 258 kts how long will it take to cover 58nm?
a. 258/60
=4.3 nm/min
58/4.3
13 mins
78
28. RWY 34, wind 010/30, x/wind?
a. 010-340
=30 degrees
x/wind (50%) = 15 kts
h/wind (90%) = 27 kts
79
29. Distance around a 12nm and 15nm arc?
a. 12 nm = 5 degrees/nm
15 nm = 4 degrees/nm
80
30. After travelling 128nm in 58mins, what’s your G/S?
a. 128/58
2.2 nm/min
X 60
132 kts groundspeed
81
31. At 240kts, how far will you travel in 58mins?
a. 240/60
= 4 nm min
X 58 mins
232 nm
82
32. On a SID 3.3% (200ft/nm), what alt do you have to be at 10nm?
83
33. G/S 170kts, how far would you have travelled in 24mins?
a. 170/60
3nm min
X 24
72 nm
84
34. What does the VSI read to maintain at 120kts g/s on a SID?
85
35. How many NM between 270 and 300 radials on a 10nm arc?
a. 30 degrees
30/6
=5 nm
86
36. What speed will an aircraft with a single wheel pressure of 100psi hydroplane?
a. 9 x √100
=9x10
90 kts
87
37. What’s the pressure alt of an airfield if the elevation 700ft and QNH is 1003hpa?
a. 1013-1003
10
X30
300 +700
PA = 1000 ft
88
38. Add these: 1:24, 2:48, 2:27 =
89
39. Tire pressure is 64psi, at what speed will the a/c aquaplane?
a. 9 x √64
9 x 8
=72 kts
90
40. Cruising at 10000ft at 180kts, what’s you TAS?
a. ? 2% for CAS for evert 1000 ft
2x10
20% faster
180x0.2
=216 kts TAS
91
41. What’s the aquaplaning speed for a tyre pressure of 121psi?
a. 9 x √121
9 x 11
99 kts
92
42. 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. 420/60
=7nm min
480/60
= 8nm min
8+7=15 nm
225/15
=15 minutes
93
43. 8kg/nm burn rate, what’s your range with 1000kgs of gas on board?
a. 1000/8
125 nm
94
44. Hydroplaning speed of a rotating wheel at 100psi?
a. 9 x √100
9 x 10
90 kts
95
45. How many ft/nm is 2.5%, 3.3%, 4.9%?
96
46. At 400kts how many nm will you travel in 24mins?
a. 400/60
=7
24x7
168nm
