Calculation Questions Flashcards
- If a hold inbound is 300 and left-hand hold, if flying 030 what entry type?
a. Calculating entry, out bound +70 gives you sectors.
120+70
=190
HDG 030, entry is parallel
- Calculating distance on the arc. 56 degrees, 15 mile arc?
a. 56/(60/15)
=56/4
=14 nm on arc
- Inbound 270, standard hold, heading 030 what entry
a. Standard holds are right hand
Offset entry
- Inbound hold axis 270 degrees. You are tracking a heading of 180 to the hold, what is your entry procedure?
Parallel
- Draw a hold and work out 1,2,3 sector entry.
Standard hold is right hand.
For calculating entries, left holds, LADD, add 70 degrees to inbound. Right hand holds, -70 from inbound.
- ROC required with 6% and 140 kts?
a. Gradient x GS x 1.013
140kts x 6 x 1.013
850 fpm
- 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. 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)
- 6% profile on VOR approach, at 140 kts, what is the ROD required?
a. Gradient x GS x 1.013
6 x 140 x 1.013
850 fpm
- 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. ROD @ 160kts = 800ft/min
ROD @ 130kts = 650ft/min
800-650 = 150 ft
150fpm
- Flying an approach with a 6% gradient at 140kts groundspeed what ROD is required to maintain profile
a. Climb gradient x GS x 1.013
6 x 140 x 1.013
=850
- 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. 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
- 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. 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
- Climb to FL190 22min 110nm covered with planned 79kt headwind, what is the distance covered with 30 kt headwind
a. 110/22=5
5x60=300 kts (GS)
(79-30)=49kts
300+49=349
349/60 = 5.8
5.8x22
128nm
- 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. 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
- Travelling at 290kts, for 58 nm, how long will it take you
a. 290/60=4.8nm/min
58/4.8=12mins
- Travelling at 178kts, how far will you go in 7 minutes?
a. 178/60=2.9nm/min
2.9x7=20nm
- 480nm distance +60 kts headwind on the way there, 380 TAS. How long does it take?
a. 380-60=320 kts GS
320/60=5.3nm/min
480/5.3=90min
2hours 30 min
- 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?
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
- 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. 450/150=3
6 minutes behind in total
240/60=4nm minute
450/4=112minutes
ATD 1000 +112=ETA 1152
+6 minutes AETA= 1158
- 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. 430/55=7.8
7.8x2=15.6 minutes late in total
- 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. 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
- 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
b. 1023-1013=10
10x30
300ft variant, so 700 ft in separation
- Climb to FL190 22min 110nm covered with planned 70kt headwind, what is the distance covered with 30kt headwind
a. 110/22=5nm/min
5x60=300kt GS
300+(70-30)=340kts new GS
340kts/60=5.6nm minute
5.6x22minutes=124nm
- 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. 27000-16000=11000ft
11000/800=13.75minutes
256/60=4.28nm/min
4.28x13.75=59nm
59+6=65nm TOD
- 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
- 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
- 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
- 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%
- 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?)
- Elevation 371’, QNH 998 hPa. What is the pressure altitude?
a. 1013-998=15
15x30=450
450+371= 821PA
- 998hpa elevation 1171, what’s the pressure altitude?
a. 1013-998=15
15x30=450
1171+450=1621PA
- 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
- 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
- How to write ISA-18 figures ect
- 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
- 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
- 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
- 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