Questions: Navigation in Climb and Descent

---

---

► An aircraft maintaining a 5.2% gradient is at 7 NM from the runway, on a flat terrain; its height is approximately:

Answer: 2210FT

Height difference = Distance travelled (feet) x (Gradient / 100)
Height difference = (7 x 6076) x (5.2 / 100)
Height difference = 2211FT

---

► An aircraft is descending down a 12% slope whilst maintaining a GS of 540 kt. The rate of descent of the aircraft is approximately:

Answer: 6500 FT/MIN

Rate of descent = Gradient (%) x GS
12 x 540 = 6480 ft/min

---

► An aircraft at FL390 is required to descend to cross a DME facility at FL70 . Maximum rate of descent is 2500 FT/MIN. Mean GS during descent is 248 kt. What is the minimum range from the DME at which descent should be commenced?

Answer: 53 NM

(Ground distance x 60) / (Groundspeed) = Time = (Altitude difference) / (Vertical speed)

Altitude difference (32000ft) / VS (2500ft/min) = Time (12,8 )
Time (12,8 ) = Distance x 60 / 248kt
Distance = 53 NM

(Or just simply D = V x T , just remember that your time in these questions (12,8 ) is in minutes so make sure to divide by 60 or work with your degrees, minutes and seconds function on your calculator)

---

► An aircraft at FL330 is required to commence descent when 65 NM from a VOR and to cross the VOR at FL100. The mean GS during the descent is 330 kt. What is the minimum rate of descent required?

Answer: 1950 FT/MIN

(Ground distance x 60) / (Groundspeed) = Time = (Altitude difference) / (Vertical speed)

65NM x 60 / 330 = Time (11,8 )
Time (11,8 ) = altitude difference (23000ft) / VS
VS = 1950 ft/min

---

► Assuming zero wind, what distance will be covered by an aircraft descending 15000 FT with a TAS of 320 kt and maintaining a rate of descent of 3000 FT/MIN?

Answer: 26.7 NM

Altitude (15 000) / VS (3000) = 5 minutes
Distance = 5 minutes (5 / 60) x 320kt
Distance = 26,7 NM

---

► Given: ILS GP angle = 3.5 DEG, GS = 150 kt. What is the approximate rate of descent?

Answer: 875 FT/MIN

When the GP angle is 3° we can use: 5 x GS
When the GP angle is different than 3° we can use: 5 x GS x (GP angle/3)

5 x 150 x (3.5/3) = 875 FT/MIN

---

► Given: aircraft height 2500 FT, ILS GP angle 3°. At what approximate distance from THR can you expect to capture the GP?

Answer: 8.3 NM

Gradient (GP angle) = Ch altitude x 60 / Ground distance (ft)
3° = 2500ft x 60 / Ground distance (ft)
Ground Distance (ft) = 150.000 / 3
Ground Distance (ft) = 50.000ft (=8.22 NM)

---

► You are required to descend from FL 230 to FL 50 over a distance of 32 NM in 7 minutes. What is the required TAS when you expect WC-25 during the descent?

Answer: 300 kt

V = D / T
V = 32NM / 7min
V = 274 kt (GS)

274 GS (+25) = 299kt TAS

---

► You are required to descend from FL 230 to FL 50 over a distance of 32 NM in 7 minutes. What is the required rate of descent when you expect WC-25 during the descent?

Answer: 2570 ft/min

Turn it into a gradient:

Gradient (FT/NM) = Ch Alt (feet) / Ground Distance (NM)
Gradient = 18 000 / 32
Gradient = 562.5 ft/nm

Now find VS from gradient and GS (we can find GS as we did above):

VS = Gradient x (GS/60)
VS = 562.5 x (274/60)
VS = 2569

---

► You are required to descend from FL 230 to FL 50 over a distance of 32 NM in 7 minutes. What will the glideslope be when you expect WC-25 during the descend?

Answer: 5.29°

Gradient (as GP °) = Ch alt (ft) x 60 / Ground distance (ft)
Gradient = 18000 x 60 / (32 x 6080)
Gradient = 5.5 (approx 5.30°)

---

► What is the average TAS climbing from 2000ft up to FL 120 at standard temperatures, given a CAS 185 KT and QNH 1013?

Answer: 210 kt

Note: QNH is 1013 so we don't have to correct for pressure

To find the average TAS in a climb, we take it at 2/3 altitude change:

(2000ft + (2/3 x 10 000ft)) = Approx 8700ft

ISA Temperature at 8700ft is -2.4°C

Align -2.4°C with 8700 on the navigation computer. Find CAS 185 on inner scale and read TAS approx 208kt on outer scale (210 being closest answer)

---

► An aircraft is descending from FL 270 to FL 100 following MT 054° and maintaining CAS 250 KT. Given are variation 13°E, temperatures ISA-10°C, W/V 020/60. What is your GS?

Answer: 277 kt

- Find average TAS

For a descent we take it halfway so at 18 500ft and with temperature -32°C (ISA at 18.500 - 10°C)

Via the navigation computer we align 18.500ft with -32°C , look at 250kt CAS at inner scale and find TAS 325kt at outer scale

- Use TAS to find GS

Now we can use the values to calculate the GS on the wind side of the nav computer;

TT = 054 + 13 = 67°
Wind = 020/60
TAS = 325kt

We find GS (in circle) to be just below 280kt, so approx 277kt is the closest

---

► Due to pressurisation problems you are requested to descent with 1000 ft/min only from FL 120 down to FL 50 maintaining a CAS 200 kt. What descent profile will you follow at no wind conditions and standard temperature?

Answer: 2.5°

Alt Ch (7000) / VS (1000) = 7 minutes
Find TAS with CAS, altitude and temperature as we've done above using your nav computer = 228 TAS (no wind so = 228 GS)

228kt x 7min = 26.6 NM

GP ° = Ch alt (ft) x 60 / Ground Distance (ft)
GP ° = 7000ft x 60 / (26.6 x 6080)
GP ° = 2.5°

---

► You are on ILS 3° glideslope which passes over the runway threshold at 50 feet. Your DME range is 25 NM from the threshold. What is your height above the runway threshold elevation? (Use the 1 in 60 rule and 6.000ft = 1 NM)

Answer: 7.550 feet

TE = DOT/DAT x 60

DOT = distance off track
DAT = distance along track

3° = DOT / 150.000ft (6000x25NM) x 60
DOT = (3 / 60) x 150 000
DOT = 7500ft

Add the 50ft = 7550ft

---

► An aircraft at FL350 is required to descend to cross a DME facility at FL80. maximum rate of descent is 1800 ft/min and mean GS for descent is 276 kts. The minimum range from the DME at which descent should start is:

Answer: 69 NM

Ch alt (27 000) / VS (1800) = 15 minutes

15min x 276kt GS = 69 NM

---

► By what amount must you change your rate of descent given a 10 knot increase in headwind on a 3° glideslope:

Answer: 50 feet per minute decrease

Just do an example with a random chosen GS

Basic formula for a 3° Glide: VS = 5 x GS

5 x 200kt GS = 1000ft/min

Now consider the 10kt HWC:

5 x 190kt GS = 950ft/min (a 50ft/min decrease)

---

► An aircraft at FL370 is required to commence descent when 100 NM from a DME facility and to cross the station at FL120. If the mean GS during the descent is 396 its, the minimum rate of descent required is approximately:

Answer: 1650 ft/min

Turn into a Gradient:

Gradient (ft/nm) = Ch alt (ft) / Ground distance (nm)
Gradient = 25 000ft / 100nm
Gradient = 250 ft/nm

Find Vs:

VS = Gradient x (GS/60)
VS = 250 x (396/60)
VS = 1650 ft/min

---

► The outer marker of an ILS with a 3° glide slope is located 4,6 NM from the threshold. Assuming a glide slope height of 50ft above the threshold, the approximate height of an aircraft passing the outer marker is (use the 1:60 rule):

Answer: 1450ft

3° = Distance off track (DOT) / Distance along track (DAT) x 60

DAT should be in feet. In this Question they don't state what number to use to convert from NM to ft so I would just use the 1NM = 6080ft which gives the most accurate result

3° = DOT / 27968ft x 60
DOT = (3/60) x 27968
DOT = 1398ft

Add the 50ft = 1448ft (1450 being the closest answer)

---

► During approach the following data are obtained:
DME: 12,0 NM , altitude 3000ft
DME: 9,8 NM , altitude 2400ft
TAS: 160kt
GS: 125kt
The rate of descent is:

Answer: 570 ft/min

We lost 600ft during 2.2 NM

Gradient (ft/nm) = 600 / 2.2
Gradient = 273 ft/nm

VS = Gradient x GS/60
VS = 273 x (125/60)
VS = 569 ft/min

---

► An aircraft at FL360 is required to descent to FL120. The aircraft should reach FL120 at 40 NM from the next waypoint. The rate of descent is 2000ft/min. The average GS is 420 kt. The minimum distance from the next waypoint at which descent should start is:

Answer: 124 NM

Ch alt (24 000ft) / VS (2000ft/min) = 12 minutes

12min x GS 420kt = 84NM

BUT we should reach it 40 NM BEFORE the next waypoint so we should start the descent further away from that waypoint, 84 + 40 = 124 NM

---

► An aircraft is departing from an airport which has an elevation of 2.000ft and the QNH is 1003 hPa. The TAS is 100 kts. The head wind component is 20 its and the rate of climb is 500 ft/min. Top of climb is FL050. At what distance from the airport will this be achieved?

Answer: 7,2 NM

Difference QNH and 1013 = 10hPa
10hPa x 30 = 300ft difference
2.000ft + 300ft = 2300ft

From 2300ft to 5000ft te aircraft has to climb 2700ft at 500ft/min

2700 / 500 = 5.4 minutes

5.4 minutes x GS 80 = 7,2 NM

---

► You are departing from an airport which has an elevation of 1500ft. The QNH is 1003 hPa. 15 NM away there is a waypoint you are required to pass at an altitude of 7500ft. Given a groundspeed of 120 its, what is the minimum rate of climb?

Answer: 800ft/min

Question is not speaking about FL this time so we can assume both were taken from the same QNH.

The aircraft has to climb 6000ft (Ch Alt)

Gradient (ft/nm) = 6000 ft / 15 NM
Gradient = 400ft/nm

VS = Gradient x (GS/60)
VS = 400 x (120/60)
VS = 800 ft/min

---

► Given:
A descending aircraft flies in a straight line to a DME.
DME 55,0 NM , altitude 33.000ft
DME 43,9 NM , altitude 30.500ft
M=0.72 , GS=525 its, OAT=ISA
The descent gradient is:

Answer: 3,7%

Change in altitude = 2500ft for a distance covered of 11.1 NM

Gradient (%) = Ch alt (feet) / Ground distance (feet) x 100
Gradient = 2500ft / (11.1 x 6080) x 100
Gradient = 3.7

---

► Given:
TAS: 197 kt
True course: 240°
W/V: 180/30kts
Descent is initiated at FL220 and completed at FL40. Distance to be covered during descent is 39 NM. What is the appropriate rate of descent?

Answer: 1400ft/min

We can use the quick formula to find wind component:
Wind direction (180) - true course (240) = -60
COS(-60) = 0.5 (positive value so HWC)
0.5 x 30kt (wind speed) = 15kt

= 15kt headwind component

TAS 197 - 15 = 182kt GS
Distance (39NM) / GS (182) = 13 minutes

Altitude need to lose (18 000) / 13 minutes = 1385 (1400 closest answer)