atpl practice navigation exam 6 · pdf fileq16. refer to scan of nav computer below. steps: 1....

ATPL Practice Navigation Exam 6 Working.

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VOR

HD

G 2

80

Wind

HD

G 0

80

Q1.

Q2.

Q3.

Q4.

No turning errors in east/west headings. Turning errors are greatest on North/South headings.

Acceleration errors greatest on East/West headings using floating magnetic compass. Rule is “SAND” - South apparent turn if Accelerating, North apparent turn for Deceleration. The gyro com-pass (DG) can be assumed to not be affected by acceleration/deceleration errors.

E125 Meridian 230620 LMT

Less Arc/Time conversion (refer AIP) - 0820

Time in Greenwich 222200 UTC

Less flight time - 0200

Earliest departure time 222000

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Q5.

Curtin

ED

P (B

) ET

P (

D)

To

Dar

win

82 nm

82 nm

Distance EDP to either Alternate TAS

X Wind speed

81 nm 180 kt

= X 40 kt = wind vector length of 18 nm.

Simplified working diagram (not to scale).

Gibb River

18 n

m

1. Find the mid-point between the two alternates (Curtin and Gibb River in this case). Position “A” 2. Draw a line up at 90° to where it hits the Curtin - Darwin planned track line (refer dashed blue line) Posi-

tion “B”. This is the equi-distant point (EDP) - dashed blue circle. From the planned track line (dashed blue circle) it is the same distance from EDP “B” to either return to Curtin or divert off track to the right to Gibb River. In nil wind it takes the same time to fly to either of these airports as the groundspeeds are of course the same.

3. The ETP is always into wind from the EDP so draw a line up into wind from position “A” representing the 300°M wind (not sure how long yet - we will find out in step 4). Nearly finished !

4. To find out how far into wind use the formula below every time.

A

Wind line length formula is

5. Measure 18 scale miles (if using a WAC ruler in this case with scale being 1 to 3 million it is 6 WAC miles). 6. Draw a line up to the track line so it passes through the end of the wind line (step 5 - position “C” ) up to your Curtin - Darwin planned track. This line must be at 90° to the line that connects Curtin and Gibb River →→ es-sentially parallel to line “A” - “B” you drew in step 2. Where this 90° line intersects the Curtin - Darwin planned track line is the ETP (position “D”). With the two differ-ent groundspeeds that will occur on the diversion legs, ETP to Gibb River and ETP back to Curtin from the ETP it will take the same time to divert off track to the right to Gibb River as it will to divert back to Curtin. This plotting method is accurate to within half a minute depending on your plotting accuracy. Okay tolerance for CASA exam where answers are typically about 15-20 nm apart or more. Do not waste our time check if flight time is the same as you do not have time in the exam. Just take the marks and move on to the next question !

C

Off track ETP question.

Win

d 0

30°M


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Q6.

Q7.

Q8.

Q10.

Q9.

It is not possible to get a position fix when overhead Leonora in this case as it is too far (more than legal 30 nm maximum for NDB/NDB) from Laverton and Leinster NDB’s. Similarly using Kalgoorlie NDB. Can not use Kal-goorlie VOR in combination with Laverton because rated coverage of VOR at 10000 ft is only 90 nm and Leon-ora is 115 nm from Kalgoorlie.

1. The CP (ETP) always moves into wind from half way point. How far it moves depends on aircraft TAS, and wind component and the distance between the two airport under consideration.

2. Greater headwind would move the CP further into wind (closer to the destination airport.

1. The CP (ETP) always moves into wind from half way point. How far it moves depends on aircraft TAS, and wind component and the distance between the two airport under consideration.

2. Greater TAS reduces your exposure time to the wind and so the CP is closer to the half way point. Not so far into wind.

1. The position of the PNR depends lot on your TAS. Small TAS produces shorter distance to the PNR. 2. As the 1 engine Inop TAS is slower than the two engine TAS the return flight time will be longer and as

you have a fixed amount of fuel the 1 Inop PNR must lie before the two engine PNR (reduced aircraft performance on the home leg with an engine out).

The PNR must be at or beyond the CP (ETP) or else you did not have sufficient fuel for the flight I the first place.

Q11.

A. Ground Wave. Yes NDB operates in MF band which uses ground waves. VLF omega nav uses skywaves but accuracy depends on height of ionosphere which is higher at night than through the day which causes errors. Called “Terminator error”.

B. Ground and Sky wave. No, mixing of these reduces overall signal strength and with it usable range. Hap-pens at night with NDB (night affect).

C. Direct wave. No, direct wave is feature of VOR (VHF band) and DME (UHF band), not NDB (MF band). D. Sky waves. Is the basis on which VLF Omega nav works but the enemy of range for NDB.





Q12.

Q13.

Tilt error (due to magnetic dip) affects the floating magnetic compass if not operating at the equator where tilt is zero. To compensate for this so we can reliably operate at other latitudes the pivot point is deliberately placed a distance from the compass centre of gravity. This reduces tilt error but brings with it acceleration errors and turn-ing errors.

Aircraft 1 0200 UTC TAS 240 kt GS 180 kt Dist nm

BRAVO

60 kt

SNAGS

200 nm

Aircraft 1 0205 UTC TAS 500 kt GS 440 kt Dist nm

25 nm

ALPHA

1. Difference in the 2 grounds speeds is 260 kt. So this is the rate of closure speed. 2. Aircraft 1 (the slow one) arrived at SNAGS 5 minutes before aircraft 2, so by the time aircraft 2 reaches

SNAGS aircraft 1 will be 5 minutes past SNAGS. At a GS of 180 kt (3 nm per minute that is 15 nm). 3. Simply divide the 15 nm by the closure speed of 260 kt to get the minutes it will take for the faster aircraft

(aircraft 2) to catch up to aircraft 1. It is about 3.5 minutes. 4. In 3.5 minutes aircraft 2 at a GS of 440 kt will have travelled 25 nm. So it will pass aircraft 1 about 25 nm

past SNAGS. This is 225 nm past airport ALPHA where the question asked you to measure from. 5. The 30 kt wind between ALPHA and SNAGS is not required - just a CASA diversion.

Working diagram.




Q14.

Q15.

16.7 nm

1420 UTC

1420 UTC 1415 UTC

TAS 300 kt

GS 340 kt Plane Logic

TAS 240 kt

GS 200 kt Plane Logic

40 kt

300 nm

Flight Profile

Working: 1. Move aircraft #1 to a 1420 UTC position. I this case 5 minutes at 200 kt = 16.7 nm. 2. Divide miles separating the aircraft at 1420 UTC (283.3 nm) by the closure speeds (combined GS’s of

540 kt), and then multiply by 60 to get minutes to crossing. The answer is about 31.5’ (rounded slightly). 3. Move aircraft 2 a further 31.5minutes at 200 kt GS to get 105 nm past the 1420 UTC position. Add the

16.7 nm to the 105 nm to get miles from Airport “A. In this case 121.7 nm.

Airport “A” Airport “B”

283.3 nm

31’ @ 200 kt = 105 nm

Aircraft #2

Aircraft #1

Working: 1. Using your Nav computer or the speed of sound formula on a calculator (as True OAT quoted) find TAS

for M0.65 is about 395 kt. 2. Headwind component is 52 kt so ground speed is 343 kt. 3. Again using Nav computer or calculator find that at a GS of 343 kt it will take 40.25 min to get to BRAVO. 4. SO ETA BRAVO is 1545Z + 40.25 min = 1625.25Z.



Q16. Refer to Scan of Nav computer below.

Steps: 1. Set the Mach number in the Mach window → refer blue dashed circle. 2. Set curved green line so that it passes through the –27°C TAT line where it intersects the thick black line

that runs approximately horizontally through the window → refer green circle. Not the straight line though. 3. Follow the curved green line through the TAS window toward the centre of the wheel to the “Temperature

rise” window and read off about 28 degrees of ram air temperature rise → refer red solid circle. 4. The true (Static air temperature) is always colder than that “indicated” in the cockpit because of the in-

crease in temperature due to a combination of friction heating and compression heating within the Total Air Temperature (TAT) probe. So true OAT (abbreviated TOAT) is 28° cooler than the –27°C indicated in the cockpit TAT gauge. So TOAT is –55°C.

5. ISA for FL330 is about –51°C, so today it is ISA –4 outside. ISA-5 answer is closest. Point to note:

• CASA sometimes refer to Total Air Temperature (TAT) as Indicated Outside Air Temperature (IOAT).

• CASA sometimes refer to True Outside Air Temperature (TOAT) as simply OAT, or Static Air Tempera-ture (SAT). The terms are interchangeable and mean the same thing.

• Be careful not to go the wrong way with ISA deviation. Easy to think of ISA minus when it should be ISA plus.




Q17.

Q18.

Working: 1. This is a 1 Engine Inop CP, so the normal TAS or wind data is not applicable. Use the 1 Inop level wind

and 1 Inop TAS only. You are trying to find the position along track where it takes the same time to go on at the 1 Inop TAS/FL as it takes to go back 1 Inop to the airport behind you. Normal ops does not feature.

2. The TAS to use is 410 kt. The wind component is that half way between the two levels (FL230 Home and FL240 on) so use FL235. It is as it happens a full 45 kt headwind on and 45 kt tailwind home.

Distance between the alternate airports x GS HOME GS ON + GS HOME

=

CP distance from departure airport is

600 nm x 455 kt 365 nm + 455 kt

= 333 nm from departure airport (Alpha).

3.

4. The question asked for distance from the destination (Bravo) which is 600 nm - 333 nm = 267 nm. 270 nm answer is closest. Note the CP has moved 33 nm into wind from the halfway point. Note also that the time ON 1 Inop of 44’ Is the same as the time to fly 1 Inop HOME. Truly this is the equi-time point.

Pressure error (also called position error) is created as an aircraft moves through the air. It can af-fect the static and pitot ports though they are positioned to reduce this error. It affects all air data instruments. Anything that uses air data as an input.

Q19. To get a “positive” (valid) radio fix the position lines must cross at 45° or more. Refer AIP.

HOME 1 Inop TAS 410 kt

GS 455 kt Time HOME 44 min

BRAVO

45 kt @ FL235

333 nm 267 nm

ALPHA

Working diagram.

600nm

ON 1 Inop TAS 410 kt

GS 365 kt Time ON 44 min

CP 1 Inop

FROM ALPHA Two engines



Q20. WORKING (Refer ERC L4)

1. Find time from fix to destination (Mt Isa), and time from destination to alternate (Townsville). See below.

Longreach

Mt Isa

ETAS 255 kt GS 228 kt

Distance 314 nm ETI 82.5’

Townsville

ETAS 252 kt

GS 252 kt

Distance 420 nm

ETI 100’

Rough LPSD formula = Time to Dest x End Available Time to Dest + Time to Alternate

82.5’ x 110’ 82.5’ + 100’

= = 49.7’ @ GS228 kt = 189 nm from fix. →→ plot on ERC L4 chart. This is just to get a starting point. Sometimes it nails the LPSD first time and some-times not. All depends on the angular relationships.

2. Find track/dist/time from rough LPSD to alternate (Townsville). See below.

Longreach

Mt Isa



Townsville

ETAS 252 kt

GS 244 kt

Distance 337 nm

ETI 82.9’ Rough LPSD

Trk 059°M

ETI to Rough LPSD 49.7’

Rough LPSD to Townsville 82.9’

Total Fuel required 132.6

Fuel Available 110.0

Must reduce total flight time by 22.6’

With the angle of the diversion track at the Rough LPSD being about 135° then reducing the time from the fix to the LPSD by 16 minutes will affect the diversion time by less than this → about the remaining 6.6 minutes. This will give a ‘tweaked’ LPSD. So reduce fix (Longreach) to LPSD time back to 33.7 minutes (49.7’ - 16’) @228 kt = 128 nm past LR and plot this position on ERC L4 chart. See diagram on next page →→→





Longreach

Mt Isa



Townsville

ETAS 252 kt

GS 240 kt

Distance 306 nm

ETI 76.5’

Tweaked (final) LPSD

Trk 050°M

Q20 working continued...

ETI to tweaked LPSD 33.7’

Rough LPSD to Townsville 76.5’

Total Fuel required 110.2

Fuel Available 110.0

Shortfall of fuel available at tweaked PSD

0.2’

So at the tweaked LPSD we have 0.2 minutes of fuel spare (well within CASA tolerance so just leave it at this position). As the diversion from the tweaked position now involves about a 90 degree turn then adding or sub-tracting a few minutes to the Longreach to Mt Isa track will not change the diversion leg distance to Townsville much. This is because we are essentially on a DME arc and roughly abeam Townsville. You will find more about this in the LPSD module of the training textbook/online course. The question asked or the distance from Mt ISA so it will be 314 nm -128 nm = 186 from Mt Isa. 190 nm answer best. Being conservative, for multi-choice answers CASA will always round DOWN if asked to measure from the airport behind you (120 nm from Longreach), and being complimentary to this would round UP if measuring from the airport aheads as here 190 nm from Mt Isa.

Step 3.

Q21.

• NDB is not reliable enough for INS/IRS updates. Can not update either manually or automatically.

• You can update over a ground feature (e.g. Ayres Rock) when no more than 5000 ft above the feature.

• You can manually update the INS/IRS when you get station passage overhead a VOR (“TO” flag chang-ing to “FROM”).




Q22.

LRE VOR

Winton NDB

0835Z Air Plot

0835Z Ground Plot (Fix)

0815Z fix

Trk

made g

oo

d L

RE

288 r

adia

l

WTN 205°M FROM

Win

d 15

5°M

@ 4

8 kt

16 n

m

HD

G 2

80°M

WORKING: 1. On ERC L4 plot 280°M track out of Longreach VOR using your protractor centred on the LRE VOR.

Measure up this line 20’ @ 240 kt TAS = 80 scale miles. This is the “air plot” which is simply where you would have been if there was no wind at all.

2. Again with your protractor centred on the LRE VOR draw in the track made good (quoted as 288°M). 3. The bearing TO Winton NDB is quoted as 025°M from the aircraft (this is not a relative bearing so be

careful). The bearing from the WTN NDB will be the reciprocal of 025°M which is 205°M. Plot this line with your protractor centred this time on the Winton NDB up to where it crosses the 288 bearing from LRE. Where these two position lines cross is where you actually are at 0835 Z → The “ground plot”.

4. Now simply draw a line connecting the air plot and the ground plot. This is the direction the wind has been coming from between the fix positions.

5. Measure the length of the line in step 4 and you should get about 16 scale miles. 6. In 20 minutes the wind line is 16 nm long so in one hour it would be 3 times this (48 nm). This is the wind

speed. 7. For wind direction place your protractor over the ground plot. You should get about 155°M. 8. So the wind affecting the aircraft between 0815Z and 0835Z has been 155°M@ 48 kt.



Q23.


1. Find time from Moree fix to St George and get the planned endurance at that time → it is about 26.4’. So safe endurance at St George will be about 38.6 minutes.

2. Flight time to Roma works out to be 19.8’. The diversion time from Roma to Oakey is about 29.5’. 3. Based on the planned safe endurance at St George (38.6’) use the Rough LPSD formula as below.

Moree

Roma

TAS 290 kt GS 248 kt


OAKEY



Rough LPSD formula = Time to Dest x End Available Time to Dest + Time to Alternate

19.8’ x 38.6’ 19.8’ + 29.4’

= = 15.5’ @ GS276 kt = 71 nm from St George. →→ plot on ERC L4 chart. This is just to get a starting point. Sometimes it nails the LPSD first time and sometimes not. All depends on the angular relationships.

St George

ETAS 285 kt

GS 276 kt

Distance 91 nm

ETI 19.8’

1320Z fix Safe End 65 min

Roma

OAKEY

ETAS 290 kt

GS 340 kt

Distance 161 nm

ETI 28.4’

St George Safe End 38.6’

ETAS 285 kt

GS 276 kt

Distance 71 nm

ETI 15.5’

Rough LPSD Trk 093°M

ETI to Rough LPSD 15.5’

Rough LPSD to Oakey 28.4’

Total Fuel required 43.9’

Fuel Available @ St George 38.6’

Over endurance by 5.3 min

4. Plot the rough LPSD and measure track/distance from there to Oakey. Then get diversion time → see below.


5. As the rough LPSD is almost abeam Oakey we can safely reduce the St George to Roma time by the full 5.3 minutes that we are over by because the diversion distance to Oakey will not vary much as a result. 6. At a groundspeed on this section of 276 kt that equates to a distance of 24 nm. So the distance past St George of the “tweaked” LPSD is 47 nm. This is 44 nm from Roma (91 nm - 47 nm). Being conservative we round UP if measuring from destination as in this case, so the 50 nm answer is best !

Q23 working continued...


Q24.


As the aircraft nears to abeam position the groundspeed reduces because the rate of closure with the DME is reducing. Abeam the ground station the groundspeed would momentarily be zero, counting up again after the aircraft passes the abeam position. Remember DME measures “slant distance”. The rate of groundspeed reduction would reduce as it gets closer to the abeam position. See diagram below.

END OF NAVIGATION PRACTICE EXAM 6 WORKING FILE.

DME

Groundspeed decreasing at an ever reducing RATE here.

Groundspeed increasing at an increasing RATE here.


atpl practice navigation exam 6 · pdf fileq16. refer to scan of nav computer below. steps: 1....

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