General Nav Problems
TightYorksherMan
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General Nav Problems
If anyone could help me I would be much appreciated.
Many thanks,
1) Given:
Compass Heading = 090º
Deviation = 2ºW
Variation = 12ºE
TAS = 160kts
Whilst maintaining a radial 070 from a VOR station, the aircraft flies at a ground distance of 14nm in 6min. What is the wind velocity?
Answer = 160 degrees/ 50knots
2) An aircraft at FL140, IAS 210kt, OAT – 5C and wind component minus 35 knots, is required in order to cross a reporting point 5 MIN later than planned. Assuming that flight conditions do not change, when 150nm from the reporting point the IAS should be reduced to?
Answer = 20kts
Distance = 150nm
Groundspeed = 210 – 35 = 175kts
So 150nm @ 175kts = 51mins
51 + 5mins = 56mins
56mins @ 150nm = 159kts
159 + 35kts = 194.5kt
194.5kt – 210kt = 15.4kts – why is answer 20kts? What have I done wrong?
3) At 0422, an aircraft at FL370, GS 320kts, is on the direct track to VOR ‘X’ 185nm distant. The aircraft is required to cross VOR ‘X’ at FL80. For a mean rate of descent of 1800 ft/min at a mean groundspeed of 232kt, the latest time to commence descent is?
Answer = 0445
Height to Loose = 37,000 – 8000 = 29,000ft
29,000 / 1800 = 16mins to descend
Any Ideas where I go from here????
4) The two standard parallels of a conical lambert projection are at N10º 40’N and N41º 20’. The constant of the cone is approximately?
Answer given = 0.44
5) An aircraft departing A (N40º 00’ E080º 00’) flies a constant true track of 270º at a groundspeed of 120kt. What are the coordinates of the position after 6hrs?
120 x 6 = 720nm x cos 40 = 551nm / 60 = 9º 11’ – E080º = 70º 49’E – what have I done wrong?
6) Given: A is N55º 000º, B is N54º E010º. The average true course of the great circle is 100º. The true course of the rhumbline at point A is =
Answer given = 100º
7) At latitude 60N the scale of a Mercator projection is 1: 5,000,000. The length of the chart between ‘C’ N60 E008 and ‘D’ N60 W008 is?
Answer given = 17.8cm
Many thanks,
1) Given:
Compass Heading = 090º
Deviation = 2ºW
Variation = 12ºE
TAS = 160kts
Whilst maintaining a radial 070 from a VOR station, the aircraft flies at a ground distance of 14nm in 6min. What is the wind velocity?
Answer = 160 degrees/ 50knots
2) An aircraft at FL140, IAS 210kt, OAT – 5C and wind component minus 35 knots, is required in order to cross a reporting point 5 MIN later than planned. Assuming that flight conditions do not change, when 150nm from the reporting point the IAS should be reduced to?
Answer = 20kts
Distance = 150nm
Groundspeed = 210 – 35 = 175kts
So 150nm @ 175kts = 51mins
51 + 5mins = 56mins
56mins @ 150nm = 159kts
159 + 35kts = 194.5kt
194.5kt – 210kt = 15.4kts – why is answer 20kts? What have I done wrong?
3) At 0422, an aircraft at FL370, GS 320kts, is on the direct track to VOR ‘X’ 185nm distant. The aircraft is required to cross VOR ‘X’ at FL80. For a mean rate of descent of 1800 ft/min at a mean groundspeed of 232kt, the latest time to commence descent is?
Answer = 0445
Height to Loose = 37,000 – 8000 = 29,000ft
29,000 / 1800 = 16mins to descend
Any Ideas where I go from here????
4) The two standard parallels of a conical lambert projection are at N10º 40’N and N41º 20’. The constant of the cone is approximately?
Answer given = 0.44
5) An aircraft departing A (N40º 00’ E080º 00’) flies a constant true track of 270º at a groundspeed of 120kt. What are the coordinates of the position after 6hrs?
120 x 6 = 720nm x cos 40 = 551nm / 60 = 9º 11’ – E080º = 70º 49’E – what have I done wrong?
6) Given: A is N55º 000º, B is N54º E010º. The average true course of the great circle is 100º. The true course of the rhumbline at point A is =
Answer given = 100º
7) At latitude 60N the scale of a Mercator projection is 1: 5,000,000. The length of the chart between ‘C’ N60 E008 and ‘D’ N60 W008 is?
Answer given = 17.8cm
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Heres a couple...
1) Magnetic heading = 090 - 2 (deviation) = 088 degrees
Magnetic trk = 070 (radial from VOR)
therefore drift angle = 088 - 070 = 18 degrees (BIG!)
GS = 14 / 6 * 60 = 140 knots
Now to jepp flight computer on wind side....
turn TAS (with white triangle above) pointing to 160
trun green numbers to trk = 070 degrees mag
Now on 2nd most outer scale look at 18 degrees corresponds to 50 knots of x-wind which is to (edit : FROM) the right
for drift angles greater than 10 degrees must consult the black degrees scale next to TAS arrow: 18 degrees corresponds to our new TAS of 152 knots.
therefore headwind = 152 - 140 = 12 knots.
Now draw mark on grid at 50 knots right x-wind and 12 knots headwind.
turn it around to vertical wind = 147 / 51
147 degrees + 12 (variation) = 159 degrees true
answer 159 / 51 (pretty close)
2) On jepp flight computer line up CAS 210 (assume IAS = CAS) and FL140.
Mach number = M0.41
Mach index arrow in TAS window onto -5 degrees
Inner scale M0.41 outer scale 261 KTAS
GS = 261 - 35 = 226 kts
150nm @ 226 kts ~ 40 minutes
Must extend to 40 + 5 minutes = 45 minutes
New groundspeed required = 150 * 60 / 45 = 200 kts
Headwind = 35 therefore TAS must = 200 + 35 = 235 knots
back to jepp computer tas 235 outer scale with mach index on -5 degrees gives mach no. ~ M0.37
Mach no. onto M0.37 read 189 knots next to FL140
Answer 210 - 189 = 21 kts close enough
The problem you have here is assuming 210kts IAS is equal to the aircraft's true airspeed. Have a look at some past posts to work find the relationship between indicated airspeed (IAS), calibrated airspeed (CAS), equivalent airspeed (EAS) and true airspeed (TAS).
3) Descent time 16 minutes as you say
16 minutes @ 232 kts = 61.9 NM from station (TOPD)
185 - 61.9 = 123.1 NM = present postion to TOPD
123.1 NM @ 320 kts = 23 mins.
Present time 0422 + 0023 = 0445
If i've messed any of these up please point it out...
1) Magnetic heading = 090 - 2 (deviation) = 088 degrees
Magnetic trk = 070 (radial from VOR)
therefore drift angle = 088 - 070 = 18 degrees (BIG!)
GS = 14 / 6 * 60 = 140 knots
Now to jepp flight computer on wind side....
turn TAS (with white triangle above) pointing to 160
trun green numbers to trk = 070 degrees mag
Now on 2nd most outer scale look at 18 degrees corresponds to 50 knots of x-wind which is to (edit : FROM) the right
for drift angles greater than 10 degrees must consult the black degrees scale next to TAS arrow: 18 degrees corresponds to our new TAS of 152 knots.
therefore headwind = 152 - 140 = 12 knots.
Now draw mark on grid at 50 knots right x-wind and 12 knots headwind.
turn it around to vertical wind = 147 / 51
147 degrees + 12 (variation) = 159 degrees true
answer 159 / 51 (pretty close)
2) On jepp flight computer line up CAS 210 (assume IAS = CAS) and FL140.
Mach number = M0.41
Mach index arrow in TAS window onto -5 degrees
Inner scale M0.41 outer scale 261 KTAS
GS = 261 - 35 = 226 kts
150nm @ 226 kts ~ 40 minutes
Must extend to 40 + 5 minutes = 45 minutes
New groundspeed required = 150 * 60 / 45 = 200 kts
Headwind = 35 therefore TAS must = 200 + 35 = 235 knots
back to jepp computer tas 235 outer scale with mach index on -5 degrees gives mach no. ~ M0.37
Mach no. onto M0.37 read 189 knots next to FL140
Answer 210 - 189 = 21 kts close enough
The problem you have here is assuming 210kts IAS is equal to the aircraft's true airspeed. Have a look at some past posts to work find the relationship between indicated airspeed (IAS), calibrated airspeed (CAS), equivalent airspeed (EAS) and true airspeed (TAS).
3) Descent time 16 minutes as you say
16 minutes @ 232 kts = 61.9 NM from station (TOPD)
185 - 61.9 = 123.1 NM = present postion to TOPD
123.1 NM @ 320 kts = 23 mins.
Present time 0422 + 0023 = 0445
If i've messed any of these up please point it out...
Last edited by ROB-x38; 5th Jun 2004 at 06:40.
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Just had an idea re: quesiton six
6) Rhumb line track = (initial track + final track) / 2
(initial track + final track) / 2 I suppose is the same as "average true course" therefore 100 degrees is also the rhumb line track
6) Rhumb line track = (initial track + final track) / 2
(initial track + final track) / 2 I suppose is the same as "average true course" therefore 100 degrees is also the rhumb line track
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Q.4 Cionstant of the cone refers to the sin of the mean latitude on a Lamberts, add N10º 40’N and N41º 20’ together and divide by 2, giving 26º. Taking the sine of this gives 0.438
Q.5 Departure = Chg Long x Cos Lat, we know departure is 6hrs x 120kts so 720nm and latitude is 40, so to find chg long:
chg long (mins) = 720/cos 40, now divide chg long by 60 to find degrees and minutes then subtract this from 80 to get your answer of about 064º 20’E
Q.6 Your answer here is wrong, it is 104º (check with who you got the questions from but I have this exact question and answer is 104), to get 104:
A great circle is always convex to the nearer pole, as in the north hemisphere, we can draw this out and as such we will see that the initial rhumb line (straight line) track will be greater than the great circle track. The difference between the 2 tracks is conversion angle:
CA = 1/2 chg long (in degrees) x sin mean lat
= 0.5 x 10 x sin 54.5
= 4.07º
therefore rhumb line angle = 100 + 4.07 = 104.7º
Q.7 first you need to find departure:
dep = chg long x cos lat
= (16x60) x cos 60
= 480nm
now convert 480nm to km (x 1.852) = 888.96, convert this to meters (x 1000) then to centimeters (x100) = 88,896,000cm
This is the distance on the ground covered, so to find distance on the chart divide this by the denominator of the scale (5,000,000) which gives 17.7792cm
Wow, I actually know some Gan Nav, handy with my JAA on Wednesday... Revision is so boring
BigAir
Q.5 Departure = Chg Long x Cos Lat, we know departure is 6hrs x 120kts so 720nm and latitude is 40, so to find chg long:
chg long (mins) = 720/cos 40, now divide chg long by 60 to find degrees and minutes then subtract this from 80 to get your answer of about 064º 20’E
Q.6 Your answer here is wrong, it is 104º (check with who you got the questions from but I have this exact question and answer is 104), to get 104:
A great circle is always convex to the nearer pole, as in the north hemisphere, we can draw this out and as such we will see that the initial rhumb line (straight line) track will be greater than the great circle track. The difference between the 2 tracks is conversion angle:
CA = 1/2 chg long (in degrees) x sin mean lat
= 0.5 x 10 x sin 54.5
= 4.07º
therefore rhumb line angle = 100 + 4.07 = 104.7º
Q.7 first you need to find departure:
dep = chg long x cos lat
= (16x60) x cos 60
= 480nm
now convert 480nm to km (x 1.852) = 888.96, convert this to meters (x 1000) then to centimeters (x100) = 88,896,000cm
This is the distance on the ground covered, so to find distance on the chart divide this by the denominator of the scale (5,000,000) which gives 17.7792cm
Wow, I actually know some Gan Nav, handy with my JAA on Wednesday... Revision is so boring
BigAir
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ROB-x38
This is new to me. Is it a jepp-only trick, or can you do it on the CRP5 too?
Can you expand, please?
On jepp flight computer line up CAS 210 (assume IAS = CAS) and FL140. Mach number = M0.41
Can you expand, please?
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Hi keg167l,
> This is new to me. Is it a jepp-only trick, or can you do it on
> the CRP5 too?
>
> Can you expand, please?
You beat me to it! I was going to ask the same question as I would have worked it out thus:
Assume CAS = IAS
CAS is 210 knots
Align -5 C with 14,000 on the Pressure Window of the CRP5
Read off 264 kts TAS on the outer scale against 210 kts CAS on the inner.
No Compressibility Error calculation involved as TAS is under 300 kts.
No need to get involved with Mach numbers with this one... or am I wrong?
This is how I have been going through this type of ATPL Gen Nav question.
Best wishes,
Charlie Zulu.
> This is new to me. Is it a jepp-only trick, or can you do it on
> the CRP5 too?
>
> Can you expand, please?
You beat me to it! I was going to ask the same question as I would have worked it out thus:
Assume CAS = IAS
CAS is 210 knots
Align -5 C with 14,000 on the Pressure Window of the CRP5
Read off 264 kts TAS on the outer scale against 210 kts CAS on the inner.
No Compressibility Error calculation involved as TAS is under 300 kts.
No need to get involved with Mach numbers with this one... or am I wrong?
This is how I have been going through this type of ATPL Gen Nav question.
Best wishes,
Charlie Zulu.
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1) Given:
Compass Heading = 090º
Deviation = 2ºW
Variation = 12ºE
TAS = 160kts
Whilst maintaining a radial 070 from a VOR station, the aircraft flies at a ground distance of 14nm in 6min. What is the wind velocity?
quick way to do this one on the CRAP 5
work out the true heading using C D M V T making sure you get you get your sign code right.
C D M V T
090 2W 088 12E 100
now convert the 070 on the VOR into true direction, giving 082 giving a drift on 100 true heading of 18 port.
put the blue TAS circle onto 160 and then find the 18 port drift position that corresponds to the groundspeed (140) and mark it.
turn the slide til you have the mark on the wind numbers line, make sure you read downwards (wind down method). voila 160/50 kts.
hope this helps
Compass Heading = 090º
Deviation = 2ºW
Variation = 12ºE
TAS = 160kts
Whilst maintaining a radial 070 from a VOR station, the aircraft flies at a ground distance of 14nm in 6min. What is the wind velocity?
quick way to do this one on the CRAP 5
work out the true heading using C D M V T making sure you get you get your sign code right.
C D M V T
090 2W 088 12E 100
now convert the 070 on the VOR into true direction, giving 082 giving a drift on 100 true heading of 18 port.
put the blue TAS circle onto 160 and then find the 18 port drift position that corresponds to the groundspeed (140) and mark it.
turn the slide til you have the mark on the wind numbers line, make sure you read downwards (wind down method). voila 160/50 kts.
hope this helps
Last edited by Chintito; 5th Jun 2004 at 17:50.
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keg167l, Charlie Zulu
Either way would work fine for this question. As you mention CZ there's no need for a compressibility correction at the lower speeds.
I just use the mach no. method because I find it is the most accurate and the majority of sample questions i've faced have been in the transonic region.
Either way would work fine for this question. As you mention CZ there's no need for a compressibility correction at the lower speeds.
I just use the mach no. method because I find it is the most accurate and the majority of sample questions i've faced have been in the transonic region.
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Jinkster
The ansers you've received so far are all good except, I fancy, Big Air's response to Q6. The clue here is that you are given the average GC TC; this can be taken as the GS TC measured at the mid-point of the track, where the GC is parallel to the RL! - that, is the RL and GC values are identical at 100. As, by definition, the RL course never changes, the RL course at A is also 100.
Feel free to PM me if you want any more on the other Qs.
The ansers you've received so far are all good except, I fancy, Big Air's response to Q6. The clue here is that you are given the average GC TC; this can be taken as the GS TC measured at the mid-point of the track, where the GC is parallel to the RL! - that, is the RL and GC values are identical at 100. As, by definition, the RL course never changes, the RL course at A is also 100.
Feel free to PM me if you want any more on the other Qs.
