A Couple of GNav Problems.
TightYorksherMan
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A Couple of GNav Problems.
Thanks in advance for the help,
1) The constant of the cone on a lambert chart where the convergence angle between longitiudes 010E and 030degW is 30deg, is:
a) 0.50
b) 0.64
c) 0.75 <--answer
d) 0.40
2) The constant of the cone of a lambert conformal conic chart is quoted as 0.3955. At what latitude on the chart is earth convergency correctly represented?
a) 68deg 25'
b) 21deg 35'
c) 23deg 18' <---answer
d) 66deg 42'
3) in order to maintain an accurate vertical using pendulous system, an aircraft inertial platform incorporates a device:
a)without damping and a period of 84.4sec
b) with damping and a period of 84.4sec
c) with damping and a period of 84.4min <---answer
d) without damping and a period of 84.4min
This question is definately 84.4min but damping or not damping - I am getting confilcting info
When accelerating on a westerly heading in the northern hemisphere the compass card of a direct reading magnetic compass will turn:
a) clockwise giving an apparent turn to the north
b) clockwise giving an apparent turn towards the south
c) anti-clockwise giving an apparent turn to the north <--answer
d)anti-clockwise giving an apparent turn towards the south
Thanks for the help, Jinkster
1) The constant of the cone on a lambert chart where the convergence angle between longitiudes 010E and 030degW is 30deg, is:
a) 0.50
b) 0.64
c) 0.75 <--answer
d) 0.40
2) The constant of the cone of a lambert conformal conic chart is quoted as 0.3955. At what latitude on the chart is earth convergency correctly represented?
a) 68deg 25'
b) 21deg 35'
c) 23deg 18' <---answer
d) 66deg 42'
3) in order to maintain an accurate vertical using pendulous system, an aircraft inertial platform incorporates a device:
a)without damping and a period of 84.4sec
b) with damping and a period of 84.4sec
c) with damping and a period of 84.4min <---answer
d) without damping and a period of 84.4min
This question is definately 84.4min but damping or not damping - I am getting confilcting info
When accelerating on a westerly heading in the northern hemisphere the compass card of a direct reading magnetic compass will turn:
a) clockwise giving an apparent turn to the north
b) clockwise giving an apparent turn towards the south
c) anti-clockwise giving an apparent turn to the north <--answer
d)anti-clockwise giving an apparent turn towards the south
Thanks for the help, Jinkster
Jet Blast Rat
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1. convergence = ch. long x constant of the cone
rearrange to give:
constant of the cone = convergence / ch. long.
2. Convergence is correct at the parallel of origin. Constant of the cone is sine (parallel of origin), therefore parallel of origin is inverse sine (constant of the cone). Use the 'inv' or '2ndF' buttong on your calculator and the 'sin' button for inverse sine.
3. With damping as far as I can find out.
4. Indicates (falsely) a turn toward the nearer pole, i.e. toward the north or a right turn. To indicate a right turn when not turning the compass card must turn left, anti-clockwise (as viewed from above).
rearrange to give:
constant of the cone = convergence / ch. long.
2. Convergence is correct at the parallel of origin. Constant of the cone is sine (parallel of origin), therefore parallel of origin is inverse sine (constant of the cone). Use the 'inv' or '2ndF' buttong on your calculator and the 'sin' button for inverse sine.
3. With damping as far as I can find out.
4. Indicates (falsely) a turn toward the nearer pole, i.e. toward the north or a right turn. To indicate a right turn when not turning the compass card must turn left, anti-clockwise (as viewed from above).
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Q1 You get 30º of inclination of the meridians (ie, chart convergence) in 40º of change of longitude.
Chart convergence = ch long x sin parallel of origin
30 = 40 x 'n'
where 'n' is the sin of the parallel of origin, or 'constant of the cone'
Q2 Chart convergence will be the same as earth convergence on the parallel of origin.
The arc sine of the constsnt of the cone is the parallel of origin.
Q3 The question has been removed. I complained to the CAA about it and they took it out. I will give a more detailed explanation if anybody requests it on this thread.
Your feedback is out of date.
Q4 In the northern hemisphere, the red end of the magnet hangs down because of the Z component of earth magnetism. Pendulous suspension is an attempt to correct this, but it still leaves the centre of pivotage (ie, the bit where the airframe is attached to the compass needle pivot) over the red end, not over the centre of gravity.
As the aircraft accelerates westward, the point of pivotage is dragged by the accelerating airframe westwards. This point is north of the centre of gravity. The resistance to the acceleration (ie, the inertia, as described by Newton's Third Law) will act through the centre, setting up a couple, which will turn the magnet anti-clockwise (draw it out, if you don't like it in words). If the needle turns anti-clockwise, it goes to the 'eleven o'clock position', not the 'one o'clock' position. This means that, measuring clockwise, you measure a BIGGER angle, ie, you get an apparent turn towards the north.
Can't your instructor answer these, Jinkster? If not, what are you supposed to be paying him for?
Chart convergence = ch long x sin parallel of origin
30 = 40 x 'n'
where 'n' is the sin of the parallel of origin, or 'constant of the cone'
Q2 Chart convergence will be the same as earth convergence on the parallel of origin.
The arc sine of the constsnt of the cone is the parallel of origin.
Q3 The question has been removed. I complained to the CAA about it and they took it out. I will give a more detailed explanation if anybody requests it on this thread.
Your feedback is out of date.
Q4 In the northern hemisphere, the red end of the magnet hangs down because of the Z component of earth magnetism. Pendulous suspension is an attempt to correct this, but it still leaves the centre of pivotage (ie, the bit where the airframe is attached to the compass needle pivot) over the red end, not over the centre of gravity.
As the aircraft accelerates westward, the point of pivotage is dragged by the accelerating airframe westwards. This point is north of the centre of gravity. The resistance to the acceleration (ie, the inertia, as described by Newton's Third Law) will act through the centre, setting up a couple, which will turn the magnet anti-clockwise (draw it out, if you don't like it in words). If the needle turns anti-clockwise, it goes to the 'eleven o'clock position', not the 'one o'clock' position. This means that, measuring clockwise, you measure a BIGGER angle, ie, you get an apparent turn towards the north.
Can't your instructor answer these, Jinkster? If not, what are you supposed to be paying him for?
TightYorksherMan
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What Id like to know from any instructor is why do I need to know any of this bollocks in the first place.
When you go to an airport and see a shiny new aeroplane maybe it be a 737 or 747 taking you on your holidays and think "I really really do want to fly one of those" then that is why you have to learn all this "bollocks"
I stopped saying why and just got on with it last year sometime - I am now more than half way through my ATPL ground exams.
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Only time I want to be in a 737 or 747 is when its is taking me somewhere warm for my hols.
Been flying helicopters " Commercially " for 8 years for the Queen. God bless her. I have 1500 hrs (H)and 250 (A) and FAA CPL H & A IR and never come across this stuff before.
So I ask the question again.
Been flying helicopters " Commercially " for 8 years for the Queen. God bless her. I have 1500 hrs (H)and 250 (A) and FAA CPL H & A IR and never come across this stuff before.
So I ask the question again.
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Because the ATPL(H) exams are not yet written (a former colleague of mine is onto the case, so soon I hope) and you have to take the ATPL (A) General Navigation exam if you want an ATPL (H). Therefore you have to learn what is, to you, pointless junk. If you only want a CPL (H), then take the CPL (H) exams. This is not then included.
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If you only want a CPL (H), then take the CPL (H) exams. This is not then included.
Im doing the CPL (H) distance learning now and its all in there along all the other mind numbing and totally irrelevant subjects that have nothing to do with helicopter flying.
Floppy
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I don't teach the CPL course, only the ATPL, but I am surprised if the convergence is covered. If it is in your notes, I suggest you check the learning objectives.
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I have checked checked the JAR FCL and Floppy thing is right.
The only exception in the CPL(H) is:
Polar, transverse and oblique general properties:
I fully agree with Floppy: Too much of this JAR stuff is irrelevant. In an ideal world all questions should be relevant to what pilots do and need to know.
Who cares about the resonant freqency of a Cap / Coil parallel cct!!!
The only exception in the CPL(H) is:
Polar, transverse and oblique general properties:
I fully agree with Floppy: Too much of this JAR stuff is irrelevant. In an ideal world all questions should be relevant to what pilots do and need to know.
Who cares about the resonant freqency of a Cap / Coil parallel cct!!!
