Indigo Call letters for Freshers
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some more problems...
Q 1.a.Aircraft Height= 2500ft which = 0.4111Nm (approx of 1Nm=6080ft)
b. Using 1 in 60 rule we have, Required distance= (0.4111 Nm / 3° ) X60
c. Required D = 8.22 Nm
Note = you can also use trigonometry but you will get answer of 7.8Nm
Q.2.Convergency = change of Longitude X sine of Lat (mean Lat)
= 18 X Sine 60 = 15.5°
Convergent Angle = 15.5/2 which Equals 7.8°
Q.3 Convergency between 030°W & 025°W = 4.33 CA= 2.2
departure= {(Ch"long X60) X Cos Lat} Departure = {(5°X60) x Cos 60}= 150Nm
Using 1 in 60 rule Find distance off track : 2.2 = {Dot / 150 } X 60 = 5.5Nm
Therefore 5.5Nm/60 =5°30' which equates 60°60'S
Q4.departure @ 53° N = 12999Nm ,
Scale @53°N ={133/12999}=18100000 aprx
Scale @ 30 N = {18100000 X cos 30}/Cos 53 = 26000000(apprx to 25000000)
Q.5 convergence = Dlong X sinLat
30 = 40 X sin lat
Sinlaat (constant of cone) = 30/40= 0.75
b. Using 1 in 60 rule we have, Required distance= (0.4111 Nm / 3° ) X60
c. Required D = 8.22 Nm
Note = you can also use trigonometry but you will get answer of 7.8Nm
Q.2.Convergency = change of Longitude X sine of Lat (mean Lat)
= 18 X Sine 60 = 15.5°
Convergent Angle = 15.5/2 which Equals 7.8°
Q.3 Convergency between 030°W & 025°W = 4.33 CA= 2.2
departure= {(Ch"long X60) X Cos Lat} Departure = {(5°X60) x Cos 60}= 150Nm
Using 1 in 60 rule Find distance off track : 2.2 = {Dot / 150 } X 60 = 5.5Nm
Therefore 5.5Nm/60 =5°30' which equates 60°60'S
Q4.departure @ 53° N = 12999Nm ,
Scale @53°N ={133/12999}=18100000 aprx
Scale @ 30 N = {18100000 X cos 30}/Cos 53 = 26000000(apprx to 25000000)
Q.5 convergence = Dlong X sinLat
30 = 40 X sin lat
Sinlaat (constant of cone) = 30/40= 0.75
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Question
Answer
NPA should be 90, not NAP, right?
My understanding is that we need to find the true course at B(which isn't really angle NBA). You have done a great job ITAviator.
Since it is a Polar Stereographic Chart(PSC), Chart convergency = dLong = 102 deg.
Also, in a PSC, above 70N, Great Circles are straight lines. So, APB is a Great Circle. True course of this GC at A can be found out since we have already calculated angle NAP.
True course of GC at A is 360-55=305.
Convergency is equal to the difference between the GC directions between two given meridians. So, difference between true course of GC at A and B is convergency i.e. 102. Also, westerly in N.H., GC direction decreases.
So, true course of GC at B=305-102= 203
Correct me if wrong, please.
Originally Posted by KKAARR
Q. Two positions plotted on a Polar Stereographic A 80N000E/W B 70N102W are joined by a Straight line whose highest lattitude is reached at 035W. At point B, the true course is ?
a) 230
b) 203
c) 023
d) 030
a) 230
b) 203
c) 023
d) 030
Originally Posted by ITAviator
Diagram has to be drawn, which i cannot here..
A = 80N000E/W lies on smaller circle of 80N intersecting at 000 E/W
B = 70N102W lies on bigger circle of 70N intersecting at 102W
Centre of circle is North Pole (N)
From given data: we get a triangle ANB, Angle ANB=102 (Long diff between A and B),
Required is angle NBA, which is True Course from B to A.
Highest lat=35W, a line drawn from N perpendicular
to AB at point P.
Hence Angle ANP = 35 deg.
In triangle ANP, Ang ANP=35, and NAP=90, hence ang PAN = 180-(90+35) = 55
Now with this data, in triangle ANB, ang ANB=102, ang PAN/BAN = 55, hence ang NBA = 180-(102+55) = 023
A = 80N000E/W lies on smaller circle of 80N intersecting at 000 E/W
B = 70N102W lies on bigger circle of 70N intersecting at 102W
Centre of circle is North Pole (N)
From given data: we get a triangle ANB, Angle ANB=102 (Long diff between A and B),
Required is angle NBA, which is True Course from B to A.
Highest lat=35W, a line drawn from N perpendicular
to AB at point P.
Hence Angle ANP = 35 deg.
In triangle ANP, Ang ANP=35, and NAP=90, hence ang PAN = 180-(90+35) = 55
Now with this data, in triangle ANB, ang ANB=102, ang PAN/BAN = 55, hence ang NBA = 180-(102+55) = 023
My understanding is that we need to find the true course at B(which isn't really angle NBA). You have done a great job ITAviator.
Since it is a Polar Stereographic Chart(PSC), Chart convergency = dLong = 102 deg.
Also, in a PSC, above 70N, Great Circles are straight lines. So, APB is a Great Circle. True course of this GC at A can be found out since we have already calculated angle NAP.
True course of GC at A is 360-55=305.
Convergency is equal to the difference between the GC directions between two given meridians. So, difference between true course of GC at A and B is convergency i.e. 102. Also, westerly in N.H., GC direction decreases.
So, true course of GC at B=305-102= 203
Correct me if wrong, please.
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@aditya104
Thanks for your method, I usually opt trigo for solving as I find it easier, although your approach is great.
Although the last step should be: 203-180 = 023
Reason: N is North, the True Course at A for B is 305, hence the corresponding course from B to A has to approx be less than 180, its not practical otherwise. You can try drawing a general Lambert's figure for same to get a better idea.
Note: The angle NBA is the true course at B. True course is always, if not mentioned otherwise, a GC track. Here the line B to N i.e. BN is True North and BA is GC, hence NBA is True GC track at B for A.
Although the last step should be: 203-180 = 023
Reason: N is North, the True Course at A for B is 305, hence the corresponding course from B to A has to approx be less than 180, its not practical otherwise. You can try drawing a general Lambert's figure for same to get a better idea.
Note: The angle NBA is the true course at B. True course is always, if not mentioned otherwise, a GC track. Here the line B to N i.e. BN is True North and BA is GC, hence NBA is True GC track at B for A.
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Originally Posted by KKAARR
Q. Two positions plotted on a Polar Stereographic A 80N000E/W B 70N102W are joined by a Straight line whose highest lattitude is reached at 035W. At point B, the true course is ?
a) 230
b) 203
c) 023
d) 030
a) 230
b) 203
c) 023
d) 030
If it is-What is the value of initial straight line track from B to A? Then ITAviator's answer 023 is correct.
If it asked-What is the value of the final straight line track from A to B? Then, my answer of 203 is correct.
We are both correct.
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Originally Posted by captaa
Q4
The total length of the 53°N parallel of latitude on a direct Mercator chart is 133 cm. What is the approximate scale of the chart at latitude 30°S?
A)1 : 30 000 000 B)1 : 18 000 000 C)1 : 21 000 000 D)1 : 25 000 000
The total length of the 53°N parallel of latitude on a direct Mercator chart is 133 cm. What is the approximate scale of the chart at latitude 30°S?
A)1 : 30 000 000 B)1 : 18 000 000 C)1 : 21 000 000 D)1 : 25 000 000
Originally Posted by captaa
how is dep at 53N is 12999?
Departure
=dLong(in mins) x cos Lat
=360 x 60 x cos53
=12,999nm
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@ flexiflyer
Paper was easy for those who read Human performance in depth..
60 % of paper was from Human Factors ( Good Q.s ) ,
10 % from Air Regs,
15% from Instruments & Radio Aids,
& Remaining 15 % from Gen Nav, Performance & Theory of Flight !!
23-25 people made on Day 1 (including me) !!
60 % of paper was from Human Factors ( Good Q.s ) ,
10 % from Air Regs,
15% from Instruments & Radio Aids,
& Remaining 15 % from Gen Nav, Performance & Theory of Flight !!
23-25 people made on Day 1 (including me) !!
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@KKAARR
Can u plz list down some questions??????
Did they asked abt visibility mimina's, marshalling or something in else in regs????
And wht questions were asked frm performance topic?????
Did they asked abt visibility mimina's, marshalling or something in else in regs????
And wht questions were asked frm performance topic?????
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So on the 17th the exam was pretty easy. Honestly not tough. If youve really studied, you would clear.
I'm going to try and put down as many questions as i can.
Human performance, in my exam was the key, so that, you have to do thoroughly, there were a bunch of questions so you have to really do that book inside out, again this was for my exam I've heard in the past that many questions haven't come from this topic.
QUESTIONS FROM MY EXAM
REGS
emergency codes, highjack radio comm failure.
ifr altitudes to be maintained inside and outside controlled airspace.
light signals double cross at an airport - that is glider flying in progress.
MET
tropopause altitude at the equator and at poles.
windshear, is it caused due to updrafts or downdrafts below a cb.
a metar question where they gave a code it was a strange one it was numbers, atleast 8 of them and they asked what is the state of the runway.
raining, snow, broken, or braking coefficient which was .28 not sure about the answer.
HUMAN PERFORMANCE
Tons of questions, hypoxia, hyperventilation, SHELL model, psychological processes. Honestly I dont remember them all. BUT DO THIS TOPIC WELL.
INSTRUMENTS- not much
Magnetism- directive force.
DG.
Mach No, speed of sound, TAS calculations.
GENERAL/ BASIC NAV
descent
lots of calculations on G/S determining wind speeds etc - Flight computer can be used for most.
ISOGRIVES - not sure what that was but it might be lines connecting places of equal grivation. Thats what I marked atleast.
RADIO NAV
Nothing came from here. Surprisingly.
OK IMPORTANT ONE HERE. - FLIGHT PLANNING.
atleast 5 questions came from here. cant recall them all, but there were some regarding particular items in a flight plan. item no.7, item no. 9, wake turbulence categories.
I hadnt done anything from flight planning, so it was a bit of an issue.
AERODYNAMICS
10 Questions towards the end of the paper
mainly about stalling, what affects stall speed, stall speed in a 60 degree turn 1.41 percent etc.
Position of CP in relation to Aerodynamic center in a symmetrical aerofoil.
CL= 0 for a positively cambered aerofoil at what angle of attack positive, negative, neutral.
PERFORMANCE
Nothing from here as far as i remember.
OK GUYS SO AGAIN THIS IS WHAT MY PAPER WAS LIKE. So any advice that i may have offered may not be what you have to study, because the exam varys all the time. If I could give any useful advice, well then just study hard I know everybody says this, but it does pay off, no shortcuts.
Hope this helps. If I do remember anything else I will keep you'll posted.
A special shoutout to Gambl3, really appreciated the responses to almost every question that I posted on this thread. Thanks alot.
I'm going to try and put down as many questions as i can.
Human performance, in my exam was the key, so that, you have to do thoroughly, there were a bunch of questions so you have to really do that book inside out, again this was for my exam I've heard in the past that many questions haven't come from this topic.
QUESTIONS FROM MY EXAM
REGS
emergency codes, highjack radio comm failure.
ifr altitudes to be maintained inside and outside controlled airspace.
light signals double cross at an airport - that is glider flying in progress.
MET
tropopause altitude at the equator and at poles.
windshear, is it caused due to updrafts or downdrafts below a cb.
a metar question where they gave a code it was a strange one it was numbers, atleast 8 of them and they asked what is the state of the runway.
raining, snow, broken, or braking coefficient which was .28 not sure about the answer.
HUMAN PERFORMANCE
Tons of questions, hypoxia, hyperventilation, SHELL model, psychological processes. Honestly I dont remember them all. BUT DO THIS TOPIC WELL.
INSTRUMENTS- not much
Magnetism- directive force.
DG.
Mach No, speed of sound, TAS calculations.
GENERAL/ BASIC NAV
descent
lots of calculations on G/S determining wind speeds etc - Flight computer can be used for most.
ISOGRIVES - not sure what that was but it might be lines connecting places of equal grivation. Thats what I marked atleast.
RADIO NAV
Nothing came from here. Surprisingly.
OK IMPORTANT ONE HERE. - FLIGHT PLANNING.
atleast 5 questions came from here. cant recall them all, but there were some regarding particular items in a flight plan. item no.7, item no. 9, wake turbulence categories.
I hadnt done anything from flight planning, so it was a bit of an issue.
AERODYNAMICS
10 Questions towards the end of the paper
mainly about stalling, what affects stall speed, stall speed in a 60 degree turn 1.41 percent etc.
Position of CP in relation to Aerodynamic center in a symmetrical aerofoil.
CL= 0 for a positively cambered aerofoil at what angle of attack positive, negative, neutral.
PERFORMANCE
Nothing from here as far as i remember.
OK GUYS SO AGAIN THIS IS WHAT MY PAPER WAS LIKE. So any advice that i may have offered may not be what you have to study, because the exam varys all the time. If I could give any useful advice, well then just study hard I know everybody says this, but it does pay off, no shortcuts.
Hope this helps. If I do remember anything else I will keep you'll posted.
A special shoutout to Gambl3, really appreciated the responses to almost every question that I posted on this thread. Thanks alot.
Last edited by thearsenal; 19th Apr 2012 at 06:07.
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Radio Aids Ques
Can Some body please solve this for me :
What is the slant range error for an aircraft flying at 9000 feet absolute altitude above a DME located at elevation 2000 ft, when the slant range is 12 NM?
A)0,09 NM.
B)0,57 NM.
C)1,42 NM.
D)0,31 NM.
What is the slant range error for an aircraft flying at 9000 feet absolute altitude above a DME located at elevation 2000 ft, when the slant range is 12 NM?
A)0,09 NM.
B)0,57 NM.
C)1,42 NM.
D)0,31 NM.
Last edited by buddyaviator; 18th Apr 2012 at 12:07.
What is the slant range error for an aircraft flying at 9000 feet absolute altitude above a DME located at elevation 2000 ft, when the slant range is 12 NM?
A)0,09 NM.
B)0,57 NM.
C)1,42 NM.
D)0,31 NM.
A)0,09 NM.
B)0,57 NM.
C)1,42 NM.
D)0,31 NM.
The difference between slant range and true horizontal range can be tested using the following equation:
True Range = √(Slant Range squared – Aircraft Height squared)
The wording of the question is rather curious in that it states that the “an aircraft at 9000 feet absolute altitude above a DME located at 2000 feet elevation”. This should be interpreted as meaning that the height of the aircraft is 9000 feet above the DME.
Multiplying by 1 nm / 6080 feet converts the 9000 feet into 1.48 nm
With the aircraft at 1.48 nm above the DME at slant range of 12 nm we have the following:
True Range = √(Slant Range squared – Aircraft Height squared)
True Range = √(12 nm squared – 1.48 nm squared) = 11.91.
So the slant range is 12 nm and the true range is 11.91 nm, giving an error of 0.09 nm.
An aircraft flying at flight level 250 wishes to interrogate a DME beacon situated 400 ft above mean sea level.
What is the maximum range likely to be achieved?
A)198 nm.
B)222 nm.
C)175 nm.
D)210 nm.
What is the maximum range likely to be achieved?
A)198 nm.
B)222 nm.
C)175 nm.
D)210 nm.
Because radio waves travel in straight lines, the maximum range of DME is limited by the height of the aeroplane, the height of the transmitter and the curvature of the earth.
Maximum range can be calculated using the standard equation:
Max range = 1.25 x (√ aircraft height + √ ground station height)
The aircraft is at FL250 which is 25000 feet.
The DME ground station is at 400 feet so:
Max range = 1.25 x (√ 25000 ft + √ 400 ft) = 222.64 nm.
The closest option is 222 nm.
Last edited by keith williams; 18th Apr 2012 at 13:33.