Indigo Call letters for Freshers
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1 act QDM 330° act HDG 060° req QDM 350° Intercept HDG?
QDM 330
QDR = 330 - 180 = 150 (Seg AC)
So a/c on radial 150 (Point C)
req QDM = 350, QDR = 170 (Seg AD)
interception angle 45 deg (if angle is not mentioned). *answer to question 3*
angle CED = 45
AD parallel to BC
CD perpendicular to AD and BC
In triangle EDC, CED = ECD = 45 deg and EDC = 90 deg
DAC = BCA = 20 deg (QDR 170 - QDR 150)
angle BCD = BCA + ACE + ECD
90 = 20 + ACE + 45
ACE = 25 deg
So if the a/c at point C on seg AC (QDM 330/QDR 150) and wants to intercept seg AD (QDM 350/QDR 170) at 45 deg, it will have to turn left to a heading of 330 - ACE = 330 - 25 = 305.
2 .act QDM 140° act HDG 020° req QDM 110° First turn?
Similarly
45 deg intecept
act QDM 140 + 15 = 155
ans : Right hdg 155 (dont know how 'Left')
QDM 330
QDR = 330 - 180 = 150 (Seg AC)
So a/c on radial 150 (Point C)
req QDM = 350, QDR = 170 (Seg AD)
interception angle 45 deg (if angle is not mentioned). *answer to question 3*
angle CED = 45
AD parallel to BC
CD perpendicular to AD and BC
In triangle EDC, CED = ECD = 45 deg and EDC = 90 deg
DAC = BCA = 20 deg (QDR 170 - QDR 150)
angle BCD = BCA + ACE + ECD
90 = 20 + ACE + 45
ACE = 25 deg
So if the a/c at point C on seg AC (QDM 330/QDR 150) and wants to intercept seg AD (QDM 350/QDR 170) at 45 deg, it will have to turn left to a heading of 330 - ACE = 330 - 25 = 305.
2 .act QDM 140° act HDG 020° req QDM 110° First turn?
Similarly
45 deg intecept
act QDM 140 + 15 = 155
ans : Right hdg 155 (dont know how 'Left')
Last edited by gAMbl3; 12th Apr 2012 at 13:48.
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@gambl3
cheers for the response man.
one more for you or anybody
An aircraft is over flying a VOR at 30.000 ft, at a groundspeed of 300 kt. The maximum time during which no usable signals will be received (in minutes and seconds) is:
A) 4.40
B) 1:40
C) 2.25
D) 0.50
cheers for the response man.
one more for you or anybody
An aircraft is over flying a VOR at 30.000 ft, at a groundspeed of 300 kt. The maximum time during which no usable signals will be received (in minutes and seconds) is:
A) 4.40
B) 1:40
C) 2.25
D) 0.50
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1.Aircraft Height = 30,000 Ft which= 4.9NM (taking 6080 ft = 1 Nm)
2.Assuming VOR elevation angle of 40 degrees (default)
3.Using Trigonometry : (right angled Triangle)
a. Tan 40 Degrees = 4.9 NM / distance traveled (D)
b. D= 4.9 / Tan 40 Degrees therefore D= 5.8 NM
c. find Time taken At G/S of 300 Kts to cover 5.8 NM ?
d. time taken = (5.8 NM / 300 NM / hr ) x 2 = 2.32 approx to 2.25
2.Assuming VOR elevation angle of 40 degrees (default)
3.Using Trigonometry : (right angled Triangle)
a. Tan 40 Degrees = 4.9 NM / distance traveled (D)
b. D= 4.9 / Tan 40 Degrees therefore D= 5.8 NM
c. find Time taken At G/S of 300 Kts to cover 5.8 NM ?
d. time taken = (5.8 NM / 300 NM / hr ) x 2 = 2.32 approx to 2.25
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An aircraft is over flying a VOR at 30.000 ft, at a groundspeed of 300 kt. The maximum time during which no usable signals will be received (in minutes and seconds) is:
As per ICAO, VOR shall provide signal up to a minimum elevation angle of 40 deg.
Point B is directly overhead VOR at height 30,000 ft.
So angle AOB = 90 - 40 = 50 deg
AB x 2 = total distance in which there will no usable signal.
AB = OB x tan 50 = 30,000 x tan 50 ft.
AB x 2 = 60,000 x tan 50 ft = (60,000 x tan 50) / 6080 nm = 11.76 nm
groundspeed = 300 kts
Time = 11.76 / 300 = 2 min 21 sec
As per ICAO, VOR shall provide signal up to a minimum elevation angle of 40 deg.
Point B is directly overhead VOR at height 30,000 ft.
So angle AOB = 90 - 40 = 50 deg
AB x 2 = total distance in which there will no usable signal.
AB = OB x tan 50 = 30,000 x tan 50 ft.
AB x 2 = 60,000 x tan 50 ft = (60,000 x tan 50) / 6080 nm = 11.76 nm
groundspeed = 300 kts
Time = 11.76 / 300 = 2 min 21 sec
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@ KKAARR:bro do we count Groundspeed fpr Still air gradient ? ( instead of TAS )
I agree, there is a confusion here..
Its my understanding that gradient will be the same any which way. But if the wind is no longer still, we use the same gradient and GS. TAS can be used only in still air conditions, but in the example, we clearly have 50kts headwind..
Consider we have to clear an obstacle during T/O, and have a authorized gradient for still wind, but have a tailwind component. We cannot calculate the ROC using TAS.
Hence, for the required problem, we would need to use GS.. Its of course debatable.
Its my understanding that gradient will be the same any which way. But if the wind is no longer still, we use the same gradient and GS. TAS can be used only in still air conditions, but in the example, we clearly have 50kts headwind..
Consider we have to clear an obstacle during T/O, and have a authorized gradient for still wind, but have a tailwind component. We cannot calculate the ROC using TAS.
Hence, for the required problem, we would need to use GS.. Its of course debatable.
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Can anybody give me the right formula for approximate maximum range for VORs? I have ROOT 1.5 x height in feet. I tried a couple of sums but couldnt get the answer.
An aircraft at FL 100 should be able to receive a VOR ground station at 100 FT above MSL at an approximate maximum range of:
A) 123 NM
B) 135 NM
C) 142 NM
D) 130 NM
Either one of these is the correct answer. Dont know how though.
An aircraft at FL 100 should be able to receive a VOR ground station at 100 FT above MSL at an approximate maximum range of:
A) 123 NM
B) 135 NM
C) 142 NM
D) 130 NM
Either one of these is the correct answer. Dont know how though.
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1.23 root h1 + 1.23 root h2 = max range
h1 = height of VOR station
h2 = height of a/c
1.23 root 100 + 1.23 root 10000 = 12.3 + 123 = 135.3
If only Flight level is given and height of station = 0 then use 12 root FL i.e. 12 root 100 (if in the above question h1 = 0)
h1 = height of VOR station
h2 = height of a/c
1.23 root 100 + 1.23 root 10000 = 12.3 + 123 = 135.3
If only Flight level is given and height of station = 0 then use 12 root FL i.e. 12 root 100 (if in the above question h1 = 0)
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@gambl3
cheers man, you're the go to guy here. really appreciate all the responses man, really nice of you.
The time interval between the transmission of a given DME interrogation pulse and the reception of the appropriate response pulse at the aircraft is 2 milli seconds. The slant range is:
A) 162 nm
B) 92 nm
C) 73 nm
D) 323 nm
cheers man, you're the go to guy here. really appreciate all the responses man, really nice of you.
The time interval between the transmission of a given DME interrogation pulse and the reception of the appropriate response pulse at the aircraft is 2 milli seconds. The slant range is:
A) 162 nm
B) 92 nm
C) 73 nm
D) 323 nm
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The time interval between the transmission of a given DME interrogation pulse and the reception of the appropriate response pulse at the aircraft is 2 milli seconds. The slant range is:
Time interval = 2 mili sec
Speed of radio wave = 3 x 10^8 m/s
Distance traveled by the pulse in 2 mili sec = (2 x 10^-3) x (3 x 10^8) m = 6 x 10^ 5 m = (6 x 10^5) / 1852 nm = 323.9 nm
In 2 mili sec the pulse has traveled from the a/c to the ground station and back to the a/c.
323.9 nm = distance traveled by the pulse from the a/c to the ground station and back to the a/c.
So distance (slant range) of a/c from ground station = 323.9/2 = 161.9 nm
Time interval = 2 mili sec
Speed of radio wave = 3 x 10^8 m/s
Distance traveled by the pulse in 2 mili sec = (2 x 10^-3) x (3 x 10^8) m = 6 x 10^ 5 m = (6 x 10^5) / 1852 nm = 323.9 nm
In 2 mili sec the pulse has traveled from the a/c to the ground station and back to the a/c.
323.9 nm = distance traveled by the pulse from the a/c to the ground station and back to the a/c.
So distance (slant range) of a/c from ground station = 323.9/2 = 161.9 nm
Last edited by gAMbl3; 13th Apr 2012 at 17:31. Reason: Typo
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Distance traveled by the pulse in 2 mili sec = (2 x 10^-6) x (3 x 10^8) m = 6 x 10^ 5 m = (6 x 10^5) / 1852 nm = 323.9 nm
CAN U PLEASE EXPLAIN Y DID U TAKE 10^-6 instead of 10^-3. CANT UNDERSTAND.
CAN U PLEASE EXPLAIN Y DID U TAKE 10^-6 instead of 10^-3. CANT UNDERSTAND.
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2 millisec = 2/1000 sec = 2x10*-3 seconds = Time
Speed = Speed of sound = Speed of Radio wave = 300000000 m/s =3x10*8 m/s
Distance = Speed x Time
= (2x10*-3) x (3x10*8) = 6,00,000 m = 600 km
1nm = 1.85 km
Hence, 600km = 600/1.85 = 323.9 nm
This distance is to and fro distance
Hence, 323.9/2 = 161.95 nm
Speed = Speed of sound = Speed of Radio wave = 300000000 m/s =3x10*8 m/s
Distance = Speed x Time
= (2x10*-3) x (3x10*8) = 6,00,000 m = 600 km
1nm = 1.85 km
Hence, 600km = 600/1.85 = 323.9 nm
This distance is to and fro distance
Hence, 323.9/2 = 161.95 nm
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some problems...
Q1
Given:
aircraft height 2500 FT,
ILS GP angle 3°.
At what approximate distance from THR can you expect to capture the GP?
a)8.3 NM b)7.0 NM c)13.1 NM d)14.5 NM
Q2
The angle between the true great-circle track and the true rhumb-line track joining the following points: A (60° S 165° W) B (60° S 177° E), at the place of departure A, is:
A)7.8° B)9° C)15.6° D)5.2°
Q3
"Given: Waypoint 1. 60°S 030°W
Waypoint 2. 60°S 020°W
What will be the approximate latitude shown on the display unit of an inertial navigation system at longitude 025°W?"
A)060°11'S B)059°49'S C)060°00'S D)060°06'S
Given:
aircraft height 2500 FT,
ILS GP angle 3°.
At what approximate distance from THR can you expect to capture the GP?
a)8.3 NM b)7.0 NM c)13.1 NM d)14.5 NM
Q2
The angle between the true great-circle track and the true rhumb-line track joining the following points: A (60° S 165° W) B (60° S 177° E), at the place of departure A, is:
A)7.8° B)9° C)15.6° D)5.2°
Q3
"Given: Waypoint 1. 60°S 030°W
Waypoint 2. 60°S 020°W
What will be the approximate latitude shown on the display unit of an inertial navigation system at longitude 025°W?"
A)060°11'S B)059°49'S C)060°00'S D)060°06'S
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some more problems...
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
Q5
The constant of the cone, on a Lambert chart where the convergence angle between longitudes 010°E and 030°W is 30°, is:
A)0.50 B)0.64 C)0.75 D)0.40
THX......
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
Q5
The constant of the cone, on a Lambert chart where the convergence angle between longitudes 010°E and 030°W is 30°, is:
A)0.50 B)0.64 C)0.75 D)0.40
THX......