Ground school questions
Thread Starter
Join Date: Apr 2010
Location: Bangkok
Age: 40
Posts: 96
Likes: 0
Received 0 Likes
on
0 Posts
Ground school questions
Hi,
I have a few questions relating to the one in sixty rule and critical point. I solved the problem on one of the questions but I got stuck on another. Just wondering if anyone can check my work.
1.Critical point is usually used when
a. making sea crossing
b. flying over developed country
c. making remote area crossing
d. A and B
I think the correct answer is A and C because ETP is used to determined the time to go on and to go home when crossing the ocean or remote area whereas the PNR is used for sea crossing.
2.An aircraft obtains an on track fix "x" at 0210 utc. A constant heading of 150M is maintained until 0222 utc when a pinpoint is obtained 2 miles right of track and 15 miles from "x" if heading is altered to 138M at 0222 utc
a. be regained at 0234 utc
b. be regained at 0246 utc
c. be regained at 0258 utc
I worked out the Track error to be 8 degree and got the speed at 75 knots then totally got stuck
3.An aircraft obtains a pinpoint at "a", 2 nm left of flight path at 0515 utc. At 0530 utc, after maintaining a constant heading another pinpoint is obtained at "b", 3.5 nm right of flight path and 26 nm. From "a" the alteration of heading necessary at "b" to make good the flight planned track at a dr position "c", 38 nm from "b" and the ETA at that dr position are respectively
my answer is: 19 degree left and 0552 utc
3a.The alteration to heading at the dr position to maintain the flight path is
a. 6 degree left
b. 9 or 10 degree right
c. 6 degree right
d. 9 or 10 degree left
Since the closing angle from the previous question is 6 degree so the correct answer is 6 degree right to maintain the track.
I have a few questions relating to the one in sixty rule and critical point. I solved the problem on one of the questions but I got stuck on another. Just wondering if anyone can check my work.
1.Critical point is usually used when
a. making sea crossing
b. flying over developed country
c. making remote area crossing
d. A and B
I think the correct answer is A and C because ETP is used to determined the time to go on and to go home when crossing the ocean or remote area whereas the PNR is used for sea crossing.
2.An aircraft obtains an on track fix "x" at 0210 utc. A constant heading of 150M is maintained until 0222 utc when a pinpoint is obtained 2 miles right of track and 15 miles from "x" if heading is altered to 138M at 0222 utc
a. be regained at 0234 utc
b. be regained at 0246 utc
c. be regained at 0258 utc
I worked out the Track error to be 8 degree and got the speed at 75 knots then totally got stuck
3.An aircraft obtains a pinpoint at "a", 2 nm left of flight path at 0515 utc. At 0530 utc, after maintaining a constant heading another pinpoint is obtained at "b", 3.5 nm right of flight path and 26 nm. From "a" the alteration of heading necessary at "b" to make good the flight planned track at a dr position "c", 38 nm from "b" and the ETA at that dr position are respectively
my answer is: 19 degree left and 0552 utc
3a.The alteration to heading at the dr position to maintain the flight path is
a. 6 degree left
b. 9 or 10 degree right
c. 6 degree right
d. 9 or 10 degree left
Since the closing angle from the previous question is 6 degree so the correct answer is 6 degree right to maintain the track.
Thread Starter
Join Date: Apr 2010
Location: Bangkok
Age: 40
Posts: 96
Likes: 0
Received 0 Likes
on
0 Posts
critical point definition
Critical point is usually used when
a. making sea crossing
b. flying over developed country
c. making remote area crossing
d. A and B
a. making sea crossing
b. flying over developed country
c. making remote area crossing
d. A and B
Join Date: Apr 2011
Location: NL
Age: 38
Posts: 91
Likes: 0
Received 0 Likes
on
0 Posts
I'll try q2:
with the 1 in 60 rule the drift is 2NM times 4 (60/15 NM) is 8 degrees. So making this 8 degree heading correction will parallel the planned track. (150 deg.- 8deg. is 142 deg.)
If another 8 degree correction is made (134 deg.), the plane will cross the planned track after the same time it took to fly from x to pinpoint.
However, in the question the total correction is only 1,5 times that required. 150 deg. - 138 deg. is 12, so minus 8 leaves only 4 degrees for track regaining.
I guess this simply means it will take twice as long to regain planned track, so instead of 12 minutes, 24 minutes, hence answer B.
with the 1 in 60 rule the drift is 2NM times 4 (60/15 NM) is 8 degrees. So making this 8 degree heading correction will parallel the planned track. (150 deg.- 8deg. is 142 deg.)
If another 8 degree correction is made (134 deg.), the plane will cross the planned track after the same time it took to fly from x to pinpoint.
However, in the question the total correction is only 1,5 times that required. 150 deg. - 138 deg. is 12, so minus 8 leaves only 4 degrees for track regaining.
I guess this simply means it will take twice as long to regain planned track, so instead of 12 minutes, 24 minutes, hence answer B.
