Hi,
Just wondering if someone can guide me to the right direction.
1. 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. Since I don't know the rest of the distance then how can I find the closing angle?

2. 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. 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
I found the closing angle to be 6 degree therefore in order to maintain the flight path I have to turn right by the same amount of the closing angle which is 6 degree.
Is this correct?

3. Given: HDG 085 M, OBS 255
TO/FROM TO, CDI 2 dots left
The aeroplane is on the
a. 071 radial
b. 075 radial
c. 255 radial
d. 251 radial
I take 255-180 = 075. 2 dots left is 4 degrees less. therefore 075-004 = 071

4. Track required 090M, HDG 102M, ADF(destination) 000R, DME(destination) 21NM
According to the "one in sixty rule", the aeroplane is
a. 12 NM right of the required track
b. 4 NM right of the required track
c. 12 NM left of required track
d. 4 NM left of required track
I have no clue how to solve this question.
Cheers

G'day,

I'm just about to duck off, but I might as well make a start on your questions

Here's question #1:

There has been a HDG change of 12 degrees right. You got the TE of 8 degrees which means we must be working on a CA of 4 degrees. Think about how long it will take us to close the 2 nm if we use a closing angle of half the track error. You can also shuffle the 1-in-60 formula to find out the distance to the intercept point and using your calculated GS work out an ETA.

Here's the solution - click the spoiler to see it.

Got to dash. Other answers later, unless someone else wants to have a crack at them

Cheers,

Rich

Ok, now for the other questions:

Question #2:

You need to close the 3.5nm of off-track distance in the 38 nm from B to the DR point at C. This is a CA of close enough to 6 degrees. You were right of track at B so once you have reached the DR point at C you will turn 6 degrees right to track along the flight planned track.


Question #3:

With the OBS on 255 TO and the CDI centred, you would be on the 075 degree radial. You have 2 dots left deflection on the CDI though which, as you said, equates to a difference of 4 degrees. You are 4 degrees to the right of the 255 degree track TO the station which is the same as being 4 degrees to the left of the 075 degree radial FROM the station. Therefore you must be on the 071 degree radial. These situations become clearer if you draw a quick situation diagram.

Question #4:

To solve this question, put all your information together and draw a situation diagram. Your FPT to the destination is 090. You are heading 102 and the NDB at the destination is right on the nose since your ADF is reading 000R. You'd have to assume this relative bearing has been constant which means your TMG is 102 degrees and you are experiencing zero drift. If you had drift the ADF would not be reading 000R.

If you are actually tracking 102 and are heading directly inbound for the destination then you must be to the West-Northwest of the destination. You really wanted to be tracking 090 inbound so you are North (i.e. left) of your FPT. If this difficult to follow, again, draw a diagram to see how it all fits together.

The DME says you are 21 nm out and you have a 12 degree difference between your FPT and the TMG so rearranging the 1-in-60 formula gives us:

TE = (d x 60) / D ====> (TE x D ) / 60 = d

Putting in our values: (12 x 21) / 60 = 4.2.

We are 4.2 nm off track so the answer is (d) 4 nm left of track.

Cheers,

Rich

P.S. I'm on an iPad so I can't upload any situation sketches. If the answers make no sense, I'll prepare some diagrams and upload them when I'm back at the office.