I have one more question to ask you. Could you please check to see if I am on the right track?
Given: Track required left 090M, HDG 102M
ADF (destination) 000R DME (destination) 21 NM
According to the "one in sixty rule" the aeroplane is
a. 12 NM right of required track
b. 4 NM right of required track
c. 12 NM left of required track
d. 4 NM left of required track

The difference between heading and track is 12 degree to the right so the drift is 12 degree left which means that the wind is coming from the right so the airplane is on the left of the track. Am I on the right track? According to the one in sixty rule, every mile off track along a sixty miles along track is equal to one degree of track error. Now what is the next step?
Cheers

Could someone kindly guide me to the right direction?
Thanks

So the correct answer is d. latitude, longitude, date, cloud cover, reduced visibility, and terrain?

What is the origin of the question?
Under whose syllabus are you being examined?

I ask because (for example) the JAA syllabus has no such term as 'time of last light' in the learning objectives. Does this term have a specific definition or is it just being used by the author to describe 'when it's too dark to see'.

BTW folks there is a big glaring RTFQ in here. (C) wouldn't even be correct for an 'end of twilight' definition because the question states UTC not LMT. So we need Lat, Long and Date for starters. Only (D) has all 3.

Problem is, discussions about multiple choice questions really don't work well on a bulletin board like Pprune. They're more for a tutor at a training provider.

There are a great many people here with a lot of experience, much of which just adds, not reduces, complexity in any given problem. Multiple choice exam questions, whilst inevitable, are often very simplistic.

Nobody on Pprune does simplistic very well!

G

Your last question is very unclear.

Maybe this helps:

The 1 in 60 rule in formula form is:

(1 degree) / 60 = (1 NM) / (60 NM)

Finding an ETA with 1 in 60
 

An aircraft obtains a pinpoint at "A", 2 NM left of flight path at 0515 UTC. At 0530 UTC, after maintaining a constant heading (and TAS) another pinpoint is obtained at "B", 3.5 NM right of the flight path and 26 NM. From "A", the alteration of heading necessary at "B" (to make good the FPT at a DR position "C", 38 NM from "B" and the ETA at that DR position and respectively.

I worked out the problem and I got 19 degree left but I could not figure out the ETA time. Does anyone know how to work this out?

Cheers

I agree with 18 left (close enough huh), but for the ETA I went with:

Time = 38 x 15 divided by 26 = 22 mins

So an extra 22 mins for the remaining 38 miles at same speed.

Added to 0530 = 0552 UTC.

Any thoughts?

Tuuapache

I echo Genghis's comments!

These are questions more suited for your tutor than a bulletin board because the spectrum of experience here really is very wide and the answers will vary in accuracy accordingly!

This thread will be locked in 24 hours!

Halfwayback

Convert GMT to Local time
 

Does anyone know how to convert GMT to local time?

GMT 0200 hrs, 30 June, local longitude 68 degrees 21 minutes, E. What is the local time? (conversion factor is +5)
What I did was I converted the arc to time. 68 degrees 21 minutes E is 4 HR 33 MIN then I got stuck. What am I supposed to do next?
Cheers

Are you serious? Just in case you are, do you know what 2+5=?

what about the longitude 68 degrees 21 minutes, E?
What does this have anything to do with the question?