Hey guys.
I don’t get this one, and I have seen several explanations for it, among others right here, but in another thread, but I haven’t considered them to be satisfactory, so I’ll give it another go.
Given: Waypoint 1 is 60°S 030°W, waypoint 2 is 60°S 020°W. What will be the approximate latitude on the display of an inertial navigation system at longitude 025°W?
a. 60°06’S
b. 59°49’S
c. 60°00’S
d. 60°11’S
The correct answer is supposed to be, a. 60°06’S.
My solution was:
0.5 x 10° x sin(60) = 4.33°, conversion angle.
5° x cos(60) x 60 =150 nm, departure.
Then do some trig-stuff, or 1-60 rule, whichever, to come up with roughly 11 nm, which would make d. an appropriate answer in my opinion.
Most of the explanations I have seen for this question seem intent on halving the conversion angle, or rather getting the conversion angle from the ch.long between 030°W and 025°W, to get 2.2°-ish, and obviously that would present an answer more consistent with the correct one. But why is this done? I don’t see how that’s relevant as it would imply that we are flying from 60°N 030°W to 60°N 025°W and then on to 60°N 020°W, in which case I would fully agree with the answer, but that would mean that 025°W was an additional waypoint… Unless I'm missing something else.
-Anders