General Navigation Problems

[Jinkster]

Having a few problems with General Navigation

If, on a lamberts chart the scale at the PO is 1:2,000,000. What is the approximate scale at the northern edge of the chart if the SPs are 16° Apart

An aircraft is flying at FL310 with an ISA deviation of –10C and a cruising mach no of 0.83. The distance between A and B is 800nm and ATC required the aircraft to slow down at 360nm from B to achieve a delay in ETA of 3mins. To what Mach number must the aircraft reduce to achieve the delay?

Answer = 0.78M

Find ISA temp 31 x 2 – 15 = -47 +10 = -57C
Find TAS using whiz wheel ISA temp and 0.83M# = 457kts

Therefore new TAS is ???? –then convert M# using whiz wheel



A route over the North Atlantic has three waypoints (WP). WP1 is at 60N 030W, WP2 is at 60N 020W and WP3 is at 60N 010W. Assuming still air conditions the course change at WP2 will be:

8° to the left

Ch Long x sin mean lat = convergence angle
Ch long x sin PO = 10 x sin 60 = 8.6 (not sure which direction to turn)

Thanks all for the help

Jinkster

[Send Clowns]

There is no requirement in the syllabus to calculate the scale change across a Lambert's Chart. You should however be aware that the chart is at minimum scale at the parallel of origin, maximum at the northerly and southerly edges. With SPs 16 degrees apart the scale variation from the nominal (SP) scale is about 1%, so the answer should be about 2% larger, or 1:1,960,000. The last bit is not relevant to the course.

Second one we need a wind to calculate accurately. Still-air the time for 360 nm at 457 kts would be about 47 minutes, so 50 minutes with the delay. 360 nm in 50 minutes is 430 kts TAS. Try that (do not have my CRP-5 here.

I'll be back - have to go teach.

Send Clowns
BCFT

[Send Clowns]

Apologies for the delay in further response. I suppose since you are getting for free what I usually charge £25 an hour for I don't really need to apologise ;)

The final one is not a convergence question, although you are right, the answer is 8.66 degrees, closer to 9. One report of the actual exam question gave it at 64N rather than 60N, giving pretty much 9 degrees as the answer.

Now for technique. The rhumb-line track from (1) to (2) is 090, as is the rhumb-line track (2) to (3). They are all on the same parallel, so the rhumb line between them must be that parallel, which runs East-West.

Draw a diagram with the three meridians (30W, 20W and 10W) and the 60N parallel. The diagram should look basically like a Lambert's chart graticule - meridians converging to the north, parallel a distinct curve concave to the north. Label WPTs (1), (2) and (3). These are the intersections of the meridians and the parallel.

Join (1) to (2) with a straight line, extending the line beyond (2) but remaining straight (should be south of the parallel beyond (2)). That represents the great-circle track, ans the original question mentions FMS or INS, either of which will direct you on a great circle. Draw a second straight line from (2) to (3), again representing a great circle.

It now becomes obvious that at (2) the aircraft must turn left to fly towards (3). By how much? To turn left onto 090 it would have to turn left by the conversion angle (CA) between (1) and (2), as the current course is the great circle and 090 the rhumb line - CA is th difference between the two.

CA = 1/2 x ch.long. x sin mean lat = 1/2 x 10 x sin 60 = 4.33

To turn from 090 to the new track is again CA, this time between (2) and (3), as 090 is the rhumb line and the new track the great circle. since the ch.long is the same as is the latitude, this is again 4.33 degrees. So the total left turn is

4.33 + 4.33 = 8.66 degrees

This is far easier with a diagram so if you want a better explanation send me a PM with your email address and I will draw and email one.

[Alex Whittingham]

Jinkster, I'd be interested to hear where your first question came from. I've never seen anything like it in the feedback we have.