Q.4 Cionstant of the cone refers to the sin of the mean latitude on a Lamberts, add N10º 40’N and N41º 20’ together and divide by 2, giving 26º. Taking the sine of this gives 0.438
Q.5 Departure = Chg Long x Cos Lat, we know departure is 6hrs x 120kts so 720nm and latitude is 40, so to find chg long:
chg long (mins) = 720/cos 40, now divide chg long by 60 to find degrees and minutes then subtract this from 80 to get your answer of about 064º 20’E
Q.6 Your answer here is wrong, it is 104º (check with who you got the questions from but I have this exact question and answer is 104), to get 104:
A great circle is always convex to the nearer pole, as in the north hemisphere, we can draw this out and as such we will see that the initial rhumb line (straight line) track will be greater than the great circle track. The difference between the 2 tracks is conversion angle:
CA = 1/2 chg long (in degrees) x sin mean lat
= 0.5 x 10 x sin 54.5
= 4.07º
therefore rhumb line angle = 100 + 4.07 = 104.7º
Q.7 first you need to find departure:
dep = chg long x cos lat
= (16x60) x cos 60
= 480nm
now convert 480nm to km (x 1.852) = 888.96, convert this to meters (x 1000) then to centimeters (x100) = 88,896,000cm
This is the distance on the ground covered, so to find distance on the chart divide this by the denominator of the scale (5,000,000) which gives 17.7792cm
Wow, I actually know some Gan Nav, handy with my JAA on Wednesday... Revision is so boring
BigAir