CPL/ATPL Navigation Exam - Scale Calculation 1

[Brand Wessels]

Hoping to assist those lucky ones busy preparing for their CAA examination. The solution for this type of question, often escape some students:
A course of 120°(T) is drawn between ’X’ (61°30’N) and ’Y’ (58°30’N) on a Lambert Conformal conic chart with a scale of 1:1 000 000 at 60°N.
The chart distance between ’X’ and ’Y’ is:

Possible Answers given:

a) 66.7 cm
b) 33.4 cm
c) 38.5 cm
d) 36.0 cm

Prior knowledge required:

• Scale formula, i.e. Scale = Chart Length/Earth Distance
• Knowing that earth distance on a longitude, between two consecutive latitudes = 60nm.
• Knowing that longitudes and latitudes always cross perpendicular (90˚).
• Knowing that any track that does not run exactly East/West or North/South, likely involves basic right-angled trigonometry. (The Sin-, Cos-, Tan
ratios as well as Pythagoras)
• Knowing the conversion factors for earth distance (Nautical Miles, Kilometres and Statute Miles) to chart length (mm, cm or inches).
• Knowing how Longitudes/Latitudes are numbered and when to add or subtract to establish difference in Longitude (DLong) and Latitude (DLat).
• Knowing how to use your scientific calculator.
• Basic mathematics skills (rewriting a formula to calculate a variable, inverse operation, supplementary angles)

Solution to the question:

What do they want? They want Chart Length – which means they must have given us Scale and Earth Distance. Scale of 1:1000 000 is obvious. The Earth Distance is hidden – so we have to find that first.

We will need trigonometry for this - so we need one angle and one line length of a right-angled triangle, to calculate the hypotenuse (Track distance).

Then we use the Scale formula to calculate the Chart Length.

[V5 - LEO]

...your knowledge is incredible!

Also why I never became a CPL - dumm as mud