Ch. 27: Vertical Navigation

Question 1

What is the approximation of “1 NM = 6000 ft” from?

Answer 1

1:60 rule

Question 2

For descent problems, which wind vector and TAS is used?

Answer 2

The wind vector / TAS corresponding to the altitude halfway through the descent.

Question 3

For climb problems, which wind vector and TAS is used?

Answer 3

The wind vector / TAS corresponding to the altitude 2/3 through the climb.

Question 4

Equation for average gradient (degrees)? (ft-NM)

Answer 4

(Vertical distance (ft) / 100) / (ground distance (NM))

Question 5

Average gradient (%) equation?

Answer 5

(Vertical distance (ft) / 60) / (ground distance (NM))

Question 6

Average gradient (°) equation? (Both terms ft)

Answer 6

Arctan (altitude difference (ft) / ground distance (ft))

Question 7

ROD (ft/min) equation, i. using GP°, and ii. using GP%?

Answer 7

i. GP° * GS (NM/min) * 100
ii. GP% * GS (kt)

Question 8

Vertical speed equation? Kt, ft/NM

Answer 8

(GS (kt) x gradient (ft/NM)) / 60

Question 9

Why is important to determine the position of an aircraft accurately before commencing descent?

Answer 9

To ensure safe ground clearance

Question 10

What do most aircraft operating handbooks supply to calculate climb and descent problems?

Answer 10

Graphical material