Navigation in Climb and Descent - ATPL 061 — General Navigation

So far the module's legs have been flat. This closing lesson tilts them: a climb or descent is a leg with a third dimension, and its navigation adds two things to everything already built. First, angles and rates: how steep a path is, and how fast it takes altitude away. Second, honest averages: a climbing aircraft's TAS and wind change all the way up, so the still-air arithmetic needs to know at which altitude to sample them. The payoff is the profile question every jet flies daily: how far out, how fast down, to cross the fix at the cleared level.

Gradients

A path's steepness is its gradient, and the syllabus wants it three ways, all built on the 1:60 rule and its companion approximation that 1 NM is about 6,000 ft:

gradient in per cent = (vertical distance (ft) / 60) / ground distance (NM);
gradient in degrees = (vertical distance (ft) / 100) / ground distance (NM);
gradient in degrees = arctan(altitude difference (ft) / ground distance (ft)), the exact version, the two above being its rules of thumb.

One descent measured all three ways: 1,800 ft lost over 6 NM. In per cent: 1,800 over 60 is 30, over 6 NM is 5 per cent. In degrees: 1,800 over 100 is 18, over 6 NM is 3°. Exactly: 6 NM is about 36,000 ft, and 1,800 over 36,000 is the tangent of about 2.9°. The 3°/5 per cent pair is worth recognising on sight: it is the standard glide path, and half the numbers in this lesson orbit it.

Rates of climb and descent

A gradient becomes a rate the moment ground speed is known, and the syllabus's rules of thumb come as a matched set:

ROD (rate of descent, ft/min) = glide-path degrees × GS (NM/min) × 100;
ROD (ft/min) = glide-path per cent × GS (kt).

Run all three on the standard case and they agree. A 3° path at GS 180 kt (3 NM per minute): 3 × 3 × 100 is 900 ft/min. As 5 per cent: 5 × 180 is 900. As a 300 ft/NM gradient: 180 × 300 over 60 is 900. The vertical-speed formula also runs backwards: a 900 ft/min limit against a 300 ft/NM gradient allows a GS of 900 × 60 over 300, which is 180 kt, and 900 ft/min at GS 180 is a gradient of 900 × 60 over 180, which is 300 ft/NM. For the 3° path specifically the syllabus adds two purpose-built shortcuts:

ROD = (GS × 10) / 2, five times the ground speed;
ROD = SF × glide-path angle × 100 (SF: the speed factor of the last lesson, the speed in question divided by 60, here the GS).

Both give the same 900 at 180 kt, and the first is worth saying aloud: on a three-degree path, the rate of descent is five times the ground speed.

Planning the climb and descent

A climb has no single TAS or wind: both change with altitude all the way up. The planning convention picks one honest sampling altitude for each phase:

for climb problems, the average TAS and the W/V (wind velocity) are taken at 2/3 of the cruising altitude, because the aircraft climbs slowly through the fast, high part and spends most of the climb's time up there;
for descent problems, the average TAS and the W/V are taken at 1/2 of the descent altitude.

With the samples taken, the flat-leg machinery runs unchanged: the average climb or descent GS is the sampled TAS corrected by the wind component at the sampling altitude, resolved along the true track. Climbing to FL 360 with a sampled TAS of 240 kt at 2/3 height and 30 kt of headwind component there, the average climb GS is 210 kt, and every time and distance below uses it. For the numbers themselves, most aircraft operating handbooks supply graphical material for climb and descent problems: time, fuel and distance charts against weight and temperature, the same answers this lesson computes by rule of thumb.

The three questions of the profile

Every climb and descent problem is one of three questions, and all three are the same proportion read from a different corner. First, time and distance from a rate: minutes = altitude to change / rate, and distance = minutes × the average GS in NM/min. A 24,000 ft climb at an average 1,200 ft/min takes 20 minutes, and at an average GS of 240 kt (4 NM/min) it covers 80 NM. Second, the rate to make a gate: the minutes available are the distance over the GS, and the rate required is the altitude over those minutes. Third, the distance required at a chosen rate: minutes from altitude over rate, then miles from minutes.

Worked from the other corners with the same machinery: with a chosen ROD of 800 ft/min and 12,000 ft to lose, the descent takes 15 minutes and, at GS 240 kt, needs 60 NM; with 6,000 ft to lose and only 24 NM available at GS 180 kt (8 minutes), the required rate is 750 ft/min. Underneath all three questions sits one sober fact the syllabus states plainly: before commencing descent, the aircraft's position must be determined accurately, because safe ground clearance depends on knowing where the descent begins. A perfect profile from the wrong starting point is a perfectly flown mistake.

What the exam asks

The gradient and rate rules of thumb are pure marks for anyone who has run them once, and the 2/3 and 1/2 sampling altitudes are asked verbatim. The overview, following the learning objectives:

Estimate gradients in per cent and degrees by the rules of thumb and by arctan

Calculate ROD from glide-path angle or per cent, and vertical speed from GS and a ft/NM gradient

State the 2/3-of-cruise and 1/2-of-descent sampling altitudes for average TAS and W/V; calculate the average climb/descent GS; state that handbooks supply graphical climb/descent material

Calculate time and distance from a rate, the rate to make a position at an altitude, and the distance a chosen rate requires, using the 3°-path formulae where they apply; state why position must be known before descent

Gradients

Rates of climb and descent

Check you understood

Planning the climb and descent

The three questions of the profile

What the exam asks