Level Flight, Climb and Descent - ATPL 081 — Principles of Flight

The method of this whole module fits in one sentence: draw the forces, resolve them along and perpendicular to the flight path, and set each direction's sum to zero — that is what steady means. This first lesson applies it to the three straight cases: level flight, the climb and the descent, ending with the formula the exam asks most often, the climb angle from thrust, weight and the lift-to-drag ratio.

Straight, horizontal, steady flight

The four forces, with their addresses:

Lift — at the wing's centre of pressure.
Weight — at the centre of gravity.
Thrust — along the engine's line.
Drag — along the airflow, opposing the motion.

Steady and level, the sums close: lift equals weight, thrust equals drag. But equal sums are not yet balance — the forces act at different points, so they form couples: lift behind the CG pairs with weight into a pitching moment, and the thrust and drag lines usually offset into another. The stability module built the answer in full: the tailplane supplies whatever moment (normally a download) closes the sum about the lateral axis. Four forces, two couples, and one tailplane closing the moment sum. (Strictly, that download makes the wing carry a little more than the weight: the stall module's forward-CG point; "lift equals weight" counts wing and tail together.)

The climb: geometry first

Tilt the flight path up and one angle rules everything: the flight-path angle γ (gamma), between the flight path and the horizontal. It completes a triangle the course has been assembling since the first lessons: with zero wind and zero bank, the pitch attitude is the flight-path angle plus the angle of attack — attitude = γ + α. The nose points where the path goes, plus the extra the wing needs to make its lift. Confusing the three is the classic error this module exists to prevent:

attitude is where the nose points.
γ is where the aircraft goes.
α is the difference the wing flies at.

The climb: forces

Resolve the forces about the climbing path. Perpendicular to it: lift balances only the weight component across the path, $L = W \cos\gamma$ — so in a steady climb, lift is less than weight, a result students resist until they see that part of the weight is carried along the path instead. Parallel to it: thrust must carry both the drag and the weight component sliding back down the path:

This is why thrust exceeds drag in a climb: the surplus is not accelerating the aircraft (the climb is steady); it is paying the weight component, lifting the aircraft's mass up the slope of the path. Divide the parallel equation by W and apply the small-angle habit (sin γ ≈ γ, and L ≈ W so D/W ≈ D/L):

The formula reads like a budget: T/W is what the engines offer, 1/(L/D) is what the drag charges, and the climb angle is what remains. It explains climb performance in one line (more thrust or less drag steepens the climb, more weight flattens it), and it is the exam's favourite quick calculation.

One dynamic case closes the climb (and recalls a warning from the stall module): a climb flown at constant vertical speed with a constant thrust setting. As altitude increases, the engines' excess over drag shrinks, but the vertical-speed demand does not. The aircraft gives up the only thing it has left: speed. IAS decreases, and it decays fast enough that the TAS, the speed that feeds the vertical-speed equation, falls with it; to keep the lift, α increases; and since the vertical speed is the TAS times sin γ, the slowing aircraft needs a steeper γ to hold the same rate. Left alone, the sequence walks toward the stall — the very trap that the stall module's closing lesson pinned on modes that command a trajectory without protecting speed.

The descent

Tilt the path down and the same resolution runs in the aircraft's favour. Perpendicular: $L = W \cos\gamma$ again — lift below weight, for the same reason as in the climb: only the across-path share of weight needs carrying. Parallel: the weight component now points down the slope, with the motion, so the engines need supply only what drag demands beyond it:

Thrust is less than drag in a steady descent: gravity has taken over part of the propulsion. Steepen the descent far enough and the weight component covers the drag entirely — thrust to idle, and the aircraft is gliding, which is the next lesson's whole subject.

What the exam asks

The two parallel-force formulas and the climb-angle calculation carry this topic's marks. The overview, following the learning objectives:

Describe the four forces in level flight, where they act, and how they balance including the tailplane's function

Define γ, and relate pitch attitude, γ and α (zero wind and bank)

Describe the climb's forces; name the parallel and perpendicular sets; apply T = D + W sin γ; explain thrust > drag and lift < weight

Explain and use the small-angle climb formula relating γ, thrust, weight and L/D

Explain how IAS, α and γ change in a constant-vertical-speed, constant-thrust climb

Describe the descent's forces; apply T = D − W sin γ; explain lift < weight and thrust < drag