Climb Performance - ATPL 032 — Performance

A climb is level flight with a surplus. The engines produce more than the drag bill, and the excess is spent on going up: spent one way it buys angle, spent the other it buys rate, and the two purchases peak at two different speeds. This lesson resolves the forces, extracts the two climb speeds from the curves, and then walks through everything that erodes them, from a hot day to a failed engine, ending at the one calculation the syllabus demands.

The forces in a steady climb

Tilt the flight path up at a climb angle γ and the weight no longer sits neatly opposite the lift. Resolving along and across the path: along it, thrust must carry the drag plus a slice of the weight, W·sin γ; across it, the lift only has to balance W·cos γ. Two consequences are exam favourites. First, in a steady climb the lift is slightly less than the weight: the thrust is carrying the rest. Second, the climb angle is set by how much thrust remains after drag is paid:

For the small angles of transport flying, sin γ and the gradient are interchangeable: a 3° climb is close to a 5 per cent gradient (the angle is measured in degrees, the gradient in per cent, height gained over horizontal distance; that distinction is LO material from module 1 and returns here as a one-line conversion). The formula pair above is the lesson's spine: excess thrust buys climb angle; excess power buys rate of climb. A steady climb holds both books balanced at once.

Vx and Vy from the curve gaps

Plot available over required and the two climb speeds fall out as the widest gaps. The speed for the best angle of climb, Vx, sits where the thrust gap is widest; for a jet, whose thrust line is flat, that is simply the bottom of the drag bucket, so a jet's Vx is at Vmd. The speed for the best rate of climb, Vy, sits where the power gap is widest, always faster than Vx. A propeller aeroplane shifts both down the scale: its thrust is greatest at low speed, so its Vx is very low, just above the minimum safe speed, and its Vy sits near the minimum-power speed, well below a jet's. One more definition rides with this LO: the critical engine is the engine whose failure would most adversely affect the aeroplane's performance or handling qualities.

Reading the climb from the curves

The curves answer more than the two named speeds. At any speed, the vertical gap in the thrust picture divided by the weight is the climb gradient available at that speed, and the gap in the power picture divided by the weight is the rate of climb. Where the gaps close, at the intersections, climbing stops being possible and level flight is all that remains. Between the intersections the aeroplane can trade: fly at Vx when a mountain or an obstacle demands the most height per mile, fly at Vy when the goal is the most height per minute, and fly faster than both when the schedule matters more than the altimeter.

What erodes the climb

Weight — more weight raises the required curves and divides the surplus by a bigger W: climb angle and rate of climb both fall, and the best-angle and best-rate speeds (as IAS) both rise.
Altitude and temperature — thinner or hotter air sinks the available curves: angle and rate both fall. With altitude, Vx (IAS) creeps up while Vy (IAS) falls, and the two meet at the absolute ceiling. Climbing into an inversion means suddenly warmer air and a step-change loss of climb performance.
Configuration — flaps and gear add parasite drag: angle and rate fall, and both Vx and Vy move to lower speeds as the buckets shift left.
Engine failure — a twin loses half its thrust but far more than half its climb: the windmilling engine adds drag, the rudder and bank needed to hold it straight add more, and the excess collapses. Vx and Vy settle slightly lower.

The gradient calculation

The syllabus closes this lesson with arithmetic it will reuse throughout class A: given thrust, drag and weight, produce the climb gradient. The recipe is the formula from the top of the lesson, with the small-angle shortcut that sin γ times 100 is the gradient in per cent.

Resolve the forces in a steady climb and state the lift/weight relationship

Define the critical engine, Vx and Vy, and explain climb angle from the thrust picture and rate of climb from the power picture

State the difference between climb angle and climb gradient

Interpret the available and required curves in a steady climb

Explain the effects of weight, altitude and temperature (including an inversion), configuration and engine failure on climb angle, rate of climb, Vx and Vy

Calculate all-engine and one-engine-out climb gradients from thrust, drag and weight

The forces in a steady climb

Check you understood

Vx and Vy from the curve gaps

Reading the climb from the curves

What erodes the climb

The gradient calculation