Turn excess energy into height

Why is best angle different from best rate?

In a steady climb, T − D = W sin γ and L = W cos γ. For small angles, air gradient is approximately (T − D)/W × 100. Rate of climb is (Pavailable − Prequired)/W. Best angle VX maximises excess thrust per weight; best rate VY maximises excess power per weight. Their speeds need not coincide.

Resolve the forces along the path

In a steady climb, T − D = W sin γ and L = W cos γ. For small angles, air gradient is approximately (T − D)/W × 100. Rate of climb is (Pavailable − Prequired)/W. Best angle VX maximises excess thrust per weight; best rate VY maximises excess power per weight. Their speeds need not coincide.

Angle is measured in degrees; gradient is tan γ × 100. A 5% gradient is about 2.86°, not 5°. For obstacle work use ground-relative distance and wind. In a propeller aircraft VX often rises and VY often falls in IAS toward the ceiling; real schedules depend on the engine and configuration.

Spend the surplus carefully

Extra mass raises required lift and drag, lowers both climb gradient and ROC, and generally raises characteristic IAS. Hot/high conditions reduce available thrust or power and increase TAS for a given IAS. An inversion can produce an abrupt temperature and thrust penalty as the aircraft climbs into warmer air. Gear, inappropriate flap and icing add drag; bleed extraction can remove thrust.

The critical engine is the engine whose failure most adversely affects performance or handling. Losing one of two engines is not losing half the excess climb power: extra asymmetric and windmilling drag appear while much of the original power was already needed for level flight. The remaining surplus can become very small or negative.

What can you trade when the engine cannot hold height?

In a steady descent, D − T = W sin |γ|; lift is W cos γ. In a glide T is approximately zero, so tan |γ| ≈ D/L and still-air glide ratio is approximately L/D. Excess power required appears as a rate of height loss. Adding thrust reduces that deficit and can flatten a constant-speed descent.

Gravity pays the energy bill

In a steady descent, D − T = W sin |γ|; lift is W cos γ. In a glide T is approximately zero, so tan |γ| ≈ D/L and still-air glide ratio is approximately L/D. Excess power required appears as a rate of height loss. Adding thrust reduces that deficit and can flatten a constant-speed descent.

At best L/D, increasing mass does not change the ideal still-air glide angle in the same configuration, but requires higher speed and produces a higher sink rate. Increasing altitude at a fixed IAS increases TAS and sink rate approximately together in a simple aerodynamic model. Neither result creates a better L/D.

Wind and configuration change the useful path

A headwind shortens ground distance from the same height; a tailwind lengthens it. The best ground-range speed shifts with wind. Drag from gear, flap, windmilling propellers or a damaged aircraft reduces L/D. Flying above or below the correct best-glide schedule also reduces range.

In a powered descent, thrust, speed, mass, altitude and configuration affect both angle and sink rate. A speed increase may first consume height to accelerate, then settle to a new drag and power balance. Do not combine a transient speed change with a steady-state descent formula without accounting for kinetic energy.

Can you make the performance decision?

Four original questions. Work from the stated conditions and review the reasoning.