Thrust required, power required and energy

Multiply by airspeed to obtain power

Drag is a force. The rate of doing work against drag is drag multiplied by TAS.

The two curves answer different questions

In steady level flight, thrust required equals drag. Propulsive power required is D × TAS. A force curve in kN and a power curve in kW cannot be compared without converting at the same speed.

For D = A/V² + BV², power required is A/V + BV³. Its minimum occurs at a lower speed than minimum drag because multiplying drag by speed gives extra weight to the high-speed side.

Use propulsive and shaft power carefully

For a propeller, propulsive power is η times shaft power, with η varying across operating conditions. A constant shaft-power engine does not deliver constant thrust over all speeds.

For a jet over a limited low-speed range, approximately constant available thrust gives available propulsive power that rises with speed. These are simplified tendencies, not substitutes for engine data.

One operating point, two units

At a selected speed of 60 m/s, suppose the drag curve requires 5 kN. The required propulsive power is 300 kW. If a propeller has 75% efficiency at this point, shaft power must be 300/0.75 = 400 kW. Comparing the 5 kN directly with a 400 kW engine rating would be a comparison between incompatible quantities.

At a different speed, both drag and propeller efficiency may change. Even if the shaft-power rating stays 400 kW, available thrust is not automatically the same. First convert shaft to propulsive power using the applicable efficiency, then divide by TAS when a thrust comparison is required.

Switch the laboratory between drag and power axes without changing the selected speed. The selected point represents the same physical flight condition but has a different numerical vertical coordinate. Find each curve’s minimum separately. Explain the displacement using the additional V factor rather than assuming that every performance optimum must occur at the best-L/D speed.

Excess power changes total mechanical energy

The same energy can become height, speed or a mixture of both.

Specific excess power

With thrust aligned to the path, (T − D)V equals the rate of change of potential plus kinetic energy, neglecting other losses. At constant speed, rate of climb is excess power divided by weight.

If speed increases during a climb, some excess power builds kinetic energy and less remains for height. An instantaneous pitch-up can trade speed for height even without positive sustained excess power.

Read the condition before using a maximum

Maximum excess thrust corresponds to a strong path-angle tendency in the simplified steady case. Maximum excess power corresponds to the best rate of energy gain and, at constant speed, best rate of climb.

The laboratory holds a fictional polar and engine model fixed so the distinction can be isolated. Real best-angle and best-rate speeds vary with altitude, mass, configuration and engine state.

Do not spend the excess power twice

A 10,000 kg aeroplane has 200 kW of excess propulsive power at a particular instant. If its speed is constant, ROC = 200,000/(10,000 × 9.81) = 2.04 m/s. If it is simultaneously accelerating at 0.20 m/s² while travelling at 60 m/s, the kinetic-energy rate is mVdV/dt = 120 kW.

Only 80 kW then remains for potential-energy gain, giving ROC about 0.82 m/s. Assigning the full 200 kW to climb while also claiming that acceleration would count the same energy twice. Conversely, decelerating can temporarily add height gain by releasing kinetic energy, without proving sustainable climb capability.

Use the laboratory to choose a positive excess-power point and calculate the constant-speed result. Then state how an acceleration demand would change the answer, even though the laboratory holds speed steady. This keeps the useful steady equation attached to its assumption and prepares you to interpret zoom climbs and energy-limited manoeuvres.

Can you explain and apply the mechanism?

Four original questions. Work from the stated facts, then compare your reasoning with the explanation.