Climb and Descent Performance Questions and Answers

Answer: B

The new rate of climb is lower by more than 10%.

Rate of climb depends on excess power: \(R/C = (T - D)V/W\). At the same speed and weight, drag is unchanged, so only the thrust term moves.

A 10% drop in thrust removes \(0.1T\) from the excess thrust \(T - D\). Because \(T - D\) is smaller than \(T\), losing \(0.1T\) is more than 10% of the excess. For example, if \(D = 0.6T\), the excess falls from \(0.4T\) to \(0.3T\), a 25% drop in rate of climb.

'Exactly 10%' would only be true if there were no drag, and 'less than 10%' gets the direction of the effect wrong.

Answer: C

A dive is initiated by generating a nose down pitch rate.

To enter a dive the flight path must be bent downward, which needs the aircraft to rotate nose down. The pilot pushes the stick forward, the elevator produces a nose-down pitching moment and the resulting pitch rate lowers the nose and the flight path angle.

Reducing thrust alone makes the aircraft slow down or descend gently at its trim speed, not dive. Reducing angle of attack alone is a consequence of the pitch change, not a way of starting it, and increasing thrust tends to make the aircraft climb or accelerate.

Answer: A

The pitch angle equals the angle of attack in wings level constant altitude flight.

The pitch angle is the sum of the angle of attack and the flight path angle: \(\theta = \alpha + \gamma\). In level flight at constant altitude the flight path is horizontal, so \(\gamma = 0\) and \(\theta = \alpha\).

In an unaccelerated climb \(\gamma \gt 0\), so the pitch angle exceeds the angle of attack. In an unaccelerated descent, and on a landing approach, \(\gamma \lt 0\), so the pitch angle is less than the angle of attack.