Philip Cassady writes. I must respond to Art Delagrange's reply (TR March 2002) to my solution of 2000 N/D 3 (TR October 2001). The required power scales with the cube of the velocity and not linearly with the velocity as suggested by Delagrange and not as the square of the velocity as proposed in the original solution to this problem (TR May 2001).

We need to consider the power required to propel the aircraft through the air, regardless of the means by which that power is provided. When we take this approach, we see that the aerodynamic drag force on the aircraft varies as the square of the velocity (Drag=3D1/2 r V2 CD A, where r is the air density, V is the aircraft velocity, CD is the drag coefficient, and A is a reference area). In straight and level flight enough thrust is required to balance this drag. Therefore the required thrust varies as the square of the velocity. If a rocket provides a fixed thrust then it will propel the aircraft at exactly that speed at which the drag equals this fixed thrust. If we want to fly at a different speed then the thrust will need to be changed proportional to the square of the velocity.

The power required to overcome the aerodynamic drag thus varies as the
cube of the velocity. This is true regardless of the source of this
power, whether it be a propeller, a jet, or a rocket. The fuel
consumption per unit time, to a first approximation, is proportional
to the output power for any type of engine. This can be seen if we
consider the engine as a closed system into which fuel flows at a
given rate (joules or liters per second, since each liter contains a
fixed number of joules) and both heat and work flow out (joules per
second). If we assume the engine efficiency is constant in this first
approximation, the rate of work flow out (power) is proportional to
the rate of fuel flow in (liters per second). Therefore the required
fuel flow rate is proportional to the cube of the velocity, as well.