Joseph R. Badick

Flight Theory and Aerodynamics


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be no unbalanced moments acting on the body. Mathematically, ∑M = 0.Moments at the fulcrum in Fig. 1.8 are 5000 ft‐lb clockwise and 5000 ft‐lb counterclockwise. The weight (force) of A is 100 lb and is located 50 inches (″) to the left of datum (fulcrum), thus 100 lb × −50″ = −5000 lb‐in. The weight of B is 200 lb and is located 25 inches to the right of datum, thus 200 lb × 25″ = 5000 lb‐in. So, ∑M = 0.

      Sir Isaac Newton summarized three generalizations about force and motion. These are known as the laws of motion.

      Newton’s First Law

      In simple language, the first law states that a body at rest will remain at rest and a body in motion will remain in motion, in a straight line, unless acted upon by an unbalanced force. The first law implies that bodies have a property called inertia. Inertia may be defined as the property of a body that results in its maintaining its velocity unchanged unless it interacts with an unbalanced force. For example, an aircraft parked on the ramp would not even need chocks unless an unbalanced force (such as wind, or gravity if parked on a slope) acted on it. The measure of inertia is what is technically known as mass.

      Newton’s Second Law

      The amount of the acceleration a is directly proportional to the unbalanced force, F, and is inversely proportional to the mass, m, of the body. For a constant mass, force equals mass times acceleration.

      Newton’s second law can be expressed by the simple equation:

      Then, solving for a,

equation

      EXAMPLE

      An airplane that weighs 14 400 lb accelerates down a runway with a net force of 4 000 lb, what is the acceleration (a) assuming constant acceleration?

equation equation

      Newton’s Third Law

      The third law states that for every action force there is an equal and opposite reaction force. Note that for this law to have any meaning, there must be an interaction between the force and a body. For example, the gases produced by burning fuel in a rocket engine are accelerated through the rocket nozzle. The equal and opposite force acts on the interior walls of the combustion chamber, and the rocket is accelerated in the opposite direction. As a propeller aircraft pushes air backward from the propeller, the aircraft is pushed forward.

      Newton’s laws of motion express relationships among force, mass, and acceleration, but they stop short of discussing velocity, time, and distance. These are covered here. In the interest of simplicity, we assume here that acceleration is constant. Then,

equation

      where

       Δ (cap delta) means “change in”

       Vf = final velocity at time tf

       Vi = initial velocity at time ti

      If we start the time at ti = 0 and Vi = 0 (brakes locked before takeoff roll) and rearrange the above where Vf can be any velocity given, for example liftoff velocity, then

equation

      The distance s traveled in a certain time is

equation

      where the average velocity Vav is

equation equation

      which yields

      EXAMPLE

      An aircraft that weighs 15 000 lb begins from a brakes‐locked position on the runway, and then accelerates down the runway with a net force of 5000 lb until liftoff at a velocity of 110 kts. Calculate the average acceleration down the runway, the average time it takes to reach liftoff speed, and the total takeoff distance on the runway.

      First, to calculate the acceleration, we need find the force (F) and the mass of the aircraft during the takeoff roll, Eq. 1.3: F = m a

      Finding the mass: images