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Chapter 9: Problem 112

Show that for unpowered flight (for which the lift, drag, and weight forces are in equilibrium) the glide slope angle, \(\theta\), is given by \(\tan \theta=C_{D} / C_{L}\)

Short Answer

Expert verified
This problem is solved by understanding that in a state of equilibrium, the lift force is equal to the weight of the aircraft, and the force of drag opposes the motion. Represent these forces mathematically and create a ratio to define the glide slope angle, which simplifies to \(\tan \theta = C_D/C_L\) proving the given relationship.

Step by step solution

01

Understanding the Forces at Play

First, understand that in a state of equilibrium for unpowered flight, the lift force is equal to the weight of the aircraft and the force of drag opposes motion. The lift force can be represented as \(L = C_L \cdot 0.5 \cdot \rho \cdot v^2 \cdot S\) and the drag force as \(D = C_D \cdot 0.5 \cdot \rho \cdot v^2 \cdot S\), where \(C_L\) and \(C_D\) are the lift and drag coefficients, \(\rho\) is the air density, \(v\) is the velocity and \(S\) is the wing area.
02

Determining the Glide Slope Angle

The glide slope angle, \(\theta\), is the angle at which the aircraft is descending. When the aircraft is in equilibrium, the weight force equals the lift force, and the tan of the glide slope angle (\(\tan \theta\)) is the ratio of the drag force over the lift force.
03

Proving the Relationship

Rewrite the drag and lift forces in terms of the coefficients: \(\tan \theta = \frac{D}{L} = \frac{C_D \cdot 0.5 \cdot \rho \cdot v^2 \cdot S}{C_L \cdot 0.5 \cdot \rho \cdot v^2 \cdot S}\). Simplify to reveal that \(\tan \theta = C_D/C_L\), proving the given relationship.

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