Parachute Surface Area Calculation

Problem Statement: A parachute is needed to safely land a mobile radar station weighing 1000 kg after it exits an aircraft at a normal altitude, under average weather conditions. Calculate the surface area of the parachute required for safe descent.

Solution:

Step 1: Weight of the payload:

The weight of the mobile radar station is given as:

\( W = m \cdot g = 1000 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 9810 \, \text{N} \)

Step 2: Terminal velocity and drag force:

The drag force \( F_d \) acting on the parachute at terminal velocity is equal to the weight of the payload when the descent reaches a constant speed. The drag force is given by the drag equation:

\( F_d = \frac{1}{2} \cdot C_d \cdot \rho \cdot A \cdot v^2 \)

Where:

\( C_d \) = Coefficient of drag for a parachute (typical value: 1.5)
\( \rho \) = Air density at normal altitude, \( 1.225 \, \text{kg/m}^3 \)
\( A \) = Surface area of the parachute (what we are solving for)
\( v \) = Terminal velocity (assume a safe descent speed of \( 5 \, \text{m/s} \))

Step 3: Equating the drag force to the weight:

At terminal velocity, the drag force equals the weight of the payload:

\( F_d = W \)

Substitute the values into the drag equation:

\( 9810 = \frac{1}{2} \cdot 1.5 \cdot 1.225 \cdot A \cdot (5)^2 \)

Step 4: Solve for the surface area \( A \):

\( 9810 = 0.5 \times 1.5 \times 1.225 \times A \times 25 \)

Simplifying the equation:

\( 9810 = 22.97 \times A \)

Solving for \( A \):

\( A = \frac{9810}{22.97} \approx 427.09 \, \text{m}^2 \)

Summary:

The surface area of the parachute required to safely land the mobile radar station is approximately:

\( A \approx 427.09 \, \text{m}^2 \)