Report on Parachute Suitability and Minimum Size Determination
Introduction
Parachutes are essential devices used to slow down the descent of objects or people through the atmosphere. Understanding the physics of parachutes and their interaction with the air is crucial for determining their effectiveness. This report will analyze the suitability of parachutes with a drag coefficient of 1.5 and an air density of 1.22 kg/m³ and calculate the minimum size parachute necessary to achieve a safe descent.
Principles of Terminal Velocity
Terminal velocity occurs when the force of gravity acting on a falling object is balanced by the drag force acting against it. The drag force can be calculated using the formula:
\[ F_{D} = \frac{1}{2} \cdot C_d \cdot \rho \cdot A \cdot v^2 \]
Where:
- \(F_{D}\) = drag force (N)
- \(C_d\) = drag coefficient (dimensionless)
- \(\rho\) = air density (kg/m³)
- \(A\) = cross-sectional area of the parachute (m²)
- \(v\) = velocity (m/s)
At terminal velocity, the drag force equals the gravitational force:
\[ F_g = m \cdot g \]
Where:
- \(F_g\) = gravitational force (N)
- \(m\) = mass of the object (kg)
- \(g\) = acceleration due to gravity (approximately \(9.81 , \text{m/s}^2\))
Therefore, at terminal velocity:
\[ \frac{1}{2} \cdot C_d \cdot \rho \cdot A \cdot v_t^2 = m \cdot g \]
In this equation, \(v_t\) is the terminal velocity we seek to calculate.
Determining Minimum Size Parachute
To find the minimum size of the parachute, we need to rearrange the terminal velocity equation to solve for \(A\):
\[ A = \frac{2mg}{C_d \cdot \rho \cdot v_t^2} \]
Assuming we want a terminal velocity that ensures a safe landing, we can decide on a typical value. Depending on the person's preference/needs, 5 m/s is often considered a safe landing speed for parachutists.
Variables Defined
- Let's say the affected object/mass is a standard human at \(m = 80 , \text{kg}\) (average weight).
- Using \(g = 9.81 , \text{m/s}^2\) for gravity.
- Using the given values: \(C_d = 1.5\) and \(\rho = 1.22 , \text{kg/m}^3\).
- We will use \(v_t = 5 , \text{m/s}\) as a target terminal velocity.
Substituting Values
\[ A = \frac{2 \cdot (80 , \text{kg}) \cdot (9.81 , \text{m/s}^2)}{1.5 \cdot (1.22 , \text{kg/m}^3) \cdot (5 , \text{m/s})^2} \]
Calculating each part: \[ A = \frac{1569.6}{1.5 \cdot 1.22 \cdot 25} \]
Calculating the denominator: \[ 1.5 \cdot 1.22 \cdot 25 = 45.75 \]
Now substituting back: \[ A \approx \frac{1569.6}{45.75} \approx 34.3 , \text{m}^2 \]
Discussion
Suitability of the Parachute
A parachute with a drag coefficient of 1.5 and an air density of 1.22 kg/m³ is suitable for stabilization during descent. The resulting drag force can balance the gravitational force for safe terminal velocity levels if the parachute is sized appropriately.
However, a parachute with an area of approximately 34.3 square meters is quite large, which might present challenges in packing and deployment and can also increase resistance beyond desired descent rates if not managed properly.
Conclusion
For a parachute to be effective for a person weighing 80 kg to achieve a terminal velocity of approximately 5 m/s, the parachute must have a cross-sectional area of at least 34.3 m². While a parachute with a drag coefficient of 1.5, under the conditions considered, is capable of ensuring a safe descent, practical considerations about size, weight, and deployment must also be taken into account for real-world applications. Further refinements could be made by exploring different drag coefficients or design variations to achieve smaller sizes while maintaining safety.