External Ballistics: Bird's Eye View and Windage - workshop 2

Introduction to Bird's Eye View in Trajectory Analysis

• This lecture transitions from the side-on view of trajectory maths (covered in the previous workshop) to a bird's-eye view. This is relevant to shooting reconstruction and investigation, allowing analysis of scenarios from above.

• The side-on view utilizes Suvat equations, while the bird's-eye view considers environmental conditions like crosswinds on projectile flight.

• The concept of windage, the lateral movement or deviation caused by the wind, is central when viewed from above.

Scenario: Police Marksman and Crosswind

• Scenario: A police marksman needs to shoot a hostage taker. The shooter and target are on the same horizontal plane, 140 meters apart.

• Given Data:

  • Range: 140 meters.

  • Muzzle velocity: 1000 meters per second.

  • Crosswind: 37 kilometers per hour at 84 degrees from the bullet's path.

• Questions to Address:

  • How does the crosswind affect the bullet's actual velocity?

  • How far off the target horizontally must the marksman aim to compensate for the crosswind?

• The goal is to understand how much the wind deviates the bullet so the marksman can aim slightly into the wind, allowing it to push the bullet onto the target.

Setting Up the Problem

• Process: Draw a diagram, write down the data, check units, and then perform the calculations.

• Two diagrams are used:

  • One for dimensions (distances).

  • One for velocities.

• Diagram 1: Dimensions:

  • A horizontal line represents the target location.

  • The distance between the marksman and the target (SY) is 140 meters.

  • The bullet deviates from its path due to wind by an angle theta. The lateral deviation is labeled Sx.

Diagram of Dimensions:

Shooter--------------------------------------------Target

Distance between the marksman and the target (Sy) is 140 meters. Bullet deviates with an angle (\theta), and lateral deviation labeled as (Sx).

• To calculate (S_x), we need to determine (\theta).

• Diagram 2: Velocities:

  • The bullet's initial velocity (VB) is 1000 meters per second.

  • The wind velocity is 37 kilometers per hour at 84 degrees from the bullet's path.

  • X and Y axes are drawn to resolve vector components.
    Diagram of Velocities:

Shooter--------------------------------------------Target

Velocity of the bullet (V_B) is 1000 meters per second. The angle of the wind is 84 degrees, and the velocity of the wind is 37 kilometers per hour.

Key Considerations

• Distances and velocities must be kept separate in calculations.

• When dealing with crosswinds in a top-down view, Suvat equations are not applicable; this is a vector problem.

• In vector questions, the angle of deviation is consistent between the velocity and dimension diagrams.

Unit Conversion

• Convert the wind velocity from kilometers per hour to meters per second.

• V_{wind} = 37 \frac{km}{h} = 37 \frac{1000 m}{3600 s} = 10.2\overline{7} \frac{m}{s}

Vector Resolution and Calculation of the Actual Velocity

• Resolve vectors into X and Y components:

  • V{wx} = Vw \cdot cos(6^{\circ})

  • V{wy} = Vw \cdot sin(6^{\circ})

Where:*

• (V_{wx}) is the x component of wind velocity.

  • (V_{wy}) is the y component of wind velocity.

  • V_w is the velocity of the wind.

• The original bullet velocity only has a Y component.

• Calculate the total velocity in the X and Y directions by adding the respective components.
  In the X direction:

• V_{bx} = 0 (no x component for the bullet)

  • V{totalx} = V{wx} + V{bx} = V_{wx} = 10.22147 \frac{m}{s}
    In the Y direction:

• V_{by} = 1000 \frac{m}{s} (all bullets velocity in the Y)

  • V{totaly} = V{wy} + V{by} = 1.0743 + 1000 = 1001.0743 \frac{m}{s}

Calculating the Actual Velocity (VA) and Angle of Deviation (θ)

• Use the Pythagorean theorem to calculate the actual velocity (V_A)

  • VA = \sqrt{Vx^2 + V_y^2}

• Determine the angle of deviation (\theta) using the inverse tangent function:

  • tan(\theta) = \frac{Vx}{Vy}

  • (\theta) = tan^{-1}(\frac{Vx}{Vy})

• Substitute values to find (V_A) and (\theta)

Calculation of Actual Velocity and Angle of Deviation

• Calculating the Actual Velocity (VA): VA = \sqrt{(10.221)^2 + (1001.0743)^2} \approx 1001.13 m/s

• Final Answer (Actual Velocity): 1001.13 \frac{m}{s}

• Calculating the Angle of Deviation \theta: \theta = tan^{-1}(\frac{10.221}{1001.0743}) \approx 0.585^{\circ}

Calculating Lateral Deviation (Sx) Using Dimensions

• Use the calculated angle of deviation and the known distance to find the lateral deviation.

• Using trigonometric relationships:

  • tan(\theta) = \frac{Sx}{Sy}

  • Sx = Sy \cdot tan(\theta)

• Given Sy = 140 meters and (\theta) = 0.585 degrees, calculate (S_x)

Calculating Lateral Deviation

• S_x = 140 \cdot tan(0.585^{\circ}) \approx 1.43 meters.

• The marksman must aim 1.43 meters off the target to compensate for the crosswind.

• This deviation is significant, emphasizing the importance of considering environmental factors in shooting scenarios.

• Lighter bullets are more affected by crosswinds.

Introduction to Bird's Eye View in Trajectory Analysis

• This lecture transitions from the side-on view of trajectory maths (covered in the previous workshop) to a bird's-eye view, offering a comprehensive understanding of projectile motion. This perspective is particularly useful in shooting reconstruction and investigation, enabling a top-down analysis of various scenarios.

• The side-on view primarily utilizes Suvat equations to analyze motion in a vertical plane, while the bird's-eye view incorporates environmental factors such as crosswinds that affect projectile flight in a horizontal plane.

• The concept of windage, which refers to the lateral movement or deviation of a projectile caused by wind, becomes crucial when analyzing trajectories from a bird's-eye view. Understanding and compensating for windage is essential for accurate shooting.

Scenario: Police Marksman and Crosswind

• Scenario: Consider a scenario where a police marksman needs to neutralize a hostage taker. Both the shooter and the target are positioned on the same horizontal plane, with a distance of 140 meters separating them. A crosswind is present, influencing the bullet's trajectory.

• Given Data:

  • Range: 140 meters, representing the distance between the shooter and the target.

  • Muzzle velocity: 1000 meters per second, indicating the initial speed of the bullet as it leaves the barrel of the firearm.

  • Crosswind: 37 kilometers per hour at an angle of 84 degrees relative to the bullet's intended path. This crosswind introduces a lateral force that affects the bullet's motion.

• Questions to Address:

  • How does the crosswind impact the bullet's actual velocity, considering both speed and direction?

  • How far off the target, horizontally, must the marksman aim to effectively compensate for the crosswind and ensure the bullet hits the intended target?

• The primary objective is to quantify the wind's effect on the bullet's trajectory, enabling the marksman to accurately aim slightly into the wind. This technique allows the wind to naturally push the bullet back onto the intended target, achieving precise shot placement.

Setting Up the Problem

• Process: Begin by creating a detailed diagram, accurately documenting all available data, verifying the consistency of units, and subsequently performing the necessary calculations. Accurate diagrams are essential for visualizing the problem and ensuring correct solutions.

• Utilize two distinct diagrams:

  • One diagram to represent dimensions, including distances and spatial relationships.

  • Another diagram to represent velocities, indicating both magnitude and direction of relevant vectors.

• Diagram 1: Dimensions:

  • Draw a horizontal line representing the target's location, establishing a clear reference point.

  • Indicate the distance between the marksman and the target as (S*y), which is 140 meters. This notation clarifies the known distance in the scenario.

  • Illustrate the bullet's deviation from its intended path due to the wind, denoting the angle of deviation as (\theta). The lateral deviation is labeled as (S*x), representing the horizontal displacement caused by the wind.

Diagram of Dimensions:

Shooter--------------------------------------------Target

Distance between the marksman and the target (Sy) is 140 meters. Bullet deviates with an angle (\theta), and lateral deviation labeled as (Sx).

• To accurately calculate (S_x), we need to first determine the angle of deviation (\theta).

• Diagram 2: Velocities:

  • Represent the bullet's initial velocity as (V_B), which is 1000 meters per second, indicating its speed as it leaves the firearm.

  • Indicate the wind velocity as 37 kilometers per hour, acting at an angle of 84 degrees relative to the bullet's path.

  • Establish X and Y axes to facilitate the resolution of vector components, enabling individual analysis of horizontal and vertical influences.

Diagram of Velocities:

Shooter--------------------------------------------Target

Velocity of the bullet (V_B) is 1000 meters per second. The angle of the wind is 84 degrees, and the velocity of the wind is 37 kilometers per hour.

Key Considerations

• Ensure that distances and velocities are treated separately throughout the calculations to maintain accuracy and avoid confusion.

• When addressing crosswinds from a top-down perspective, traditional Suvat equations are not applicable. This scenario requires a vector-based approach to accurately account for the directional forces.

• In vector-based questions, the angle of deviation remains consistent between the velocity and dimension diagrams, providing a cohesive understanding of the impact of wind on projectile motion.

Unit Conversion

• Convert the wind velocity from kilometers per hour to meters per second to maintain consistency with other measurements.

• V_{wind} = 37 \frac{km}{h} = 37 \frac{1000 m}{3600 s} = 10.2\overline{7} \frac{m}{s}

Vector Resolution and Calculation of the Actual Velocity

• Resolve vectors into X and Y components to facilitate calculations:

  • V{wx} = Vw \cdot cos(6^{\circ})

  • V{wy} = Vw \cdot sin(6^{\circ})

Where:

• (V_{wx}) is the x component of wind velocity.

• (V_{wy}) is the y component of wind velocity.

• V_w is the velocity of the wind.

• The original bullet velocity only has a Y component.

• Calculate the total velocity in the X and Y directions by adding the respective components.

In the X direction:

• V_{bx} = 0 (no x component for the bullet)

• V{totalx} = V{wx} + V{bx} = V_{wx} = 10.22147 \frac{m}{s}

In the Y direction:

• V_{by} = 1000 \frac{m}{s} (all bullets velocity in the Y)

• V{totaly} = V{wy} + V{by} = 1.0743 + 1000 = 1001.0743 \frac{m}{s}

Calculating the Actual Velocity (VA) and Angle of Deviation (θ)

• Use the Pythagorean theorem to calculate the actual velocity (V_A)

  • VA = \sqrt{Vx^2 + V_y^2}

• Determine the angle of deviation (\theta) using the inverse tangent function:

  • tan(\theta) = \frac{Vx}{Vy}

  • (\theta) = tan^{-1}(\frac{Vx}{Vy})

• Substitute values to find (V_A) and (\theta)

Calculation of Actual Velocity and Angle of Deviation

• Calculating the Actual Velocity (VA): VA = \sqrt{(10.221)^2 + (1001.0743)^2} \approx 1001.13 m/s

• Final Answer (Actual Velocity): 1001.13 \frac{m}{s}

• Calculating the Angle of Deviation \theta: \theta = tan^{-1}(\frac{10.221}{1001.0743}) \approx 0.585^{\circ}

Calculating Lateral Deviation (Sx) Using Dimensions

• Use the calculated angle of deviation and the known distance to find the lateral deviation.

• Using trigonometric relationships:

  • tan(\theta) = \frac{Sx}{Sy}

  • Sx = Sy \cdot tan(\theta)

• Given Sy = 140 meters and (\theta) = 0.585 degrees, calculate (S_x)

Calculating Lateral Deviation

• S_x = 140 \cdot tan(0.585^{\circ}) \approx 1.43 meters.

• The marksman must aim 1.43 meters off the target to compensate for the crosswind, demonstrating the practical impact of wind on shooting accuracy.

• This deviation is significant, underlining the critical importance of considering environmental factors in shooting scenarios. Accurate wind assessment and compensation are vital for mission success.

• Lighter bullets are generally more susceptible to the effects of crosswinds due to their lower mass and ballistic coefficient. Understanding a bullet's characteristics is crucial for accurate shooting in windy conditions.