Lesson 1: Fundamentals of Projectile Motion

Introduction

External ballistics begins the moment a projectile leaves the barrel. In this lesson, we'll establish the fundamental equations that govern projectile motion in a vacuum, then build toward more realistic models.

Coordinate System

We use a right-handed Cartesian coordinate system following ballistics convention:

X-axis: Horizontal, perpendicular to flight path (for windage/lateral deflection)
Y-axis: Vertical, pointing upward
Z-axis: Horizontal, pointing downrange (direction of initial velocity)

Key Concept: The origin (0,0,0) is typically placed at the muzzle, with the rifle bore aligned along the Z-axis at the departure angle.

Basic Equations of Motion

In a vacuum (no air resistance), a projectile experiences only gravitational acceleration:

$$\vec{F} = m\vec{g}$$ Newton's Second Law applied to a projectile

This gives us the acceleration components:

$$a_x = 0$$ $$a_y = -g$$ $$a_z = 0$$ Where $g \approx 9.81 \, \text{m/s}^2$ (32.174 ft/s²)

Velocity Components

Given an initial velocity $v_0$ at departure angle $\theta$ (angle above horizontal):

$$v_x(t) = 0$$ $$v_y(t) = v_0 \sin(\theta) - gt$$ $$v_z(t) = v_0 \cos(\theta)$$ Velocity as a function of time

Position Equations

Integrating the velocity equations gives us position:

$$x(t) = 0$$ $$y(t) = v_0 \sin(\theta) \cdot t - \frac{1}{2}gt^2$$ $$z(t) = v_0 \cos(\theta) \cdot t$$ Position as a function of time

The Parabolic Trajectory

Eliminating time from the position equations gives us the trajectory equation:

$$y = z\tan(\theta) - \frac{gz^2}{2v_0^2\cos^2(\theta)}$$ The parabolic trajectory equation

Important: This parabolic equation is only valid in a vacuum. Real trajectories deviate significantly due to air resistance, especially at longer ranges.

Key Ballistic Parameters

Maximum Range (in vacuum)

$R_{max} = \frac{v_0^2 \sin(2\theta)}{g}$$ Maximum when $\theta = 45°

Maximum Height

$H_{max} = \frac{v_0^2 \sin^2(\theta)}{2g}$$ Reached at time $t = \frac{v_0\sin(\theta)}{g}

Time of Flight

$$T = \frac{2v_0\sin(\theta)}{g}$$ Total time to return to launch height

Practical Example

Consider a rifle bullet with:

Muzzle velocity: $v_0 = 850$ m/s (2789 ft/s)
Departure angle: $\theta = 0.5°$ (typical for 100m zero)

At 100 meters downrange (ignoring air resistance):

t = \frac{z}{v_0\cos(\theta)} = \frac{100}{850 \times \cos(0.5°)} \approx 0.118 \, \text{seconds}$$ $$y = 100 \times \tan(0.5°) - \frac{9.81 \times 100^2}{2 \times 850^2 \times \cos^2(0.5°)}$$ $$y \approx 0.873 - 0.068 = 0.805 \, \text{meters above bore}

Limitations of the Vacuum Model

The vacuum model fails to account for:

Air Resistance: Dramatically reduces velocity and range
Magnus Effect: Spin-induced lateral forces
Coriolis Effect: Earth's rotation (important at long range)
Wind: Lateral and vertical deflection
Atmospheric Variation: Changes in air density with altitude

Reality Check: A .308 bullet with 2700 ft/s muzzle velocity has a vacuum range of ~15,000 meters at 45°, but actual maximum range is only ~4,000 meters due to air resistance!

Summary

This lesson established the fundamental equations of projectile motion. While these vacuum equations are too simplistic for practical ballistics, they provide the foundation upon which we'll build more sophisticated models. In the next lesson, we'll introduce drag forces and see how dramatically they affect trajectory.

← Home Lesson 2: Drag Forces →