How Does A Cannon Work?

It would be an odd sight indeed to watch a medieval-era cannon wheeled onto a modern field of battle, with drones zooming about overhead and armored, motorized tanks on the ground.

However, not only was the cannon the most-feared mechanical weapon in the world for a very long time, but the physical principles governing the form of projectile motion embodied by a cannon ball also dictate those of modern guns. A cannon, really, is simply a kind of gun in which the mass of the "bullet" is very large. As such, it obeys the same laws of projectile motion, and understanding projectile physics will help you understand cannon physics.

Cannonballs are often depicted in film as exploding on impact, wreaking most of their havoc through pyrotechnics. In reality, before the mid-1800s, comparatively few projectiles were designed to explode after launch. They did their damage by blunt-force impact, making use of tremendous ​momentum​ (mass times velocity) to accomplish this.

In the 1400s, the warlords of the day produced cannonballs equipped with fuses and designed to explode in enemy territory, but this came with the grave risk of bad timing or a misfiring cannon, leading to precisely the opposite result as the one the fighting force sought.

The sizes of purposefully launched heavy objects have varied tremendously over time, but a glance at 18th-century England offers a view of what cannonballs actually looked like. The national war ministry used eight standard sizes, rising in diameter in increments of about 1/2 inch (1.27 cm).

This choice was useful because the ​volume of a sphere​ is

\(V=\frac{4}{3}\pi r^3\)

where ​r​ is the radius (half the diameter), so the masses of uniform-density objects thus rise in predictable proportion to the cube of the radius. The diameters were actually rounded to allow for exact weights of cannonballs, from 4 to 42 pounds in unequal increments.

It takes considerable might to launch a cannonball, heralded by the fact that such events are typically noisy and violent. But what is less intuitive is that at the instant a projectile leaves the device that powers its launch, ​the only force acting on it from that moment on, if air resistance is neglected, is Earth's gravity​ (assuming Earth is where this event is being staged).

This means that you can treat a projectile-motion cannon problem as two separate problems, one for constant-velocity horizontal motion imparted by the launch, and one for constant-acceleration vertical motion owing to both the object's initial upward motion (if any) and the results of gravity acting on the cannonball. The solution is found by adding these together as vector sums.

Specifically, in addition to gravity, what determines the path of a cannonball are its ​launch angle​ θ and ​launch (initial) speed​ v₀.

The initial velocity must be separated into horizontal (v_0x) and vertical (v_0y) components for solving; you can obtain these from

\(v_{0x}=v_0\cos{\theta}\text{ and }v_{0y}=v_0\sin{\theta}\)

For horizontal motion, you have

\(v_x(t) = v_{0x}\)

which can be assumed to not diminish until the object strikes something (recall there is no friction in this idealized setting). The ​horizontal​ distance traveled as a function of time ​t​ is simply

\(x(t) = v_{0x}t.\)

For vertical motion, you have

\(v_y(t) = v_{0y} – gt\)

where g = 9.8 m/s², and

\(y(t) = v_{0y}t – (1/2)gt^2\)

This shows that as the effects of gravity prevail, the vertical speed increases in the negative (downward) direction.