Physics of the Trebuchet

Equivalent forms

R_{\max} = \frac{2\eta M h}{m}\sin 2\theta

\tfrac{1}{2}mv^2 = \eta M g h

Two textbook laws bolted together — energy conservation feeding the projectile-range formula — explain a war-winning machine.

Symbol	Name	SI unit	Dimension	Range
$R$	Rangeoutput Horizontal distance the projectile travels	m	L	0 – 400
$M$	Counterweight mass Mass of the falling counterweight	kg	M	100 – 2000
$m$	Projectile mass Mass of the thrown stone	kg	M	5 – 150
$h$	Counterweight drop Vertical distance the counterweight falls	m	L	1 – 6
$θ$	Release angle Launch angle above the horizontal	rad	1	0.35 – 1.2

Unit systems

Symbol	SI	CGS	Imperial
$R$	m	cm	ft
$M$	kg	g	lb
$m$	kg	g	lb
$h$	m	cm	ft

Where it holds

Idealized energy-transfer model with drag neglected and a lumped efficiency

\eta

; a full treatment solves the coupled beam-sling-projectile Lagrangian and adds air resistance for heavy/fast shots.

Dimensional analysis

[L]

[L^{2}T^{-2}]\times [1] / [LT^{-2}] = [L]

v^{2} \sin 2\theta / g

has units

(m^{2}/s^{2})/(m/s^{2}) = m

.

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Discovery

Medieval engineers; analyzed by Mersenne & Galileo era · 1200

The counterweight trebuchet appeared around the 12th century and dominated siege warfare until gunpowder. Its physics is the projectile range law (Galileo, ~1638) fed by energy conservation from a falling mass; modern 'punkin chunkin' contests still optimize the same M/m and release-angle trade-offs.

Try this

A medieval trebuchet hurls a 90 kg stone the length of three football fields. Where does all that range come from?

A falling counterweight dumps gravitational energy into a light projectile. The energy ratio M/m sets the launch speed, and v²sin(2θ)/g sets the range.

Research status: stable

Real-world applications

Medieval siege engineering and modern replicas
Pumpkin-throwing and watermelon-launch competitions
Teaching coupled energy transfer and projectile motion
Ballistic intuition for catapult-launched drones and aircraft

Common misconceptions

A heavier projectile flies farther — for fixed counterweight, range falls as 1/m because the stone carries less speed
Steeper is always better — beyond $45^{\circ}$ range drops; $\sin (2\theta )$ is symmetric about $45^{\circ}$
All the counterweight energy reaches the stone — sling and beam losses $cap \eta$ well below 1

Experimental verification

Scale-model and full-size trebuchets at engineering contests match the M/m scaling; the 'Warwolf' replica (2005) and annual Punkin Chunkin events confirm range grows roughly linearly with the counterweight-to-projectile mass ratio.

Derivation

The counterweight M falls a height h, releasing energy Mgh.

A fraction

\eta

goes into the projectile's kinetic energy:

\tfrac{1}{2}mv^{2} = \eta Mgh

, so

v = \sqrt{2\eta Mgh/m}

.

Launched at angle

\theta

over flat ground, projectile kinematics give time of flight

t = 2v \sin \theta /g

and range

R = v \cos \theta \cdot t = v^{2} \sin 2\theta /g

.

Substituting

v^{2} = 2\eta Mgh/m

yields

R = (2\eta Mh/m) \sin 2\theta

.

Limiting cases

\theta = 45^{\circ}

⟶ Maximum range (no drag)

\sin (2\theta )

peaks at 1, so

45^{\circ}

throws farthest in vacuum; real releases

use \sim 43

–

45^{\circ}

.

M ≫ m⟶

v \approx \sqrt{2\eta Mgh/m}

growsA huge counterweight on a light stone delivers near-maximal speed; that is why trebuchets used multi-tonne boxes.

\eta \to 1

(ideal)⟶ All CW energy reaches the stoneReal machines lose

\sim 40

% to the beam and sling;

\eta \approx 0.5

–0.7 in practice.

What if…

What if you add a sling to the throwing arm?

It lengthens the effective lever and release speed, boosting $\eta and$ range — the key medieval innovation.

What if there is significant air drag?

Range saturates and the optimal angle drops below $45^{\circ}$ , since high-speed light stones bleed energy fast.

What if the counterweight could fall twice as far?

$v \propto \sqrt{h}$ , so range $R \propto h$ doubles — taller frames throw farther.

1

Classic siege shot

Given ·

M:: 1000
m:: 50
h:: 4
θ:: 0.785

Find ·

R (\eta = 0.6)

Steps

Energy to stone $= 0.6\times 1000\times 9.8\times 4 = 23520\,\mathrm{J}$
$v = \sqrt{2\times 23520/50} \approx 30.7\,\mathrm{m/s}$
$R = 30.7^{2}\times 1/9.8 \approx 96\,\mathrm{m}$

Result ·

v = \sqrt{2\times 0.6\times 1000\times 9.8\times 4/50} \approx 30.7\,\mathrm{m/s}

;

R = v^{2} \sin 90^{\circ}/g \approx 96\,\mathrm{m}

2

Halve the projectile mass

Given ·

M:: 1000
m:: 25
h:: 4
θ:: 0.785

Find · R

Steps

Same 23520 J now into 25 kg
$v = \sqrt{1881.6} \approx 43.4\,\mathrm{m/s}$
$R \approx 43.4^{2}/9.8 \approx 192\,\mathrm{m}$ : lighter shot doubles range

Result ·

v = \sqrt{2\times 23520/25} \approx 43.4\,\mathrm{m/s}

;

R \approx 192\,\mathrm{m}

— double the range

Projectile Range→Gravitational Potential Energy→Kinetic Energy→Work-Energy Theorem→Conservation of Linear Momentum→