Engineering playground · hands-on

Physics you can get hold of.

A few fundamentals from a mechanical engineering degree – as small machines, oscillations and flows to turn and tweak. Everything is calculated live in the browser, without third-party libraries. A click in the picture sets a new starting point.

↓ nine experiments, astonishing facts & a chart · drag the sliders, expand the derivations

The slider-crank

Rotation becomes stroke: crank, connecting rod and piston. The same principle is at work in every piston engine and every piston pump.

01 Slider-crank Kinematics of mechanisms
Crank angle
0°
Piston travel (from TDC)
0%
Piston speed (rel.)
0
Why it is interesting: Piston travel is not a pure sine curve – the second term (cos 2φ) comes from the finite length of the connecting rod. It is exactly this term that produces the second-order inertia forces an engine has to be balanced against.
Derivation & calculation

Wanted: the piston travel x as a function of the crank angle φ. Crank (radius r) and connecting rod (length l) form a triangle; the piston runs along the cylinder axis.

1
The crank pin is at (r·cos φ, r·sin φ). The piston sits on the axis; the connecting rod of fixed length l joins the two. Pythagoras gives x.
x = r·cos φ + √( l² − r²·sin²φ )
2
The square root is unwieldy. With λ = r/l and an approximation it becomes an easily readable sum:
x ≈ r·( cos φ + (λ/2)·cos 2φ ) + const
the cos 2φ term is the “obliquity” effect
3
Why it matters: Differentiating twice with respect to time gives the acceleration – and therefore the inertia force. The cos 2φ component oscillates at twice the rotational speed: this is the second-order inertia force, which in-line engines are specifically balanced against.

Example: at φ = 90°, cos φ = 0, but cos 2φ = −1 – so the piston is not in the middle but closer to bottom dead centre. That is exactly what the animation above shows.

Gears & gear ratio

Two meshing gears turn in opposite directions; their circumferences roll on each other without slip. The ratio of the numbers of teeth is the gear ratio.

02 Spur gear pair Machine elements
Gear ratio i = z₂/z₁
2.00
Input
100min⁻¹
Output n₂
50min⁻¹
Remember: Few teeth driving many → slower, but more torque. The power stays (ideally) the same – speed and torque trade places in the ratio i.
Derivation & calculation

Two gears mesh without slip. At the point of contact (the pitch point, marked in orange) both have exactly the same circumferential speed v.

1
Equal circumferential speed: v = ω₁·r₁ = ω₂·r₂. Because the module m (the tooth size) must be the same for both, r = m·z/2 – so the pitch circle is proportional to the number of teeth z.
2
Replace r with z and rearrange:
i = ω₁ / ω₂ = n₁ / n₂ = z₂ / z₁
3
Without friction the power is conserved, P = M·ω. If the speed drops by a factor i, the torque rises by i:
M₂ = i · M₁ slower, but stronger

Example z₁ = 14, z₂ = 28: i = 2. 100 min⁻¹ becomes 50 min⁻¹, and the torque doubles. This is exactly how a gearbox turns a fast, weak motor into a slow, strong drive.

The damped oscillation

Mass, spring, damper – the basic model of every oscillation. With little damping it keeps swinging for a long time; in the critically damped case it returns to rest fastest without overshooting.

03 Spring-mass-damper Engineering mechanics
Natural angular freq. ω₀
6.3rad/s
Damping ratio ζ
0.12
Behaviour
underdamped
ζ = c / (2·√(k·m)): below 1 it oscillates (underdamped), at exactly 1 it is critically damped (the shock-absorber ideal), above 1 it creeps back (overdamped).
Derivation & calculation

Newton for the mass m: the spring pulls back with −k·x, the damper brakes with −c·ẋ (proportional to the velocity ẋ).

1
Sum of forces m·ẍ = −k·x − c·ẋ, rearranged:
m·ẍ + c·ẋ + k·x = 0
2
Without damping (c = 0) it oscillates purely sinusoidally at the natural angular frequency:
ω₀ = √( k / m ) [rad/s]
3
How strongly it is damped is measured by the dimensionless damping ratio ζ – it compares the damping with the “critical” damping:
ζ = c / ( 2·√(k·m) )
4
ζ < 1 underdamped (oscillates and decays), ζ = 1 critically damped (fastest return without overshoot – this is how shock absorbers are designed), ζ > 1 overdamped (creeps back).

Example k = 40 N/m, m = 1 kg → ω₀ = 6.3 rad/s (≈ 1 Hz). With c = 1.5 Ns/m, ζ = 0.12 – lightly damped, it keeps oscillating for a long time.

Kármán vortex street

When fluid flows around a cylinder, vortices shed alternately from the top and bottom. This vortex street makes chimneys sway and power lines sing.

04 Flow around a cylinder Fluid mechanics
Strouhal number: The shedding frequency grows with velocity – f ≈ St·U/D, with St ≈ 0.2 over a wide range. That is why the wind whistles sometimes higher, sometimes lower.
The physics behind it

The base flow here follows the potential flow around a circular cylinder – which is why the streamlines split cleanly and close again behind the cylinder:

u = U·( 1 − R²·(x²−y²)/r⁴ ) , v = −U·2R²·x·y/r⁴
r = distance from the cylinder centre, R = radius
1
In reality the flow separates at the rear and rolls up into vortices, alternately at the top and bottom. Each one is modelled as a Rankine vortex: a rigid core inside, v ∝ 1/r outside.
2
How often a vortex is shed is given by the dimensionless Strouhal number St:
f = St · U / D with St ≈ 0.2 for the cylinder
3
Everyday life: For a thin wire D is small, so f is high → it whistles. For a thick chimney f is low – if it matches the chimney's natural frequency, it can oscillate dangerously (hence the helical strakes on tall chimneys).

Fourier: waves from circles

Every periodic function is a sum of sine waves – visually: rotating circles linked together. The tip of the chain draws the curve.

05 Epicycle synthesis Advanced mathematics
Gibbs phenomenon: At the jumps of the square wave a small overshoot remains even with many terms – a famous detail of the Fourier series.
Derivation & calculation

Idea: every periodic function is a sum of pure sine waves – a fundamental plus overtones. A rotating circle produces exactly one sine (its height above the centre).

1
A circle with radius Aₙ rotating at frequency n gives the height Aₙ·sin(n·ω·t). Link many of them together and their heights add up – the tip of the chain draws the sum.
f(t) = Σ Aₙ · sin(n·ω·t)
2
The square wave needs only the odd multiples (1, 3, 5, …), and the amplitude falls with 1/n:
Aₙ = 4/(π·n) , n = 1, 3, 5, … Sawtooth: all n · Triangle: ∝ 1/n²
3
The more terms, the squarer – but at the jumps an overshoot of ≈ 9 % remains, the Gibbs phenomenon, which never quite disappears.

Exactly this decomposition is behind MP3, JPEG and every vibration analysis: split a signal into its frequencies and leave out what does not matter.

The double pendulum

Hang two pendulums one below the other, and simple mechanics turns into chaos. Two almost identical starting angles diverge completely after a short time.

06 Chaotic pendulum Dynamics
Deterministic, but unpredictable: The motion follows exact equations but reacts extremely sensitively to the starting value – the hallmark of chaos.
Why chaos?

The upper and lower pendulums pull on each other – their motions are coupled. The Lagrangian formalism yields two linked differential equations for the angles φ₁, φ₂.

1
The angular acceleration of the upper pendulum depends on both angles and both angular velocities:
φ̈₁ = f( φ₁, φ₂, φ̇₁, φ̇₂ )
a non-linear expression with sin(φ₁ − φ₂)
2
The sin(φ₁ − φ₂) term makes the system non-linear – there is no closed-form solution. It is computed step by step (here by time integration, 8 sub-steps per frame).
3
Sensitivity: A starting angle that differs by 0.1° leads to a completely different path within seconds – the “butterfly effect”, the same mathematics as in weather.

The Otto cycle

Compress, ignite, expand, exhaust. In the p-V diagram the state runs in a loop – the enclosed area is the work gained.

07 p-V cycle Thermodynamics
Stroke
Compression
Efficiency η
60%
Peak pressure
65bar
More ε = more η – but only until the mixture ignites by itself (knocking). That is why petrol engines run at ε ≈ 10–12 and diesels (compression ignition) at ε ≈ 16–22.
Derivation & calculation

Efficiency is the share of the heat supplied that turns into work: η = W/Qin. For the ideal Otto cycle (two isentropic + two isochoric steps) almost everything cancels – what remains is a surprisingly simple formula:

1
Compression (1→2) and expansion (3→4) are isentropic, so T·Vκ−1 = const. The volume ratio between the isochores is exactly ε.
2
Heat flows only at constant volume (Q = m·cv·ΔT). Substitute the temperatures and cancel – only ε remains:
η = 1 − 1 / εκ−1 κ ≈ 1.4 for air
3
Calculation: ε = 10 → η = 1 − 1/100.4 = 1 − 0.40 = 60 % (ideal). A real engine achieves about half of that – friction, heat losses, real gas.

The area enclosed by the loop in the p-V diagram is directly the useful work per cycle. Larger area = more work.

Motor meets load: the operating point

A drive settles where motor torque and load torque are equal. As long as the motor delivers more than the load needs, the flywheel accelerates.

08 Speed-torque characteristic Drive dynamics
Speed
0%
Operating point n_B
–%
Torque M
0%
Accelerating torque: The difference motor − load drives the flywheel. When it is zero, the machine runs steadily – exactly at the intersection of the two curves.
Derivation & calculation

Everything revolves around a torque balance at the flywheel (moment of inertia J):

J · dω/dt = Mmotor(n) − Mload(n)
1
The motor characteristic (here an induction motor, Kloss equation) has a breakdown torque and drops steeply towards rated speed. The load characteristic rises – for a fan with the square of the speed, Mload ∝ n².
2
As long as Mmotor > Mload, dω/dt > 0 – the machine accelerates. The surplus shrinks as the intersection comes closer.
3
At the operating point both are equal, dω/dt = 0, and the speed stays constant:
Mmotor(nB) = Mload(nB)

A large flywheel (large J) runs up sluggishly but holds the speed steady. If the load exceeds the breakdown torque, the motor “stalls” – it never gets up to speed at all.

The jet engine

Intake, compression, combustion, exhaust. The large fan at the front pushes cold air past the hot core; the core supplies the energy.

09 Turbofan Turbomachinery
Thrust
83kN
Gas after combustor
1420°C
Pressure ratio
38:1
Bypass
10:1
The trick of modern engines: accelerating a lot of air slowly is more efficient than accelerating a little air quickly. Hence the huge fan – it produces ~85 % of the thrust and is quieter too.
Derivation & calculation

Thrust comes from a change in momentum: the engine draws in air at vin and expels it faster at vout – by Newton (actio = reactio):

F = ṁ · ( vout − vin )
ṁ = air mass drawn in per second
1
Behind this runs the Joule/Brayton cycle: intake → compression (pressure ×40) → combustion at constant pressure → expansion in turbine and nozzle. The turbine only drives the compressor – the rest of the energy becomes thrust.
2
Why the large fan? The same thrust F can be achieved with a large ṁ and a small Δv (fan) or the other way round (core only). The wasted power lies in the Δv²: accelerating a lot of air gently costs less energy.
ηV = 2 / ( 1 + vout / vflight ) propulsive efficiency
3
That is why modern engines have a bypass ratio of ~10:1 – ten times as much cold bypass air as hot core air. This saves fuel and makes them considerably quieter.

Astonishing facts

Engineering facts that are hard to believe – and that still work reliably every single day.

11 tonnes

On a finger-sized blade

Each turbine blade of a jet engine produces as much power as a Formula 1 engine – while withstanding a centrifugal force as if a fully loaded coach were hanging from it.

Jet engine
1700 °C

Hotter than its melting point

The gas in the turbine is hotter than the melting point of the blade metal (~1300 °C). The blade survives only thanks to fine cooling-air channels inside and a ceramic thermal barrier coating.

Materials
1 crystal

The blade without grain boundaries

A turbine blade is cast as a single metal crystal. Grain boundaries would “creep” under heat and centrifugal force – the single crystal holds.

Casting
×40

Hot before anything burns

Compression alone squeezes the air in the engine to ~40 times its pressure and heats it to around 600 °C – before a single drop of kerosene has burned.

Thermodynamics
1 t/s

The fan shovels air

The large, cold fan at the front moves more than a tonne of air – per second. And it provides ~85 % of the thrust, not the hot core.

Propulsion
13 t/s

Saturn V, the moon rocket

At lift-off it burned 13 tonnes of propellant per second. Its power output was roughly that of 85 Hoover Dams at once.

Spaceflight
300 km/h

The wind turbine blade tip

The rotor turns at a leisurely pace, yet the tip of a large rotor blade races through the air at around 300 km/h – hence the “whoosh” in the wind.

Wind power
15 cm

The Eiffel Tower in summer

Heat expands steel: on hot days the Eiffel Tower is around 15 cm taller than in winter. That is why every bridge needs expansion joints.

Thermal expansion
7 m

How far a wing bends

In the ultimate load test, the wing of a wide-body jet is bent upwards by several metres – up to 150 % of the maximum flight load before it fails.

Strength
125 years

Concrete that is still curing

Without built-in cooling pipes, the concrete of the Hoover Dam would still not have cooled down today – the residual heat would have taken around 125 years to dissipate.

Civil engineering

What I built with what

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