ISEGORIA / MATH ENCYCLOPEDIA
Piston engines: from crank to cycle
A piston engine turns pressure into rotation through a crank and a connecting rod, and almost everything about how it feels comes from that linkage. The rod’s tilt gives the piston a second harmonic at twice crank speed; the gas cycle sets the efficiency through the compression ratio alone; and the pulses of torque from each cylinder, and the shaking forces of each piston, either add up or cancel depending on how the crankshaft is laid out.
Before you begin: Kinematics, thermodynamics of ideal gases, and Fourier series
Predict, manipulate, then check your reasoning against the example and question. Graphs illustrate the mathematics; they do not replace a proof.
1. The crank–slider and its second harmonic
The crank turns at a steady rate, but the piston does not move as a pure sine wave, because the connecting rod tilts as it goes round. With crank radius r and rod length l, the piston’s distance from top dead centre has a second harmonic whose size is set by λ = r/l: the piston moves faster in the half of the stroke near the top and turns round more violently there. The acceleration it needs, and so the force on every bearing, grows with the square of engine speed. Turn the crank, change the rod ratio and the speed, and compare the exact acceleration with its first two harmonics.
Worked example. Take a 86 mm stroke (\(r=43\) mm) with \(\lambda=0.3\), a rod of 143 mm, at 6000 rpm, so \(\omega=628\) rad/s and \(r\omega^2=1731g\). At top dead centre the piston decelerates at \(r\omega^2(1+\lambda)=2250g\), at bottom dead centre only \(r\omega^2(1-\lambda)=1212g\). A 450 g piston and pin therefore pull on the rod with almost 10 kN at the top of every exhaust stroke, 50 times a second. The mean piston speed is 17.2 m/s, but it peaks at 28.2 m/s at 74.5°, well before the crank is at right angles.
Watch out. The piston is drawn on the cylinder axis with no offset, and the crank turns at constant speed; in a real engine the crankshaft speeds up and slows down slightly within each cycle, which the third experiment measures. The two-harmonic formula drops terms of order \(\lambda^3\), under 2% here. The drawing is to scale.
Why is the piston fastest before the crank reaches 90°?
At 90° the crank pin is moving straight along the cylinder axis, but the rod is tilted and is still swinging, so the piston is already being slowed by the rod’s rotation. Before 90° the rod’s swing adds to the pin’s motion. Differentiating \(x(\theta)\) gives \(\dot x=r\omega\sin\theta\big(1+\lambda\cos\theta/\sqrt{1-\lambda^2\sin^2\theta}\big)\), whose maximum moves to smaller \(\theta\) as \(\lambda\) grows.
2. The Otto and Diesel cycles
An ideal engine takes a kilogram of air through four processes: adiabatic compression by the compression ratio r, heat added by burning fuel, adiabatic expansion, and heat rejected with the exhaust. In the Otto cycle of a petrol engine the heat arrives at top dead centre, at constant volume; in the Diesel cycle fuel burns while the piston moves away, at roughly constant pressure. The work per cycle is the area enclosed on the pressure–volume diagram, and the efficiency depends on how far the gas expands after being heated, which is set by r. Change the compression ratio and the cycle, and press Play to follow the gas around the loop.
Worked example. With \(\gamma=1.4\) and \(r=10\), the Otto efficiency is \(1-10^{-0.4}=60.2\%\). Compressing air from 300 K and 1 bar heats it to \(300\times10^{0.4}=754\) K at 25.1 bar before any fuel burns. A Diesel at \(r=18\) with cutoff ratio 2 reaches 953 K by compression alone, enough to ignite injected fuel, and has an efficiency of 63.2%, less than the 68.5% an Otto cycle would have at the same compression, but more than a petrol engine can reach at the \(r=10\) to 13 that knock allows.
Watch out. These are air-standard cycles: air as an ideal gas with constant specific heats, instant reversible processes, no heat loss to the walls and no friction, with 1.8 MJ of heat per kilogram of air in the Otto cycle. Real engines reach about half the ideal efficiency, and real cylinder pressures peak well below the ideal constant-volume spike because combustion takes time, as the third experiment models. The gas exchange strokes, intake and exhaust, add no net work here.
Why does a higher compression ratio give a higher efficiency?
Because the heat is added to gas that is already hot and compressed, and then the expansion recovers it over a larger volume ratio. In the Otto cycle the exhaust temperature is the peak temperature divided by \(r^{\gamma-1}\), and the fraction of heat thrown away is \(T_4/T_3\)\(=r^{1-\gamma}\). Squeezing harder before burning means the gas cools further while doing work before it is let go.
3. Torque, flywheels and balance
Each cylinder delivers torque in a sharp pulse after it fires and takes torque back to compress its charge, and its piston adds an inertia torque that reverses twice a revolution. A flywheel stores the surplus and lends it back, but the more cylinders share the cycle, the smoother the sum. Separately, every piston shakes the engine along the cylinder axis with forces at crank speed and twice crank speed; arranging the cranks so that these forces, and the rocking couples they make along the crankshaft, cancel is the art of engine balance. Choose a layout and an engine speed; the torque comes from a single half-litre cylinder model with gradual combustion, repeated at each cylinder’s firing angle.
Worked example. An inline four at 3000 rpm and full load averages 258 N·m in this ideal model, with peaks of 755 N·m just after each firing. The flywheel must absorb and return \(\Delta E=286\) J each cycle, so with \(I=0.12\) kg·m² the crankshaft speed swings by \(286/(0.12\times314^2)=2.4\%\). A single cylinder of the same size averages a quarter of the torque but needs to store 788 J, a speed swing of 6.7%, which is why single-cylinder engines have heavy flywheels.
Watch out. The cylinder model has no friction, heat loss or pumping loss, so its torques are higher than a real engine’s at the same size; the combustion follows a Wiebe function starting 10° before top dead centre and lasting 50°. Balance is shown for the reciprocating masses only, with the cylinders equally spaced along an inline crankshaft; rotating masses can be cancelled with counterweights and are left out. V engines and boxers add another set of possibilities, such as the perfectly balanced flat six.
Why does an inline four shake at twice crank speed, when its pistons are paired up and down?
Because the second harmonic of piston motion does not care about the sign of \(\cos\theta\): \(\cos2\theta\) is the same for a piston at the top and one at the bottom. When pistons 1 and 4 are at top dead centre, decelerating hard, pistons 2 and 3 are at the bottom, decelerating less hard (\(1+\lambda\) against \(1-\lambda\)), so the four inertia forces add up to \(4\lambda r\omega^2m\) upwards. Half a turn later the situation repeats, so the net force oscillates at twice crank speed.