Advanced11 min read6 / 9

Crank-slider kinematics

Between the burning gas and the turning wheels sits one mechanism: the crank-slider — piston, connecting rod, crank. Its geometry looks trivial and is not. The finite length of the rod bends the piston’s motion away from a pure sine wave, and that small distortion is the origin of engine vibration, cylinder-wall wear and the rev limit itself.

The exact position equation

Fig. 1. The mechanism, drawn to the demo's true proportions (r = 43 mm, L = 143 mm) at θ = 50°. The crank pin swings on the circle; the rod leans by φ to reach the wrist pin, which slides only vertically.

Measure the crank angle $\theta$ from TDC. The crank pin sits $r\cos\theta$ up and $r\sin\theta$ sideways from the crank axis. The rod of length $L$ must span from there to the wrist pin on the cylinder axis, so by Pythagoras the wrist pin rides at height

s(\theta) = r\cos\theta + \sqrt{L^2 - r^2\sin^2\theta}

(1)

above the crank axis. It peaks at $r + L$ (TDC, $\theta=0$ ) and bottoms at $L - r$ (BDC) — stroke $= 2r$ , as expected. The displacement below TDC, which is what the demo plots, is:

d(\theta) = (r+L) - s(\theta) = r(1-\cos\theta) + L - \sqrt{L^2 - r^2\sin^2\theta}

(2)

$r$: crank radius = stroke/2 [m]
$L$: connecting-rod length (centre to centre) [m]
$\lambda$: rod ratio r/L — the small parameter of this whole subject (~0.25–0.35) [–]
$\varphi$: rod lean angle from the cylinder axis [rad]

The square root is the rod’s doing: because the crank pin swings sideways by $r\sin\theta$ , the leaning rod’s vertical reach shrinks. Let $L \to \infty$ and (2) collapses to $d = r(1-\cos\theta)$ — simple harmonic motion. Every interesting effect below is the difference between the two.

The harmonic expansion

Binomial-expand the square root in powers of $\lambda = r/L$ and (2) becomes, to excellent accuracy:

d(\theta) \approx r\left[(1-\cos\theta) + \frac{\lambda}{4}\left(1-\cos 2\theta\right)\right]

(3)

A fundamental at crank frequency plus a small wave at twice crank frequency. Differentiating twice (at constant $\omega = \dot\theta$ ) gives the acceleration the crank must impose on the piston mass:

a(\theta) \approx -r\omega^2\left(\cos\theta + \lambda\cos 2\theta\right)

(4)

The $\cos\theta$ term is the primary motion; the $\lambda\cos 2\theta$ term is the secondary. (The demo does not use this approximation — it differentiates (2) exactly — but (4) is how engineers think, and it is all the next article needs to explain engine balance.)

Fig. 2. Computed from equation (2) with the demo's geometry: the piston passes mid-stroke before 90° and hurries through the top half of its travel — the SHM curve is symmetric, the real one is not.

Fig. 3. Acceleration over one revolution. The two cosine terms add at TDC (peak ≈ (1+λ)rω²) and oppose at BDC (≈ (1−λ)rω²) — the piston is yanked hardest around the top, twice as hard as a sine wave would suggest at λ = 0.3.

Side thrust: the rod pushes sideways too

The rod can only transmit force along its own axis, and that axis leans at $\varphi$ , where $L\sin\varphi = r\sin\theta$ . So whatever axial force $F$ the gas and inertia apply, the cylinder wall must supply a lateral reaction:

F_{\text{side}} = F\,\tan\varphi, \qquad \varphi = \arcsin(\lambda\sin\theta)

(5)

Zero at both dead centres, maximum near 90° — which the piston spends skating down the thrust side of the bore under full combustion load. This is the dominant source of bore wear, and a second reason (after equation (4)) that designers like long rods. Feel both effects below:

Rod / crank ratio

L/r

3.30

\lambda = \frac{r}{L} = 0.303

\tan\varphi_{\max} = 0.318

Secondary inertia force ≈ $\lambda$ × the primary (30%), and the piston presses the cylinder wall with up to 32% of the axial force. A longer rod calms both — at the cost of a taller, heavier block.