Slider-Crank (Exact Kinematics and Gas Torque)

kinematicstorque-power

Verified build 10 relations · 6 identities proven · 3 modeling steps · 6 parity samples

Take a four-bar linkage and push one ground pivot off to infinity: the rocker that used to swing on an arc now slides in a straight line, and the four-bar becomes a slider-crank — crank, connecting rod, piston. It is the mechanism that turns the up-and-down of a piston into the round-and-round of a crankshaft (and back): every internal-combustion engine, every reciprocating compressor and pump, every old steam locomotive’s driving wheel. One degree of freedom — set the crank angle θ\theta and the whole mechanism is decided.

Unlike its four-bar parent, the in-line slider-crank assembles on a single branch. The parent’s open/crossed pair came from a quadratic; here the position problem is a right triangle — rod length ll as the hypotenuse, the crank pin’s transverse offset rsinθr\sin\theta as one leg, and the rod’s axial reach q=l2r2sin2θq=\sqrt{l^2-r^2\sin^2\theta} as the other — with only the positive root physical. The piston sits at

x=rcosθ+l2r2sin2θx = r\cos\theta + \sqrt{l^2 - r^2\sin^2\theta}

measured from the crankshaft axis: x=r+lx=r+l at top dead centre, x=lrx=l-r half a turn later, a stroke of exactly 2r2r no matter how long the rod.

Two passes: motion, then force

The page carries the two analyses an engine designer runs in turn, as its two configurations:

Each configuration hands you the knobs for that pass and holds the other driver — gas force FF in the kinematic pass, crank speed ω\omega in the force pass — at a labelled reference value, so the full quasi-static state is always on screen.

There is no material picker here, and that is the point: kinematics and the quasi-static force path are pure geometry. Steel, aluminium, or titanium, the piston follows the same path and the same gas force makes the same torque. Material re-enters the moment inertia and stiffness matter — the crank’s own whirl and torsional vibration, the rod’s buckling, the bearing loads — which is where the combined bending + torsion shaft and the torsional oscillator pick the story back up.

Try it

Inputs
Rod axial reach
Piston position (from crank axis)
Piston velocity
m/s
Piston acceleration
m/s²
Piston travel from TDC (exact)
Piston travel from TDC (two-term)
Approximation error
Connecting-rod angle (obliquity)
Connecting-rod force
Crank torque
N·m

Governing relations

l2=q2+r2sin2θl^2 = q^2 + r^2\sin^2\theta

Assumes: rigid crank and connecting rod, ideal pin joints, no clearances; in-line (non-offset) slider-crank; the piston pin slides on the cylinder axis through the crank centre. The rod length l, the crank pin's transverse offset r·sinθ, and the rod's axial projection q form a right triangle, so q = sqrt(l^2 - r^2 sin^2 theta) · Valid while: The connecting rod is no longer than the crank (l ≤ r), so the mechanism cannot assemble through a full rotation — near mid-stroke the position radical sqrt(l² − r²sin²θ) goes to zero and then imaginary. A real slider-crank always has l > r (typically l ≈ 3–4 r).

Source: Norton, R. L., Design of Machinery: An Introduction to the Synthesis and Analysis of Mechanisms and Machines, 5th ed., McGraw-Hill, 2012 — slider-crank position, velocity, and acceleration analysis (the exact closed forms and the r/l two-term series for piston motion) and engine dynamics (the gas-force torque with connecting-rod obliquity, T = F·r·sin(θ+φ)/cosφ).

x=rcosθ+qx = r\cos\theta + q

Assumes: piston position measured from the crank axis along the cylinder line: the crank pin's axial coordinate r·cosθ plus the rod's axial projection q. x = r + l at top dead centre (θ = 0) and x = l − r at bottom dead centre (θ = π); the stroke is 2r

Source: Norton, R. L., Design of Machinery: An Introduction to the Synthesis and Analysis of Mechanisms and Machines, 5th ed., McGraw-Hill, 2012 — slider-crank position, velocity, and acceleration analysis (the exact closed forms and the r/l two-term series for piston motion) and engine dynamics (the gas-force torque with connecting-rod obliquity, T = F·r·sin(θ+φ)/cosφ).

v=ωrsinθ(1+rcosθq)v = -\omega r\sin\theta\left(1 + \dfrac{r\cos\theta}{q}\right)

Assumes: constant crank speed ω, so v = dx/dt = ω·dx/dθ and nothing is integrated in time; the derivative is taken exactly and keeps q = sqrt(l² − r²sin²θ) as a shared subterm. That this expression is exactly ω·dx/dθ is re-derived by independent differentiation in the physics test

Source: Norton, R. L., Design of Machinery: An Introduction to the Synthesis and Analysis of Mechanisms and Machines, 5th ed., McGraw-Hill, 2012 — slider-crank position, velocity, and acceleration analysis (the exact closed forms and the r/l two-term series for piston motion) and engine dynamics (the gas-force torque with connecting-rod obliquity, T = F·r·sin(θ+φ)/cosφ).

a=ω2(rcosθ+r2cos2θq+r4sin2θcos2θq3)a = -\omega^2\left(r\cos\theta + \dfrac{r^2\cos 2\theta}{q} + \dfrac{r^4\sin^2\theta\cos^2\theta}{q^3}\right)

Assumes: constant crank speed ω, so a = d²x/dt² = ω²·d²x/dθ² (no dω/dt term); the second derivative is exact. For a long rod it reduces to the familiar a ≈ −rω²(cosθ + (r/l)cos2θ). Machine- checked against the second derivative of x in the physics test

Source: Norton, R. L., Design of Machinery: An Introduction to the Synthesis and Analysis of Mechanisms and Machines, 5th ed., McGraw-Hill, 2012 — slider-crank position, velocity, and acceleration analysis (the exact closed forms and the r/l two-term series for piston motion) and engine dynamics (the gas-force torque with connecting-rod obliquity, T = F·r·sin(θ+φ)/cosφ).

s=(r+l)xs = (r + l) - x

Assumes: piston travel measured DOWN from top dead centre (where x = r + l), the natural frame for the engine designer's two-term series; s runs from 0 at TDC to 2r at BDC

Source: Norton, R. L., Design of Machinery: An Introduction to the Synthesis and Analysis of Mechanisms and Machines, 5th ed., McGraw-Hill, 2012 — slider-crank position, velocity, and acceleration analysis (the exact closed forms and the r/l two-term series for piston motion) and engine dynamics (the gas-force torque with connecting-rod obliquity, T = F·r·sin(θ+φ)/cosφ).

sapprox=r(1cosθ)+r24l(1cos2θ)s_{\text{approx}} = r(1 - \cos\theta) + \dfrac{r^2}{4l}(1 - \cos 2\theta)

Assumes: the classic Fourier/binomial two-term approximation of the piston travel, exact frame s = r(1−cosθ) + (l − q) with (l − q) binomially expanded to (r²/2l)sin²θ = (r²/4l)(1−cos2θ). This is a cited modeling approximation, NOT an identity — its error is bounded and grows with r/l (proven across samples in the physics test), which is the whole point of showing it

Source: Norton, R. L., Design of Machinery: An Introduction to the Synthesis and Analysis of Mechanisms and Machines, 5th ed., McGraw-Hill, 2012 — slider-crank position, velocity, and acceleration analysis (the exact closed forms and the r/l two-term series for piston motion) and engine dynamics (the gas-force torque with connecting-rod obliquity, T = F·r·sin(θ+φ)/cosφ).

ε=sapproxs\varepsilon = s_{\text{approx}} - s

Assumes: the signed gap between the two-term approximation and the exact travel — small at engine proportions (r/l ≈ 1/4–1/3) and growing steeply as the rod shortens

Source: Norton, R. L., Design of Machinery: An Introduction to the Synthesis and Analysis of Mechanisms and Machines, 5th ed., McGraw-Hill, 2012 — slider-crank position, velocity, and acceleration analysis (the exact closed forms and the r/l two-term series for piston motion) and engine dynamics (the gas-force torque with connecting-rod obliquity, T = F·r·sin(θ+φ)/cosφ).

lsinϕ=rsinθl\sin\phi = r\sin\theta

Assumes: the connecting rod tilts from the cylinder axis by the obliquity angle φ; the rod foot and crank pin share the same transverse offset, so l·sinφ = r·sinθ, i.e. sinφ = (r/l)sinθ. Because l > r this has |sinφ| < 1 and cosφ > 0 for every crank angle — the slider-crank assembles on a single branch, no open/crossed pair · Valid while: Rod obliquity is extreme (r/l > 0.5). The connecting-rod force F/cosφ climbs steeply, the side thrust on the cylinder wall grows, and the two-term approximation degrades fast. Real engines keep l/r ≈ 3–4 (r/l ≈ 0.25–0.33) for exactly these reasons.

Source: Norton, R. L., Design of Machinery: An Introduction to the Synthesis and Analysis of Mechanisms and Machines, 5th ed., McGraw-Hill, 2012 — slider-crank position, velocity, and acceleration analysis (the exact closed forms and the r/l two-term series for piston motion) and engine dynamics (the gas-force torque with connecting-rod obliquity, T = F·r·sin(θ+φ)/cosφ).

Frod=FcosϕF_{\text{rod}} = \dfrac{F}{\cos\phi}

Assumes: quasi-static: the piston is in force balance under the gas force F (along the cylinder axis) and the connecting-rod force F_rod (along the rod, tilted by φ). The axial component F_rod·cosφ balances F, so F_rod = F/cosφ. Piston mass and friction are neglected (no inertia term — this is the force PATH, not a dynamic balance)

Source: Norton, R. L., Design of Machinery: An Introduction to the Synthesis and Analysis of Mechanisms and Machines, 5th ed., McGraw-Hill, 2012 — slider-crank position, velocity, and acceleration analysis (the exact closed forms and the r/l two-term series for piston motion) and engine dynamics (the gas-force torque with connecting-rod obliquity, T = F·r·sin(θ+φ)/cosφ).

T=Frsin(θ+ϕ)cosϕT = F r\,\dfrac{\sin(\theta + \phi)}{\cos\phi}

Assumes: moment of the connecting-rod force about the crankshaft axis: T = F·r·sin(θ+φ)/cosφ. At φ → 0 (very long rod) this becomes the elementary T = F·r·sinθ. The torque is zero at both dead centres and peaks near mid-stroke — the fluctuation a flywheel exists to smooth

Source: Norton, R. L., Design of Machinery: An Introduction to the Synthesis and Analysis of Mechanisms and Machines, 5th ed., McGraw-Hill, 2012 — slider-crank position, velocity, and acceleration analysis (the exact closed forms and the r/l two-term series for piston motion) and engine dynamics (the gas-force torque with connecting-rod obliquity, T = F·r·sin(θ+φ)/cosφ).

Derivation

Steps marked modeling step are where physics enters by citation — every other line is machine-proven to follow from them, and the cited models are independently re-derived in the test pipeline where possible. See what is and isn't machine-verified.

l2=q2+r2sin2(θ)l^{2} = q^{2} + r^{2} \sin^{2}{\left(\theta \right)}

1. Draw the mechanism as a right triangle. The connecting rod is the hypotenuse of length l; the crank pin sits a transverse distance r·sinθ off the cylinder axis, and the rod reaches a distance q along the axis to the piston pin. Rigid links and pin joints are the entire physics of the device — everything below is geometry and calculus. So l² = q² + r²sin²θ, and q = sqrt(l² − r²sin²θ). This radical is real for all θ only when l > r. — geometry: the rod-crank-axis right triangle modeling step

x=q+rcos(θ)x = q + r \cos{\left(\theta \right)}

2. The piston pin's distance from the crank axis is the crank pin's axial coordinate r·cosθ plus the rod's axial reach q. At θ = 0 the piston is at x = r + l (top dead centre); half a turn later, at θ = π, it is at x = l − r (bottom dead centre). The difference — the stroke — is 2r, independent of the rod length. — add the two axial projections

v=ωr(1+rcos(θ)q)sin(θ)v = - \omega r \left(1 + \frac{r \cos{\left(\theta \right)}}{q}\right) \sin{\left(\theta \right)}

3. The crank turns at a constant speed ω, so the piston velocity is v = dx/dt = ω·dx/dθ — a space derivative scaled by ω, never a time integral. Differentiating x and keeping q as a shared subterm gives v = −ω r sinθ (1 + r cosθ/q). It is zero at both dead centres and its peak leans toward TDC — the asymmetry a long rod suppresses and a short rod exaggerates. (The claim that this is exactly ω·dx/dθ is machine-checked by independent differentiation in the physics test — the audit surface for this page.) — differentiate once; ω constant

a=ω2(rcos(θ)r2cos(2θ)qr4sin2(θ)cos2(θ)q3)a = \omega^{2} \left(- r \cos{\left(\theta \right)} - \frac{r^{2} \cos{\left(2 \theta \right)}}{q} - \frac{r^{4} \sin^{2}{\left(\theta \right)} \cos^{2}{\left(\theta \right)}}{q^{3}}\right)

4. Differentiate again: a = d²x/dt² = ω²·d²x/dθ² (no dω/dt term, since ω is fixed). The exact acceleration carries three terms over powers of q. For a long rod (q → l) it collapses to the textbook a ≈ −rω²(cosθ + (r/l)cos2θ): a dominant once-per-rev term plus a twice-per-rev correction — the "secondary" imbalance that engine designers cannot balance with a simple counterweight. — differentiate twice; ω constant

dispapprox=r(1cos(θ))+r2(1cos(2θ))4ldisp_{approx} = r \left(1 - \cos{\left(\theta \right)}\right) + \frac{r^{2} \left(1 - \cos{\left(2 \theta \right)}\right)}{4 l}

5. Measure piston travel down from top dead centre: s = (r + l) − x = r(1 − cosθ) + (l − q). The second piece is small — rationalise it exactly, l − q = r²sin²θ/(l + q), and binomially expand, (l − q) = l − l√(1 − (r/l)²sin²θ) ≈ (r²/2l)sin²θ = (r²/4l)(1 − cos2θ), to get the two-term form every engine text carries. This is a cited APPROXIMATION, not an identity: it is exact only as r/l → 0. (The rationalised s is the solved form — subtracting x from r + l directly would lose precision near dead centre.) — binomial two-term approximation (r/l small) modeling step

err=dispapproxdispexacterr = disp_{approx} - disp_{exact}

6. Put the approximation's cost on screen: ε = s_approx − s. Rationalised, the two near-equal travels collapse to ε = −r⁴sin⁴θ / 2l(l + q)² — third order in r/l, so at engine proportions (r/l ≈ 1/4 to 1/3) it stays well under a percent of the stroke, and it grows as the cube when the rod shortens. The physics test bounds it across a sweep of r/l — the comparison IS the pedagogy, which is why the approximation is a readout and never a relation the solver leans on. — define the signed approximation error

lsin(ϕ)=rsin(θ)l \sin{\left(\phi \right)} = r \sin{\left(\theta \right)}

7. Now the force path. The connecting rod is not along the cylinder axis; it tilts by the obliquity angle φ. The rod foot and the crank pin share the same transverse offset, so l·sinφ = r·sinθ. Because l > r this always has |sinφ| < 1 and cosφ > 0 — the slider-crank assembles on a single circuit, with none of the open/crossed ambiguity of the general four-bar. — obliquity from the shared transverse offset modeling step

Frodcos(ϕ)=FF_{rod} \cos{\left(\phi \right)} = F

8. Treat the loaded piston quasi-statically (mass and friction neglected). It balances the gas force F along the axis against the connecting-rod force F_rod along the rod. The rod's axial component F_rod·cosφ carries the gas force, so F_rod = F/cosφ — always larger than F, and the cross-axis remainder F·tanφ is the side thrust the cylinder wall must take. — piston free body (quasi-static)

Tcos(ϕ)=Frsin(ϕ+θ)T \cos{\left(\phi \right)} = F r \sin{\left(\phi + \theta \right)}

9. Take moments of the rod force about the crankshaft axis. The crank torque works out to T = F·r·sin(θ + φ)/cosφ. For a very long rod (φ → 0) it is the elementary T = F·r·sinθ — the piston force times the crank's transverse offset. T is zero at both dead centres and swings to a peak near mid-stroke: the torque ripple every reciprocating engine smooths with a flywheel. — moment about the crankshaft axis

How it fails

A slider-crank, like its four-bar parent, mostly misbehaves kinematically and dynamically long before a link breaks — and several of its troubles are written directly into the obliquity angle ϕ\phi this page computes.

The torque ripple, and why the flywheel exists

The single clearest lesson of the force pass is that crank torque is not constant. Over one revolution T(θ)T(\theta) swings from zero at a dead centre to a mid-stroke peak and back — and in a real engine it goes negative over the compression stroke, when the crank drives the piston rather than the other way round. A load, meanwhile, wants steady torque. The mismatch between the engine’s lumpy delivery and the load’s steady demand is stored and released twice a revolution by a flywheel: its rotational inertia soaks up the excess near the torque peak and gives it back through the dead centres, holding the speed fluctuation to a chosen coefficient.

Sizing that flywheel means integrating the torque curve — accumulating the energy (TTmean)dθ\int(T-T_\text{mean})\,\mathrm{d}\theta between the speed extremes to get the required inertia. This page refuses to do that honestly: it evaluates TT at the crank angle you set, exactly, but it does not integrate over the cycle, because time- and angle-integration of a varying quantity is a dynamics capability the catalog has deliberately not built yet. So the flywheel story is told here and sized on the flywheel page — the torque ripple is the input that page’s energy method consumes, and wiring one THING’s output into the next is the job of the chaining work still to come. Naming the hand-off rather than faking the integral is the honest version.

Where material finally re-enters

Everything on this page is geometry and quasi-static force, so nothing here depends on what the parts are made of. The failures that do depend on material live one link away:

So the slider-crank’s own page stops at the clean, material-blind skeleton: where the piston is, how fast it moves, and what torque a given gas force makes. Every failure that needs a modulus, a density, or a yield strength is deliberately handed to the THING that carries it.

  • DC Motor (Permanent Magnet)

    The machine that turns current into torque — and its whole personality is one straight line. At fixed voltage a PM DC motor trades speed for torque along T = T_stall(1 − ω/ω₀): two datasheet numbers pin every operating point, and the peak power hides at half the no-load speed.

    • torque-power
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  • Four-Bar Linkage (Position)

    Four pinned links — ground, crank, coupler, rocker — and the oldest mechanism in the book. Spin the crank and the rocker answers through pure geometry. Every position has TWO valid assemblies (open and crossed): the first THING in the catalog where one input has two honest answers, and the widget lets you pick the circuit.

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  • Torsional Oscillator (Disk on a Shaft)

    A disk on an elastic shaft is a torsion pendulum: twist it and let go and it rings at one natural frequency, ω_n = √(k_t/J_d). The pitch is set entirely by the shaft's stiffness and the disk's inertia — not by how hard you twist it — while the stress it survives is set by the amplitude.

    • dynamics
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  • Flywheel (Solid Rotating Disk)

    The machine that stores work as spin — and loads itself doing it. Centrifugal self-loading grows with ρω²R², so the energy a flywheel can hold per kilogram is capped not by its size but by one material index: strength over density.

    • energy-storage
    • stress
    • mass-cost
  • Planetary (Epicyclic) Gearset

    Three coaxial members — sun, ring, planet carrier — share one gear mesh law. With two degrees of freedom, it has no single "ratio": fix a different member and the same hardware becomes a different transmission.

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  • Axial Disk Clutch / Brake (Uniform Wear vs Uniform Pressure)

    The torque an axial plate clutch can pass depends on an assumption you cannot see: how the contact pressure is distributed across the friction annulus. A new, rigid clutch presses uniformly; a worn-in one wears until pressure ∝ 1/r, concentrating load at the inner edge. This page shows both torque predictions side by side — never picking a winner — with the worn-in model always giving the smaller (safe) number, and the r_i = r_o/√3 that squeezes the most torque from a given lining.

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Chains with

Outputs whose SI dimension and quantity kind match another THING's input — the only wires the planner's connectionLegal accepts (invariant 2, computed at build time, not hand-listed). Wire these on the chaining demo.

+ 27 more THINGs its outputs can legally feed (showing the first 8 in course order).

Sources