Mechanism design · Rotary to reciprocating

Slider-Crank Workshop

Why does a crank turning at steady speed move the slider unevenly? Because the slider follows the crank pin's position along the guide, not its angle, and the tilting rod distorts that further. Near each end the pin moves mostly across the guide, so the slider stops while the crank turns on; in between, the pin moves along the guide and the slider is fastest. A shorter rod distorts more, and an offset guide makes one stroke take more of the turn than the other.

Drive a piston along its guide with a crank and a connecting rod. Change the rod length and the guide offset, compare how long the two strokes take, and see why the piston force gives no torque on the crank at dead center.

Slider position $0$

Slider velocity $0$

Slider acceleration $0$

Acute transmission angle $0$

Required input torque $0$

Signed load power $0$

Length-closure residual $0$

Quasistatic power residual $0$

Build & turn

Change the geometry and follow the motion

Input angle

\theta

40.00\,{}^\circ

Prescribed input angular speed

\dot\theta

45.00\,{}^\circ/\mathrm s

Crank length

a

0.40\,\mathrm m

Connecting link length

b

1.20\,\mathrm m

Guide vertical offset

e

0.00\,\mathrm m

Tracer fraction along coupler

u

0.50\,

Tracer normal offset

h

0.30\,\mathrm m

Signed slider force

F_x

-5.00\,\mathrm N

Angles are measured counterclockwise from the horizontal axis pointing right, and a positive guide offset is upward. Blue is the input crank, mint the coupler, coral the output and gold the tracer point on the coupler. Drag the gold crank-pin handle in the separate plan view, or use the angle slider with the keyboard.

Playback turns the input at a constant angular speed. It stops at the next geometric limit and never jumps across a range where the linkage cannot be assembled. The Grashof inequality alone does not guarantee a full turn; which links are grounded and driven also matters. When the two circles coincide, the assembly is not unique, and the Lab says so.

The connecting rod takes the right-hand intersection of its circle with the guide. At the slider's dead centre the transmitted velocity is zero, and a finite slider force needs zero input torque. This differs from the singular position with the rod vertical, where the derivatives are undefined. A guide offset can make the two strokes take different times.

The load is a constant external force or torque with a sign. It does not always resist; on part of the cycle it can supply power. In the massless, frictionless, quasistatic limit, the input torque balances the load's virtual work. Joint reactions, inertia forces, bearing safety and motor sizing are not computed.

Lengths between pin centres and tracer coordinates are in metres. Bar widths, pin diameters, bearing blocks and the separate depth layers are schematic; the real motion is planar, with pin axes along the depth. Links may overlap. The gold curve is the computed path of the tracer at sampled angles, not a physical rail. The CSV holds 721 samples, one every half degree; entries stay blank where the linkage cannot be assembled or the derivatives are singular.

The slider is assembled on the right-hand side, the rod is rigid and the guide is ideal. Positions follow exact geometry. A constant force of either sign acts on the slider and is treated quasistatically. The gas cycle, pressure, flywheel dynamics, bearing loads and combustion are not modeled.

Start with the essentials

Focus question: Why does the slider pause at an end while the crank keeps turning?
One-sentence intuition: The connecting rod makes the slider's position change out of proportion with the crank angle, and at dead center no finite slider force can turn the crank back.

Core mathematical model

Slider position on the right-hand branch

x=a\cos\theta+\sqrt{b^2-(e-a\sin\theta)^2}

The square root must be real. When it reaches zero, the connecting rod stands vertical and the position is singular.

Velocity and acceleration

\dot x=x'(\theta)\dot\theta,\qquad\ddot x=x''(\theta)\dot\theta^2

The crank turns at constant angular speed during playback. The acceleration shown comes from the motion alone, not from a force calculation.

Input torque from virtual power

\tau_{\rm in}=-F_x x'(\theta),\qquad\tau_{\rm in}\dot\theta+F_x\dot x=0

At an ordinary dead center the displacement derivative is zero, so the required torque is zero too. A constant external force keeps its sign; it does not flip when the slider changes direction.

Common difficulties

A zero velocity is not a singularity

Typical misconceptionWhenever the output stops moving, the mechanism must be at a singular position.

Better mental modelWhen the output simply reverses, its velocity is zero but the solution stays regular. Only at a closure singularity do the differential equations fail to determine the motion uniquely.

The input speed is set, not computed

Typical misconceptionIf the animation runs, this motor could drive the real mechanism.

Better mental modelThe input speed is simply prescribed. Link inertia, friction, flexing of the links and collisions are left out, and the power balance is quasistatic.

Run the experiment

01
Turn the crank
Press play, then pause and drag the gold crank pin in the plan view.
What to observe: The 3D links and every plot move together because they all use the same input angle.
02
Change a length
Compare Inline guide, Short connecting rod and Offset guide.
What to observe: A shorter rod changes how the slider's speed varies through the cycle. An offset guide changes how many degrees of crank rotation each stroke takes.
03
Look at the limiting position
Select Dead center and read the input torque. Then export the full sweep.
What to observe: Missing and singular values are flagged on their own and never mixed up with genuine zeros.

Theory & apparatus: Carnegie Mellon · Planar Linkages · Universidad de Almería · Four-bar mechanisms