Machine kinematics · Slotted rocker

Crank-Shaper Quick Return

Why does a steady motor give the ram a fast return? Because the crank turns through unequal angles on the two strokes. The rocker reaches its ends where its slot is tangent to the crank circle, and those two points split each turn into a large arc for the cutting stroke and a small one for the return. At constant motor speed, the smaller arc takes less time. A larger crank radius makes the split more lopsided and the stroke longer.

A crank pin slides in the slot of a swinging rocker, and the rocker drives a horizontal ram through a connecting rod. Change the crank radius to change the ratio of cutting time to return time, then find which dimensions change the stroke but leave the timing alone.

Ram displacement from rod-length datum $0$

Signed output velocity $0$

Output acceleration $0$

One shaft turn $0$

Working / return time $0$

Ram stroke $0$

Return time $0$

Contact / linkage residual $0$

Turn & compare

Give the return less time.

Shaft phase

\theta

240.00\,{}^\circ

Prescribed shaft rate

\omega

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

Crank radius / fixed-pivot spacing

r/d

0.45\,

Rocker pivot to rod joint

L

2.50\,\mathrm m

Connecting rod length

c

1.60\,\mathrm m

In the plane of motion, right and up are positive, and depth points toward you. The shaft angle is measured counterclockwise from the crank pointing right. The controls use degrees; the calculations and the CSV export use radians, metres and seconds.

The two fixed pivots are 1.2 metres apart. The lower rocker pivot, the turning crank pin, the rocker tip and the connecting rod together place the ram on the right-hand branch. The rail height equals the lever length minus the pivot spacing. Negative ram velocity means the slower working stroke; positive velocity means the faster return.

The parameter range here avoids singular positions. Within it, changing the connecting rod reshapes the speed curve but leaves the reversal angles and the stroke unchanged. A longer rocker changes the stroke without changing the ratio of stroke time to return time. The output displacement is the ram coordinate minus the connecting rod length.

Playback runs through one cycle and stops. All curves and exports are exact ideal kinematics sampled over one turn; they are not measured data. Positions, velocities and accelerations come from the same model. Mass, friction, backlash, elastic deformation, real contact loads and drive safety are not modeled.

The slot and sliding shoe are real parts of the mechanism, but their widths are schematic. Here the slotted lever only swings back and forth; this is not the inversion in which the slotted arm turns all the way round. No material is actually cut.

The links are ideal, rigid and planar, the slot is open, and the ram slides in a horizontal guide. The crank turns at a fixed speed. "Working stroke" only names a direction of motion: cutting load, material removal, inertia, friction and motor dynamics are not modeled.

Start with the essentials

Focus question: Which dimension changes the return time without changing motor speed?
One-sentence intuition: The rocker swings farthest when its slot is tangent to the crank circle, and those two crank positions split each turn into unequal angles.

Core mathematical model

Crank pin and rocker direction

\mathbf A=(0,-d),\quad\mathbf P=(r\cos\theta,r\sin\theta),\quad\mathbf B=\mathbf A+L\frac{\mathbf P-\mathbf A}{|\mathbf P-\mathbf A|}

The crank pin slides along the rocker slot; the rocker tip stays at fixed distance from its pivot.

Ram position on the right-hand branch

H=L-d,\quad Q_x=B_x+\sqrt{c^2-(H-B_y)^2},\quad Q_y=H

The positive square root picks the connecting-rod arrangement that is drawn. The output displacement is this horizontal coordinate minus the rod length.

Quick-return time ratio

\alpha=\arcsin(r/d),\quad\frac{T_{\rm work}}{T_{\rm return}}=\frac{\pi+2\alpha}{\pi-2\alpha}

At constant shaft speed, the time for each stroke is proportional to the crank angle it spans.

Stroke length, velocity and acceleration

S=2L\frac rd,\quad v=\omega\frac{dQ_x}{d\theta},\quad a=\omega^2\frac{d^2Q_x}{d\theta^2}

The stroke formula holds for every allowed combination of link lengths, because over that range the ram position moves steadily one way as the rocker angle grows.

Common difficulties

A kinematic model is not a force model

Typical misconceptionIf the animation runs smoothly, a real machine can run at that speed.

Better mental modelSmooth motion only shows that the geometry works. Contact loads, inertia, friction, elastic deformation and the drive itself need their own models, and the part thicknesses are schematic.

Constant input is not constant output

Typical misconceptionThe ram cuts at a steady speed during the working stroke.

Better mental modelThe ram speed changes continuously in both directions. The quick-return ratio compares the times for whole strokes; it is not a speed ratio that holds at every instant.

Run the experiment

01
Run one turn
Choose the motion-plane view and run or drag the coral crank.
What to observe: The ram covers the same stroke both ways, but one direction takes longer.
02
Change the radius ratio
Compare mild and strong quick-return presets.
What to observe: A larger radius ratio makes the two directions differ more in time and also lengthens the stroke.
03
Separate stroke from timing
Change rocker length and connecting rod length independently.
What to observe: Rocker length changes the stroke, and rod length changes the shape of the motion curve. Neither one changes the ideal time ratio.

Geometry & apparatus: CMU · Cams · NITK Virtual Labs · Crank and Slotted Mechanism