Velocity and acceleration: polygons and Coriolis

The force starts with the acceleration

• The piston's acceleration sets the inertia force that loads the rod, the bearings and the frame. Velocity comes first, because the acceleration terms use it.
• For two points on one rigid link, $\mathbf v_C = \mathbf v_B + \boldsymbol\omega \times \mathbf r_{CB}$: the relative velocity is perpendicular to the link.

Velocities at 60°

Accelerations

• Each link adds a centripetal part $\omega^2r$ along it and a tangential part $\alpha r$ across it: the acceleration polygon.
• The crank pin has $a_B = r\omega^2 = \mathbf{39.5}$ m/s² towards O; closing the polygon gives the piston $a_C = \mathbf{14.8}$ m/s², matching $r\omega^2(\cos\theta + \cos2\theta/n)$ with $n = 4$.

The Coriolis component

• A point sliding along a rotating link gains an extra $2\omega v_{rel}$, perpendicular to the link: the Coriolis component. The rotation turns the sliding velocity, and the radius changes while the link turns.
• A block sliding at 0.6 m/s on an arm turning at 5 rad/s gets $2 \times 5 \times 0.6 = \mathbf{6.0}$ m/s². It appears only on a rotating guide, never a fixed one.

The fastest point is not at 90°

• The piston is fastest where the rod is perpendicular to the crank, because the crank pin's velocity then points straight along the rod.
• That happens at $\tan\theta = l/r = 4$, $\theta = \mathbf{76.0°}$, with $v_C = r\omega\sqrt{1 + (r/l)^2} = \mathbf{1.296}$ m/s, a little above $r\omega$.

Worked example: the inertia force on the rod

• The 0.8 kg piston at 60° accelerates at 14.8 m/s²: the rod must push $0.8 \times 14.8 = 11.8$ N to drive it.
• At top dead centre $a_C = r\omega^2(1 + 1/n) = 39.5 \times 1.25 = 49.3$ m/s², and the rod carries $\mathbf{39.5}$ N. The peak, not the 60° value, sizes the rod and its bearing.

Three polygon slips

Recap