Relative velocity method and velocity image

Principle

For two points A and B on the same rigid link, the velocity of B relative to A is perpendicular to AB, and its magnitude is $\omega \cdot AB$:

$$\vec v_B = \vec v_A + \vec v_{BA}, \qquad v_{BA} = \omega_{AB}\cdot AB \ \ (\perp AB)$$

The vector equation is solved by drawing a velocity polygon: the direction of each vector is known, and the magnitude of at least one is known.

Four-bar mechanism ABCD (crank AB with $\omega_2$, AD fixed)

1. Draw the configuration diagram (space diagram) to scale.
2. Compute the velocity of B: $v_{B} = \omega_2 \cdot AB$, perpendicular to AB. Choose a velocity scale (for example 1 cm = 0.5 m/s).
3. Choose a pole $o$ (the zero-velocity point, representing the fixed points A and D). Draw $ob$ perpendicular to AB, with length proportional to $v_B$.
4. Velocity of C relative to B: from $b$ draw a line perpendicular to BC (the direction of $v_{CB}$).
5. Velocity of C relative to D: from $o$ (as D is fixed) draw a line perpendicular to CD (the direction of $v_C$).
6. The intersection of these two lines is point $c$. Then $oc$ is the velocity of C, and $bc$ is the velocity of C relative to B.
7. Scale the lengths:

$$\omega_3 = \frac{v_{CB}}{BC} = \frac{bc}{BC}, \qquad \omega_4 = \frac{v_C}{CD} = \frac{oc}{CD}$$

Sense: $\omega_3$ is found by placing the vector $bc$ at C on the space diagram and seeing which way it turns C about B; $\omega_4$ by placing $oc$ at C and seeing which way it turns C about D.

 velocity polygon        space diagram
      b                  B-------C
     /|\                 |        \
    / | \                A         D
   o--+--c

Velocity of a point E on BC (velocity image)

The triangle $bce$ in the polygon is similar to the triangle BCE on the link (the polygon of the points of a rigid link is an image of the link, rotated by $90^\circ$). So:

1. For E on BC, mark $e$ on $bc$ so that $\dfrac{be}{bc} = \dfrac{BE}{BC}$.
2. For E off the line BC, make triangle $bce$ similar to BCE, with the same letter order.
3. Join $o$ to $e$. Then $oe$ gives the velocity of E.

Points to remember

• Fixed pivots are represented by the pole $o$.
• The image has the same letter order and the same sense as the link.
• The angular velocity of a link is the same for all its points: $\omega = \dfrac{\text{magnitude of relative velocity of two points}}{\text{distance between them}}$.