---
title: "Relative velocity method and velocity image"
subject: "Theory of Mechanism and Machine I"
chapter: 7
key: "semester_6/theory_of_mechanism_and_machine_i:relative-velocity-method-velocity-polygon"
source: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/relative-velocity-method-velocity-polygon
site: BCT Engineering
---

# Relative velocity method and velocity image

Theory of Mechanism and Machine I · Chapter 7: Kinematic Analysis of Mechanisms

- Asked in: year not given, 6 marks
- Video: [Velocity analysis of Four bar chain by Relative velocity method](https://www.youtube.com/watch?v=MKrfyubQGF0) — Dr Joji Thomas
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/theory_of_mechanism_and_machine_i/relative-velocity-method-velocity-polygon
- Whole chapter: https://www.bctengineer.com/md/semester_6/theory_of_mechanism_and_machine_i/chapter-7

## Question

Explain the relative velocity method for finding velocities in a mechanism. Show, with the help of a four-bar mechanism ABCD (AD fixed), the steps for drawing the velocity polygon, finding the angular velocities of the coupler BC and follower CD, and obtaining the velocity of a point E on the coupler using the velocity image principle.

## Short answer

Relative velocity method: solve v_B = v_A + v_{BA} by drawing a velocity polygon. For four-bar ABCD, draw ob perpendicular to AB (v_B). From b, line perpendicular to BC; from pole o, line perpendicular to CD; intersection c gives v_C and v_{CB}. Angular velocities: ω_BC = bc/BC, ω_CD = oc/CD. Velocity image: triangle bce similar to BCE on coupler, rotated 90°, gives velocity of point E.

## Answer

#### 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.

```text
 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: $\omega = \dfrac{\text{magnitude of relative velocity of two points}}{\text{distance between them}}$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For two points A and B on the same rigid link, the velocity of B relative to A is:
   A. Along AB with magnitude omega x AB
   B. Perpendicular to AB with magnitude omega x AB
   C. Zero
   D. Equal to v_A
   Answer: B. Points on a rigid link cannot approach each other, so v_BA is perpendicular to AB with magnitude omega x AB. (Key point: Relative velocity of two points)

2. What does the pole o represent in the velocity polygon?
   A. The crank pin B
   B. The coupler midpoint
   C. Points of zero velocity, such as fixed pivots A and D
   D. The centre of mass of the mechanism
   Answer: C. Fixed pivots have zero velocity, so they are drawn at the pole. (Key point: Pole of velocity polygon)

3. How is point c located in the four-bar velocity polygon?
   A. Intersection of a line from b perpendicular to BC and a line from o perpendicular to CD
   B. A line from b parallel to BC and from o parallel to CD
   C. Midpoint of ob
   D. Along ob extended
   Answer: A. v_CB is perpendicular to BC (from b) and v_C is perpendicular to CD (from o); their intersection is c. (Key point: Finding point c)

4. How is the velocity of a point E on the coupler BC found?
   A. Measure E directly on the space diagram
   B. Use oe = ob x BE/BC
   C. Take e at the pole
   D. Locate e on bc so that be/bc = BE/BC (image similar to link, same letter order), then oe is v_E
   Answer: D. The velocity image of a rigid link is similar to the link, so be/bc = BE/BC and oe gives the velocity of E. (Key point: Velocity image)

## Related in Theory of Mechanism and Machine I

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