In the first tutorial of this series concerning slider and crank mechanisms we firstly found from geometry an expression for displacement x of the slider as a function of crank angle θ and the ratio n (= L/R) and then differentiated with respect to time to obtain expressions for velocity and linear acceleration also as functions of θ and n.

This analysis gave the required information for the slider but not the velocity and acceleration of the connecting rod, the motion of which is a combination of translational and rotational motion. This information can be obtained from velocity and acceleration diagrams. We will explore these methods in this tutorial.

As an introduction we review some general principles and techniques.

Velocity and acceleration diagrams are drawn to scale using known quantities and the unknown values then measured directly from the completed diagrams. It is also possible to obtain unknown values from the geometry of the diagrams and we adopt this approach below. (In the third tutorial in the series we obtain the equivalent information from vector equations).

A key point when constructing velocity and acceleration diagrams is the
distinction between ** absolute **and

To create the velocity diagrams, v_{a} and v_{b }
are connected at their common reference point **o** and
the diagram closed by a third arrow. In (i) the arrow
direction represents the **relative velocity** v_{b/a}
meaning the velocity of point **b** relative to point
**a**. In (ii) the arrow v_{a/b}
represents the velocity of point **a **relative to
point **b**.

Construction of a velocity diagram is similar to, but not exactly the same, as conventional vector diagrams shown below. In both cases the magnitude and direction of the resulting velocities are identical.

In (i) vector *v*_{a} is subtracted from
vector *v*_{b} giving the resultant vector
*v*_{b/a}
which represents the velocity of point** b** relative to
the velocity of point **a**. In
(ii) vector *v*_{b} is subtracted from
vector* v*_{a} giving the
resultant vector* v*_{a/b}
which represents the velocity of point **a** relative
to the velocity of point **b**.

Acceleration diagrams are constructed in a similar way to velocity diagrams bearing in mind that directions of velocity and acceleration for a point are not necessarily the same. Acceleration diagrams must also take into account that rotational motion of any point always has a radial component of acceleration directed towards the centre of rotation and a tangential component of acceleration when the point is subject to angular acceleration. The diagram below shows an acceleration diagram for the crank pin of a rotating crank arm.

In this diagram point A rotates about point O with angular
acceleration α. a_{Ar} is the radial (centripetal)
acceleration of point A relative to point O with direction from A
towards O. a_{At} is the tangential acceleration of
point A relative to point O resulting from angular acceleration α.
a_{A} is the resultant acceleration of point A relative to
point O. Note that the diagram represents a snapshot for a
specific crank angle and angular velocity ω, which is constantly
changing on account of angular acceleration α .

In this example the acceleration diagram is the same as a vector diagram as all accelerations are relative to a single point O. As we shall see below this is not the case for the complete crank mechanism.

When we examine the motion of the crank arm and connecting rod in a crank mechanism we are dealing with the kinematics of a rigid body moving in a two dimensional plane. Classic theory approaches this as a combination of a longitudinal translation and a rotation represented in the example below.

The diagram is a snapshot at one instant of velocity vectors at multiple points along the longitudinal axis of a rigid bar moving in a two dimensional plane. The longitudinal component of velocity at all points along this axis must have the same magnitude, otherwise the body would be expanding or contracting. If the absolute velocities at two points on the axis are known (in this example end points A and B) the corresponding transverse velocities perpendicular to the line at points A and B define a pure rotational motion with centre of rotation on the longitudinal axis at point C*. This principle applies to any arbitrary line on a rigid plane body. We choose the longitudinal axis here simply for convenience.

* The point C defined here must not be confused
with the *instantaneous centre of
rotation* or *velocity pole* described below although as
we shall see the two points are related. I adopt the
description "centre of rotation on the longitudinal axis" on
the basis that for any practical example this axis is clearly
defined.

Note also that the magnitude of the rotational velocity for the centre of rotation on the longitudinal axis derived from this construction is the same whatever arbitrary line is taken for the longitudinal axis on the rigid body.

It follows from the principle of combined translational and
rotational motion of a rigid body that there will be a point in the
plane of motion where the tangential velocity arising from rotation
is equal and opposite to the longitudinal velocity. This point
of zero velocity must therefore be a centre of rotation for that
instant, with every point in the body being in pure rotation around this
point. The point is called the ** instantaneous centre of rotation**
or

The velocity pole can be found in two ways.

- An axis is extended perpendicular to the longitudinal axis on the body through the centre of rotation C. The longitudinal velocity (blue vector) is a constant at all positions on this perpendicular axis. The tangential velocity arising from rotation of the body (red vector) is also perpendicular to this axis but in the opposite direction and varies in magnitude. The point on this perpendicular axis where longitudinal and tangential velocities are of equal magnitude and thus cancel is the velocity pole.
- If directions of the absolute velocities of two points on the body are known, lines perpendicular to these directions can be drawn from the respective points. The point of intersection of these perpendicular lines is the velocity pole. This is a simple and useful construction as we shall shortly see. The diagram above clearly shows how the absolute velocity of each point on the body (green vector) is tangential to a line from the velocity pole, indicating pure rotation about the velocity pole at that instant.

We now construct a velocity diagram for the crank mechanism below based on a crank angle θ = 50°. The angle between the connecting rod and the horizontal axis is designated φ which we use in subsequent calculations. Displacements are in metres.

The velocity diagram has three components:

- velocity of point A relative to O (ground) designated v
_{A/O} - velocity of point B relative to point A designated v
_{B/A} - velocity of point B relative to O (ground) designated
v
_{B/O}

We know the following:

- The line of action of v
_{A/O}is perpendicular to the crank arm with magnitude ω x (OA) = (2 π) x (1) = 2 π. m/s. The crank arm rotates in an anti-clockwise direction which determines the direction of v_{A/O} - Assuming the connecting rod is rigid the motion of point B
relative to point A must be purely rotational. Thus the
line of action of v
_{B/A}is perpendicular to AB. - The motion of slider point B is constrained on a horizontal
plane which defines the line of action of v
_{B/O}

The diagram below shows the lines of action of these velocities.

We now construct the velocity diagram. Firstly, calculate
the magnitude of v_{A/O} and draw the arrow to scale along
the line of action. The head of the arrow represents point A
relative to point O at the tail. Now draw the line of action
of v_{B/A} through point A and the line of action of v_{B/O}
through point O. The point of intersection of these two lines
determines the magnitude and direction of v_{B/A}
and v_{B/O}

The completed velocity diagram is shown below. Note that the
directions of v_{B/A} and v_{B/O} are resolved by
this construction, the head of the arrow representing the velocity
at that point relative to the point at the tail.

Values scaled from well drawn velocity diagrams are sufficiently accurate for practical purposes. Alternatively, velocities can be calculated directly from the diagram.

Firstly, we calculate angle φ previously mentioned :

Show angle φ on the velocity diagram with the horizontal axis and the axis of the connecting rod extended to intersect at point P .

It is now a reasonably simple matter to find sides AB and BO of
triangle ABO given we know AO = v_{A/O} = 2π m/s and
angle φ = 14.79°

From triangle AOP angle PAO = 180° - (140° + 14.79°) = 25.21°

Thus in triangle ABO angle BAO = (90° - 25.21°) = 64.79°

and angle ABO = 180° - (40° + 64.79°) = 75.21°

gives BO = v_{B/O} = 5.88 m/s

This is the identical value for the velocity of the slider for crank angle θ = 50° obtained in the previous tutorial from differentiation with respect to time, viz.

Correspondingly we calculate v_{A/B} as follows:

gives AB = v_{B/A} = 4.18 m/s

We use this result to find the rotational velocity ω_{AB
}of the connecting rod of length 3 m at the instant where the
crank angle is 50°.

The diagram below shows the construction for the instantaneous velocity pole of the connecting rod in our example mechanism where the crank angle = 50°.

Construction of the velocity pole diagram is very
straightforward. Firstly, rod AB is drawn to scale at the
correct angular position. In this case we know the direction
of velocities v_{A/O} and v_{B/O} thus the velocity
pole P is the intersection of lines perpendicular to these directions
extended from points A and B.

Bearing in mind that the velocity pole represents the centre of
pure rotational motion of the rod the rotational velocity of the rod
ω_{AB} is found by scaling length AP, which is the
radius of rotation for point A. We know from above that v_{A/O}
= 6.28 m/s. Thus ω_{AB} = 6.28/AP and correspondingly
by scaling length BP from the diagram velocity v_{B/O} = ω_{AB}
x BP.

The radii from the velocity pole to points A and B can be easily calculated, noting that we already know angle φ on the diagram and AB = 3 m..

Using the trig identity:

confirming the value of ω_{AB} derived from the
velocity diagram in the previous section.

confirming the value of v_{B/O} derived from
the velocity diagram in the previous
section.

We derive the centre of rotation on the longitudinal axis of the connecting rod directly from the velocity pole diagram. Single subscripts indicate absolute velocities at points on the rod.

The line PD perpendicular to AB at point D extended through the
velocity pole P is the radius of rotation of point D on the
longitudinal axis of the rod. Thus the direction of velocity vector v_{D}
must align exactly with the longitudinal axis. Point D is the
instantaneous centre of rotation on this axis and is the only point
on the axis that has no transverse component of velocity.

We calculate v_{D} with reference to the diagram as
follows.

From the previous velocity pole diagram : AP = 4.51 m

5.68 m/s is the longitudinal component of velocity at every point
on the rod axis. We can check this is the case at point B
where we know v_{B} = 5.88 m/s along the horizontal axis.

The acceleration diagram for the example mechanism has the components listed below. Lines of action of these accelerations are shown in the diagram above.

- Point A has centripetal acceleration designated a
_{rA/O }acting radially along the axis of the crank arm towards its centre of rotation point O, which is also the ground reference point for the diagram. Because the crank arm rotates with constant angular velocity ω there is no tangential component of acceleration at point A. In our example a_{rA/O}= ω^{2}(AO) = (2π)^{2}(1) = 39.44 m/s^{2} - Point B has centripetal acceleration relative to point A
designated a
_{rB/A}acting radially along the axis of the connecting rod towards point A which is the centre of rotation in this case. In our example mechanism a_{rB/A}= ω^{2}_{AB}.(AB) = (1.39)^{2}(3) = 5.80 m/s^{2} - We know the rotational velocity of the connecting rod
ω
_{AB}is not constant and thus has angular acceleration α_{AB}Consequently point B has a tangential acceleration relative to point A designated a_{tB/A}with a line of action perpendicular to the rod. Magnitude and direction on the line of action are not known. - Point B has acceleration relative to point O
designated a
_{B/O}acting horizontally. Magnitude and direction on the line of action are not known.

We now construct the acceleration diagram, beginning with the known quantities.

Firstly, draw a_{A/O} from point O to a point
designated A on its line of action with direction from point A
to point O shown on the physical diagram for the mechanism at θ =
50° (see above) . It is important to note that the relative
positions of A and O on the acceleration diagram are opposite to the
positions on the physical diagram.

Next, draw a_{rB/A} on its line of action from point A
to a point designated B^{/} noting the direction of this centripetal acceleration is towards its
centre of rotation which in this case is from point B to point A on
the physical diagram. Knowing that tangential acceleration at point
B relative to point A also exists we designate the point as B^{/} which can be
considered a half-way point between point A and point B.

Now draw the line of action of a_{tB/A} through point B^{/}
and line of action of a_{B/O} through point O. The
intersection of these lines determines point B. a_{tB/A}
is drawn from point B^{/} to point B and a_{B/O} is
drawn from point O to point B. The completed acceleration
diagram is shown below.

Values of a_{B/O} and a_{tB/A} from a scaled diagram are satisfactory for practical purposes.
However. as we did for the velocity diagram. we can calculate the
value of a_{B/O} from the diagram and compare with the value
obtained previously from the expression derived from differentiation
with respect to time in a previous
tutorial.

The calculation is slightly more involved than for the velocity
diagram. We know all angles in triangles PBO and PB^{/}A
and we know lengths OA and AB^{/}. We then find side
AP in triangle PB^{/}A which gives side OP = OA + AP
in triangle PBO from which we find side OB. The calculation
gives a_{B/O} = 23.4 m/s^{2} which corresponds to
the value previously obtained.

From these triangles we also calculate a_{tB/A} = 29.72
m/s^{2}. From this value we determine the
angular acceleration of the connecting rod α_{BA} from the
relation a_{tB/A} = AB x α_{BA }
where AB = 3 m giving α_{BA} = 9.91 rad/s^{2}

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