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Velocity and Acceleration

The analysis of velocity and acceleration in a range of mechanisms including Klein's Construction for piston acceleration.

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Contents

Introduction
Analysis Of Velocity And Accelerations Components
The Velocity And Acceleration Of A Piston By Analysis
Instantaneous Centre Method For Velocities.
The Vector Method For Velocity And Acceleration
Klein's Construction For Piston Acceleration
Page Comments

Introduction

The Theory of Machines is concerned with the Motion of parts of machines and the forces which act on those parts. In most cases these forces are not constant and their calculation demands that we know the velocities and accelerations which occur in the various components.

The concept of Instantaneous Centres of Velocity was covered in the section on Mechanisms. In this section the Analysis of Velocity and Acceleration are considered with particular reference to Cranks and Pistons. Klien's Construction for Piston Acceleration is introduced and a description of the Coriolis Component is given.

Analysis Of Velocity And Accelerations Components

Velocity

In the diagram the point $P$ moves in the plane $XOY$ . The length $OP = r$ and the $\displaystyle\angle POX = \theta$ .

Velocity is the measurement of the rate and direction of change in the position of an object. It is a vector physical quantity; both magnitude and direction are required to define it.

Then:

$x = r\cos\theta$

And,

$y = r\sin\theta$

Differentiating with respect to time :

$\frac{dx}{dt} = \dot{ x} = \dot{r }\cos\theta - \dot{ r}\sin\theta \;.\;\dot{ \theta }$

(1)

$\frac{dy}{dt} = \dot{y} = \dot{r}\sin\theta + r\cos\theta \;.\;\dot{\theta }$

(2)

The radial component of Velocity $v$ ( i.e. in the direction of $OP$ ) is given by:

$v = \dot{x}\cos\theta + \dot{y}\sin\theta$

(3)

And using equations (1) and (2)

$\displaystyle v = \dot{r}$ Which is the rate of increase of $OP$

The tangential component of velocity (i.e. Perpendicular to $OP$ in the direction of $\displaystyle \theta$ increasing)

$= \dot{y}\cos\theta - \dot{x}\sin\theta \;.\;\dot{\theta }= r\dot{\theta } = r\omega$

Where $\displaystyle \omega = \dot{\theta }$ = angular velocity of $OP$

Acceleration

Acceleration is the time rate of change of velocity with respect to magnitude or direction; the derivative of velocity with respect to time.

Differentiating (1) and (2)

$\ddot{x} = \ddot{r}\cos\theta - \dot{r}\sin\theta \;.\;\dot{\theta } - \dot{r}\sin\theta \;.\;\dot{\theta } - r\cos\theta \;.\;\dot{\theta }^2 - r\sin\theta \;.\;\ddot{\theta }$

$\ddot{y} = \ddot{r}\sin\theta + \dot{r}\cos\theta \;.\;\dot{\theta } + \dot{r}\cos\theta \;.\;\dot{\theta } - r\sin\theta \;.\;\dot{\theta }^2 + r\cos\theta \;.\;\ddot{\theta }$

The Radial components of acceleration

$= \ddot{x}\cos\theta + \ddot{y}\sin\theta = \ddot{r} - r\;\dot{\theta }^2$

And from equation (3)

$= \dot{v} - r\;\omega ^2$

(4)

The Tangential component of acceleration $= \ddot{y}\cos\theta - \ddot{x}\sin\theta$

By substitution:

$= r\;\ddot{\theta } + 2\,\dot{r}\;\dot{\theta } = r\;\alpha + 2\;v\;\omega$

(5)

(Note $\displaystyle \alpha = \ddot{\theta }$ = Angular acceleration of $OP$ )

Of these four terms in equations (4) and (5)

$\displaystyle \dot{v}$ is the rate of change of radial velocity
$\displaystyle r\;\omega ^2$ is the centripetal acceleration due to the rotation of $OP$
$\displaystyle r\;\alpha$ is due to the change in angular velocity.
$\displaystyle 2\;v\;\omega$ is called the compound supplementary acceleration or Coriolis Component. Notice that the direction of this is the same as $\displaystyle r\;\omega$ when $v$ is radially outwards.

The Velocity And Acceleration Of A Piston By Analysis

In the following analysis $\displaystyle \omega$ is the uniform angular velocity of the crank. The positive direction of velocity and acceleration is away from the crankshaft.

$x = r\cos\theta + l\cos\phi$

And,

$r\sin\theta = l\sin\phi$

$\therefore \;\;\;\;\;\;\cos\phi = \sqrt{1 - \frac{\sin^2\theta }{n^2}}\;\;\;\;\;\;\;\,\;\;\;\;n = \frac{l}{r}$

From the above three equations:

$x = r\,\left (\cos\theta + \sqrt{n^2 - \sin^2\theta } \right )$

Thus Piston velocity,

$\dot{x}\;= - r\,\omega \left (\sin\theta + \frac{\sin2\theta }{2\;\sqrt{n^2 - \sin^2\theta }} \right )$

(6)

And Piston acceleration,

$\ddot{x}\;= - r\,\omega^2 \left (\cos\theta + \frac{n^2\;\cos2\theta + \sin^4\theta }{(n^2 - \sin^2\theta )^{\frac{3}{2}}} \right)$

(7)

Normally $\displaystyle \sin^2\theta$ can be neglected in comparison with $\displaystyle n^2$ and equations (6) and (7) can be reduced as follows:

$\dot{x}\;= - r\,\omega \left (\sin\theta + \frac{\sin2\theta }{2n} \right )$

$\ddot{x}\;= - r\,\omega^2 \left (\cos\theta + \frac{\cos2\theta }{n} \right )$

Instantaneous Centre Method For Velocities.

This has been covered in the section on Mechanisms.

The Vector Method For Velocity And Acceleration

The Law of addition of velocities states that:

Velocity of $B$ = Velocity of $A$ + Velocity of $B$ relative to $A$

i.e.

$\underset{OB}{\displaystyle\rightarrow} = \underset{OA}{\rightarrow} + \underset{AB}{\rightarrow}$

Absolute velocities (or accelerations) are given from $O$ to the corresponding point on the diagram.

For Velocities

The relative velocity between two points $A$ and $B$ on the same link of a mechanism must be perpendicular to the line joining the points and is equal to $\displaystyle AB\;\omega$ (Equation (2))since $r$ is constant and $\displaystyle \dot{r}$ is zero.
The relative velocity for two points sliding over one another is along the common tangents of their paths and represents the component $\displaystyle \dot{r}$ , $\displaystyle r\;\omega$ is zero since $\displaystyle r = 0$

For Acceleration

The relationships for acceleration are similar to those given for velocity:

Acceleration of $B$ = Acceleration of $A$ + Acceleration of $B$ relative to $A$
Equations (4) and (5) are the general expressions for the radial and tangential components of relative acceleration.
For two points on the same link $\displaystyle v = \dot{v} = 0$ leaving centripetal component $\displaystyle - r\,\omega ^2$ (which can be calculated when the velocities are determined) and the tangential component $\displaystyle r\;\alpha$
For a uniformly rotating Crank $\displaystyle \alpha = 0$ leaving the centripetal as the only term.
The Coriolis component arises when a point on one link is sliding along another link which is itself rotating.

If $A$ , $B$ , $C$ , are three points on the same link of a mechanism and $a$ , $b$ , $c$ , are the corresponding points on the velocity (or acceleration) diagram, it can be shown that the triangles $ABC$ and $abc$ are similar. $abc$ is called the Velocity (or acceleration) image of the link.

Klein's Construction For Piston Acceleration

The above is a diagramatic sketch of a piston, connecting rod, and crank assembly where,

$PC$ is the connecting rod with $C$ the Crank Pin.
$OC$ is the crank.
$OP$ is the line of stroke.
$P$ is the gudgeon pin

The Construction is as follows:

Extend $PC$ to meet the line through $O$ perpendicular to the line of stroke. Let the point of intersection be $N$ .
Draw a circle centre $C$ and radius $CN$ .
Draw a circle with $CP$ as diameter.
Let the common cord cut the line $CP$ at $L$ and the line of stroke $PO$ at $M$ .

Then the quadrilateral $OCLM$ represents, to a certain scale, the acceleration diagram for $OCP$ . It can be shown that this scale is $\displaystyle \omega ^2$ .

The Centripetal acceleration of the crank pin is $\displaystyle C\;O\;\times \omega ^2$
The piston acceleration is $\displaystyle M\;O\;\times \omega ^2$
$CL$ is the Centripetal component and $LM$ the Tangential component of the acceleration of $P$ relative to $C$ , so that $CM$ is the acceleration image of $CP$ .
For any point $Q$ on $CP$ draw a line parallel to $OP$ cutting $CM$ in $q$ . The acceleration of $Q$ is $q\;O\times\omega ^2$ in magnitude and direction.

Example:

[imperial]

Example - Example 1

Problem

An aeroplane $A$ flying at 180 m.p.h. in a direction $\displaystyle 30^{0}$ North of West sights another $B$ due North of $A$ . After 30 seconds flying $B$ is seen to be in a North-Easterly direction from $A$ and after a further 45 seconds $B$ is directly astern of $A$ . If $B$ is flying at a constant speed in a direction due South, find:

a) The speed of $B$
b) For how long $B$ is within 2 miles of $A$

Workings

In the following diagram the Paths of $A$ and $B$ are shown. Their three particular positions are $\displaystyle a_1\,,\,a_2\,,\,a_3$ and $b_1\,,\,b_2\,,\,b_3$ respectively.

$a_1\;a_2 = 1.5\;miles$ (30 seconds flying)

By scaling or by calculation $\displaystyle b_2\;b_3 = 2.05\;miles$ and occupies 45 seconds of flying time.

Hence, Speed of $B$ = $\displaystyle\frac{2.05}{45}\times 3600 = 164\;m.p.h.$

Relative displacement of $B$ to $A$ = displacement of $B$ - displacement of $A$

i.e. $b_1\;{b_{1}}^{'} = b_1\;b_2 + b_2\;{b_{1}}^{'}$

In the first 30 seconds $\displaystyle b_2\;{b_{1}}^{'}\;= - a_1\;a_2$

Assuming that A is at rest at $\displaystyle a_1$ , the relative path of $B$ is $\displaystyle b_1\;{b_{1}}^{'}$ produced and a circle centre $\displaystyle a_1$ and of radius 2 miles cuts this path at 1 and $m$ . Between 1 and $m$ , $B$ lies within 2 miles of $A$ .

$1\;m = 1.85\;miles$

But $b_1\;{b_{1}}^{'} = 2.47\;miles$

This represents 30 seconds of relative displacement

But, Time corresponding to 1 mile = $\displaystyle\frac{1.85}{2.47}\times 30 = 22.5\;seconds$

Solution

a) The speed of $B$ is $164\;m.p.h.$
b) $22.5\;seconds.$