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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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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.


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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


In the diagram the point \inline P moves in the plane \inline XOY. The length \inline OP = r and the \inline \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.


x = r\cos\theta
y = r\sin\theta

Differentiating with respect to time :


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

And using equations (2) and (3)

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

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

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

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


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

Differentiating (2) and (3)

\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 (4)

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

By substitution:

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

Of these four terms in equations (5) and (6)

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

The Velocity And Acceleration Of A Piston By Analysis

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


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


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,

And Piston acceleration,

Normally \inline \displaystyle \sin^2\theta can be neglected in comparison with \inline \displaystyle n^2 and equations (7) and (8) 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 \inline B = Velocity of \inline A + Velocity of \inline B relative to \inline A

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

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

For Velocities

  • The relative velocity between two points \inline A and \inline B on the same link of a mechanism must be perpendicular to the line joining the points and is equal to \inline \displaystyle AB\;\omega (Equation (3))since \inline r is constant and \inline \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 \inline \displaystyle \dot{r}, \inline \displaystyle r\;\omega is zero since \inline \displaystyle r = 0
For Acceleration

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

  • Acceleration of \inline B = Acceleration of \inline A + Acceleration of \inline B relative to \inline A
  • Equations (5) and (6) are the general expressions for the radial and tangential components of relative acceleration.
  • For two points on the same link \inline \displaystyle v = \dot{v} = 0 leaving centripetal component \inline \displaystyle - r\,\omega ^2 (which can be calculated when the velocities are determined) and the tangential component \inline \displaystyle r\;\alpha
  • For a uniformly rotating Crank \inline \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 \inline A,\inline B,\inline C, are three points on the same link of a mechanism and \inline a,\inline b,\inline c, are the corresponding points on the velocity (or acceleration) diagram, it can be shown that the triangles \inline ABC and \inline abc are similar. \inline abc is called the Velocity (or acceleration) image of the link.

Klein's Construction For Piston Acceleration


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

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

The Construction is as follows:

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

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

  • The Centripetal acceleration of the crank pin is \inline \displaystyle C\;O\;\times \omega ^2
  • The piston acceleration is \inline \displaystyle M\;O\;\times \omega ^2
  • \inline CL is the Centripetal component and \inline LM the Tangential component of the acceleration of \inline P relative to \inline C, so that \inline CM is the acceleration image of \inline CP.
  • For any point \inline Q on \inline CP draw a line parallel to \inline OP cutting \inline CM in \inline q. The acceleration of \inline Q is \inline q\;O\times\omega ^2 in magnitude and direction.
Example - Example 1
An aeroplane flying at 180 m.p.h. in a direction North of West sights another due North of . After 30 seconds flying is seen to be in a North-Easterly direction from and after a further 45 seconds is directly astern of . If is flying at a constant speed in a direction due South, find:
  • a) The speed of
  • b) For how long is within 2 miles of
In the following diagram the Paths of and are shown. Their three particular positions are and respectively.

(30 seconds flying)


By scaling or by calculation and occupies 45 seconds of flying time.

Hence, Speed of =

Relative displacement of to = displacement of - displacement of


In the first 30 seconds

Assuming that A is at rest at , the relative path of is produced and a circle centre and of radius 2 miles cuts this path at 1 and . Between 1 and , lies within 2 miles of .


This represents 30 seconds of relative displacement

But, Time corresponding to 1 mile =
  • a) The speed of is
  • b)