1.1 Cam

A cam may be defined as a rotating, reciprocating or oscillating machine part, designed to impart reciprocating and oscillating motion to another mechanical part, called a follower. Usually cam-follower contact is a line contact and together they constitute a higher pair. The contact is generally maintained with a help of a spring.

1.2 Types of cam

Fig 1.1 Classification of cams

 

 

1.3 Types of follower

Fig 1.2 Classification of followers

                                                                                                               

1.4 Radial Cam: Terminology

Fig 1.3 Radial cam terminology

a) Base circle - Smallest circle drawn from the cam center inside the cam profile.

b) Trace point - It is that point on the follower which when the cam is operated generates the pitch curve. Hence in case of knife edge follower it is the knife edge and in case of a roller follower it is the center of the roller which is called as trace points.

c) Pressure angle - - It is the angle line of follower motion and normal to the pitch curve at that point.

d) Pitch point - The pressure angle keeps changing during the cam rotation. The point where the pressure angle is maximum is called as the pitch point.

e) Pitch circle - It is a circle drawn from the center of the cam through the pitch points.

f) Pitch curve - It is the curve generated by the trace point of the follower. Hence in case of knife edge follower is the curve generated by the knife edge whereas in case of roller follower it is the curve generated by the center of the roller.

g) Prime circle - It is the smallest circle drawn from the cam center to the pitch curve. Hence for knife edge follower prime circle and base circle are same whereas for a roller follower the prime circle is more than the base circle by an amount equal to the radius of the roller.

h) Lift or stroke - It is the distance moved by the follower from cam dwell or lowest point to Rise or highest point is the called as the lift or stroke of the follower.

1.5 Motion of the follower

The follower, during its travel, may have one of the following motions.

       a) Uniform velocity
       b) Simple harmonic motion
       c) Uniform acceleration and retardation
       d) Cycloidal motion

1.6 Displacement, velocity and acceleration diagrams when the follower moves with uniform velocity

This is a case when the displacement graph of the follower will be a straight line with a constant slope. Constant slope means the velocity of the follower will not change with time. Second differentiation of displacement will lead to a zero acceleration when the velocity is constant and to infinite acceleration at the start and end of the rise or fall of the follower. This kind of motion is not desirable. Steps to draw the graphical representation is as below:

1. Divide the horizontal axis in 360° and as per the dwell 1, rise, dwell 2 and fall angle.
2. On the vertical axis take the stroke of the cam as “S”.
3. Starting from dwell 1, draw a line on the horizontal axis for the complete dwell 1 angle. This line will lie on the horizontal axis because there is no motion of the follower.
4. Draw a straight line from the end of the dwell 1 angle to the beginning of dwell 2 angle. This is the angle through which the cam rotates and raises the follower equal to the stroke length.
5. Draw the horizontal straight line at the stroke line starting from the end of the rise angle to the beginning of the fall angle. This is dwell 2 or the angle in which the follower remains in raised position.
6. Draw a straight line starting from the end point of dwell 2 to the end point of the all angle. This is the fall angle or the angle through which the cam rotates and brings back the follower from the raised position to the initial position.
7. From the displacement diagram we can say that the velocity of the follower will remain zero at both dwell 1 and dwell 2 and will be a constant value during the rise and fall of follower. This is shown in the velocity graph.
8. In the velocity diagram we see that the velocity of the follower from an initial condition (of rest or motion) is suddenly changed from zero to a value or from a value to zero. This is possible only if we have infinite acceleration of the follower. This is shown in the acceleration diagram.

 

Fig 1.4 Follower in uniform velocity

1.7 Displacement, velocity and acceleration diagram when the follower moves with a simple harmonic motion

If P is a point on the circumference of a circle with a diameter “S” and if P moves at a uniform speed “ω” on the circumference of the circle then the projection of P on the diameter of the circle will execute a simple harmonic motion. Let the projected point be called as P’. The general equation of SHM is

X(t) = S*Cos(ωt) ,

differentiating, we get,

dx/dt = -S * ω * Sin(ωt),

differentiating again, we get

$\frac {d^2 x}{dt^2} = -S \times \omega^2 \times cos\omega t$

Steps to draw the graphical representation of the follower motion is as below

1. Divide the horizontal axis in 360° and as per the dwell 1, rise, dwell 2 and fall angle.
2. On the vertical axis take the stroke of the cam as “S”.
3. Draw a semi-circle on the vertical axis with “S” as diameter.
4. Divide the semi-circle into any number of even parts, say 8.
5. Divide the angular displacement of the cam during the rise and fall stroke into same number of equal parts.
6. Project the points on the semicircle on the vertical line.
7. Join the projected point by a smooth curve as shown. This is the displacement diagram.
8. Since the displacement diagram is essentially a Cosθ curve hence velocity which is a differential of the displacement will be a -Sinθ curve.
9. This will be a second differential of the displacement diagram and will be a -Cosθ curve. The acceleration of the follower is positive maximum in the beginning, drops to zero in the midpoint and is a negative maximum at the end.

 

Fig 1.5

Simple harmonic motion (Fig.- 3.5)

Let S = Stroke of the follower,
$\theta_f \quad and \quad \theta_r$ = Angular displacement of the cam during fall and rise motion respectively, and
$\omega$ = Angular velocity of the cam in rad/s.
∴ Time required for the rise motion of the follower in seconds, $t_r={\theta_r \over \omega}$, and time required for the fall motion of the follower in seconds,$t_f={\theta_f \over \omega}$
An analogy of body in S.H.M can also be imagined as below: Let a point P move in circle with angular velocity of ω rad/s. then the projection of the point P on the diameter of the circle (say P’) executes an SHM.

Fig 1.6 Follower in SHM

∴ Peripheral speed of the point P′

$V_p= \frac {\pi S}{2 . t_r} \hbox{ or } V_p= \frac {\pi S\omega}{2 . \theta_r}$,

This is also equal to the maximum velocity of the follower and is same for both rise and fall motion of the follower.

The centripetal acceleration of the point P is 

$a_p = \frac {V_p^2}{\frac {S}{2}} \hbox{ or } a_p= \frac {\pi^2 \omega^2 S^2}{2 . {\theta_r}^2}$

This is also equal to the maximum acceleration of the follower and is same for both rise and fall motion of the follower.

1.8 Displacement, Velocity and Acceleration Diagrams when the Follower Moves with Uniform Acceleration and Retardation

Since the follower moves with a uniform acceleration and retardation say some value “C”, the velocity of the follower will be an integral of C, a straight line as Cx+a, and the displacement will be an integral of the velocity, a parabolic curve Cx2 + K. The method to draw this parabolic curve is as below

a) Divide the horizontal axis in 360° and as per the dwell 1, rise, dwell 2 and fall angle.
b) On the vertical axis take the stroke of the cam as “S”.
c) Divide the angular displacement of the cam during the rise into even number of equal parts, say 8. Draw vertical lines through these points.
d) Divide the stroke of the follower into same number of parts.
e) Draw straight lines from the origin. These new straight lines intersect the vertical lines drawn earlier. Join these points of intersection to obtain the parabolic displacement diagram for the rise angle.
f)Similarly draw the parabolic displacement diagram for the fall angle also.
g)Since the acceleration and retardation is uniform, the velocity diagram is linear and varies directly with time.
h)The acceleration diagram will be uniform.

Fig 1.7 Follower in uniform acceleration and retardation

S = Stroke of the follower

Let

θf and θr = Angular displacement of the cam during fall and rise motion respectively, and

ω = Angular velocity of the cam in rad/s.

∴ Time required for the rise motion of the follower in seconds,$t_r={\theta_r \over \omega}$

and time required for the fall motion of the follower in seconds,$t_f={\theta_f\over \omega}$

Mean velocity of the follower during rise is = $S\over t_r$

and mean velocity of the follower during the fall motion is =$S\over t_f$

Now, during the rise motion we see that the follower has to first accelerate from 0 to Vmax and then decelerate to reduce the velocity to zero level again. This is possible only when the time to accelerate and decelerate is same. The same is also true for the fall motion of the follower. Hence 

Maximum velocity of the follower during Rise motion is

$\frac {2 . S}{t_r} => \frac {2 . ω . S}{\theta_r}$

and the maximum velocity of the follower during the fall motion is

$\frac {2 . S}{t_f} => \frac {2 . ω . S}{\theta_f}$

Maximum acceleration of the follower during the rise motion is

$a_r = \frac {V_r}{\frac {t_r}{2}} => \frac {4 . \omega S}{\omega_r ^2}$

the same will be true for the fall motion of the follower.

1.9 Displacement, velocity and acceleration diagrams when the follower moves with cycloidal motion

A cycloid is a curve traced by a point on a circle when the circle rolls without slipping on a straight line. We can see the curve being traced in such a manner in the animation. The method to draw this cycloidal curve is as below

a) the horizontal axis in 360° and as per the dwell 1, rise, dwell 2 and fall angle.
b) On the vertical axis take the stroke of the cam as “S”.
c) Draw a circle of S/2π with point “A” as centre. Divide the circle into even number of parts, say 6.
d) Project the points on the circumference of the circle to the circle centreline.
e) Divide the rise stroke into same number of even parts and draw vertical lines.
f) Join AB
g) Draw line parallel to AB
h) Trace a graph joining the point of intersection.
i) The velocity diagram is made as shown.
i) The acceleration diagram is made as shown.

Fig 1.8 Follower in cycloidal motion

Let

θ = Angle through which the cam rotates in time t second, and
ω = Angular velocity of the cam
We know That displacement of the follower after time t seconds,

$x= S \bigg[ \frac {\theta}{\theta_o}-\frac {1}{2 \pi}sin \bigg( \frac{2 \pi \theta}{\theta_o} \bigg) \bigg]$

Therefore, velocity of the follower after time t seconds

$\frac {dx}{dt} = S \bigg[ \frac {1}{\theta _o} \times \frac {d \theta}{dt} - \frac {2 \pi}{2 \pi \theta_o}cos \bigg( \frac {2 \pi \theta}{\theta_o} \bigg)\frac {d \theta}{dt} \bigg]$

[Differentiating  equation for the displacement 'x'.

$\frac {S}{\theta_o} \times \frac {d\theta}{dt} \bigg[1-cos \bigg( \frac {2 \pi \theta}{ \theta_o} \bigg)\bigg] = \frac {ω.S}{\theta_o} \bigg[ 1-cos \bigg( \frac {2 \pi \theta}{\theta_o} \bigg) \bigg]$

The velocity is maximum, when

 $cos \bigg( \frac {2 \pi \theta}{\theta_o} \bigg)= -1 \qquad or \qquad \frac {2 \pi \theta}{\theta_o}= \pi \qquad or \qquad \theta= \theta_o/2$

Substituting $\theta ={ \theta_0 \over 2}$ , We have maximum velocity of the follower during outstroke,

$v_o = \frac {ω.S}{\theta_o}(1+1)= \frac {2ω.S}{\theta_o}$

Similarly, maximum velocity of the follower during return stroke,

 $v_r = \frac {2ω.S} {\theta_R}$

Now, acceleration of the follower after time t sec,

 $\frac {d^2x}{d t^2} = \frac {ω.S}{\theta_o} \bigg[ \frac {2\pi}{\theta_o} sin \bigg( \frac {2 \pi \theta}{\theta_o} \bigg) \frac{d\theta}{dt} \bigg]$$= \frac {2 \pi ω^2 .S }{ (\theta_o)^2 } sin \bigg( \frac {2 \pi \theta}{\theta_o} \bigg)$

The acceleration is maximum, when

 $sin \bigg( \frac {2 \pi \theta}{\theta_o} \bigg)=1 \qquad or \qquad \frac {2 \pi \theta}{\theta_o} = \frac {\pi}{2} \qquad or \qquad \theta = \theta_o/4$

Substituting $\theta = \theta_o/4$ in equation, We have maximum acceleration of the follower during outstroke,

$a_o = \frac{2 \pi ω^2.S} {(\theta_o)^2}$

Similarly, maximum acceleration of the follower during return stroke,

$a_r = \frac{2 \pi ω^2.S} {(\theta_R)^2}$

The velocity and acceleration diagrams are shown in figure respectively

1.10 Construction of cam profile for a radial cam

In order to draw the cam profile for a radial cam, first of all the displacement diagram for the given motion of the follower is drawn. Then by constructing the follower in its proper position at each angular position, the profile of the working surface of the cam is drawn.

Fig 1.9 Graphical method of generating cam profile

In constructing the cam profile, the principle of kinematic inversion is used, i.e. the cam is imagined to be stationary and the follower is allowed to rotate in the opposite direction to the cam rotation.

Construction of various types of CAM is dealt in the Application for design of CAM.

1.11 Special cams

Special Cams

Based of flank, cams are classified into two groups,

1. Circular arc cam
2. Straight edged cam or Tangent cam

Circular arc cam

When the flank of the cam connecting the nose and base circles are of convex circular arc, such cams are referred as circular arm cam

Fig 1.10 Circular arc cam

Tangent cam

When the flanks between the nose and base circles are of straight and tangential to both the circle, then, the cams are called tangent cam

Fig 1.11 Tangent cam

These are usually symmettric about the centre line of the cam. Generally, the following combinations of cam and follower are used.

1. Circular arc cam with flat faced follower
2. Tangent cam with reciprocating roller follower

1.12 Circular arc cam with flat faced follower

The fig 1.12 represents various main dimentions of circular arc cam.

Fig 1.12

Expression for determining the displacement, velocity and acceleration of the follower when flat face of the follower has contact on the circular flank

Fig 1.13

Let,
r1 = OB = Least base circle radius 
r2 = Nose circle radius
R = QD = Flank circle radius
d = Distance between the centre of cam and nose circle 
α = Angle of ascent
φ = Angle of contact on circular flank

Displacement:

$X = BC = OC - OB = DE - r_1$

$= (QD - QE) - r_1$ =$(R - OQ cos \theta) - r_1$= $R - (R - r_1) cos \theta - r_1$

Thus,
 \(X = (R - r_1)(1 - cos \theta)\)

Velocity:

\(V = \frac {dx}{dt} = \frac {dx}{d\theta}\times \frac {d\theta}{dt}\)= \(\omega(R - r_1)(sin \theta)\)= \(V = \omega (R - r_1)sin \theta\)

From this equation, it is evident that, at the beginning of the ascent, the velocity is zero (when &theta = 0) and it increases with θ. It will be maximum when the follower is just shift from circular flank to circular nose.

$V_{max} = \omega (R - r_1)sin \phi$

Acceleration:

$a = \frac {dv}{dt} = \frac {dv}{d\theta}\times \frac {d\theta}{dt}$

$V = \frac {dx}{dt} = \frac {dx}{d\theta}\times \frac {d\theta}{dt} \quad and \quad hence\quad a = \omega^2 (R - r_1)cos \theta$

It is obvious from the above equation that, at the beginning of the ascent when θ = 0, acceleration is maximum and it goes on decreasing and is maximum when $\theta =\phi$

$a_{max} = \omega^2 (R - r_1)$ and ,

$a_{min} = \omega^2 (R - r_1)cos \phi$

Expression for determining the displacement, velocity and acceleration of the follower when flat face of the follower has contact on the nose

Fig 1.14

Let,

$r_1=$ OB =least base circle

$r_2$= nose circle radius

R=QD=flank circle radius

d= distance between centers of cam and nose circle

$\alpha$= angle of ascent

$\phi$=angle of contact on cicular flank.

Displacement

$x=BC=OC-OB=DE-r_1$

= $DP+PE-r_1$$=r_2+OP cos(\alpha-\theta)-r_1$, thus 

$x=r_2+dcos(\alpha -\theta)-r_1$

Velocity

$v={dx\over dt}={dx\over d\theta}.{d\theta \over dt}$,

therefore

the veclocity,

 $v= \omega dSin(\alpha -\theta)$,

the velocity is minimum when $\alpha = \theta$, this happens when the follower is at the apex of the circular nose and it is maximum when $(\alpha -\theta)$ is maximum and it is so when the contact chnaes from circluar flank to circlular nose i.e when  $\alpha -\theta =\phi$

Acceleration

$a={dv\over dt} ={dv\over d\theta}.{d\theta \over dt}$ =$-\omega^2d cos(\alpha -\theta)$,

negative sin indicates retardation. It is maximum when $\alpha -\theta =0,$, i.e when the follower is at the apex of the nose and minimum when $(\alpha -\theta)$ is maximum, ie when the follower changes the contact from circular flank to circular radius.

$a_{max}= -\omega^2d$ and $a_{min} = -\omega^2dcos\phi$

1.13 Tangent cam with roller reciprocating follower

Tangent cams are made with straight flank. A tangent cam is shown below. The flanks AB and IH are straight line and tangent to the base circle at 'A' and 'I'. The centre of the circular nose is 'P'. The path of the centre of the roller is shown by dotted line.

Fig 1.15

Expression for determination of displacement, velocity and acceleration of the roller follower when in contact on the straight flank

Fig 1.16

Let,
r1 = Least base circle radius 
r2 = Roller radius 
r3 = Nose circle radius
d = Distance between the centre of cam and nose circle 
L = (r1 + r3)
α = Angle of ascent
φ = Angle of contact of cam with straight flank

Displacement:

$X = OG - OB$ $= \frac {OB}{cos \theta} - OB$ = $OB \bigg (\frac {1}{cos \theta} - 1 \bigg )$

\(X = (r_1 + r_3 )\ ( \frac {1}{cos \theta} - 1 )\)

 Velocity:

$V = \frac {dx}{dt} = \frac {dx}{d\theta}\times \frac {d\theta}{dt}$ = $= \bigg(r_1 + r_3 \bigg)\bigg( \frac {sin\theta}{cos^2 \theta} - 0 \bigg) \times \omega$ 

Thus ,

 $V = \omega \bigg(r_1 + r_3 \bigg)\bigg( \frac {sin\theta}{cos^2 \theta}\bigg)$

from this equation, it is evident that, as θ increases, sin θ also increases whereas cos θ decreases. with that, the velocity increases, Velocity will be maximum when θ is maximum. It happens when point of contact is just leaving a straight flank ie., when θ = φ

$V_{max} = \omega \bigg( r_1 + r_3 \bigg) \bigg(\frac {sin \phi}{cos^2 \phi} \bigg)$

Acceleration:

\(a = \frac {dv}{dt} = \frac {dv}{d\theta}\times \frac {d\theta}{dt}\)

$a=\omega \bigg(r_1 + r_3 \bigg)\left[ \frac {cos^2 \theta . cos \theta - sin \theta . 2cos \theta . - sin \theta}{cos^4 \theta} \right]\frac {d\theta}{dt}$

$a=\omega^2 \bigg(r_1 + r_3 \bigg)\left[ \frac {cos^2 \theta - 2sin^2 \theta}{cos^3 \theta} \right]$ = $\omega^2 \bigg(r_1 + r_3 \bigg)\left[ \frac {cos^2 \theta + 2(1 - cos^2 \theta)}{cos^3 \theta} \right]$=$\omega^2 \bigg(r_1 + r_3 \bigg)\left[ \frac { 2 - cos^2 \theta}{cos^3 \theta} \right]$

Acceleration is maximum when, $\left[ \frac { 2 - cos^2 \theta}{cos^3 \theta} \right]$ is minimum. This is possible when $2 - \cos^2 \theta$ is minimum and $cos^3\theta$is maximum. It is so, when $\theta =0$ ,or the roller is at the beginning of its lift along the straight flank. Acceleration is maximum when the roller shifts from flank to nose circle ie., when $\theta = \phi$.

$a_{min} = \omega^2 (r_1 + r_3)$

$a_{max} = \omega^2 (r_1 + r_3)\left[\frac {2 - cos^2 \phi}{cos^2 \phi} \right]$

1.14 Expression for determination of displacement, velocity and acceleration of the roller follower when in contact with nose

Let,
r1 = Least base circle radius 
r2 = Roller radius 
r3 = Nose circle radius
d = Distance between the cam and nose circle 
L = (r1 + r3)
α = Angle of ascent
φ = Angle of contact of cam with straight flank

Fig 1.17

Displacement:

$X = OJ - OG=(OP + PJ) - (OE + EG)= (d + L) - (OP cos \theta_1 + PG cos \beta)$,

 $= (d + L) - (d cos \theta_1 + L cos \beta)=L + d - d cos \theta_1 - L cos \beta$

From right angle triangles, OEP and GEP

$EP = GP sin \beta = OP sin \theta_1 \quad or \quad L sin \beta = d sin \theta_1$

Squaring from both sides

$L^2 sin^2 \beta = d^2 sin^2 \theta_1$, or  $L^2 ( 1 - cos^2 \beta ) = d^2 sin^2 \theta_1$,

=$L^2 - L^2 cos^2 \beta = d^2 sin^2 \theta_1$

=$L cos \beta = {(L^2 - d^2 sin^2 \theta_1)}^{\frac {1}{2}}$

Substituting above value in equation (i)

$X = L + d - d cos \theta_1 - {(L^2 -d^2 sin^2 \theta_1)}^{\frac {1}{2}}$

Velocity:

$V = \frac {dx}{dt} = \frac {dx}{d\theta 1}\times \frac {d\theta 1}{dt}$

=$= -d -sin \theta_1 \frac {d \theta 1}{dt} - \frac {1}{2}{(L^2 -d^2 sin^2 \theta_1)}^{-\frac {1}{2}} (-d^2 2sin \theta_1 cos \theta 1 )\frac {d \theta 1}{dt}$

$= d sin \theta_1 \frac {d \theta 1}{dt} - \frac {1}{2}{(L^2 -d^2 sin^2 \theta_1)}^{-\frac {1}{2}} (-d^2 sin2 \theta_1 )\frac {d \theta 1}{dt}$

$V = \omega d \left[sin \theta_1 + \frac {d sin^2\theta 1}{{(L^2 -d^2 sin^2 \theta_1)}^{\frac {1}{2}}} \right]$

Acceleration:

$a = \frac {dv}{dt} = \frac {dv}{d\theta 1}\times \frac {d\theta 1}{dt}$

$a = \omega d \left[cos \theta_1 + \frac {2{(L^2 -d^2 sin^2 \theta_1)}^{-\frac {1}{2}}(dcos2 \theta_1)-d sin2\theta_1 \times \frac {1}{2}{(L^2 -d^2 sin^2 \theta_1)}^{-\frac {1}{2}}(-d^2 2sin^2 \theta_1 . cos \theta_1)}{2{(L^2 -d^2 sin^2 \theta_1)}^{\frac {1}{2}}} \right]$

Multiply numerator and denominator with $\sqrt {L^2-d^2sin^2\theta_1}$, we get

a=$\omega^2 d \left[cos \theta_1 + \frac {{(L^2 -d^2 sin^2 \theta_1)}(dcos2 \theta_1)-d sin2\theta_1 + \frac {1}{2}{(d^3 sin^2 \theta_1)}}{2{(L^2 -d^2 sin^2 \theta_1)}^{\frac {3}{2}}} \right]$$= \omega^2 d \left[cos \theta_1 + \frac {L^2 2d cos2 \theta_1 -2d^3sin^2 \theta_1 . cos2 \theta_1 + \frac {1}{2}{d^3 {(2sin \theta_1 cos2 \theta_1)}^2}}{2{(L^2 -d^2 sin^2 \theta_1)}^{\frac {3}{2}}} \right]$

$= \omega^2 d \left[cos \theta_1 + \frac {2L^2 d cos\theta_1 -2d^3sin^2 \theta_1 . (1 - sin2 \theta_1) + 2d^3 sin^2 \theta_1 ( 1- sin^2 \theta_1)}{2{(L^2 -d^2 sin^2 \theta_1)}^{\frac {3}{2}}} \right]$,

thus acceleration,

$a = \omega^2 d \left[cos \theta_1 + \frac {L^2 d cos2\theta_1 + d^3sin^4 \theta_1}{{(L^2 -d^2 sin^2 \theta_1)}^{\frac {3}{2}}} \right]$