Cams

The Cam shown in the diagram rotates about $O$ at a uniform speed of 500 r.p.m. and operates a follower attached to a roller of centre $A$. The path of $Q$ is a straight line passing through $O$.

Draw the time lift diagram for the roller centre $Q$ on a base of 1 inch to 0,01 seconds and to a vertical scale four times full size, for a movement of $180^{0}$ from the position shown; determine the maximum velocity of the roller centre $Q$ and the cam angle at which it occurs.

• Leaving the cam "stationary" draw a series of radial lines $OQ_1\;\;OQ_2\;\;OQ_3....$ at equal angles of $10^{0}$ from the position at which lift starts.

• The points $\displaystyle\Q_1\;\;Q_2\;\;Q_3....$ are such that circles of the roller radius , with these points as centres, just touch the cam. Note the point of contact is not in general on the line of stroke.
• Thus $\displaystyle OQ_1$ = lift + Minimum Cam Radius + Roller Radius

The construction of the required time lift diagram is as follows:

• Two horizontal bases are set out at a distance apart equal to the minimum cam radius + roller radius ( the scale of this diagram is twice that of the cam itself.
• Mark off verticals from the lower base line equal to $\displaystyle OQ_1\;\;OQ_2\;\;OQ_3...$ at cam angles of $10^{0}\;\;20^{0}\;\;30^{0}\;.\;.\;.$ These give the points on the displacement time diagram

• The slope of this graph is proportional to the velocity and hence the maximum velocity occurs at a cam angle of $\displaystyle 125^{0}$ and by drawing the tangent at this point its magnitude is found to be $3.6 ft./sec.$

• The cam angle is $\displaystyle 125^{0}$.

The diagram shows a cam which is used to raise and lower the head of a loom. The line of stroke of the follower passes through the centre of the camshaft and the lift is 2.25 in.The follower dwells for a third of a revolution at the top and bottom of the stroke and the rise and fall each occupy a sixth of a revolution. The minimum cam radius is 1.25 in. and the roller on the follower is 2 in. in diameter. Complete the profile to give harmonic motion during rise and fall. Scale half full size.

If the shaft rotates at 30 r.p.m. determine the maximum velocity and the maximum acceleration of the follower.

The minimum distance from the cam axis to the centre of the roller = 1.25 + 1 = 2.25.in.

Set off this distance vertically as is on diagram (a) and above this the lift of 2.25 in. is drawn.

Simple harmonic motion is obtained by the projection from a semi circle onto a diameter. In this case the circumference is divided into 12 equal arcs.

The angle of lift is also divided into 12 equal parts. See diagram (b)

Mark of the distances $oa$, $ob$, $oc$, $...$ at successive intervals of the cam at $OA$, $OB$, $OC$, $...$ The points $A$, $B$, $C$, $...$ are the roller centres as the line of stroke moves round the cam.

From each of the centres an arc of 1 in radius is drawn to represent the roller surface.

The contour of the cam can now be drawn to touch each of the above arcs.

Since the cam turns through $\displaystyle 60^{0}$ during the lift stage $\beta = \displaystyle\frac{\pi }{3}$ and the equation of motion where $\theta$, the cam angle is $h = 1.125\left (1 - \cos 3\theta \right )$

Differentiating, Maximum velocity $= 1.125\times 3\times \displaystyle\frac{d\theta }{dt}$
(At $3\theta = 90^{0}$) $= 1.125\times 3\times 30\times \displaystyle\frac{2\pi }{60} = 10.6\;in.\;sec^{-1}$

Differentiating again gives Maximum acceleration

(at $\theta = 0^{0}$)

• The maximum velocity is $10.6\;in.\;sec^{-1}$
• The maximum acceleration is $100\;in.\;sec.^{-2}$

A cam turns at a uniform speed of 180 r.p.m. and gives an oscillating follower, 2.5 in. long, an angular displacement of $\displaystyle 30^{0}$ on each stroke. The follower is fitted with a roller 2 in. in diameter, which makes contact with the profile of the cam. The outward and return displacements each take place with uniform acceleration and retardation whilst the cam turns through $\displaystyle 60^{0}$ and there is a period of dwell in the outward position whilst the cam turns through $\displaystyle 15^{0}$.

If the axis of the follower is 3.5 in. from the axis of the cam and the least distance of the roller axis from the cam is 2.5 in., draw the outline of the cam.

Find the maximum angular velocity and the maximum angular acceleration of the follower.

If $O$ is on the cam axis and the centre line is vertical then it is possible to draw a circle of radius 3.5 in. and centre $O$. The axis of rotation of the follower must lie on this circle.

In the central lowest position, the roller axis is at $Y$ ($OY = 2.5 in.$) and the follower axis is at $X$ ($YX = 3.5 - 1 = 2.5 in.$) $\displaystyle \angle XOY = 46^{0}$

When the follower is rotated by $\displaystyle 30^{0}$ the roller axis moves to $Z$ and the relative position is then given by $\displaystyle \angle XOZ = 40^{0}$

Mark off $OS$ equal to $OZ$ and at $\displaystyle 7\frac{1}{2}^{0}$ to the vertical. $S$ is the position of the roller axis at the beginning of dwell.

The follower axis is now at $I$ where $\displaystyle \angle IOS = \angle XOZ$

Lift commences with the follower axis at $A$ where $\displaystyle \angle AOI = 60^{0}$

Divide the arc $AI$ into eight parts at $B$, $C$, $D$, etc and join these points to $O$.

For uniform acceleration and retardation $\displaystyle (\alpha)$ the angular displacement of the follower:

The acceleration occurs up to $\displaystyle \theta = 30^{0} = \frac{\pi }{6}\;rad.$ when the follower displacement will be $\displaystyle 15^{0} = \frac{\pi }{12}\;rad.$

$\therefore \;\;\;\;\;\;\;\frac{\pi }{12} = \frac{1}{2}\alpha \times \left (\frac{\pi }{6} \right )^2\left (\frac{60}{180\times 2\pi } \right )^2$

Thus, $\alpha = 677\;rad.\;sec.^{-2}$

Note. The displacement is proportional to the square of the cam angle. At the four stages $B$, $C$, $D$, and $E$ it must be :

$\displaystyle 15\times\left (\frac{1}{4} \right )^2\;;\;\;15\times \left (\frac{1}{2} \right )^2\;;\;\;15\times\left (\frac{3}{4} \right )^2$ and $15$ repectively

i.e. $\displaystyle\frac{15}{26}^{0}\;;\;\;3.75^{0}\;;\;\;8.4375^{0}$ and $15^{0}$

Similarly, for the stages $F$, $G$, $H$, and $I$ the angular displacement will be 30 - 8.4375, 30 - 3.75, 30 - 0.9375 and 30 i.e. $\displaystyle 21.5625^{0},\;\;26.25^{0},\;\;29.0625^{0}$ and $30^{0}$

These angles are added in turn to angle $OXY$ and henced the lines $BK$, $CL$, $DM$, etc are drawn (i.e. $\angle OBK = \angle OXY + 0.9375^{0}\;\;\;:\;\;\;\angle OCL = \angle OXY + 3.75^{0}$, etc.) . The positions $K$, $L$, $M$, $N$, ....of the roller axis are at a distance of 2.5 in. along these lines.

With $K$, $L$, $M$, $N$.... as centres arcs are drawn to represent the roller surfaces and the cam profile is drawn to touch these arcs.

The maximum velocity occurs at the end of the acceleration period and is given by: $\alpha \times \displaystyle\frac{\theta }{\omega }\;\;\;\;\;\;\;\left\;\;\;\displaystyle\frac{\theta }{\omega }$ = time of lift

i.e. $677\times \displaystyle\frac{\pi }{6}\left (\displaystyle\frac{60}{180\times 2\pi } \right ) = 18.9\;rad.\;sec.^{-1}$

• The maximum angular velocity is $\displaystyle 15^{0} = \frac{\pi }{12}\;rad.$
• The maximum angular acceleration is $18.9\;rad.\;sec.^{-1}$

A circular cam of 4 in. diameter and an eccentricity of 1.5 in. rotates about a centre $O$ as shown in the diagram.

The cam rotates at 100 r.p.m. in a clockwise direction and operates a lever follower pivoted at $B$.

When the cam has rotated by $\displaystyle 120^{0}$ from the position shown, find the angular velocity and acceleration of the follower.

The position required is shown in (a) . The direction $CD$ of the follower is the common tangent to the cam and to an arc of 3 in.radius and centre $B$.

If $D$ is the point of contact between the follower and the cam, $AD$ is of fixed length and will always be parallel to $BC$. $AD$ can therefore be replaced by an arm rigidly attached to a block sliding in a slot in the follower, the other end pinned to a link $OA$ rotating with uniform angular velocity.

If $EA$ is drawn parallel to $CD$, the cranked lever $BEA$ will have the same angular motion as the follower and will be driven by the crank $OA$ with a pined block at $A$ sliding in a slot $EA$.

The "equivalent" mechanism velocity diagram is shown as diagram (b).

• $o\;a_1$ = Velocity of "Crank Pin" $A$ = $15.7\;in.\;sec.^{-1}$

• $a_1\;a_2$ = Velocity of sliding parallel to the slot $EA$

• $b\;a_2$ = Velocity due to the rotation of $BEA$ and is perpendicular to $BA$

From this diagram the angular velocity of the follower can be found from $\displaystyle \frac{b\;a_2}{BA} = 1.95\;rad.\;sec.^{-1}$

The "equivalent" mechanism acceleration diagram is shown as (c) above.

• $o\;a_1$= velocity of crank pin = $164\;in.\;sec.^{-2}$

• $a_1\;a_2'$ = Coriolis component, perpendicular to slot

= 2 $X$ velocity of sliding $a_1\;a_2$ $X$ angular velocity of $BEA$ $= 5.7\times 9.3\times 1.95 = 36.2\;in.\;sec.^{-2}$

• $b\;a_2''$ = centripetal acceleration of $A$ to $B$, along $AB$

$= 5.7\times 1.95^2 = 21.6\;in.\;sec.^{-2}$

The diagram is completed by drawing $\displaystyle a_2''\,a_2$ perpendicular to $AB$ and $\displaystyle a_2'\,a_2$ along $EA$. The former gives the angular acceleration, $\displaystyle \frac{a_2''\,a_2}{BA}$ i.e. $14\;rad.\;sec.^{-2}$ and the latter represents the acceleration of the "block" $A$ along the slot.

• The angular velocity is $1.95\;rad.\;sec.^{-1}$
• The acceleration is $14\;rad.\;sec.^{-2}$

A cam is to cause a slider weighing 1.5 lbs. to move 2 in. in $\displaystyle \frac{1}{12}$sec. from rest to rest. Compare the maximum velocity reached and the maximum force required if:

• The slider is given simple harmonic motion.

• The slider is uniformly accelerated and then uniformly retarded.

• The motion is produced by half a revolution of a circular disc of 4 in. diameter and 1 in. eccentricity, the slider ends in a roller of 1 in. diameter and the line of stroke passes through the axis of the cam.

The displacement can be written in the form

$h = \frac{l}{2}\left (1 - \cos\,\omega \,t \right ) = 1 - \cos\,12\,\pi \,t$ ( where $l = 2\;in.$ and $\omega = 12\,\pi \;\;rad.\;sec^{-1}$ )

Differentiating, $v = 12\;\pi \;\sin\,12\,\pi \,t$

Differentiating again, $a = 144\;\pi^2 \;\cos\,12\,\pi \,t$

From which:

Maximum velocity = $12\;\pi = 37.7\;in,\;sec^{-1}$

The maximum inertia force = $144\;\pi ^2\times \displaystyle\frac{1.5}{32.2\times 12} = 5.52\;lbs.$

Note. For those of you not used to the Imperial system: The weight of the slider is in lb.. and needs to be expressed in Slugs, a unit of mass. This is done by dividing by 32.2 is the acceleration due to gravity in ft and the 12 expresses $g$ into $\displaystyle in.\;sec.^{-2}$ .

• Uniform acceleration and retardation

Using equation (2), Displacement, $h = \displaystyle\frac{1}{2}\;a\;t^2$

Substituting values: $1 = \displaystyle\frac{1}{2}\left (\displaystyle\frac{1}{24} \right )^2$ at half lift

Thus, $a = 1152\;in.\;sec.^{-2}$

Maximum Velocity = $1152\left (\displaystyle\frac{1}{22} \right ) = 48\;in.\;sec.^{-1}$

Maximum Inertia Force = $1152\times \left (\displaystyle\frac{1.5}{32.2\times 12} \right ) = 4.47\;lb.$

( Part three of question)

The lines $OA$ and $AB$ are equivalent to the links of a slider crank chain mechanism with a ratio of $\displaystyle n = \frac{AB}{OA} = 4.5$

Using equations from "Theory of Machines Velocity and Acceleration,

$v = 1\times \omega \left (\sin\theta + \displaystyle\frac{\sin2\theta }{2\times 4.5} \right )\)$ (where $\omega = 12\pi \;rad.\;sec.^{-1}$)

And, $a = \omega ^2\;\left (\cos\theta + \displaystyle\frac{\cos2\theta }{4.5} \right )$

The maximum velocity occurs when the acceleration is zero i.e.when:

$\cos2\theta = -4.5\cos\theta = 2\cos^2\theta - 1$

Solving the quadratic. ( There can only be one answer)

$\theta = 78^{0}\;15'$

Hence $\hat{v} = 12\,\pi \left (0.979 + \displaystyle\frac{0.399}{9} \right ) = 38.5\;in.\.sec.^{-1}$

The maximum Inertia Force will occur when $\displaystyle \theta = 0$

i.e. $\hat{F} = 144\,\left (1 + \displaystyle\frac{1}{4.5} \right )\displaystyle\frac{1.5}{32.2\times 12} = 6.75\;lb.$

A flat ended valve tappet is operated by a symmetrical cam with circular arcs for flank and nose profiles. The straight line path of the tappet passes through the cam axis. The total angle of action is $\displaystyle 150^{0}$, the lift is 0.25 in., the base circle diameter is 1.25 in. and the period of acceleration is half that of the deceleration during lift. The cam rotates at 1250 r.p.m.

Determine:

• a) The nose and flank radii.
• b) The maximum acceleration and deceleration whist lifting.

In the diagram $O$ is the centre of the base circle and $P$ the centre of the nose circle.

If $Q$ is the centre of the flank arcs then the limits of the flank arc are where $QO$ and $QP$ produced, cut the base and nose circles respectively.

By symmetry, the angle of lift is half the total angle of action i.e.$\displaystyle 75^{0}$ and $\displaystyle \therefore \angle POQ = 105^{0}$

Acceleration takes place on the flank and the period of acceleration is half that of deceleration during lift. Hence $\displaystyle \angle POQ = \frac{1}{3}\times 75^{0} = 25^{0}$

If $R$ in. is the flank radius and $r in.$ is the nose radius then, $PQ = R - r$ ; $OQ = R - 0.625$; $PO = 0.625 + lift - r = 0.875 - r$

• a) Calculation of nose and flank radii.

Applying the sine rule to the triangle $OPQ$

$\frac{R - 0.625}{\sin50^{0}} = \frac{0.875 - r}{\sin25^{0}} = \frac{R - r}{\sin105^{0}}$

From which:

And

Eliminating $R$ from from equations (5) and (6), $r = \displaystyle\frac{0.807}{2.02} = 0.40\;in.$

And from equation (5), $R = 2.208 - 0.724 = 1.484 in.$

• b) To find the maximum acceleration and deceleration.

On the Flank, let $G$ be the contact point for a cam angle of $\displaystyle \theta$

Dispalcement, $h = OH - 0.625$

And since OHGM is a rectangle:

$h = QGT - QM - 0.625 = 1.484 - (1.484 - 0.625)\cos\theta - 0.625$

Thus, $h = 0.859(1 - \cos\theta )$

Differentiating twice to obtain the acceleration, gives: $a = 0.859\;\omega ^2\;\cos\theta$

This will have its maximum value when $\displaystyle \theta = 0$

$\therefore \;\;\;\;\;\;\hat{a} = 0.859\times \left (1250\times\frac{2\pi }{60} \right )^2 = 14,700\;in.\;sec.^{-2}$

Let the contact on the nose be $G$'

Displacement $h = OH' - 0.625$

And Since $H$'$G$'$P$ $N$ is a rectangle, $h = ON + PG' - 0.625$

$= (0.875 - 0.40)\times \cos\;(75^{0} - \theta ) + 0.40 - 0.625$

$\therefore \;\;\;\;\;\;\;h = 0.475\times \cos\;(75^{0} - \theta ) - 0.225$

Differentiating twice for acceleration: $a\;= - 0.475\;\;\omega ^2\times \cos\;(75^{0} - \theta )$

By inspection it can be seen that this will be a maximum when the cos term equals one i.e.$\displaystyle \theta \:=\:75^{0}$.

$\displaystyle \hat{a} - 0.475\times \left ( 1250\times\frac{2\pi }{60} \right )^2 = 8120\;in.\;sec.^{-2}$

• a) $r = 0.40\;in.$, $R = 1.484 in.$
• b) $a = 0.859\;\omega ^2\;\cos\theta$

A cam of base circle diameter $D$ in. has tangent flanks and operates a follower through a roller of radius $R$ in., the path of the roller centre being a straight line passing through the camshaft axis. The follower acts against a spring of stiffness $S\;lb./in.$,and the initial compression is $x\;in.$ The total effective mass of the follower is $M$ and the spring mass is $m$.

Obtain an expression for the torque exerted on the camshaft when it is rotating at $\displaystyle \omega\;rads.per\;sec.$ and the cam has turned through an angle $\displaystyle \theta$ from the point at which the roller makes contact with the flank. Neglect the effects of friction.

Please refer to Example 6 for the geometry of a tangent cam with a roller follower.Displacement,

Differentiating w.r.t.time,Velocity,

Differentiating again for the acceleration.Acceleration,

The Total force exerted upon the follower $\displaystyle = S(x + h) + (M + \frac{m}{3})\,a$. Note that it is normal practice to add one-third of the mass of the spring to the mass of a vibrating mass in calculating the modulus of rigidity . This assumption will be justified in "General Dynamic Problems: Example 11. ( Hopefully this will be published in the not too distant future!)

Using the principle of work and if $\displaystyle \tau$ is the torque on the camshaft then:-

(where $F$ represents the Force and $v$ the Velocity)

Substituting from equations (7) (8) and (9) in (10)

$\tau = \left \{S\left [x + \left (\frac{D}{2} + R \right )\left (\frac{1}{\cos\theta } \right ) \right ] +$ $+\left (M + \frac{m}{3} \right )\left (\frac{D}{2} + R \right )\left (\frac{2 - \cos^2\theta }{\cos^3\theta } \right )\omega ^2 \right \}\times\left (\frac{D}{2} + R \right )\times \frac{\sin\theta }{\cos^2\theta }$

• The expression for the torque is given by:

$\tau = \left \{S\left [x + \left (\frac{D}{2} + R \right )\left (\frac{1}{\cos\theta } \right ) \right ] +$ $+\left (M + \frac{m}{3} \right )\left (\frac{D}{2} + R \right )\left (\frac{2 - \cos^2\theta }{\cos^3\theta } \right )\omega ^2 \right \}\times\left (\frac{D}{2} + R \right )\times \frac{\sin\theta }{\cos^2\theta }$

Show that the reaction on a shaft of radius a from a cylindrical bearing in which it rotates with line contact is tangential to a circle of radius $\displaystyle a\;\sin\phi$ called the friction circle, where $\displaystyle \phi$ is the friction angle at contact.

The diagram shows a circular cam of radius $r$ and eccentricity $e$. The radius of the camshaft is $a$. It operates a flat ended follower with a straight line passing through thee cam axis. The coefficient of fiction between the cam and follower and between the cam and bearings is $\displaystyle \mu$.

Find an expression for the turning moment $T$ required to rotate the shaft when it has turned an angle $\displaystyle \theta$ from the commencement of lift, the force $P$ on the follower being parallel to the axis and the guide friction neglected.

If $r = 1 in.$, $a = 0.5 in.$, $e = 0.25 in.$, $\displaystyle \mu = 0.2$, and $P = 5 lb.$, sketch the curve showing $T$ over one revolution.

A discussion on the first part of the problem will be found in "Theory of Machines. Mechanisms Friction circle.

The Forces between the follower and the cam are $P$ Normal and $\displaystyle \mu \,P$ tangential. This produces a moment about the camshaft axis of :-

The Total force on the bearings = The resultant of $P$ and $\displaystyle \mu P$ = $P\;\sqrt{1 + \mu ^2}$

The bearing friction moment = $P\;\sqrt{1 + \mu ^2}\times a\,\sin\phi$

But since $\tan\phi = \mu$

The bearing friction moment

The Turning Moment $T$ required to rotate the shaft is given by the addition of equations (11) and (12)

Hence, $T = \mu \,P\,(a + r) + P\,e\,(\sin\theta - \mu \;\cos\theta )$

The graph shows $T$ plotted against $\displaystyle \theta$. Note that $T$ has a mean value of $\displaystyle \mu \,P\,(a + r) = 1.5\;lb.\,in.$ The fluctuating term is obtained from the following table. The alternative signs refer to the angles in brackets.

• The expression for the turning moment $T$ is $T = \mu \,P\,(a + r) + P\,e\,(\sin\theta - \mu \;\cos\theta )$.