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Today we continue our discussion on synthesis
of cam profile for a flat-face oscillating
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follower.
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Let the base circle of the cam be given and
when the follower, flat-face follower is in
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contact with the base circle, that is the
lowest position of the follower and the oscillating
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follower is hinged at this location.
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This dimension we called ‘e’; the location
of the follower hinge from the cam shaft was
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given by these two dimensions, ‘a’ and
‘b’.
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The angle that the follower face makes with
the horizontal at this lowest position, we
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denote it by psib that is the angle e makes
with the vertical or the follower face makes
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with the horizontal.
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In our last lecture, we got an expression
for this psib which was, base circle radius
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rb, sin inverse rb plus e divided by the distance
O2O3, which is nothing but a square plus b
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square minus tan inverse b by a.
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Now when the cam rotates from this position
by an angle theta, let the follower rotates
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by an angle, delta theta.
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The follower has rotated by an angle, delta
theta and this is the follower.
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So this is the cam surface, cam profile.
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We denote the distance of this contact point
on the cam, let me call it C.
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This is the common normal within the cam profile
and the follower surface.
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And lets say this distance is l, which keeps
on changing as the cam rotates, because the
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contact point shifts from this cam surface.
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As we have done for the translating follower
situation, we take our X and Y axis passing
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through O2, this is the X-axis, this is Y-axis,
this is the X-axis.
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Our objective is to find the co-ordinates
of the contact point, namely xc and yc.
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To find this xc and yc, I will need this quantity
l.
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So, as a first step, I try to find what is
this quantity l.
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Towards that end, I drop a perpendicular from
O2 to this common normal, from O2 I drop a
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perpendicular to this common normal and let
this point I call ‘M’.
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The follower position is at this instant,
again measured from the horizontal is given
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by psi.
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And as we noted in our last class, psi is
nothing but psib plus delta theta.
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Psib we have already got psib; delta theta
is prescribed for us so I can find psi.
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To find this distance l, we consider a point
C1 on the follower surface, but at this instant
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coincident with C. C belongs to the point
on the cam and C1 belongs denotes the point
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on the follower surface which is in contact.
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To maintain the contact, velocity of this
point C1 and this point C along this the direction
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of n-n must be same.
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Otherwise, the contact will be either lost
or one point will get into the other point,
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that means, because they are rigid body and
that cannot happen.
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To maintain the contact, velocity of the point
C1 along the direction n-n must be velocity
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of the point C along the same direction.
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Now because M and C are two points on the
cam which is a rigid body, so velocity of
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C and M along the line CM that is along n-n
must be same.
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So [VC]n-n is nothing but velocity of this
point M in the n-n direction, because the
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two points on the same rigid body, so this
distance CM does not change, that means, the
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velocity along this direction must be same.
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And velocity of C and Cnn C1 along n-n must
be same; otherwise the contact will be lost.
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Now to find the velocity of the point M along
n-n is nothing but O2M into the angular velocity
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of the cam.
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So this I can write is O2M into angular velocity
of the cam d theta dt and that is velocity
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in this direction.
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If omega d theta dt is counter-clockwise,
the velocity of M is along n-n, and O2M into
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d theta dt.
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Now velocity of C1 along n-n is nothing but
l times psi dot.
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So the vC1, the velocity of the point C1 along
n-n, where N1 belongs to the follower surface
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is nothing but l into d psi dt, where psi
is measured clockwise.
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Because these two are same, I can immediately
write O2M is nothing but l into d psi d theta.
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Once I get this O2M equal to l into d psi
d theta, then I can write, as I see if this
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angle from from the horizontally is psi and
this line is perpendicular to this line, this
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line is also perpendicular to this line.
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So, these two lines are parallel, this angle
is also psi.
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So I can write O2M is a cos psi, a cos psi
minus b sin psi; e is perpendicular to M,
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so there is no component and then minus l.
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O2M is, I am going from O2 to M in this way:
O2 to this point, to this point, to this point,
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to this point and this point.
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So O2M, if I add up these vectors, the components
if I take, l will give me l cos psi, b will
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give me minus b sin psi, e will not give anything,
l will again give me cos psi, but with a negative
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sign, l will give again l, but in this direction,
so minus l and O2M is l psi prime.
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So this is a cos psi minus b sin psi minus
l.
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So that gives me l equal to a cos psi minus
b sin psi divided by one plus psi prime.
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So I get this quantity l, this distance l,
which changes with theta because psi is a
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function of theta.
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Now, we are in a position to write the coordinate
of xc, xc is a minus, this angle is psi so
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e sin psi minus l cos psi. x co-ordinate of
this point C is a; there is no projection
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of b on the horizontal direction, horizontal
projection of e is e sine psi, but in this
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direction, projection of l is l cos psi, and
in this direction, so that is the co-ordinate
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xc.
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Similarly, we can write yc is b minus e cos
psi plus l sin psi; out of which l I get from
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this expression.
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So I have got xc and yc everything in terms
of theta, once these parameters a, b, e are
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given to us.
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Rest is as before.
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Once I get the contact point, I can write
rc is square root of xc squared plus yc squared,
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as we see this is a function of theta.
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And from any line fixed on the cam, any line
which I fixed on the cam then if I measure
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the polar angle from here, if I call it say
some alphaC prime, then alphaC prime is theta
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plus tan inverse xc by yc.
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So this is also a function of theta.
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So this is the parametric equation of the
cam profile in polar co-ordinates with the
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origin here.
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rC which we can get for various values of
theta, alphaC prime which I can again get
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for various values of theta, where theta represents
the cam rotation.
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So varying theta from 0 to 2 pi, I can get
rC and alphaC for various values of theta.
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Then, plotting those values of rC and alphaC
prime, I get the cam profile as we have done
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in case of translating follower.
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This portion is exactly as before.
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One point I would like to emphasize at this
stage that once we get the rC that the distance
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from the cam shaft axis and the polar angle
for the contact point
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for all types of follower, translating or
oscillating does not matter.
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If I summarize this analytical method for
synthesis of cam profile, we are getting rC
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as a function of cam rotation and a polar
angle which I denoted by phiC prime or psiC
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prime or alphaC prime in various cases.
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Let me write one of them.
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I just call theta C prime, which is theta
plus tan inverse xc by yc and this rc is nothing
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but square root of xc squared plus yc squared.
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So these two quantities I can calculate for
all values of theta.
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And now, to draw the cam profile, after getting
this series of values what I can decide that
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I will draw the cam profile, when the cam
has rotated say by an angle theta0.
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So what I do, I take an origin which is O2,
draw a line which is vertical, and then draw
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a line at an angle theta0.
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Then, take these values of rC and thetaC prime,
but thetaC prime I measure in the clockwise
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direction from this line.
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So say for some values of theta, I get rC
and thetaC prime this point, then this point,
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then this point like that, and joining these
points, I will get the cam profile when the
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cam has rotated by an angle theta0.
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So this is where we end our discussion on
analytical method of cam profile synthesis.
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Next, we will do how to do it, the same thing
synthesis of cam profile for various types
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of follower by graphical method.
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Now that we have finished discussion on the
analytical method of cam profile synthesis,
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we will do the same problem through graphical
method.
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Graphical method of cam profile synthesis.
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Of course, we assume all the geometrical parameters
like base circle radius, offsets, all these
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quantities are known.
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And only thing, instead of y the follower
movement being given in the form of an analytical
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expression, now we have it in the form of
a displacement diagram.
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It is given in the form of a displacement
diagram.
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Now given the displacement diagram, how can
we obtain the cam profile through a graphical
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method.
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So as before, we start with translating follower.
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First we discuss, if if the follower is flat-face
type, then we will discuss if the follower
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is a roller follower.
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So what we do?
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We first draw the base circle, this is the
cam shaft.
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The offset is also given.
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So let this be the follower axis and this
is the follower face at the lowest position.
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This is the offset.
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In the graphical method, this total rotation
of theta from 0 to 2 pi, we divide in certain
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number of station points.
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Say this is zero.
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Then we take various values of theta like
theta1, theta2, theta3, theta4, these are
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called station points 0, 1, 2, 3, 4.
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How many divisions we take is the question,
but obviously more number of points we take,
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more accurate will be our drawing.
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But obviously, we cannot take too many points,
because then it will be very time-consuming.
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I will just illustrate by taking 2, 3 station
points and rest will be same; it can be easily
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followed.
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This is the follower axis, this line represents
the follower axis.
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Obviously this is corresponds to station point
0, when the follower is at its lowest position.
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Now we draw the displacement diagram at the
same level.
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Whatever is the displacement diagram, I draw
it at the same level.
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Now say this is the station point 1, this
is station point 2.
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So wat we do, we project these points onto
this line, on the follower axis.
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Lets say this is 1, this is 2.
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So here again I mark this as 1, this as 2.
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To obtain the follower cam profile, as we
have done in case of analytical method, we
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hold the cam fixed and make a kinematic inversion,
such that this fixed link rotate in the opposite
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direction.
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Suppose this is the rotation of the cam for
station point 1, the cam rotation is theta1
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in the counter-clockwise direction.
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So the follower axis has to rotate, for the
same station point the kinematic inversion
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is that the follower axis must rotate by same
value of theta1, but in the clockwise direction.
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If I draw a circle with radius e, then this
I make theta1 and draw a tangent.
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So this follower axis has rotated by theta1
in the clockwise direction, corresponding
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to this station point.
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That is the kinematic inversion that I hold
the cam fixed and allow the follower axis
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of the fixed link to rotate by the same amount,
but in the opposite direction.
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Now the distance of this point, corresponding
station point 1 from O2 is here, so this is
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the distance.
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But now this line represents the follower
axis at this distance; so I rotate this radius
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along this circular arc and intersect here.
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So this is what I call, give it a name say
1 prime and the follower, this is the follower
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axis, so the follower face is perpendicular
to that, but at this distance.
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So I draw a line perpendicular to the follower
axis.
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So this line represents the follower face
after kinematic inversion corresponding to
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station point 1.
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Similarly, corresponding to station point
2, I rotate it by theta2.
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And this particular point of the follower
axis should be at this location, that is so
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much distance from O2, so I rotate O2 and
this is what I call 2 prime.
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Then, the follower face is perpendicular to
this line; so this is the follower face corresponding
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to station point 2.
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This way, I can draw the inverted position
of the follower face corresponding to all
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the station points 1, 2, 3, and so on.
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And the cam profile is nothing but a curve,
which is an envelope to this family of straight
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line representing the follower faces.
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I draw an envelope to this straight line,
which represents the follower face after kinematic
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inversion and then, draw an envelope to this
family of straight lines and that is nothing
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but the cam profile.
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So this is the cam profile which is always
tangent to the follower face and these lines
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represents the follower faces corresponding
to station point 1 and 2.
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And the same construction can be continued
for all the station points and we have to
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draw a curve, which is tangent to this family
of straight line representing the follower
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faces.
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So the basic thing is the kinematic inversion.
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The kinematic inversion is simple, as if we
to draw these lines we draw tangent to the
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circle, which is called.
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In some books it is mentioned as offset circle,
which is a circle of radius e with center
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at O2.
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But the simplest thing is, to rotate this
line O2a1 by an angle theta1, then O22 by
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an angle theta2.
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It is very easy to show from the triangle
law that this distance is same as this distance,
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both of them are e.
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Then this distance, this distance if I consider
station point 1, this distance is same as
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this distance.
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So this particular right angled triangle is
same as this right angled triangle.
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So this whole right angled triangle has been
rotated by an angle theta1 in the clockwise
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direction.
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So I can just as well say O2a1 rotated by
theta1, I get this point, O22 rotated by theta2,
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I get this point and once I get this point,
then I can draw the follower face perpendicular
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to the follower axis.
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Next, I will take up a roller follower, but
translating.
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Let me now explain the graphical method for
a translating roller follower.
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For the roller follower, first let me draw,
the prime circle.
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This is the cam center O2 and this is the
prime circle.
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00:26:29,060 --> 00:26:52,720
So, the radius is rp, which is rb plus rR
and lets say, the offset is e, this is the
215
00:26:52,720 --> 00:26:53,910
offset, e.
216
00:26:53,910 --> 00:27:00,090
So this is as we have discussed in case of
analytical method, this is the lowest point
217
00:27:00,090 --> 00:27:03,250
of the roller center, that is the trace point.
218
00:27:03,250 --> 00:27:07,980
The displacement of the follower will be measured
from this level.
219
00:27:07,980 --> 00:27:19,760
So I draw a horizontal line
and draw whatever is the displacement diagram
220
00:27:19,760 --> 00:27:20,790
here.
221
00:27:20,790 --> 00:27:37,710
So as before, I consider few station points
namely 1, 2, 3; this is the plot of y versus
222
00:27:37,710 --> 00:27:43,480
theta, this is zero.
223
00:27:43,480 --> 00:27:56,250
So now I know, that corresponding to the station
point 1, the trace point should be here.
224
00:27:56,250 --> 00:28:08,050
Similarly, for the station point three, the
trace point should be here and for two the
225
00:28:08,050 --> 00:28:12,700
station point is here.
226
00:28:12,700 --> 00:28:21,350
So to get the inverted position of the station
points, we know this offset means, the cam
227
00:28:21,350 --> 00:28:25,090
is rotating in the counter-clockwise direction.
228
00:28:25,090 --> 00:28:31,050
Corresponding to station point 1, the cam
rotation is known which is theta1.
229
00:28:31,050 --> 00:28:38,630
So I get the inverted positions of this trace
point corresponding to one, by rotating O2a1
230
00:28:38,630 --> 00:28:47,090
to an angle theta1, but in the clockwise direction.
231
00:28:47,090 --> 00:28:59,760
So if I rotate this by theta1, this I call
inverted position of the station point 1,
232
00:28:59,760 --> 00:29:02,510
which is 1 prime.
233
00:29:02,510 --> 00:29:25,320
Similarly, for this 2, I rotate about O2,
this line O22 to an angle theta2 in the clockwise
234
00:29:25,320 --> 00:29:26,500
direction.
235
00:29:26,500 --> 00:29:36,670
So theta2 and I get the inverted position
of this trace point as 2 prime.
236
00:29:36,670 --> 00:29:40,870
Similarly I do for all the station points
namely 1, 2, 3, 4 like that.
237
00:29:40,870 --> 00:29:44,200
These are the inverted position of the trace
point.
238
00:29:44,200 --> 00:29:45,290
What is the trace point?
239
00:29:45,290 --> 00:29:49,210
Trace point is the roller center and the roller
radius is given.
240
00:29:49,210 --> 00:29:58,640
So everywhere this was the, corresponding
to 0, this is 0; so I draw the roller whose
241
00:29:58,640 --> 00:29:59,870
radius is given.
242
00:29:59,870 --> 00:30:08,180
Here again, this is the inverted, with this
point as center, I draw the roller; with this
243
00:30:08,180 --> 00:30:12,700
point as center, I draw the same roller.
244
00:30:12,700 --> 00:30:23,400
So this way, I keep on drawing roller, this
is the say three prime, the inverted position
245
00:30:23,400 --> 00:30:30,320
of this point, trace point corresponding to
third station point.
246
00:30:30,320 --> 00:30:32,160
Now what is the cam profile?
247
00:30:32,160 --> 00:30:39,210
Cam profile is nothing but the family of curve,
cam profile is nothing but the curve which
248
00:30:39,210 --> 00:30:45,110
is envelope to this family of circles representing
the roller after kinematic inversion.
249
00:30:45,110 --> 00:30:50,460
So I have to draw a curve which is tangent
to the series of circles.
250
00:30:50,460 --> 00:30:59,130
If I draw a curve which is tangent to all
these circles, that will give me the cam profile.
251
00:30:59,130 --> 00:31:07,090
So the methodology is just the same as the
analytical method; I make a kinematic inversion
252
00:31:07,090 --> 00:31:13,710
with the cam fixed, get the inverted position
corresponding to that trace point, draw the
253
00:31:13,710 --> 00:31:19,520
family of straight lines or circles representing
the flat-face of the roller follower.
254
00:31:19,520 --> 00:31:25,460
Then, the cam profile is nothing but the envelope
to this family of circles, in case of roller
255
00:31:25,460 --> 00:31:30,340
followers or in case of flat-face follower
to the family of straight lines, representing
256
00:31:30,340 --> 00:31:34,470
the follower face at the inverted positions.
257
00:31:34,470 --> 00:31:39,870
So this principle of kinematic inversion is
very important that we hold the cam fixed
258
00:31:39,870 --> 00:31:43,420
and allow the fixed link to rotate in the
opposite direction.
259
00:31:43,420 --> 00:31:48,540
The cam is rotating in the counter-clockwise
direction; so this follower axis for translating
260
00:31:48,540 --> 00:31:51,940
roller follower is rotated in the clockwise
direction.
261
00:31:51,940 --> 00:32:01,030
That rotation can be done very easily geometrically,
by rotating this line O2a1, O22, O23 by theta1,
262
00:32:01,030 --> 00:32:04,330
theta2, theta3 in the clockwise direction.
263
00:32:04,330 --> 00:32:09,400
That gives me the location of this trace point
on the roller centers.
264
00:32:09,400 --> 00:32:14,340
Once I get the roller centers, I can draw
the rollers and cam profile is enveloped to
265
00:32:14,340 --> 00:32:18,040
this family of circles representing the roller
fixed.
266
00:32:18,040 --> 00:32:29,270
In fact, if we can imagine, I can draw another
envelope touching the circles on the outer
267
00:32:29,270 --> 00:32:30,270
side.
268
00:32:30,270 --> 00:32:32,820
This was touching the circles in the inner
side.
269
00:32:32,820 --> 00:32:37,890
Then, I can use this as the groove in the
cam face where I do not need a spring.
270
00:32:37,890 --> 00:32:43,980
As I told you earlier that without the spring
for returning the follower, I can use a groove
271
00:32:43,980 --> 00:32:48,240
in the cam face of the cam and put the roller
in that groove.
272
00:32:48,240 --> 00:32:53,790
And then cam will always pull it, push it
up and pull it down, there is no need to have
273
00:32:53,790 --> 00:32:54,790
a spring.
274
00:32:54,790 --> 00:33:01,600
Actually, this curve is nothing but what we
call prime curve, pitch curve.
275
00:33:01,600 --> 00:33:09,380
This is what we call pitch curve.
276
00:33:09,380 --> 00:33:14,650
So this is the cam profile, if we have a spring
and this groove is the groove where I have
277
00:33:14,650 --> 00:33:18,690
to put the roller, when I do not need to use
the spring using a roller follower.
278
00:33:18,690 --> 00:33:25,440
Next, we will show how to use the same principle
to obtain the cam profile for a for an oscillating
279
00:33:25,440 --> 00:33:28,950
follower either roller or flat-face.
280
00:33:28,950 --> 00:33:34,690
Let me now explain the graphical method of
cam profile synthesis with reference to an
281
00:33:34,690 --> 00:33:37,730
oscillating follower.
282
00:33:37,730 --> 00:33:51,780
Here again, we shall discuss both roller follower
and flat-face follower.
283
00:33:51,780 --> 00:34:02,240
So first let me take a roller follower.
284
00:34:02,240 --> 00:34:07,500
Of course, we assume all the required geometric
dimensions like rb, rR - all these things
285
00:34:07,500 --> 00:34:15,980
are already known and the displacement function
or displacement diagram is given to us.
286
00:34:15,980 --> 00:34:24,920
So for the roller follower, let me first draw
the prime circle.
287
00:34:24,920 --> 00:34:44,530
This is the prime circle radius which is as
we have noted earlier, base circle radius
288
00:34:44,530 --> 00:34:46,470
plus roller radius.
289
00:34:46,470 --> 00:34:51,470
When the roller center or the trace point
is in contact with the prime circle, that
290
00:34:51,470 --> 00:34:56,200
is the extreme position or the lowest position
of the follower, from where we measure the
291
00:34:56,200 --> 00:34:58,810
movement of the follower.
292
00:34:58,810 --> 00:35:07,810
This point is called O2.
293
00:35:07,810 --> 00:35:30,130
Suppose this is the roller center at the lowest
most position, this is the roller and this
294
00:35:30,130 --> 00:35:43,980
roller follower is say hinged here to the
fixed link, this point we call O3.
295
00:35:43,980 --> 00:35:53,300
When the cam rotates in the counter-clockwise
direction, lets say this follower also goes
296
00:35:53,300 --> 00:36:01,700
in the counter-clockwise direction and this
rotation psi as a function of theta is known.
297
00:36:01,700 --> 00:36:13,760
So the trace point will obviously go on this
circle with radius, center at O3 and this
298
00:36:13,760 --> 00:36:14,760
as radius.
299
00:36:14,760 --> 00:36:23,030
This will move on this circle.
300
00:36:23,030 --> 00:36:30,350
So if we take station points corresponding
to theta1, theta2, theta3, then we will know
301
00:36:30,350 --> 00:36:39,970
the corresponding values of psi theta1, psi
theta2, and so on.
302
00:36:39,970 --> 00:36:53,071
So corresponding to station point 1, lets
say the trace point is here, that means, this
303
00:36:53,071 --> 00:36:57,500
rotation of the follower is psi theta1.
304
00:36:57,500 --> 00:37:06,140
Similarly, I can get all the station points
corresponding, all the trace points corresponding
305
00:37:06,140 --> 00:37:11,980
to station point 1, station point two, and
so on.
306
00:37:11,980 --> 00:37:14,930
Now as before, we make a kinematic inversion.
307
00:37:14,930 --> 00:37:27,380
To draw the cam profile, we hold the cam fixed,
which means this fixed link O2O3 has to rotate
308
00:37:27,380 --> 00:37:31,330
in the clockwise direction.
309
00:37:31,330 --> 00:37:50,100
So corresponding to station point 1, the inverted
position of O3, this is the circle with O2
310
00:37:50,100 --> 00:37:52,540
as center and O2O3 as radius.
311
00:37:52,540 --> 00:37:59,230
And after the inversion, kinematic inversion
with the cam fixed, this O3 will go on this
312
00:37:59,230 --> 00:38:00,230
circle.
313
00:38:00,230 --> 00:38:04,960
So corresponding to station point 1, when
the cam has rotated theta1 in the counter-clockwise
314
00:38:04,960 --> 00:38:14,010
direction, this fixed link will rotate in
the clockwise direction by an angle theta1.
315
00:38:14,010 --> 00:38:22,110
So this I can say O3 corresponding to one
and the inverted position.
316
00:38:22,110 --> 00:38:27,690
Now the distance of the trace point from this
hinge does not change.
317
00:38:27,690 --> 00:38:37,590
So distance of O31 is should be same as O3
prime 1 prime and the distance of one from
318
00:38:37,590 --> 00:38:43,780
O2 must be such that this required or desired
follower movement is reproduced.
319
00:38:43,780 --> 00:38:55,800
So what I do, to get the inverted position
of one, I draw a circular arc with O2 as center
320
00:38:55,800 --> 00:38:58,020
and O11 as radius.
321
00:38:58,020 --> 00:39:10,330
And from O31 prime, I draw a circle with O31
as radius so
322
00:39:10,330 --> 00:39:16,410
and where these two circles intersects; that
is nothing but the station point, inverted
323
00:39:16,410 --> 00:39:20,430
position corresponding to station point 1,
that is the roller center.
324
00:39:20,430 --> 00:39:30,340
So at this point, I can again draw the roller
whose radius is given to us.
325
00:39:30,340 --> 00:39:41,131
Let me again illustrate it with corresponding
to station point two, we draw a circle with
326
00:39:41,131 --> 00:39:44,280
O2 as center and O22 as radius.
327
00:39:44,280 --> 00:39:58,610
2 prime will lie on this circle and O32 prime,
the inverted position of O3 corresponding
328
00:39:58,610 --> 00:40:18,060
to station point two is here, on this circle
such that this angle, this angle is theta2.
329
00:40:18,060 --> 00:40:26,610
Now O32 is so much, so from O32 prime, I take
the same radius as O32 and draw a circular
330
00:40:26,610 --> 00:40:27,610
arc.
331
00:40:27,610 --> 00:40:36,570
So that let it be here.
332
00:40:36,570 --> 00:40:50,830
O32, O32, this distance is same as O32 prime
2 prime.
333
00:40:50,830 --> 00:40:56,500
This distance is same as this distance and
O22 is same as O22 prime.
334
00:40:56,500 --> 00:41:00,960
So this is the inverted position of the trace
point, that is the roller center.
335
00:41:00,960 --> 00:41:12,160
So here again, I draw
the roller of same radius.
336
00:41:12,160 --> 00:41:24,260
And now if I draw a curve which is tangent
to this family of circles, that gives me the
337
00:41:24,260 --> 00:41:26,570
cam profile.
338
00:41:26,570 --> 00:41:44,980
In fact, very simple geometrical consideration
can show that this point we call 1 prime,
339
00:41:44,980 --> 00:41:59,060
that 1 prime I can easily get if I rotate
O2a1 O21 in the clockwise direction by theta1,
340
00:41:59,060 --> 00:42:01,290
I get to 1 prime.
341
00:42:01,290 --> 00:42:14,840
Similarly, if I rotate O22 through an angle
theta2 in the clockwise direction, I get to
342
00:42:14,840 --> 00:42:15,840
2 prime.
343
00:42:15,840 --> 00:42:21,150
This can be very easily proved from congruence
of triangles, two equal triangles.
344
00:42:21,150 --> 00:42:30,680
We consider a triangle O2O31, and O311 prime
O2; we can show that very simply.
345
00:42:30,680 --> 00:42:32,610
These I leave for you to do.
346
00:42:32,610 --> 00:42:38,490
That instead of going through this by drawing
two circular arcs, I can as well rotate O21
347
00:42:38,490 --> 00:42:44,500
by theta1 in the clockwise direction to get
1 prime; O22 by theta2 in the clockwise direction
348
00:42:44,500 --> 00:42:50,290
to get 2 prime and then, draw the rollers
and then draw an envelope to this family of
349
00:42:50,290 --> 00:42:51,290
circles.
350
00:42:51,290 --> 00:42:53,390
This I did to explain the kinematic inversion.
351
00:42:53,390 --> 00:42:59,650
It is actually the fixed link which is rotating,
that means O3 is going along this circle in
352
00:42:59,650 --> 00:43:02,190
the clockwise direction, when I hold the cam
fixed.
353
00:43:02,190 --> 00:43:08,490
But, from congruence of triangle, geometric
construction can be much simplified, by just
354
00:43:08,490 --> 00:43:12,640
not doing all these; only rotating O21 through
this theta1 and O22 through theta2.
355
00:43:12,640 --> 00:43:19,980
Next, we will take up the case of a flat-face
follower.
356
00:43:19,980 --> 00:43:26,540
So at the end, let me now explain this graphical
method for an oscillating follower, which
357
00:43:26,540 --> 00:43:30,940
is flat-face type.
358
00:43:30,940 --> 00:43:54,370
If it is a flat-face follower, lets say this
is the base circle radius
359
00:43:54,370 --> 00:44:00,360
and the flat-face is tangential to this base
circle radius at the lowest point.
360
00:44:00,360 --> 00:44:13,180
And this is where the follower is hinged.
361
00:44:13,180 --> 00:44:18,170
In the analytical method, this dimension is
what we call e.
362
00:44:18,170 --> 00:44:23,660
As we see, as this thing rotates this angle,
this is one rigid body and this angle always
363
00:44:23,660 --> 00:44:37,630
remains 90 degrees, which means the flat-face
is always perpendicular to this line.
364
00:44:37,630 --> 00:44:43,480
Let me use this particular contact point as
the trace point and as the cam rotates in
365
00:44:43,480 --> 00:44:49,810
the counter-clockwise direction, as the follower
oscillates in the counter-clockwise direction,
366
00:44:49,810 --> 00:45:03,670
this particular trace point of the follower
moves on a circle with O3 as center, this
367
00:45:03,670 --> 00:45:05,030
as radius and O3 as center.
368
00:45:05,030 --> 00:45:09,370
This trace point moves on this side and where
will it be?
369
00:45:09,370 --> 00:45:25,980
It will be decided by the rotation of this,
this is what we call O2.
370
00:45:25,980 --> 00:45:35,290
So once this is prescribed, the movement of
the follower again I consider station points
371
00:45:35,290 --> 00:45:38,000
theta1, theta2, theta3, so on.
372
00:45:38,000 --> 00:45:43,740
This situation is shown for theta equal to
zero.
373
00:45:43,740 --> 00:45:46,240
This is the situation when theta is zero.
374
00:45:46,240 --> 00:45:52,410
Let for theta, when it is theta1, the cam
has rotated by theta1, this trace point goes
375
00:45:52,410 --> 00:45:53,410
here.
376
00:45:53,410 --> 00:46:01,730
So I call it as before one, similarly for
two, it comes here at two.
377
00:46:01,730 --> 00:46:08,860
To draw the cam profile, I make the same use,
make use of the same principle of kinematic
378
00:46:08,860 --> 00:46:13,010
inversion holding the cam fixed.
379
00:46:13,010 --> 00:46:20,940
As I explained earlier for the roller follower,
the inverted position of one can be easily
380
00:46:20,940 --> 00:46:33,510
obtained by rotating O21 through an angle
theta1 in the clockwise direction, because
381
00:46:33,510 --> 00:46:35,720
this O2O3 moves in the clockwise direction.
382
00:46:35,720 --> 00:46:40,920
So this is what we call 1 prime.
383
00:46:40,920 --> 00:46:49,140
The next question is how I draw the flat-face
passing through this point 1 prime, this flat-face
384
00:46:49,140 --> 00:46:57,210
in the inverted position.
385
00:46:57,210 --> 00:47:20,220
For that, O2O3 I rotate as before the O3 inverted
position is say here, which I call O31 prime.
386
00:47:20,220 --> 00:47:33,100
Now here I draw a circle of radius e with
this point as center.
387
00:47:33,100 --> 00:47:45,610
This is of radius e and the trace point is
here.
388
00:47:45,610 --> 00:47:54,470
And I know the follower face will be always
perpendicular to this offset e, which means
389
00:47:54,470 --> 00:48:15,930
this will be tangential to this circle, offset
e is same and from 1 prime, I draw a tangent,
390
00:48:15,930 --> 00:48:16,930
this is e.
391
00:48:16,930 --> 00:48:21,880
So this represents the follower face after
kinematic inversion.
392
00:48:21,880 --> 00:48:40,300
Similarly, I can rotate O22 through an angle
theta2 to get the inverted position 2 prime.
393
00:48:40,300 --> 00:48:53,510
Then I go to O32 wherever that is, by rotating
O2O3 through theta2 in the clockwise direction.
394
00:48:53,510 --> 00:48:59,160
Then, again I draw this circle of radius e
and from 2 prime I draw a tangent to that
395
00:48:59,160 --> 00:49:04,940
circle and then that tangent, I am not drawing
everything, it becomes cumbersome.
396
00:49:04,940 --> 00:49:12,230
Procedure is same; I go from 2 to 2 prime
by rotating O22 through an angle theta2 in
397
00:49:12,230 --> 00:49:16,010
the clockwise direction; that gives me this
point.
398
00:49:16,010 --> 00:49:22,700
I rotate O2O3 through theta2 in the clockwise
direction that gives me O32 prime.
399
00:49:22,700 --> 00:49:28,550
Then, with this point as center and of e radius,
I draw a circle.
400
00:49:28,550 --> 00:49:37,740
From this point, I draw a tangent to that
circle and that tangent let it be this, that
401
00:49:37,740 --> 00:49:40,630
represents the inverted follower face.
402
00:49:40,630 --> 00:49:48,380
Now a curve which is tangent to all these
inverted positions of the follower face, an
403
00:49:48,380 --> 00:49:54,090
envelope to this inverted positions of the
follower face, this was corresponding to one,
404
00:49:54,090 --> 00:49:58,600
this was corresponding two, this is corresponding
to zero.
405
00:49:58,600 --> 00:50:02,800
So once I get the inverted positions of the
follower faces which is a family of straight
406
00:50:02,800 --> 00:50:09,110
lines, then draw an envelope to this family
of straight lines, that is nothing but the
407
00:50:09,110 --> 00:50:15,410
cam profile.
408
00:50:15,410 --> 00:50:21,220
What we should note that here, the contact
point is one, but here the contact point is
409
00:50:21,220 --> 00:50:27,090
not 1 prime, the contact point has shifted
on the follower face as we have seen in case
410
00:50:27,090 --> 00:50:29,310
of while we discuss the analytical method.
411
00:50:29,310 --> 00:50:34,460
Similarly, this 2, this is the 2 prime inverted
position, but the contact point is here.
412
00:50:34,460 --> 00:50:41,010
So if we measure what is the maximum distance
of the contact point from this trace point
413
00:50:41,010 --> 00:50:46,230
on this side, that will give me the required
follower which on the right hand direction.
414
00:50:46,230 --> 00:50:52,680
To the right of this contact point, how much
the contact point moves, so that I must have
415
00:50:52,680 --> 00:50:53,890
the required follower face.
416
00:50:53,890 --> 00:51:01,340
Similarly, if we continue, you will see, now
it will be to the left of this also; right
417
00:51:01,340 --> 00:51:02,340
now it is to the right.
418
00:51:02,340 --> 00:51:05,921
For these three positions, 0 it is alright;
1 it is here, but the contact point is here;
419
00:51:05,921 --> 00:51:11,110
2 the point is here, but the contact point
is here.
420
00:51:11,110 --> 00:51:16,170
So, this is the shift of the contact point
on the follower face in position two.
421
00:51:16,170 --> 00:51:22,020
If we complete it, you will see that this
contact point shifts in one direction, then
422
00:51:22,020 --> 00:51:26,820
returns and then shifts in the other direction
and the maximum distance that is needed for
423
00:51:26,820 --> 00:51:32,461
the contact point to move on the follower
face, that decides the minimum follower face
424
00:51:32,461 --> 00:51:36,619
width, minimum width of the follower face.
425
00:51:36,619 --> 00:51:42,420
So that brings us to the end of cam-follower
mechanisms and let me summarize what we did.
426
00:51:42,420 --> 00:51:48,950
We first explained the different terms which
are needed to describe the cam follower mechanism.
427
00:51:48,950 --> 00:51:55,510
After that, we described what are the typical
follower movements, basic follower movements.
428
00:51:55,510 --> 00:52:01,880
We had both the analytical expressions and
the graphical construction for the displacement
429
00:52:01,880 --> 00:52:02,930
diagram.
430
00:52:02,930 --> 00:52:14,960
Follower motion, either y expressed as a function
of theta or through displacement diagram.
431
00:52:14,960 --> 00:52:26,740
Then, we showed how to get the basic dimensions
like rb, e, etc., so that, certain constants
432
00:52:26,740 --> 00:52:35,270
like minimum radius of curvature or maximum
value of the pressure angle, all these conditions
433
00:52:35,270 --> 00:52:36,270
are satisfied.
434
00:52:36,270 --> 00:52:41,290
We would always obviously like to go for the
smallest value of rb that we can get away
435
00:52:41,290 --> 00:52:43,900
with, so that the cam size is small.
436
00:52:43,900 --> 00:52:50,869
And after getting these basic dimensions,
we got both the analytical and graphical methods
437
00:52:50,869 --> 00:52:57,000
to obtain the cam curve or the cam profile
and that is the synthesis problem.
438
00:52:57,000 --> 00:53:02,690
Of course, given the cam curve, we can also
find what should be the follower movement;
439
00:53:02,690 --> 00:53:08,290
that is what we call analysis of the follower
movement.
440
00:53:08,290 --> 00:53:12,990
But that is purely a geometrical problem which
is quite simple; even we can do analytically
441
00:53:12,990 --> 00:53:37,619
or graphically and we are not going to discuss
that in this course.
442
00:53:37,619 --> 00:53:48,589
3