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We finished our last lecture with the geometrical
construction, which expressed the pressure
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angle for a translating roller follower which
is given by this expression.
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Today, we will take up an example to show
how from this geometrical construction, we
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can get the optimum values of this prime circle
radius and the offset.
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Of course, the roller radius rR is assumed
to be given and if I get rp the minimum value,
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then rb the base circle radius minimum value
I can get from (rp) minimum value minus rR.
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Towards this end I take an example say, the
follower displacement y theta is given by
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25 into 1 minus cosine theta so many millimeters
and this expression is valid for all values
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of theta that is the entire cam rotation 0
to 2 pi.
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If this is y theta, then differentiate it
once with respect to theta, I get y prime
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theta which is 25 sine theta.
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From these two expressions of y theta and
y prime theta it is very easy to see that,
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y theta minus 25 whole squared plus y prime
theta whole squared is 25 square.
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So the plot of y theta versus y prime theta
will be a circle of radius 25 and centre at
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y equal to 25 and y prime equal to 0.
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If we remember x minus a whole squared plus
y minus b whole squared, r squared is the
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equation of a circle of radius r with centre
at a and b.
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So if I plot, y theta versus y prime theta,
satisfying this equation and that equation
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is the circle with centre at y equal to 25
and y prime equal to 0 and radius equal to
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25.
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So this is a circle.
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Now let us say, phi max is prescribed as 30
degree, during the rise.
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We are not bothered about the pressure angle
during return, during the rise phase phi max
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must be limited to 30 degree maximum value
of phi should not go beyond 30 degree.
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In this diagram, as we see from O to this
point A, O to A, y prime is positive, that
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is the rise phase and this is the return phase
when y prime is negative.
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So during this rise phase, the value of phi
should never exceed 30 degree.
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To keep that maximum value 30 degree, what
we do, we draw a tangent to this circle such
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that, the tangent is inclined to the vertical
by an angle 30 degree.
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This line is inclined to the vertical by 30
degree that is this angle is 60 degree and
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from here, I again draw a line at 30 degree
to the vertical
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and these two lines intersect at the point
C. Then if I draw a line vertical line through
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C, this gives me the optimum amount of offset
and OC gives me the minimum value of rp, because
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if we recall the last lecture, we any point
we take on this curve then join it with C,
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then that line gives you, the inclination
of that line from the vertical gives you the
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pressure angle.
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So as we see from this point to all other
points, the inclination to the vertical never
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goes beyond 30 degree.
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Similarly, in this phase and this phase, whatever
lines I draw the inclination to the vertical
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never exceeds 30 degree.
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This is also 30 degree.
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So we get this angle as 60 degree.
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Let me now name these points: this point,
vertical line and the y prime axis, this point
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let me call E and this point, let me call
D and this point of tangency, let me call
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B and the centre let me call F. As we see,
this angle is 120 degree.
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This angle is 90 degree.
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This angle is 90 degree.
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So this angle also is 60 degree.
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Thus we get, if I draw this line just for
the sake of construction OFB, OF is FB.
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In triangle OFB, I get OF is FB is equal to
the radius of this circle which was 25 millimeters
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and this angle is 60 degree.
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So this is an equilateral triangle.
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So I can write OB also 25 millimeters.
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Now if I drop a perpendicular from here, it
is easy to see that OB is nothing but twice
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of OD into, this angle is 30 degree, this
is 120 these two are equal, so 30 degree and
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30 degree, so twice OD cosine 30 degree.
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So that tells me OD is cos 30 is root 3 by
2, so 25 by root 3 millimeters.
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Now here we see, this is 60 degree, this is
60 degree and this is also 60 degree.
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So this is again, again an equilateral triangle
OCD, triangle OCD is equilateral.
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All sides are equal because all the angles
are 60 degree.
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So OC, that is (rp) minimum or optimum turns
out to be same as OD, which is 25 by root
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3 millimeter.
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And what is the optimum amount of offset?
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That is OE, OE I can easily see, this is 30
degree this is OC.
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So this is OC sine 30 degree, so half of that.
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So, e optimum e is 25 by root 3 into sine
of 30 degree which is half so we can write
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12.5 by root 3 millimeters.
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So we get both (rp) minimum and e, and once
I get (rp) minimum, I can easily get the base
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circle radius minimum value which is (rp)
minimum minus the roller radius which have
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to be specified.
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So this is given and also we have to give
the roller radius.
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So this is a simple diagram versus y theta
versus y prime theta, we get a circle.
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So everything I could drew, I could calculate
analytically, but if y theta is a complicated
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function, it changes during rise and then
dwell then again return, then rise, dwell,
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whatever it is, I can always calculate y prime
for every value of theta and get a closed
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curve and do the same construction.
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Draw a tangent here and take a line exactly
at phimax with the vertical.
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Now if we talk of a very theoretical situation,
that if phimax is more than 45 degree, which
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normally never is, but suppose if phimax
is more than equal to 45 degree then (rp)minimum
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calculation would have been a little different,
what I would have done is, if this is the
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closed curve, then I draw a, if phimax is
less than 45, greater than 45 degree, so the
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tangent I draw at an angle phimax to the vertical,
so this angle is pi by 2 minus phimax.
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Then to get minimum value of OC, that is rp,
I just need to draw a perpendicular to this
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line.
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If this is O, this point would have been C,
because in this case the pressure angle during
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this part of the rise and this part of the
rise is automatically guaranteed to be less
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than 45 degree.
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I have to ensure only that during this phase,
phimax does not go beyond 45 degree because
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if this angle is pi by 2 minus phimax so this
angle is phimax and if this is less than 45
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degree and this is 90 degree so this angle
is.
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If phimax is more than 45 degree then this
angle will be automatically less than 45 degree.
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If this angle is pi by 2 minus phimax, then
this angle is phimax and if this phimax is
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more than 45 degree, then this angle is automatically
less than 45 degree because this is pi by
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2 minus phimax.
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So during this phase and this phase, pressure
angle is ensured to be less than 45 degree.
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But this is only theoretical because phimax
we never take so such a large value.
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For phimax less than 45 degree, I have to
draw a tangent which is inclined to the vertical
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by phimax and from O, I draw a line which
is inclined to the vertical by phimax and
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wherever they intersect, that determines the
point C and OC gives me the minimum value
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of rp and the vertical line if I draw the
distance between the Y-axis and this vertical
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line gives me the optimum offset.
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So thus we have finished our discussion to
get the basic dimensions for cam profile for
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translating follower, both roller and flat
face.
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Our next task will be to design or synthesize
the cam profile, when we are given or determine
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rb and e.
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To generate a particular follower motion,
what should be the cam profile?
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That is the discussion that we will take up
next.
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Now that we know how to size the cam that
is, how to choose the value of the base circle
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radius and the proper amount of offset, the
next task is to synthesize the cam profile
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to generate the desired follower motion given
by y equal to y theta.
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We start with a analytical procedure.
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For the time being, we are restricting our
discussion to translating follower and first
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we discuss if we have a flat face follower.
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This is the flat face follower.
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When suppose the cam has rotated by theta,
this is the cam with the cam shaft here.
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This is what we call the offset, cam is rotating
counter clockwise.
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If we draw the base circle of this cam, this
is the base circle radius rb.
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So when the follower touches the base circle
that is the lowest position of the follower,
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this tangent at the top most point of the
base circle that is the lowest position of
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the follower.
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This is the contact point, when the cam has
rotated by an angle theta
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and consequently, the follower has moved by
this distance which is y theta.
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The contact point let me call C and through
this cam shaft O2, let us take X-axis horizontal
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and Y-axis vertical.
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First of all, our objective is to find the
polar coordinates of this contact point C.
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What is the polar coordinate?
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To find the polar coordinate that is O2C and
say I measure the angle from Y-axis in the
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clockwise direction that is this angle which
we call phiC.
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So the polar coordinates of this contact point
rc is O2C and the other coordinate is phiC.
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As we see phiC is measured from the Y-axis
in the clockwise direction.
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To find this rc, we will first find what are
X and Y coordinates of the point C.
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This is the distance which we called eccentricity
of the driving effort epsilon and this point
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let me call A. So yC is AC and that is easy
to see that AC is nothing but rb plus y theta.
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If we recall yesterday’s lecture, this distance
O2A we found out which is e plus epsilon which
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is xC was shown to be y prime theta.
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The x co-ordinate which is O2A.
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It can be shown very easily that this distance
is y prime theta.
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Suppose, I use that Arnold- Kennedy theorem
of 3 centers this is the fixed link.
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There is a guide 1, 1, cam is 2 and the flat
face follower is 3.
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So it is a three link mechanism consisting
of the fixed link, cam, which I have number
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2 and this follower which have number 3.
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Out of which 12 is at O2 because there is
a revolute pair, the cam can rotate with respect
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to the fixed link about this axis.
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So this O2 is nothing but 1 2, 2 3 is on this
common normal or that is on this vertical
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line.
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So this line contains 2 3 and because 3 is
in vertical translation with respect to 1,
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1 3 is on the horizontal direction at infinity,
1 3 in the horizontal direction at infinity.
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So if I draw a horizontal line through O2
that is X- axis, this contains 1 3 at infinity
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and now I know these 3 relative instantaneous
centers, namely 1 2, 1 3 and 2 3 are collinear
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by Arnold-Kennedy theorem.
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So 1 2 is here 1 3 is at infinity, 2 3 is
on this line.
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So this point A is nothing but 2 3.
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That is the velocity of this point A, if I
consider it to be a point on the cam, that
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is on body 2, whatever velocity I get, it
will be the same velocity at this point for
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the body 3, that is the follower.
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So if I consider it to be a point on body
2, the velocity is, angular velocity of the
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cam is d theta dt into O2A in the vertically
upward direction and if I consider to be a
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point on body 3, which is the follower and
follower is in perfect translation in the
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vertical direction.
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So all points of the follower have the same
velocity.
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So this same velocity is of the follower,
which is dy dt.
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So that clearly tells me, O2A is nothing but
dy d theta which is y prime theta.
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So given y theta, I can find both the x - coordinate
of the contact point and the y - coordinate
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of the contact point, xC and yC. yC is rb
plus y theta, when rb is known, y theta is
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known, I can find yC.
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If y theta is given, I can find y prime theta
and xC is nothing but y prime theta.
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So rc is nothing but square root of xC squared
plus yC squared.
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And what is phiC?
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This angle nothing but, tan inverse xC by
yC, because this is angle measured from the
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vertical.
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So tan of this is xC by yC.
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This is xC, this is yC and this angle I am
calling phi, this angle I am calling phi.
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Do not confuse it with the pressure angle,
thats why I am writing phiC, phi is the contact
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point, C is the contact point.
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Now, this is the polar coordinates of the
contact point in the fixed frame of reference
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in this XY axis, which is fixed in space that
is, that belongs to this fixed body 1.
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But what is the cam profile?
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Cam profile is nothing but the locus of this
contact point, but it has to be expressed
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in a coordinate system fixed in the cam.
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So, I fix a coordinate system in the cam.
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What I draw, suppose this line in the cam,
I take as my fixed line.
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So first of all let me write, cam profile,
is the locus of the contact point for various
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values of theta as the cam rotates, contact
point expressed
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in a coordinate system, fixed with the cam,
fixed to the cam.
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Now how do I express this coordinate of the
point C, in a cam fixed coordinate system?
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If I fix the cam, then its a nothing but a
kinematic inversion.
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So the fixed link 1 rotates in the clockwise
direction with the same value of omega.
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If the cam is rotating with say constant angular
velocity omega and if I hold the cam fixed,
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the same kinematics is maintained.
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If the fixed link is rotated in the opposite
direction by the same amount, that is kinematic
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inversion.
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This is the relative motion of the cam rotation
with respect to the fixed link, in the counter
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clockwise direction.
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Fix the cam, that means kinematic inversion
implies the fixed link 1 will rotate in the
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clockwise direction with the same angular
velocity.
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That means this line fixed in space will rotate
in the clockwise direction by theta.
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So what is the angle of this, if I measure
it from this line fixed to the cam.
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This angle is theta and from this y axis which
is rotating, so this angle will be always
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by measured by theta, because if this line
has rotated from the fixed line by an angle
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theta in the counter clockwise direction.
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If I fix this line, the vertical axis, y axis
rotates in the opposite direction, with the
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same value of theta.
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So if I want to use the cam fixed coordinate
system, then I write rc is same as before,
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xC squared plus yC squared, but I write phiC
prime, which is the angle measured from this
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line fixed to the cam, for this line O2C,
that is phiC plus theta and it has to be measured
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in the clockwise sense.
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And we already have the expression of rc and
phiC.
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So I get the cam profile, in the polar coordinate
system, this cam profile in the polar coordinate
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system.
210
00:27:40,840 --> 00:27:46,559
I plot all these points by measuring from
the O2, this distance rc and measuring the
211
00:27:46,559 --> 00:27:53,570
angle of that line O2C, from this line fixed
in the cam, by this theta plus phiC, where
212
00:27:53,570 --> 00:27:59,250
phiC is nothing but tan inverse xC by yC which
I wrote earlier.
213
00:27:59,250 --> 00:28:16,030
PhiC is tan inverse xC by yC and if we remember
yC is nothing but rb plus y theta and xC is
214
00:28:16,030 --> 00:28:18,700
y prime theta.
215
00:28:18,700 --> 00:28:25,460
So given rb and y theta, I get xC and yC so
I get phiC and then I can get rc and phiC
216
00:28:25,460 --> 00:28:33,280
prime and to get the polar coordinate representation
of the cam profile with O2 as the origin.
217
00:28:33,280 --> 00:28:39,890
I will continue now this, for the similar
exactly same technique for the roller follower.
218
00:28:39,890 --> 00:28:45,340
Just now we have discussed the analytical
method of cam profile synthesis for a flat
219
00:28:45,340 --> 00:28:46,800
face follower.
220
00:28:46,800 --> 00:28:53,330
What we did, we obtained the polar coordinates
of the contact point and expressed those polar
221
00:28:53,330 --> 00:28:58,520
coordinates in a coordinate system fixed with
the cam and that gives you the representation
222
00:28:58,520 --> 00:28:59,840
of the cam profile.
223
00:28:59,840 --> 00:29:13,690
We will extend exactly the same technique
for a translating roller follower.
224
00:29:13,690 --> 00:29:26,210
We have been given rb, rR, e and the desired
motion of the follower y theta and we have
225
00:29:26,210 --> 00:30:09,400
to obtain, what is the curve representing
the cam surface or the cam profile?
226
00:30:09,400 --> 00:30:13,309
We first draw the cam.
227
00:30:13,309 --> 00:30:22,990
This is the cam shaft axis O2 and this circle
is as usual represents the base circle of
228
00:30:22,990 --> 00:30:24,100
radius rb.
229
00:30:24,100 --> 00:30:51,660
Suppose, the roller follower is offset that
is the roller centre is here.
230
00:30:51,660 --> 00:31:13,330
The roller radius is given to us as rR.
231
00:31:13,330 --> 00:31:37,059
This is the guide of the fixed link 1, this
is the follower 4, this is the roller 3, this
232
00:31:37,059 --> 00:31:43,340
is the cam 2.
233
00:31:43,340 --> 00:32:14,309
We have also seen that I can draw, what we
called pitch curve, and
234
00:32:14,309 --> 00:32:27,640
this is the prime circle, rp prime circle
radius is rp.
235
00:32:27,640 --> 00:32:41,840
As before, we take X and Y-axis through O2,
measure phi in the clockwise direction.
236
00:32:41,840 --> 00:32:45,880
Let me not confuse this phi as the pressure
angle.
237
00:32:45,880 --> 00:32:55,590
So I use a different symbol here, the polar
angle, I call psi.
238
00:32:55,590 --> 00:33:06,500
This is the roller centre say C. No this contact
point, let me call C and the roller centre,
239
00:33:06,500 --> 00:33:17,980
I call D. So our job is to find, what is the
polar coordinates of this contact point C?
240
00:33:17,980 --> 00:33:27,480
So through C and the roller centre if I draw
a line, that is the common normal and this
241
00:33:27,480 --> 00:33:37,049
angle we define as the pressure angle phi.
242
00:33:37,049 --> 00:33:45,780
This angle is the pressure angle phi.
243
00:33:45,780 --> 00:33:53,120
When the roller centre intersects the prime
circle, that is the lowest point, this point
244
00:33:53,120 --> 00:34:02,620
is the lowest point of the follower and this
is the movement of the follower, when the
245
00:34:02,620 --> 00:34:04,350
cam has rotated to theta.
246
00:34:04,350 --> 00:34:13,599
So this is y (theta), and this angle is theta.
247
00:34:13,599 --> 00:34:28,919
So let me first, try to get the x and y-coordinates
of this contact point C.
248
00:34:28,919 --> 00:34:39,940
For the roller centre, I can write coordinates
of yD and xD.
249
00:34:39,940 --> 00:34:42,099
First let me get to the roller centre.
250
00:34:42,099 --> 00:34:48,070
xD is very simple, that is nothing but the
amount of offset.
251
00:34:48,070 --> 00:34:53,230
This distance is e.
252
00:34:53,230 --> 00:35:00,900
So xD is e and yD is this distance plus y
theta.
253
00:35:00,900 --> 00:35:06,670
This distance as we discussed in our last
lecture, is nothing but square root of rp
254
00:35:06,670 --> 00:35:07,970
squared minus e squared.
255
00:35:07,970 --> 00:35:13,870
This is a right angle triangle with hypotenuse
rp and this horizontal side e.
256
00:35:13,870 --> 00:35:18,180
So the vertical side is square root of rp
squared minus e squared.
257
00:35:18,180 --> 00:35:24,870
So this is the vertical side plus the roller
centre has moved from here to there by amount
258
00:35:24,870 --> 00:35:26,040
y theta.
259
00:35:26,040 --> 00:35:37,330
So yD is square root of rp squared minus e
squared plus y theta.
260
00:35:37,330 --> 00:35:49,270
Now if we get to the coordinate of D, then
that of C, I can easily see xC is e plus rR
261
00:35:49,270 --> 00:35:52,960
cosine phi.
262
00:35:52,960 --> 00:36:07,070
As this angle is phi and this is rR, so extra
horizontal distance is rR sine phi, rR sine
263
00:36:07,070 --> 00:36:14,349
phi and y coordinate is below that level D
and by rR cos phi.
264
00:36:14,349 --> 00:36:19,340
So yC is yD minus rR cos phi.
265
00:36:19,340 --> 00:36:33,330
So square root of rp squared minus e squared
plus y theta minus rR cosine phi, where phi
266
00:36:33,330 --> 00:36:43,140
is the pressure angle and we already got an
expression for tan of phi is, y prime minus
267
00:36:43,140 --> 00:37:00,930
e divided by square root of rp squared minus
e squared plus y theta.
268
00:37:00,930 --> 00:37:09,790
I get the x and y-coordinate of the contact
point, in this xy coordinate system, because
269
00:37:09,790 --> 00:37:12,570
e is known to us.
270
00:37:12,570 --> 00:37:18,040
Given the value of, given the function y,
I can calculate this and I get the value of
271
00:37:18,040 --> 00:37:19,710
phi for values of theta.
272
00:37:19,710 --> 00:37:25,180
Once I know that phi, I can get xC which is
e plus rR sine phi.
273
00:37:25,180 --> 00:37:31,720
Similarly, given this quantities yC, I can
get because I know the value of phi from here,
274
00:37:31,720 --> 00:37:41,260
I can calculate cos phi. rp is nothing but
rb plus rR that is known, rb plus rR.
275
00:37:41,260 --> 00:37:44,340
So I get xC and yC.
276
00:37:44,340 --> 00:37:53,430
So the polar coordinates of this point C,
rc is nothing but square root of xC squared
277
00:37:53,430 --> 00:38:03,980
plus yC squared and psiC is nothing but tan
inverse xC by yC.
278
00:38:03,980 --> 00:38:12,020
Now this is the contact point polar coordinates
in the fixed frame of reference but the cam
279
00:38:12,020 --> 00:38:16,790
profile is nothing but the locus of the contact
point in the cam fixed system.
280
00:38:16,790 --> 00:38:23,620
So psi is measured from this vertical line,
but this vertical line which is the line belonging
281
00:38:23,620 --> 00:38:28,630
to the fixed link 1, after kinematic inversion
that is if we hold the cam fixed, we will
282
00:38:28,630 --> 00:38:31,280
rotate in the clockwise direction.
283
00:38:31,280 --> 00:38:38,480
So at this, if this is the configuration at
theta, so I draw a line at an angle theta
284
00:38:38,480 --> 00:38:52,150
from the vertical and this line I hold fixed.
285
00:38:52,150 --> 00:38:55,870
This is the line fixed to the cam and I hold
this fixed.
286
00:38:55,870 --> 00:39:02,260
So as a consequence, this y axis rotates in
the clockwise direction and that angle will
287
00:39:02,260 --> 00:39:03,340
be always theta.
288
00:39:03,340 --> 00:39:05,840
So what will be the polar coordinate in this
cam fixed system?
289
00:39:05,840 --> 00:39:13,470
If the angle is measured from this line, which
I called as before psiC prime.
290
00:39:13,470 --> 00:39:28,510
So psiC prime is nothing but theta plus psiC
and rc is same, which is square root of xC
291
00:39:28,510 --> 00:39:32,150
squared plus yC squared.
292
00:39:32,150 --> 00:39:41,760
So these are the polar coordinates representation
of the contact point in a coordinate system
293
00:39:41,760 --> 00:39:43,260
fixed in the cam.
294
00:39:43,260 --> 00:39:46,740
This line, which I am holding fixed in the
cam, when the cam has rotated by an angle
295
00:39:46,740 --> 00:39:51,740
theta, I draw this line at an angle theta
counter clockwise direction from the vertical.
296
00:39:51,740 --> 00:39:57,050
I measure this psiC prime, from this line
in the clockwise direction.
297
00:39:57,050 --> 00:39:58,180
And how do I obtain these values?
298
00:39:58,180 --> 00:40:01,450
Let me go through it once more.
299
00:40:01,450 --> 00:40:09,230
I get the coordinate xC, yC because the pressure
angle phi, I can always get once y, y prime,
300
00:40:09,230 --> 00:40:12,090
e and rp, these quantities are known.
301
00:40:12,090 --> 00:40:15,980
So for various values of theta, I can get
xC and yC.
302
00:40:15,980 --> 00:40:21,310
So for various values of theta, I can get
rc, and for various values of theta, I get
303
00:40:21,310 --> 00:40:25,410
first psiC and then add theta to get psiC
prime.
304
00:40:25,410 --> 00:40:32,180
And if I go on plotting these values r and
psi, r is the distance from O2 and psi is
305
00:40:32,180 --> 00:40:36,640
measured from this line, which has been drawn
at an angle theta from the vertical line.
306
00:40:36,640 --> 00:40:42,070
Then plot all those points and the locus of
all these points will give me the cam profile.
307
00:40:42,070 --> 00:40:47,310
So this is how we get the analytical expression
for the cam profile for a translating roller
308
00:40:47,310 --> 00:40:49,280
follower.
309
00:40:49,280 --> 00:40:55,589
So far we have discussed in great details,
the determination of the basic dimensions
310
00:40:55,589 --> 00:41:01,730
and the cam profile synthesis with reference
to translating follower.
311
00:41:01,730 --> 00:41:06,570
So for the oscillating follower is concerned,
determination of basic dimension is a little
312
00:41:06,570 --> 00:41:07,570
more involved.
313
00:41:07,570 --> 00:41:13,830
So what we will do, we will assume that the
basic dimensions like base circle radius is
314
00:41:13,830 --> 00:41:14,830
determined.
315
00:41:14,830 --> 00:41:21,900
It is already given to us and the desired
follower motion is prescribed and how to synthesize
316
00:41:21,900 --> 00:41:27,510
the cam profile or how to determine the cam
profile by analytical method?
317
00:41:27,510 --> 00:41:40,440
So we continue our discussion with analytical
method: but with reference to oscillating
318
00:41:40,440 --> 00:41:43,570
follower.
319
00:41:43,570 --> 00:41:55,860
Here again, I will restrict our discussion,
only to a flat face follower.
320
00:41:55,860 --> 00:42:01,410
Exactly same methodology can be applied to
roller follower, but the algebra involved
321
00:42:01,410 --> 00:42:03,670
is a little different.
322
00:42:03,670 --> 00:42:08,310
But the methodology is just the same what
we follow for flat face follower is equally
323
00:42:08,310 --> 00:42:11,430
applicable for a roller follower.
324
00:42:11,430 --> 00:42:18,100
Now first let me discuss, how do we describe
or how do we express the follower motion incase
325
00:42:18,100 --> 00:42:27,970
of an oscillating follower?
326
00:42:27,970 --> 00:42:32,570
As we know, the size of the cam is described
by this base circle radius.
327
00:42:32,570 --> 00:42:43,940
Let’s say this is the base circle radius
of a cam, this is as usual the cam shaft axis.
328
00:42:43,940 --> 00:42:49,589
Now when the oscillating flat face follower
is in contact with the base circle that is
329
00:42:49,589 --> 00:42:54,570
the extreme position from where the follower
motion is measured or the rise of the follower
330
00:42:54,570 --> 00:42:55,570
starts.
331
00:42:55,570 --> 00:43:04,290
Let us assume that the follower face which
is in contact with the base circle radius
332
00:43:04,290 --> 00:43:12,020
is represented by this line and this follower
is hinged at this point.
333
00:43:12,020 --> 00:43:13,530
This is a oscillating follower.
334
00:43:13,530 --> 00:43:15,630
So it is hinged at one point.
335
00:43:15,630 --> 00:43:19,160
There is a revolute pair with the fixed link
and the whole follower will oscillate about
336
00:43:19,160 --> 00:43:20,260
that point.
337
00:43:20,260 --> 00:43:32,830
So I draw
a perpendicular to this follower face.
338
00:43:32,830 --> 00:43:37,820
This is 90 degree.
339
00:43:37,820 --> 00:43:44,790
So I represent the follower by these two straight
lines the follower face and these gives me
340
00:43:44,790 --> 00:43:51,119
another dimension, kinematic dimension which
is necessary.
341
00:43:51,119 --> 00:44:00,230
So here as we know the fixed link is 1 and
cam which is number 2.
342
00:44:00,230 --> 00:44:05,520
This is the base circle of the cam and this
is the follower which is link 3.
343
00:44:05,520 --> 00:44:11,530
So let me call this revolute pair at O3.
344
00:44:11,530 --> 00:44:31,170
Now due to the shape of the cam, which is
say, then as the cam rotates when the cam
345
00:44:31,170 --> 00:44:40,560
this point was here the follower was at this
position and now this will be, follower will
346
00:44:40,560 --> 00:44:57,599
be, this is the current position and if I
drop a perpendicular to this line this dimension
347
00:44:57,599 --> 00:45:00,940
remains e.
348
00:45:00,940 --> 00:45:10,960
This is moving in a circle.
349
00:45:10,960 --> 00:45:18,590
So this rigid body rotates in the clockwise
direction as the cam rotates in the anticlockwise
350
00:45:18,590 --> 00:45:20,430
direction.
351
00:45:20,430 --> 00:45:29,080
This is the basic motion geometry of the flat
face follower which is oscillating and as
352
00:45:29,080 --> 00:45:38,170
we see, this distance changes because the
contact point on the follower face moves.
353
00:45:38,170 --> 00:45:41,660
Now how do I express this particular follower
motion?
354
00:45:41,660 --> 00:45:50,130
This is the, suppose I draw a horizontal line,
this is the lowest position of the follower
355
00:45:50,130 --> 00:46:03,150
which I express this angle by psib and when
the cam has, a follower has rotated in the
356
00:46:03,150 --> 00:46:10,290
counterclockwise direction, this I called
psi has a function of theta.
357
00:46:10,290 --> 00:46:19,050
So psi is which is a function of theta is
psib plus some angle which is a function of
358
00:46:19,050 --> 00:46:20,050
theta.
359
00:46:20,050 --> 00:46:24,520
So it is this delta theta which represents
the follower movement.
360
00:46:24,520 --> 00:46:36,370
Instead of a trace point, I am talking of
this follower movement in terms of this angle
361
00:46:36,370 --> 00:46:38,480
directly.
362
00:46:38,480 --> 00:46:41,250
That is more convenient.
363
00:46:41,250 --> 00:46:45,980
We should also note, that we have already
got another kinematic dimension namely this
364
00:46:45,980 --> 00:46:52,240
e which is perpendicular drop from O3 to this
follower face which remains same because this
365
00:46:52,240 --> 00:46:53,859
is the same rigid body.
366
00:46:53,859 --> 00:46:56,510
This angle is always 90 degree.
367
00:46:56,510 --> 00:47:07,869
Now the relative position of O2 and O3, these
two revolute pair is also two important kinematic
368
00:47:07,869 --> 00:47:10,030
dimensions.
369
00:47:10,030 --> 00:47:17,849
Suppose I draw by X-axis here and Y-axis here.
370
00:47:17,849 --> 00:47:31,570
So the x and y-coordinate of O3 are two more
important dimensions, I call it a and b, x-coordinate
371
00:47:31,570 --> 00:47:41,800
of O3 is a, y-coordinate of O3 is b, where
O3 is the hinge point for the oscillating
372
00:47:41,800 --> 00:47:48,480
follower and O2 is the cam shaft, e is another
kinematic dimension which has been obtained
373
00:47:48,480 --> 00:47:53,900
by dropping a perpendicular from O3 to the
follower face and then the follower is treated
374
00:47:53,900 --> 00:47:59,180
as this rigid body consisting of two straight
lines at right angle.
375
00:47:59,180 --> 00:48:04,310
This is the lowest position when the follower
was in contact with the base circle that is
376
00:48:04,310 --> 00:48:05,579
when this point was here.
377
00:48:05,579 --> 00:48:17,800
Now the follower has rotated, sorry the cam
has rotated through an angle theta.
378
00:48:17,800 --> 00:48:22,450
As the cam has rotated through an angle theta,
the follower has rotated through an angle
379
00:48:22,450 --> 00:48:29,140
delta theta and the angle that this follower
face makes with the X-axis or horizontal line
380
00:48:29,140 --> 00:48:30,730
that I write as psi.
381
00:48:30,730 --> 00:48:37,990
At theta equal to 0, the value of psi was
psib, when delta theta is also 0.
382
00:48:37,990 --> 00:48:42,170
At theta equal to 0, delta theta is 0, and
psi is equal to psib.
383
00:48:42,170 --> 00:48:49,089
So delta theta is the follower movement which
is the function of theta that is how it prescribed.
384
00:48:49,089 --> 00:48:56,320
And our objective will be to find the polar
coordinates of this contact point C which
385
00:48:56,320 --> 00:49:02,329
will define the cam profile when I express
this coordinate in the cam fixed coordinate
386
00:49:02,329 --> 00:49:03,329
system.
387
00:49:03,329 --> 00:49:05,430
That is basically the method.
388
00:49:05,430 --> 00:49:10,050
But to do this, we need to do a little bit
of elaborate geometry as we will see just
389
00:49:10,050 --> 00:49:11,050
now.
390
00:49:11,050 --> 00:49:17,640
For an oscillating flat face follower, the
angle that the follower face makes with the
391
00:49:17,640 --> 00:49:25,060
horizontal was expressed as psib plus delta
theta.
392
00:49:25,060 --> 00:49:31,000
So before we try to get the contact point
coordinates, first let me get the expression
393
00:49:31,000 --> 00:49:37,339
for this angle psib, which the follower face
makes at its extreme position when it is contact
394
00:49:37,339 --> 00:49:39,790
with the base circle.
395
00:49:39,790 --> 00:49:51,470
So lets say this is the base circle, with
O2 as center and the follower face is tangent
396
00:49:51,470 --> 00:49:57,820
to the base circle.
397
00:49:57,820 --> 00:50:04,690
This is the point O3, where the follower is
hinged and this angle is 90 degree.
398
00:50:04,690 --> 00:50:23,970
The
399
00:50:23,970 --> 00:50:29,720
angle that this follower face makes with the
horizontal, that we call psib.
400
00:50:29,720 --> 00:50:35,829
So if I draw a line which is horizontal, this
angle is psib.
401
00:50:35,829 --> 00:50:48,070
Now the line O2O3 this is a horizontal line
this distance we call a and this distance
402
00:50:48,070 --> 00:50:50,140
we call b.
403
00:50:50,140 --> 00:50:54,079
This is 90 degree.
404
00:50:54,079 --> 00:50:55,619
This is the X-axis.
405
00:50:55,619 --> 00:51:02,369
Now this angle is clearly seen to be tan inverse
b by a and this angle is nothing but this
406
00:51:02,369 --> 00:51:03,400
angle.
407
00:51:03,400 --> 00:51:17,770
So this angle is tan inverse let me call this
angle as some lambda.
408
00:51:17,770 --> 00:51:23,600
Sorry lambda is the angle which this line
is making with the horizontal.
409
00:51:23,600 --> 00:51:26,370
So this line is the horizontal line.
410
00:51:26,370 --> 00:51:29,530
So this angle is same as this angle.
411
00:51:29,530 --> 00:51:31,470
This is lambda.
412
00:51:31,470 --> 00:51:46,140
So lambda is tan inverse b by a and psib that
is this angle I can get, if I find this angle
413
00:51:46,140 --> 00:51:47,930
and then subtract lambda from there.
414
00:51:47,930 --> 00:51:58,910
So to find this angle, let me join this line
this is at 90 degree, this is at 90 degree
415
00:51:58,910 --> 00:52:06,340
and this angle is same as this.
416
00:52:06,340 --> 00:52:22,820
This angle if we give a name say chi, then
psib is seem to be chi minus lambda and to
417
00:52:22,820 --> 00:52:31,190
get chi I can easily get it, if I write this
as the base circle radius rb, this dimension
418
00:52:31,190 --> 00:52:42,780
is e and this point let me name A, the point
of intersection of the flat face at the lowest
419
00:52:42,780 --> 00:52:45,260
position with O2AO3.
420
00:52:45,260 --> 00:52:47,600
So this angle is equal to this angle.
421
00:52:47,600 --> 00:52:48,900
This angle is 90 degree.
422
00:52:48,900 --> 00:52:50,490
This angle is 90 degree.
423
00:52:50,490 --> 00:53:13,620
So from here, I can write sine of chi is rb
by O2A and from here I can write sine of the
424
00:53:13,620 --> 00:53:16,670
same angle chi is e by O3A.
425
00:53:16,670 --> 00:53:20,950
So this is same as e by O3A.
426
00:53:20,950 --> 00:53:35,599
Using this relation, I can write, this is
also equal to rb + e divided by O2A + O3A
427
00:53:35,599 --> 00:53:50,609
which is same as rb + e divided by O2O3 and
O2O3 is nothing but square root of a squared
428
00:53:50,609 --> 00:53:53,530
plus b squared, because this angle is 90 degree.
429
00:53:53,530 --> 00:54:13,119
So chi I get sine inverse rb plus e divided
by square root of a squared plus b squared
430
00:54:13,119 --> 00:54:19,510
and psib is nothing but chi minus lambda,
where lambda is this angle.
431
00:54:19,510 --> 00:54:24,849
The angle O2O3 makes with X axis that is tan
inverse b by a.
432
00:54:24,849 --> 00:54:31,200
So once these geometric dimensions are given
like a, b, e and rb, I am in a position to
433
00:54:31,200 --> 00:54:34,400
find the value of psib.
434
00:54:34,400 --> 00:54:38,750
In our next lecture I will take off from this
point and I will assume this value of psib
435
00:54:38,750 --> 00:54:43,300
which we can obtain given the values of a,
rb, e, b, all that.
436
00:54:43,300 --> 00:54:48,940
Then I will try to get the coordinates of
the contact point between the flat face and
437
00:54:48,940 --> 00:54:54,360
the cam profile and that will give me the
cam profile in polar coordinates as we did
438
00:54:54,360 --> 00:55:11,940
in case of translating follower.
439
00:55:11,940 --> 00:55:28,170
3