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In our last lecture, we have discussed displacement
analysis through examples using only graphical
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method. Today, we continue our discussion
on displacement analysis but through the other
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method namely, the analytical method. Now,
when do we use analytical method and when
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do we use graphical method?
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As we see, the graphical method, there is
an inherent limitation on the accuracy, because
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of the scale of the figure and your drawing
inaccuracies. So, analytical method is preferred
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when we want higher accuracy.
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Or, if the displacement analysis has to be
carried on for a very large number of configurations
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and in the graphical method, the picture becomes
really cumbersome.
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The other advantage of this analytical method
is that, it is amenable to computer programming.
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Now, before we get into the details of the
analytical method and solve examples, showing
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the power of the analytical method, let us
go through the basic methodologies, that is
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followed in this analytical method. As we
know, a mechanism consists of closed kinematic
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loops.
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So, first task is to identify all the independent
closed loops that exist in the mechanism.
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Next, we express all the kinematic dimensions
(like link lengths, offsets) and also the
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slider displacement by planar vectors. Then
the third step is to express for every close
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loop, what we call as a loop closure equation
in terms of these vectors that we have just
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now said. Now, each such vector in 2D, that
is a planar linkage is equivalent to two scalar
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equations that means, if a vector equation
is there, that is equivalent to two scalar
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equations and two unknown quantities can be
solved.
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Now, once all these vector equations are generated
using all the independent closed loops, one
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has to solve these equations to determine
the unknown quantities that is relevant to
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your particular problem. One has to remember
that in general, these equations are non-linear
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algebraic equations and can be solved only
numerically. However, in some simpler cases,
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just like a 4R mechanism or if there is a
four link closed loop, then we can show that
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these non-linear algebraic equations reduces
to quadratic equation and can be solved analytically.
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So, we start with an example that is the simplest
one. Suppose, we are given all the link-lengths
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of a 4R-linkage, then determine the orientation
of the coupler and the output link, when the
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orientation of the input link is prescribed.
So this is a problem of displacement analysis.
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We are given all the kinematic dimensions
and the position of the input link and the
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task is to determine the position of all other
moving links for that particular configuration.
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This diagram shows, a 4R linkage O2ABO4. As
we see, the fixed link O4O2 suppose has a
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length l1 and O2A - the second link has a
link length l2, AB - the coupler has a link
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length l3 and O4B is the fourth link with
link length l4. The problem is, suppose this
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four link-lengths are prescribed and the orientation
of one of the links say, input link that is
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O2A is given by this angle theta 2. Our objective
is to determine the orientation of the coupler
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and the fourth link, that is O4B.
To do this, first we set up a Cartesian coordinate
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system XY, with O2 as the origin. The first
task as I said, we have identified a closed
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loop namely - O2ABO4O2. Then the link lengths
l1, l2, l3, l4 are represented by four vectors
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namely - l1, l2, l3 and l4. These vectors
are shown by these arrows in this diagram.
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Now the angle that these vectors make with
the X-axis, we are using as the reference
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angle to determine the orientation of these
vectors, that is theta 2, theta 3, and theta
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4. Theta 1, that is the angle made by O4O2,
that is the vector l1 with X-axis is obviously
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0. We can express this loop closure equation
very simply as, l1 plus l2 (these are all
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vectors) plus l3 minus l4 is equal to 0. This
is what we say as the loop closure equation,
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expressed in terms of link vectors.
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To analyze this equation, we write, l2 vector
in terms of complex exponential notation,
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that is l2 e to the power i theta 2. Same,
for l3 vector as l3 e to the power i theta
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3 and l4 vector as l4 e to the power i theta
4. Since, theta 1 is 0, the vector l1 is simply,
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link length l1 e to the power i 0 which is
1. So, the loop closure equation, let us substitute
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these complex exponential notation, so you
get, l1 plus l2 e to the power i theta 2 plus
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l3 e to the power i theta 3 minus l4 e to
the power i theta 4 is equal to 0. We know,
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e to the power i theta is cos theta plus i
sin theta. So, first we substitute for all
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these angles, e to the power i theta 2, i
theta 3 and so on, in terms of cosine and
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sine and equate the real and the imaginary
parts of this complex equation separately
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to 0. This is nothing but really adding the
X components to 0 and adding the Y components
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to 0, because it is a closed loop. As a result,
we get, l1 plus l2 cosine theta 2 plus l3
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cosine theta 3 minus l4 cosine theta 4 is
0. And adding the imaginary parts, that is
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the sine components we get l2 sine theta 2
plus l3 sine theta 3 minus l4 sine theta 4
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is equal to 0. Thus, a vector equation, because
these vectors are two-dimensional vectors
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from one vector equation, we get two scalar
equations such as equation one and equation
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two. Each vector equation is equivalent to
two scalar equations if the vectors are two
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dimensional.
Now from these two equations, we can eliminate
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one of the unknowns. Say, either theta 4 to
solve theta 3 or we can eliminate theta 3
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to solve for the unknown theta 4, if l1, l2,
l3, and l4, that is the link lengths are given
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and the input angle theta 2 is specified.
Now, let me carry out this algebra to determine
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say, theta 4. To determine theta 4, we write
l3 sine theta 3 cos theta 3 sorry ok I can
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write this first sine theta 3 is l4 sine theta
4 minus l2 sine theta 2. And from the first
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equation, we write l3 cosine theta 3 equal
to l4 cosine theta 4 minus (l1 plus l2 cosine
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theta 2). From these two equations, if we
square and add up, obviously from the left-hand
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side I get only l3 square. So I get, square
and add these two equations we get, l3 square
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is equal to l4 square plus l1 square plus
l2 square minus twice l2l4 sine theta 4 sine
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theta 2 minus twice l4 cosine theta 4 into
l1 plus l2 cosine theta 2 minus twice l1l2
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cosine theta 2. Let me carry out this algebra
a little further and now onwards I will write
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‘C’ for cos and ‘S’ for sine for brevity.
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We can get l3 square is equal to l1 square
plus l2 square plus l4 square plus twice l1l2
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cosine theta 2 minus twice l2l4 sine theta
4 sine theta 2 minus twice l1 plus l2 cosine
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theta 2 into cosine theta 4 l4. If we divide
now both sides of this equation by twice l2l4.
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Here I missed out one l4 so let me write out
here. Now divide all the terms of this equation
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divide by twice l2l4. As a result, we will
get a simple equation a sine theta 4 plus
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b cosine theta 4 is equal to c, where a is
given by sin theta 2, that is coming from
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this term. From this term, b will get l1 by
l2 plus cosine theta 2 and c is l1 square
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plus l2 square minus l2 square minus l3 square
plus l4 square divided by twice l2l4 plus
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l1 by l4 cosine theta two. So you see, if
theta 2 and all the link lengths are given,
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these quantities namely: a, b and c are completely
known and the only unknown is this theta 4
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which has to be solved. But as we see, this
is not a linear equation in theta 4, because
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it involves sine and cosine. To solve this
angle theta 4 unambiguously, that is we need
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not judge the quadrant after we get the value,
in what quadrant the angle theta 4 lies, it
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is always better to replace sine and cosine
by the tan of the corresponding half angles.
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That means, we write sine theta 4 as twice
tan of theta 4 by 2 divided by 1 plus tan
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square theta 4 by 2 and cosine theta 4 we
write 1 minus tan square theta 4 by 2 divided
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by 1 plus tan square theta 4 by 2. Substituting
this, in that a sine theta 4 plus b cos theta
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4 equal to c equation, we get a quadratic
equation in tan theta 4 by 2 as follows: (b
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plus c) tan square theta 4 by 2 minus 2 a
tan theta 4 by 2 plus c minus b equal to 0.
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So, finally we get this quadratic equation
in tan theta 4 by 2, where a, b and c are
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all known quantities in terms of the link
lengths and the given input angle theta 2.
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Now, this is where tan theta 4 by 2 is useful,
I get a quadratic equation and I get two roots
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namely: tan theta 4 by 2 is equal to a plus
minus square root of a square plus b square
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minus c square divided by b plus c.
So, you get two roots for tan theta 4 by 2
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and the right-hand side may come out to be
positive or negative, of course, it can never
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be imaginary once the kinematic diagram is
okay, that is a 4R linkage has been made,
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a close loop has been obtained, a, b, c will
be of such values that this square root can
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never be of a negative quantity. So you may
get two positive, two negative or may be one
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positive and one negative.
Now, tan theta 4 by 2 as we know, if theta
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4 is between 0 and 180 degree and then theta
4 by 2 will be between 0 to 90 degree and
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tan will be positive. And if theta 4 is more
than 180 degree, then theta 4 by 2 will be
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more than 90 degree and tan theta 4 will be
negative. So, depending on the positive and
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negative values of tan theta 4 by 2, I can
easily determine, uniquely the value of theta
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4. Suppose we are also interested in finding
the orientation of the coupler, that is the
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angle theta 3.
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Then if we look at these two equations, we
see l3 and l4 are appearing exactly the same
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way, only l3 is appearing on the left-hand
side and l4 is appearing on the right-hand
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side. It is even more clearer, if we look
at these equations number one and two. You
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can see l3 and l4 are appearing exactly similar
way, only thing plus and minus. So, when we
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have eliminated theta 3 and obtain theta 4.
Following exactly the same procedure, I can
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eliminate theta 4 and obtain theta 3. We have
to remember that l4 has to be replaced by
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minus l3, rest of the solutions remains as
it is.
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So following exactly a similar method to eliminate
theta 4, we can get a theta 3, a very similar
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equation like tan theta 3 by 2 will be equal
to: a plus minus square root of a square plus
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b square minus c prime square divided by b
plus c prime. As we might have remembered,
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that the quantities, a and b did not involve
l4 or l3. a was sine theta 2, b was cosine
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theta 2 plus l1 by l2. However, the quantity
c involved l3 and l4. That is why, we have
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to replace c by c prime and expression of
c prime will also be very similar to c and
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we have to interchange l4 by minus l3 and
vice versa.
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That is, we will get c prime as l1 square
plus l2 square minus l3 square plus l4 square
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divided by twice times l2l3 plus l1 by l4
cosine theta 2. The thing to note is, for
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the same set of given length and the same
value of theta 2, we have obtained two values
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of theta 4 and two values of theta 3. This
is nothing surprising, if we remember that
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even by graphical method, we would have got
two solutions for the given link lengths and
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the input angle. This aspect is very clear
if we draw and we solve the same problem by
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graphical methods.
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Suppose, we have given l1 that the length
O4O2, the link length l2 that is O2A, and
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the angle theta 2, all these are prescribed.
So, you reach from O4, O2 to A. The link lengths
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l3 and l4 are given. So, I can draw a circular
arc with A as center and l3 as radius, and
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we get this circle. So the point B must lie
on this circle. Similarly, if I draw a circular
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arc with O4 as center and l4 as radius, we
get another circle, that is this circle. So,
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the B must lie on this circle, if I come from
O4 and B must lie on this circle, if I come
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from A. These two circles as I said, in general,
will intersect at two points namely: B1 and
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B2. So, we get two different configurations
for the same values of link lengths and theta
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2, one is O4, O2, A, B1 and the other is O4,
O2, A, B2. Two configurations are shown distinctly,
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one by the firm lines and the other by the
dashed lines.
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Now, we have got two values of theta 3 and
theta 4 which are indicated in this diagram
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as well. For the firm line, that is when the
diagram is above the line O4O2, this is theta
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3 one and this is theta 3 two. Corresponding
to theta 3 one, I have theta 4 one and corresponding
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to theta 3 two, I have theta 4 two. Now that
we have seen how to obtain various values
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of theta 4 as theta 2 changes. Now, we will
show you the typical variations of theta 4
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versus theta 2 for various types of 4-bar
linkages.
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For example here we see, what we call a crank-rocker
linkage, that means it is obviously a Grashofian
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linkage and a crank-rocker and theta 2 goes
from 0 to 2 pi and theta 4 because theta 4
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is the rocker angle, it only goes from a minimum
value to a maximum value and there are two
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branches, one is here and the other is there.
Because it is a rocker, again it rocks either
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from this minimum value to that maximum value
or from this minimum value to that maximum
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value. As we said earlier, that Grashofian
crank-rocker has two distinct branches, two
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distinct modes of assembly, that is why these
two curves never cross each other. If we are
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starting on one of the curves you will always
remain there, either on this curve or on this
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curve. You can never go from one curve to
the other because each curve corresponds to
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a particular mode of assembly.
Suppose we have a Grashofian double-crank,
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this is the typical theta 4 verses theta 2
characteristics of a double crank. As you
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see, theta 2 goes from 0 to 2 pi and theta
4 also goes from 0 to 2 pi, because both the
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input and the output lengths are capable of
making complete rotations. But because it
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is a Grashofian linkage, there are two distinct
modes of assembly as represented by this firm
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line and the other mode of assembly is by
this curved line and these two curves can
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never intersect, that is each curve corresponds
to a particular mode of assembly.
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If we have a Grashofian double-rocker, then
again we have two loops, one as we see here
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and the other there. Because this is double-rocker
neither theta 2 nor theta 4 can make complete
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rotation. So, they are limited between two
extreme values for example, for theta 2 from
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here to there and for theta 4 it is here to
there. Again, there are two distinct loops
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for two different modes of assembly either
you are following this curve or you are following
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this curve, you can never pass on from one
mode of assembly to the other.
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Now, let us see what happens in a non-Grashof
linkage. If it is a non-Grashof linkage then
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it has to be a double-rocker. Theta 4 versus
theta 2 plot for a non-Grashof double-rocker
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looks like this, here again neither theta
2 nor theta 4 can make complete rotation.
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Theta 2 is restricted within this region.
Theta 4 is restricted within this region,
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but there is only one curve, because there
is only one mode of assembly and as you see,
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this curve shows that the mirror image configuration
can be taken up by the same assembly. You
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can generate the entire curve by one mode
of assembly.
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Just now we have seen, how to do the displacement
analysis by analytical method for a very simple
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mechanism that is a 4R linkage which consists
of only a single loop. Now, let me show how
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we can use the kinematic analysis, the same
analytical method for a little more complicated
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mechanism, like the slotted lever quick-return
mechanism, which we did last time by graphical
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method.
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This figure shows the kinematic diagram of
the slotted lever quick-return mechanism.
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It has six links namely, the fixed link 1,
input link 2, which is the line O2A, then
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this block which is link 3 and the slotted
lever link 4, the tool holder link 6 and the
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link connecting the slotted lever with the
tool holder, that is link number 5. So there
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are six links with two sliding pairs, one
between 1 and 6, and the other between 3 and
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4. So to do the kinematic analysis of such
a mechanism by analytical method, let us first
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identify that there are more than one closed
loop.
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We draw all the link lengths by the corresponding
vectors, such as: l2 is the vector which represents
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the second link O2A, l1 is the other vector
which represents the fixed link O2O4 and O4A,
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that is the slider displacement we use another
vector namely - s4. Similarly, the slider
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displacement we represent by another vector
s6 and O4B is the other link 4, that is the
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link vector. l5 is another vector representing
the link BC, that is the fifth link. So, here
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what we see, that there are link length vectors
like l1, l2, l5, l4 and l6, that is the distance
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of the slider movement from the point O2.
So, these five, l1, l2, l4 and l6, these five
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link length vectors are constants, they are
not changing with time. Whereas there are
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two other vectors, namely s4, denoting the
distance O4A is changing with time. Same is
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true for s6 which represents the movement
of the cutting tool, here it is DC represented
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by the vector s6. And all these vectors, the
orientation of these vectors, I represent
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by angle theta, measured from this horizontal
line in the anticlockwise direction. So, we
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can have one close loop O2AO4O2, which means
l2 is equal to l1 plus s4. This is one loop
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closure equation, in terms of link lengths
vector and sliding displacement vector.
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I can have another close loop namely: O2ABCDO2.
There is an independent loop and corresponding
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to this independent loop, I can write l2 plus
l4 plus l5 is equal to l6 plus s6. So, this
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is a second vector equation. So, we have two
vector equations, which as I said earlier,
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are equivalent to four scalar equations, that
is equating the real and imaginary parts or
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X and Y components, however you said it, from
these two vector equations, I get four scalar
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equations. So, I can solve for four unknows.
Now, let us see in this problem, what are
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the unknows? What is prescribed to us are;
all the link length vectors, that is l1, l2,
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l4, l5 and l6 and the input angle theta 2,
that is the angle that this link 2 makes with
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the reference this line, that is the X-axis.
So, these are given and we have to find out
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the orientation of all other links and also
the slider displacement that is s4 and s6.
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So, the number of unknowns is: theta 4, that
is the orientation of link 4, that is the
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00:33:23,820 --> 00:33:34,410
line O4B, theta 5, that is the orientation
of the fifth link, that is BC, that is this
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00:33:34,410 --> 00:33:46,950
angle theta 5. This angle, as I said earlier
is theta 4 and also this distance O4A, that
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00:33:46,950 --> 00:33:56,150
is s4 and the distance BC that is s6. So,
these are the four scalars unknowns which
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00:33:56,150 --> 00:34:03,270
can be solved using these two vector equations.
That is how we carry out the kinematic analysis
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00:34:03,270 --> 00:34:09,770
to analytical method. Later on we will solve
some more different types of problem and show
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00:34:09,770 --> 00:34:17,460
the use of this loop closure equation and
the vector representations of link lengths.
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As a second example of this kinematic analysis
to the link vectors, let us look at the model
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of this mechanism which is a copying mechanism
or plagiograph. What we see here, that this
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is a five-link mechanism, the fixed link and
there are four moving links: 2, 3, 4 and 5.
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00:34:40,021 --> 00:34:48,550
The dimensions of these links are such that
this forms a parallelogram. This link length
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00:34:48,550 --> 00:34:55,090
is equal to this, this link length is equal
to this. And these two triangles are similar
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00:34:55,090 --> 00:35:00,900
triangles, that is this angle is same as this
angle and this angle is also same as this
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00:35:00,900 --> 00:35:07,650
angle. Now in this mechanism, we will show
later that whatever is the movement of this
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00:35:07,650 --> 00:35:14,579
point, that is reproduced by this point, only
thing there will be a change in the scale
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00:35:14,579 --> 00:35:23,570
and the orientation. For example, if I move
this mechanism such that this point follows
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this vertical line of this letter l then this
point is also following a straight line. As
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00:35:38,521 --> 00:35:53,180
we go around this letter l, this point will
also draw a l, to a bigger scale and a different
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00:35:53,180 --> 00:35:59,160
orientation. The question is that, if we given
all the link lengths and the angle of these
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two similar triangles, then can we find out
what is the scale of the reproduction, that
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00:36:04,730 --> 00:36:10,120
is what is the change in the ratio of the
size of these two l’s. And what is the angle
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00:36:10,120 --> 00:36:14,470
of these two l’s? what is the angle, that
is the change in the orientation of this letter
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00:36:14,470 --> 00:36:20,790
l to this drawn l?
As we see here the l is okay and here it has
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00:36:20,790 --> 00:36:30,050
been rotated by 90 degree almost in the clockwise
direction. The question is for this given
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00:36:30,050 --> 00:36:34,510
mechanism, that is the link lengths are given
and the angles are given, can we find out
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00:36:34,510 --> 00:36:42,990
that, what is the ratio of these two figures
and the orientation between these two figures?
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00:36:42,990 --> 00:36:46,210
That will be our second example.
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00:36:46,210 --> 00:36:52,140
This diagram shows the kinematic diagram of
the model of the mechanism which we have just
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now seen, that is the copying mechanism or
plagiograph. As we see in this mechanism,
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we have 5 links. One is the fixed link, O2A
is the second link, O2C is the third link,
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00:37:11,930 --> 00:37:23,060
BC is the fourth link and AB is the fifth
link. So, it is a five-link mechanism. So,
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00:37:23,060 --> 00:37:38,230
let us say n is equal to 5. Now let me count
the number of hinges j. We see there are three
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simple hinges at A, B and C. So j1 is 3. At
O2, three links are connected, namely: the
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00:37:52,850 --> 00:38:00,310
fixed link 1 and link number 2 and 3. Since
three links are connected at O2, this is a
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00:38:00,310 --> 00:38:10,070
second order hinge, that is of the type of
j2, so j2 is 1, So we get j equal to 4. So,
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00:38:10,070 --> 00:38:20,290
if we calculate the degree of freedom F which
we have seen, 3 times n minus 1 minus 2j,
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00:38:20,290 --> 00:38:26,990
sorry this is, I have made a mistake here,
j2 is 1, but j2 is 2j2, so this should be
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00:38:26,990 --> 00:38:30,630
counted as 2 into 1, that is 2, and j is 5.
So, if we calculate the degree of freedom
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F which we have seen, 3 times n minus 1 minus
2j. So F is 3 into 4 minus 2 into 5, which
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00:38:41,790 --> 00:38:50,130
is 2. Thus, we have a 2 degree of freedom
mechanism and this point D of this link, can
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00:38:50,130 --> 00:38:55,980
be taken anywhere arbitrarily on this plane,
because it has 2 degrees of freedom, it can
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00:38:55,980 --> 00:39:01,790
take any x value and any y value, this point
D can draw any figure.
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00:39:01,790 --> 00:39:16,400
Now let us consider, this O2ABC to be a parallelogram,
that is link length O2A is same as BC and
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00:39:16,400 --> 00:39:29,040
link length AB is same as O2C. Thus, O2ABC
is a parallelogram. Not only that, the triangle
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00:39:29,040 --> 00:39:38,510
ABD and the triangle BCE are similar, as indicated
that this angle alpha is same as this angle
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00:39:38,510 --> 00:39:45,580
alpha and this angle beta is same as this
angle beta. So, these two triangles namely
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00:39:45,580 --> 00:39:55,410
BCE and ABD are similar. Now, we can show
that the movement of the point D and the point
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00:39:55,410 --> 00:40:01,860
E will be very similar, that means the movement
of the point D will be copied by the movement
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00:40:01,860 --> 00:40:09,700
of the point E. Accordingly, E will also draw
a similar figure as D. The only change will
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00:40:09,700 --> 00:40:16,150
be the size and the orientation of the figure.
For example here, if the point D writes this
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00:40:16,150 --> 00:40:23,910
letter H, then E will write this letter H
but the height of this leg of the H is capital
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00:40:23,910 --> 00:40:31,210
L and here it is small l. Similarly, this
H is the normal way, that is vertical and
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00:40:31,210 --> 00:40:37,330
this H is almost horizontal that it has been
rotated through an angle sigma.
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00:40:37,330 --> 00:40:44,140
Our task is, given the link lengths, let us
find: what is the scale of the reproduction?
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That is what is the value of L by l and what
is this change in orientation? That is, what
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00:40:51,990 --> 00:41:00,230
is the angle sigma once we are given the link
completely, that is the link lengths and the
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00:41:00,230 --> 00:41:07,630
angles alpha and beta. To do this, we use
the link lengths as vectors. Suppose the position
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00:41:07,630 --> 00:41:14,670
of the point D, that is represented by this
vector X and the position of the point E,
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00:41:14,670 --> 00:41:25,010
that is represented by the vector Y. What
we can see? We can easily write vector X is
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00:41:25,010 --> 00:41:43,000
vector O2A plus vector AD. Similarly, vector
Y, we can write as vector O2E, which is vector
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00:41:43,000 --> 00:41:58,920
O2C plus CE. So you can write, O2C plus CE.
Now we see, that the vector O2C is same as
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00:41:58,920 --> 00:42:06,070
the vector AB, because AB and O2C are parallel
and are of equal length. So, you write O2C
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00:42:06,070 --> 00:42:17,560
is same as a vector AB. And the vector CE,
as I see is a vector which is rotated from
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00:42:17,560 --> 00:42:24,310
the vector CB through an angle alpha in the
counter clockwise sense and magnified from
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00:42:24,310 --> 00:42:38,590
CB to CE. So this I can write, CE by CB into
the vector CB, which takes care of the magnification
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00:42:38,590 --> 00:42:44,310
by a factor CE by CB and to represent the
change in orientation between the vector CB
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00:42:44,310 --> 00:42:51,950
and CE through an angle alpha, I have to multiply
it by e to the power of i alpha.
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00:42:51,950 --> 00:43:03,619
Now, vector AB, I can write in terms of vector
AD exactly the same way, the magnitude has
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00:43:03,619 --> 00:43:14,760
been magnified by AB by AD and the vector
AD and the vector AB is at an angle alpha
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00:43:14,760 --> 00:43:20,050
in the counter clockwise direction from the
vector AD, so, I multiply this by e to the
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00:43:20,050 --> 00:43:31,360
power i alpha plus this term. So, I have the
vector Y as, AB by AD into vector AD e to
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00:43:31,360 --> 00:43:38,450
the power i alpha plus CE by CB into vector
CB e to the power i alpha. Now because these
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00:43:38,450 --> 00:43:54,100
two triangles ABD and BCE are similar triangles,
it is easy to see that ratio AB by AD is same
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00:43:54,100 --> 00:44:12,820
as, CE by BC. So, CE by CB is same as AB by
AD, due to the similarity of the triangles
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00:44:12,820 --> 00:44:21,110
ABD and BCE. So I can take this ratio AB by
AD and this factor e to the power i alpha
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00:44:21,110 --> 00:44:23,090
common.
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00:44:23,090 --> 00:44:37,670
Consequently, what we get is that, vector
Y is equal to AB by AD into e to the power
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00:44:37,670 --> 00:44:54,970
i alpha, which we take common and we get AD
plus CB which is same as O2A. And AD plus
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00:44:54,970 --> 00:45:06,000
O2A is nothing but our vector X. So, AB by
AD into e to the power of i alpha into the
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00:45:06,000 --> 00:45:13,950
vector X. So you see that the position of
these two vectors, that is the point E and
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00:45:13,950 --> 00:45:19,730
the point D is given by point Y and X and
they are related through this equation. Y
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00:45:19,730 --> 00:45:25,850
is changing with time, X is changing with
time but the link lengths AB, AD and the angle
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00:45:25,850 --> 00:45:34,240
alpha of that triangle are not changing with
time. So, this factor is a constant, independent
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00:45:34,240 --> 00:45:49,360
of time. So, movement of the point dY, I can
write AB by AD into e to the power i alpha
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00:45:49,360 --> 00:45:58,181
into dX. Where dX represents the movement
of the vector X and dY represents the movement
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00:45:58,181 --> 00:46:04,730
of the vector Y.
So you see, the movements of X and Y are very
300
00:46:04,730 --> 00:46:17,930
similar, but for this factor. That means the
movement is magnified by a factor AB by AD,
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00:46:17,930 --> 00:46:25,110
which is the scale of reproduction, which
I wrote earlier L by l and there is a rotation
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00:46:25,110 --> 00:46:30,810
through an angle alpha in the counter clockwise
sense so sigma that is the change in orientation
303
00:46:30,810 --> 00:46:33,460
of the reproduction is equal to alpha.
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00:46:33,460 --> 00:46:47,650
To make it clear, let me go back to the previous
slide, AB by AD that is the factor of reproduction,
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00:46:47,650 --> 00:46:54,730
that is this length by this length, that is
this letter H is reproduced to a bigger scale
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00:46:54,730 --> 00:47:01,530
namely by the factor: AB by AD. Not only that,
this vertical side, which is vertical like
307
00:47:01,530 --> 00:47:07,230
this has been oriented through an angle sigma
and we have already shown the sigma is nothing
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00:47:07,230 --> 00:47:14,100
but the angle alpha. So, this is the copying
mechanism that any arbitrary figure, drawn
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00:47:14,100 --> 00:47:21,430
by the point B or followed by the point D,
can be reproduced at the point E at a different
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00:47:21,430 --> 00:47:27,230
size and different orientation.
Let me now summarize what we have learned
311
00:47:27,230 --> 00:47:33,910
today. We have shown, how we carry out the
displacement analysis of planar linkages by
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00:47:33,910 --> 00:47:40,310
analytical method. In the analytical method,
we represent all the kinematic dimensions
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00:47:40,310 --> 00:47:50,220
like link lengths, off-sets or slider displacements
by planar vectors. Using the closed loop that
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00:47:50,220 --> 00:47:56,250
exists in the mechanism, we generate a loop
closure equation and then by solving this
315
00:47:56,250 --> 00:48:04,350
loop closure equation, we can finish the kinematic
or displacement analysis of planar linkages.
316
00:48:04,350 --> 00:48:10,830
We have given the example of a simple 4R linkage
and also a six-link mechanism, where you have
317
00:48:10,830 --> 00:48:18,190
more than one closed loop. Finally, we have
shown an example that shows the power of this
318
00:48:18,190 --> 00:48:46,609
vector representation for link lengths to
analyze the copying mechanism or the plagiograph.
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00:48:46,609 --> 00:48:58,359
12