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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, in 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 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 are
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followed in this analytical method. As we
know, all 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 slider
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displacement by planar vectors. Then the third
step is to express for every close loop, what
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we call as a loop closure equation in terms
of these vectors that we have just now said.
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Now, each such vector in 2D, that is a planar
linkage is equivalent to two scalar equations
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that means, if a vector equation is there,
that is equivalent to two scalar equations
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and two unknown quantities can be solved.
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Now, once these entire vector equations are
generated using all the independent closed
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loops, one has to solve these equations to
determine the unknown quantities that is relevant
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to 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 such 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 the
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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 subscript
2. Our objective is to determine the orientation
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of the coupler and the fourth link, that is
O4B. To do this, first we set up a Cartesian
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coordinate system XY, with O2 as the origin.
The first task as I said, we have identified
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a closed loop namely - O2ABO4O2. Then the
link lengths l1, l2, l3, l4 are represented
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by four vectors namely - l1, l2, l3 and l4.
These vectors are shown by these arrows in
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this diagram. Now, the angle that these vectors
make with the X-axis, we are using as the
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reference angle to determine the orientation
of these vectors, that is theta subscript
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2, theta subscript 3, and theta subscript
4. Theta subscript 1, that is the angle made
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by O4O2, that is the vector l1 with X-axis
is obviously 0. We can express this loop closure
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equation very simply as, (these are all vectors)
l1 plus l2 plus l3 minus l4 is equal to 0.
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This is what we say as the loop closure equation,
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 l2e to the power i theta subscript
2. Same, for l3 vector as l3e to the power
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i theta subscript 3 and l4 vector as l4e to
the power i theta subscript 4. Since, theta
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subscript 1 is 0, the vector l1 is simply,
link length l1e to the power i0 which is 1.
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So, the loop closure equation, let us substitute
these complex exponential notation, so you
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get, l1 plus l2e to the power i theta subscript
2 plus l3e to the power i theta subscript
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3 minus l4e to the power i theta subscript
4 is equal to 0. We know, e to the power i
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theta is cos theta plus i sin theta. So, first
we substitute for all these angles, e to the
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power i theta subscript 2, i theta subscript
3 and so on in terms of cosine and sin and
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equate the real and the imaginary parts of
this complex equation separately to 0. This
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is nothing but really adding the X components
to 0 and adding the Y components to 0, because
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it is a closed loop. As a result we get, l1
plus l2 cosine theta subscript 2 plus l3 cosine
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theta subscript 3 minus l4 cosine theta subscript
4 is 0. Adding the imaginary parts, that is
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the sin components we get l2 sin theta subscript
2 plus l3 sin theta subscript 3 minus l4 sin
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theta subscript 4 is equal to 0. Thus, in
the vector equation, because these vectors
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are two dimensional vectors from one vector
equation, we get two scalar equations such
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as equation one and equation two. Each vector
equation is equivalent to two scalar equations
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if the vectors are two dimensional.
From these two equations, we can eliminate
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one of the unknowns. Say, either theta subscript
4 to solve theta subscript 3 or we can eliminate
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theta subscript 3 to solve for the unknown
theta subscript 4, if l1, l2, l3, and l4,
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that is, the link lengths are given and the
input angle theta subscript 2 is specified.
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Now, let me carry out this algebra to determine
say, theta subscript 4. To determine theta
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subscript 4, we write l3 sin theta subscript
3 is equal to l4 sin theta subscript 4 minus
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l2 sin theta subscript 2. From the first equation,
we write l3 cosine theta subscript 3 equal
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to l4 cosine theta subscript 4 minus l1 plus
l2 cosine theta subscript 2. From these two
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equations, if we square and add up, obviously
from the left hand side I get only l3 square.
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So, we get l3 square is equal to l4 square
plus l1 square plus l2 square minus twice
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l2l4 sin theta subscript 4 sin theta subscript
2 minus twice l4 cosine theta subscript 4
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into l1 plus l2 cosine theta subscript 2 minus
twice l1l2 cosine theta subscript 2. Let me
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carry out this algebra a little further and
now onwards I will write capital C for cos
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and capital S for sin for gravity.
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We can get l3 square is equal to l1 square
plus l2 square plus l4 square plus twice l1l2Cs
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theta subscript 2 minus twice l2l4Stheta subscript
4Stheta subscript 2 minus
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twice l1 plus l2Ctheta subscript 2 into C
theta subscript 4 l4. Now divide all the terms
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of this equation by twice l2 l4. As a result,
we will get a simple equation a sin theta
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subscript 4 plus b cosine theta subscript
4 is equal to c, where a is given by sin theta
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subscript 2, that is coming from this term.
b will get l1 divided by l2 plus cosine theta
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subscript 2. c is equal to l1 square plus
l2 square minus l2 square minus l3 square
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plus l4 square divided by twice l2l4 plus
l1 by l4 cosine theta. So you see, if theta
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subscript 2 and all the link lengths are given,
these quantities namely: a, b and c are completely
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known and the only unknown is this theta subscript
4 which has to be solved. But as we see, this
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is not a linear equation in theta subscript
4, because it involves sin and cosine. To
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solve this angle theta subscript 4 unambiguously,
that is we need not judge the quadrant after
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we get the value, in what quadrant the angle
theta subscript 4 lies, it is always better
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to replace sin and cosine by the tan of the
corresponding half angles.
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That means, we write sin theta subscript 4
as twice tan of theta subscript 4 by 2 whole
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divided by 1 plus tan square theta subscript
4 by 2 and cosine theta subscript 4 we write
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1 minus tan square theta subscript 4 by 2
whole divided by 1 plus tan square theta subscript
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4 by 2. Substituting this, in that a sin theta
subscript 4 plus b cos theta subscript 4 equal
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to c equation, we get a quadratic equation
in tan theta subscript 4 by 2 as follows:
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b plus c tan square theta subscript 4 by 2
minus 2 a tan theta subscript 4 by 2 plus
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c minus b equal to 0. So, finally we get this
quadratic equation in tan theta subscript
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4 by 2, where a, b and c are all known quantities
in terms of the link lengths and the given
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input angle theta subscript 2. Now, this is
where tan theta subscript 4 by 2 is useful,
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I get a quadratic equation and I get two roots
namely: tan theta subscript 4 by 2 is equal
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to a plus minus square root of a square plus
b square minus c square whole divided by b
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plus c.
So, you get two roots for tan theta subscript
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4 by 2 and the right hand side may come out
to be positive or negative, of course, it
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can never be imaginary once the kinematic
diagram is okay, that is a 4R linkage has
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been made, a close loop has been obtained,
a, b, c will be of such values that this square
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root of can never be of a negative quantity.
So you may get two positive, two negative
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or may be one positive and one negative.
Now, tan theta subscript 4 by 2, as we know,
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if theta subscript 4 is between 0 and 180
degree and then theta subscript 4 by 2 will
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be between 0 to 90 degree and tan will be
positive. If theta subscript 4 is more than
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180 degree then theta subscript 4 by 2 will
be more than 90 degree and tan theta subscript
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4 will be negative. So, depending on the positive
and negative values of tan theta subscript
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4 by 2, I can easily determine, uniquely the
value of theta subscript 4. Suppose we are
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also interested in finding the orientation
of the coupler, that is the angle theta subscript
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3.
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If we look at these two equations, we see
l3 and l4 are appearing exactly the same way,
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only l3 is appearing on the left hand side
and l4 is appearing on the right hand side.
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It is even clearer, if we look at these equations
number one and two. You can see l3 and l4
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are appearing exactly similar way, only thing
plus and minus. So, when we have eliminated
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theta subscript 3 and obtain theta subscript
4, following exactly the same procedure, I
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can eliminate theta subscript 4 and obtain
theta subscript 3. We have to remember that
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l4 has to be replaced by minus l3, rest of
the solutions remains as it is.
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So following exactly a similar method to eliminate
theta subscript 4, we can get a theta subscript
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3, a very similar equation like tan theta
subscript 3 by 2 will be equal to: a plus
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minus square root of a square plus b square
minus c prime square whole divided by b plus
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c prime. As we might have remembered, the
quantities, a and b did not involve l4 or
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l3. a was sin theta subscript 2, b was cosine
theta subscript 2 plus l1 by l2. However,
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the quantity c involved l3 and l4. That is
why, we have to replace c by c prime and expression
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of c prime will also be very similar to c
and we have to interchange l4 by minus l3
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and vice versa.
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We will get c prime as l1 square plus l2 square
minus l3 square plus l4 square divided by
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twice times l2l3 plus l1 by l4 cosine theta
subscript 2. The thing to note is, for the
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same set of given length and the same value
of theta subscript 2, we have obtained two
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values of theta subscript 4 and two values
of theta subscript 3. This is nothing surprising,
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if we remember that even by graphical method,
we would have got two solutions for the given
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link lengths and the input angle. This aspect
is very clear if we solve the same problem
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by graphical methods.
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Suppose, we are given l1 - the length O4O2,
the link length l2 that is O2A, and the angle
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theta subscript 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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subscript 2, one is O4, O2, A, B1 and the
other is O4, O2, A, B2. Two configurations
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are shown distinctly, one by the formed lines
and the other by the dashed lines.
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Now, we have got two values of theta subscript
3 and theta subscript 4, which are indicated
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in this diagram as well. For the formed line,
that is, when the diagram is above the line
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O4 O2, this is, theta subscript 3 one and
this is theta subscript 3 two. Corresponding
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to theta subscript 3 one, I have theta subscript
4 one and corresponding to theta subscript
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3 two, I have theta subscript 4 two. Now,
we have seen how to obtain various values
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of theta subscript 4 as theta subscript 2
changes. Now, we will show you the typical
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variations of theta subscript 4 versus theta
subscript 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 Grashof
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linkage and a crank-rocker and theta subscript
2 goes from 0 to 2 pi. Theta subscript 4 is
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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 Grashof crank-rocker
has two distinct branches, two distinct modes
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of assembly, that is why these two curves
never cross each other. If we are starting
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on one of the curves you will always remain
there, either on this curve or on this curve.
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You can never go from one curve to the other
because each curve corresponds to a particular
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mode of assembly.
Suppose we have a Grashof double-crank,
this is the typical theta subscript 4 verses
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theta subscript 2 characteristics of a double
crank. As you see, theta subscript 2 goes
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from 0 to 2 pi and theta subscript 4 also
goes from 0 to 2 pi, because both the input
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and the output lengths are capable of making
complete rotations. Because it is a Grashof
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linkage, there are two distinct modes of assembly
as represented by this formed line and the
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other mode of assembly is by this curved line
and these two curves can never intersect,
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that is each curve corresponds to a particular
mode of assembly.
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If we have a Grashof double-rocker, then again
we have two loops, one as we see here and
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the other there. Because this is double-rocker
neither theta subscript 2 nor theta subscript
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4 can make complete rotation. So, they are
limited between two extreme values for example,
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for theta subscript 2 from here to there and
for theta subscript 4 it is here to there.
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Again, there are two distinct loops for two
different modes of assembly either you are
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following this curve or you are following
this curve, you can never pass on from one
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mode of assembly to the other.
Now, let us see what happens in a non-Grashof
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linkage, if it is a non-Grashof linkage then
it has to be a double-rocker. Theta subscript
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4 versus theta subscript 2 plot for a non-Grashof
double-rocker looks like this, here again
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neither theta subscript 2 nor theta subscript
4 can make complete rotation. Theta subscript
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2 is restricted within this region. Theta
subscript 4 is restricted within this region
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but there is only one curve, because there
is only one mode of assembly. 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, such as 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. 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, link length
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vectors are constants, they are not changing
with time. Whereas there are two other vectors,
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namely S4, denoting the distance O4A is changing
with time. Same is true for S6 which represents
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the movement of the cutting tool, here it
is DC represented by the vector S6. All these
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vectors, the orientation of these vectors,
I represent by angle theta, measured from
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this horizontal line in the anticlockwise
direction. So, we can have one close loop
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O2AO4O2, which means l2 is equal to l1 plus
s4. This is one loop closure equation, in
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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 say it, from
these two vector equations, I get four scalar
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equations. So, I can solve for four unloads.
Now, let us see in this problem, what are
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the unloads? What is I prescribe us are, all
the link length vectors, that is l1, l2, l4,
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l5 and l6 and the input angle theta subscript
2, that is the angle that this link 2 makes
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with the reference this line, that is the
X-axis. So, these are given and we have to
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find out the orientation of all other links
and also the slider displacement that is s4
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and s6.
So, the number of unknowns is theta subscript
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4, that is the orientation of link 4, that
is the line O4B, theta subscript 5, that is
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the orientation of the fifth link, that is
BC, this angle theta subscript 5. This angle,
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as I said earlier is theta subscript 4 and
also this distance O4A, that is s4 and the
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distance BC that is s6. So, these are the
four scalars unknowns which can be solved
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using these two vector equations. That is
how we carry out the kinematic analysis to
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analytical method. Later on we will solve
some more different types of problem and show
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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.
There are four moving links: 2, 3, 4 and 5.
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The dimensions of these links are such that
this forms a parallelogram. This link length
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is equal to this, this link length is equal
to this. These two triangles are similar triangles,
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that is this angle is same as this angle and
this angle is also same as this angle. Now
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in this mechanism, we will show later that
whatever is the movement of this point, that
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is reproduced by this point, only thing there
will be a change in the scale and the orientation.
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For example, if I move this mechanism such
that this point follows this vertical line
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of this letter l then this point is also following
a straight line. As we go around this letter
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l, this point will also draw a l, to a bigger
scale and a different orientation. The question
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is that, if are given all the link lengths
and the angle of this two similar triangles,
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then can we find out what is the scale of
the reproduction, that is what is the change
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in the ratio of the size of these two l. What
is the angle of these two l? what is the angle,
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that is the change in the orientation of this
letter l to this drawn l?
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As we see here the l is okay and here it has
been rotated by 90 degree almost in the clockwise
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00:36:24,410 --> 00:36:31,410
direction. The question is for this given
mechanism, that is the link lengths are given
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and the angles are given, can we find out
what is the ratio of these two figures and
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the orientation between these two figures?
That will be our second example.
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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
tigeograph. As we see in this mechanism, we
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have 5 links. One is the fixed link, O2A is
the second link, O2C is the third link, BC
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is the fourth link and AB is the fifth link.
So, it is a five-link mechanism. So, let us
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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 j is equal
to 3 At O2, three links are connected, namely:
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00:37:52,670 --> 00:37:59,670
the fixed link 1, link 2 and 3. Since three
links are connected at O2, this is a second
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order hinge, that is of the type of j2, so
j2 is 1, but j2 is 2j2, so this should be
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counted as 2 into 1, that is 2, and j is equal
to 5. So, if we calculate the degree of freedom
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F which we have seen, 3 times n minus 1 minus
2 j. So F is 3 into 4 minus 2 into 5, which
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is 2. Thus, we have a 2 degree freedom mechanism
and this point D of this link, can be taken
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anywhere arbitrarily on this plane, because
it has 2 degrees of freedom, it can take any
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x value and any y value, this point D can
draw any figure.
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Now let us consider, this O2ABC to be a parallelogram,
that is link length O2A is same as BC and
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link length AB is same as O2C. Thus, O2ABC
is a parallelogram. Not only that, the triangle
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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 --> 00:39:45,510
alpha and this angle beta is same as this
angle beta. So, these two triangles namely
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BCE and ABD are similar. Now, we can show
that the movement of the point D and the point
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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 --> 00:40:08,860
of the point E. Accordingly, E will also draw
a similar figure as D. The only change will
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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 --> 00:40:23,150
letter H, then E will write this letter H
but the height of this league of the H is
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00:40:23,400 --> 00:40:30,400
capital L and here it is small l. Similarly,
this H is the normal l, that is vertical and
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this H is almost horizontal that it has been
rotated through an angle sigma.
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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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is the angle sigma once we are given the link
completely, that is the link lengths and the
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00:41:00,240 --> 00:41:07,240
angles alpha and beta. To do this, we use
the link lengths as vectors. Suppose the position
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of the point D, that is represented by this
vector x and the position of the point E,
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that is represented by the vector Y. What
we can see? We can easily write vector X is
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vector O2A plus vector AD. Similarly, vector
Y, we can write as vector O2E, which is vector
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O2C plus CE. So you can write, O2C plus CE.
Now we see, that the vector O2C is same as
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the vector AB, because AB and O2C are parallel
and are of equal length. So, you write O2C
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00:42:06,770 --> 00:42:13,770
is same as a vector AB. The vector CE, as
I see is a vector which is rotated from the
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00:42:17,620 --> 00:42:24,620
vector CB through an angle alpha in the counter
clockwise sense and magnified from CB to CE.
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So this I can write, CE by CB into the vector
CB, which takes care of the magnification
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00:42:38,600 --> 00:42:44,530
by a factor CE by CB and to represent the
change in orientation between the vector CB
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00:42:44,530 --> 00:42:51,530
and CE through an angle alpha, I have to multiply
it by e to the power of i alpha.
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00:42:51,910 --> 00:42:58,910
Now, vector AB, I can write in terms of vector
AD exactly the same way, the magnitude has
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been magnified by AB by AD and the vector
AD and the vector AB is at an angle alpha
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00:43:14,770 --> 00:43:20,070
in the counter clockwise direction from the
vector AD. So, I multiply this by e to the
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00:43:20,070 --> 00:43:27,070
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 --> 00:43:38,360
the power i alpha plus CE by CB into vector
CB e to the power i alpha. Because these two
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00:43:38,510 --> 00:43:45,510
triangles ABD and BCE are similar triangles,
it is easy to see that ratio AB by AD is same
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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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ABD and BCE. I can take this ratio AB by AD
and this factor e to the power i alpha common.
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00:44:23,100 --> 00:44:30,100
Consequently, what we get is that, vector
Y is equal to AB by AD into e to the power
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00:44:37,690 --> 00:44:44,690
i alpha, which we take common and we get AD
plus CB which is same as O2A and AD plus O2A
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00:44:56,350 --> 00:45:03,350
is nothing but our vector X. So, AB by AD
into e to the power of i alpha into the vector
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00:45:07,370 --> 00:45:14,340
X. So you see that the position of these two
vectors, that is the point E and the point
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00:45:14,340 --> 00:45:21,340
D is given by point Y and X and they are related
to this equation. Y is changing with time,
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00:45:21,570 --> 00:45:27,210
X is changing with time but the link lengths
AB, AD and the angle alpha of that triangle
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00:45:27,210 --> 00:45:34,210
are not changing with time. So, this factor
is a constant, independent of time. So, movement
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00:45:36,760 --> 00:45:43,760
of the point dY, I can write AB by AD into
e to the power i alpha into dX. Where dX represents
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00:45:53,500 --> 00:46:00,500
the movement of the vector X and dY represents
the movement of the vector Y.
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00:46:00,690 --> 00:46:07,690
So you see, the movements of X and Y are very
similar, but for this factor, that means the
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00:46:11,410 --> 00:46:18,410
movement is magnified by a factor AB by AD,
which is the scale of reproduction, which
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00:46:20,030 --> 00:46:27,030
I wrote earlier L by l. There is a rotation
through an angle alpha in the counter clockwise
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00:46:27,440 --> 00:46:33,470
sense so sigma that is the change in orientation
of the reproduction is equal to alpha.
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00:46:33,470 --> 00:46:40,470
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 --> 00:46:54,650
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 --> 00:47:01,540
namely by the factor: AB by AD. Not only that,
this vertical side, which is vertical like
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00:47:01,540 --> 00:47:07,240
this has been oriented through an angle sigma
and we have already shown the sigma is nothing
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00:47:07,240 --> 00:47:14,110
but the angle alpha. So, this is the copying
mechanism that any arbitrary figure, drawn
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00:47:14,110 --> 00:47:21,110
by the point B or followed by the point D,
can be reproduced at the point E at a different
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00:47:21,450 --> 00:47:28,020
size and different orientation. Let me now
summarize what we have learned today.
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00:47:28,020 --> 00:47:34,520
We have shown, how we carry out the displacement
analysis of planar linkages by analytical
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00:47:34,520 --> 00:47:41,260
method. In the analytical method, we represent
all the kinematic dimensions such as, link
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00:47:41,260 --> 00:47:48,260
lengths, off-sets or slider displacements
by planar vectors. Using the closed loop that
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00:47:50,230 --> 00:47:56,260
exists in the mechanism, we generate a loop
closure equation and then by solving this
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00:47:56,260 --> 00:48:03,260
loop closure equation, we can finish the kinematic
or displacement analysis of planar linkages.
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00:48:04,350 --> 00:48:10,820
We have given the example of a simple 4R linkage
and also a six-link mechanism, where you have
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00:48:10,820 --> 00:48:17,820
more than one closed loop. Finally we have
shown an example that shows the power of this
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00:48:18,190 --> 00:48:25,190
vector representation for link lengths to
analyze the copying mechanism or plagiograph.
323