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Brij Bhooshan Asst. Professor B.S.A College of Engg. & Technology, Mathura (India)

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In This Chapter We Cover the Following Topics

Art. Content Page

2.1 Displacement Analysis 3

2.2 Algebraic Position Analysis 7

2.3 Concept of Transmission Angle 9

2.4 General Plane Motion

Motion of a Link

Rubbing Velocity

Plane Motion of a Particle

Plane Motion of a Rigid Body

Motion Difference between Two Instantaneously Coincident Points

Brief solution of Coriolis component of acceleration

11

11

12

12

13

14

15

2.5 Instantaneous Centre of Velocity 17

2.6 Ardnhold-Kennedy Theorem of Three Centres

Types of Instantaneous Centres

Angular Velocity Ratio Theorem

19

19

20

2.7 Velocity and Acceleration Images 20

2.8 Velocity and Acceleration Analysis (Graphical)

Slider-Crank Mechanism

Four-bar Chain Mechanism

Four-bar Chain Mechanism (with off-set)

Shaper (Crank and Slotted Lever) Mechanism

22

22

23

23

26

2.9 Goodman's Indirect Method 27

2.10 Velocity and Acceleration Analysis (Analytical) 29

2.11 Klein's Construction

Single Slider Crank-Mechanism

Four Bar Mechanism

30

30

31




2 Chapter 2: Kinematic Analysis of Plane Mechanism

References:

1. Bevan, T., The Theory of Machines, CBS Publishers and Distributors, 1984.

2. Shigley, J.E., Uicker (Jr.), J.J. and Pennock, G.R. Theory of Machine and Mechanism,

Oxford University Press, New York, 2003.

3. Mallik, A. K., Ghosh, A., Theory of Mechanism and Machines, Affiliated East-West

Press (P) Ltd., New Delhi, 2004.

4. Martin, G.H., Kinematics and Dynamics of Machines, MaGraw-Hill, New York, 1982.

5. Rao, J.S., Dukkipati, R.V., Mechanism and Machine Theory, New Age International

Publishers, New Delhi, 2006.

Please welcome for any correction or misprint in the entire manuscript and your

valuable suggestions kindly mail us [email protected].




3 Theory of Mechanism and Machines By Brij Bhooshan

The objective of kinematic analysis is to determine the kinematic quantities such as

displacements, velocities, and accelerations of the elements in a mechanism when the

input motion is given. Conversely, the objective may be to determine the input motion

required to produce a specified motion of another element. In short, kinematic analysis

establishes the relationship between the motion of the various components or links of a

mechanism. In this chapter, we shall consider only the mechanisms with lower pairs.

The kinematic analysis of a higher-pair mechanism can also be carried out. This may be

done by converting the mechanism into an equivalent mechanism consisting of only the

lower pairs. Both the graphical and analytical methods can be used for kinematic

analysis. However, the graphical methods, providing better insight and visualization,

still occupy a prominent place in planar kinematics. Unlike in an analytical method, the

accuracy of the solution provided by a graphical method is often limited.

2.1 DISPLACEMENT ANALYSIS

When the kinematic dimensions and/the configuration(s) of the input link(s) of a

mechanism are prescribed, the configurations (linear and angular) of all the other links

are determined by displacement analysis.

Graphical Method

In a graphical method of displacement analysis, the mechanism is drawn to a

convenient scale and the desired unknown quantities are determined through suitable

geometrical constructions and calculations. No generalized approach can be discussed so

far as the graphical methods are concerned; the solution technique will vary from

problem to problem. We shall demonstrate the basic features of the methods through a

few examples. Some of these features which play a major role in displacement analysis

are as follows:

(i) The configuration of a rigid body in plane motion is completely defined by the

locations of any two points on it.

(ii) Two intersecting circles have two points of intersection and one has to be

careful, when necessary, to choose the correct point for the purpose in hand.

(iii) The use of a tracing paper, as an overlay, is very convenient and very often

provides an unabiguous and quick solution.

(iv) The graphical method of displacement analysis fails if no closed loop with four

links exists in the mechanism.

Application 2.1: As shown in Diagram 2.1(a) a six-link, Whitworth quick-return

mechanism used in slotting machines. Link 2 (= O2A) rotates with a constant angular

speed in the counter-clockwise (CCW) direction. The sliding link 6 represents the cutting

tool whose cutting motion is to the left and the return (idle) motion is to the right.

Determine the quick-return ratio which is defined as the ratio of the time intervals

taken by the tool to complete its cutting and idle motions, respectively.

Solution: Referring to Diagram 2.1b, the circles kA and kC are the paths of the points A

and C, respectively. The centres of these circles are at O2 and O4 and their radii are O2A

and O4C, respectively. The tool occupies its extreme right and extreme left positions (DR

and DL) when links 4 and 5 become collinear as shown in the diagram, i.e., OuCRDR and

OuCLDL. The distance DRDL is the stroke of the tool (= 2O4C). Since the points A, O4, and





C always remain collinear (on link 4), the locations of A (on kA) corresponding to the

extreme positions are now easily obtained as AR and AL, respectively. Thus, the rotations

of link 2 corresponding to the forward and return motions of the tool are obtained as θf

and θr, respectively.

From measurement, θf = 239°. Since link 2 rotates at a constant angular speed, the

quick-return ratio (qrr) is obtained as

Analytical Method

Let us consider a 4R linkage (Diagram 2.2) of given link lengths, viz., li, i = 1, 2, 3, and 4.

The configuration of the input link (2) is also prescribed by the angle θ2, and we have to

obtain the configurations of the other two links, namely, the coupler and the follower,

expressed by the angles θ3 and θ4. From Diagram 2.2, all links are denoted as vectors,

viz., l1, l2, l3, and l4. All angles are measured CCW from the x-axis which is along the

fixed vector l1 rendering θ1 = 0.

Diagram 2.2 Four bar mechanism

Considering the closed loop O2O4BAO2, we can write

Using complex exponential notation with θ1 = 0, then above equation is (2.1) can be

written as

Equating the real and imaginary parts of this equation separately to zero, we get

Thus, the two unknowns, namely, θ3 and θ4, can be solved from the two equations (2.2a)

and (2.2b) as now explained.

Rearranging (2.2a) and (2.2b), we get

B A

(a)

6 5

4

3 2

C

D

A (b)

Diagram 2.1





Squaring both sides of these two equations and adding, we obtain

or

where

It may be noted that with the prescribed data (i.e., link lengths and θ2), the coefficients

a, b, and c of (2.3) are known.

To solve for θ4, from (2.3),

Substitute these values in Eq. (2.3), then we get

which gives

Thus, for a given position of the input link, two different values of θ4 are obtained as

follows:

These two values correspond to the two different ways in which the 4R mechanism can

be formed for any given value of θ2, as explained in Diagram 2.3 where the same

problem has been solved by a graphical method.

Diagram 2.3 Four bar mechanism

To solve for the coupler orientation θ3, we can eliminate θ4 from (2.2a) and (2.2b), we get

where

It is quite obvious that we can get c' from the expression for c by interchanging l3 and l4

but attention may be drawn to the roots of θ3 and θ4 which pertain to the same

configuration.

B A





Since the 4R linkages are very useful in practice, it is instructive to go into further

details of the output-input (i.e., plots of θ4 vs. θ2) relationship of these linkages. The

typical nature of this relationship for various types of 4R linkages is indicated in

Diagram 2.4a-d. For crank-rocker and double-crank mechanisms, the output-input

characteristics are as shown in Diagram 2.4a and 2.4b, respectively. The same

characteristics for double-rocker mechanisms are like those shown in Diagram 2.4c or

Diagram 2.4d, depending on whether the linkage is Grashof type or non-Grashof type.

Diagram 2.4

The following distinctive features of various plots shown in Diagram 2.4 should be

noted:

(i) For all Grashof-type linkages, there are two disconnected branches, each

corresponding to one mode of assembly. The assembly modes are mirror image of

each other, with the mirror placed along the fixed link. The linkage shown in

Diagram 2.3 is a crank rocker, therefore the dashed configuration can never be

obtained from the configuration O2ABO4. The dashed configuration will be

obtained if initially the mechanism is assembled as the mirror image of O2ABO4

and link 2 then driven so as to coincide with O2A.

(ii) For a non-Grashof linkage, the plot of θ4 vs. θ2 is a single closed loop, implying

that, once assembled, it can take up the mirror image configuration without being

dismantled.

(iii) Except at the ends of the swing, all rocking links pass through the same position

twice the different orientations of the other link connected to the frame.

Of all the possible variations of a 4R mechanism, crank rocker is most commonly used in

practice. In general, for a uniform angular speed of the crank, the rocker takes a

different time interval during its forward and return motions. It is useful (at this stage)

to note the relationship between the link lengths that ensures equal time for the

forward and return strokes of the rocker.

Diagram 2.5

If the rotation of the crank has to be exactly equal to corresponding to both forward

and return strokes of the rocker, then the link lengths should be such that the outer and

inner dead-centre configurations are like those shown in Diagram 2.5, where the points

A1, A2, B1, B2, and O2 are all collinear. From the triangles O2O4B1 and O2O4B2, we can

write

(a)

(b)

(c)

(d)





These two equations, after simplification, result in

where, l1, l2, l3, and l4 are the lengths of frame, crank, coupler, and follower, respectively.

Thus, (2.6) constitutes the condition required for equal time interval during the forward

and return motions of the follower (link 4) when the crank (link 2) rotates with a

constant angular speed. Furthermore, we should note, from Diagram. 2.5, that for such

a 4R mechanism without quick return,

or

where is the swing angle of the rocker. Now onwards, such a 4R linkage will be

designated as one without the quick-return effect.

2.2 ALGEBRAIC POSITION ANALYSIS

In this section we present the classical approach used in the position analysis of the

slider-crank mechanism. Diagram 2.6 shows the offset version that has been chosen for

analysis. By making the offset equal to zero, the same equations can be used for the

centered or symmetrical version. The two problems that occur in the position analysis of

the slider-crank mechanism are:

Problem 1: Given the input angle θ2, find the connecting rod angle θ3 and the position xB.

Problem 2: Given the position xB, find the input angle θ2 and the connecting rod angle θ3.

Diagram 2.6

Now, starting with problem 1, we define the position of point A with the equation set

Next, we note that

From the geometry of Diagram 2.6, we see that

Then using the trigonometric identity we have,

from Eq. (2.9),

We arbitrarily select the positive sign, which we see from Diagram 2.6 corresponds

solution with the piston to the right of the crank pin.

4

2 A

B





Now using Eq. (2.11) and Eq. (2.10), we get

Thus, with the angle θ2 given, the unknowns θ3 and xB can be obtained by solving Eqs.

(2.10) and (2.12).

Problem 2 requires that, given xB, we solve Eq. (2.12) for the angle θ2. So, now we apply

well-known Newton-Raphson method. This method can be explained by reference to

Diagram 2.7. This figure is a graph of some function f(x) versus x. Let xn be a first

approximation or a rough estimate of the root where f(x) = 0 which we wish to find. A

tangent line drawn to the curve at x = xn intersects the x axis at xn + 1, which is a better

approximation to the root. The slope of the tangent line is the derivative of the function

at x = xn and is

Diagram 2.7

Solving for xn + 1,

Now solving the single slider mechanism with the help of Newton-Raphson method.

Suppose in Eq. (2.22), replace the angle θ2 with and let r2, r3, e, and xB be given

constants. Then

and

Now squaring and adding the above equations, then we get

Solving for θ2 gives

The solution of problem 1 for the centered version is, of course, obtained directly from

Eq. (1.12) by making e = 0.

The Crank-and-Rocker Mechanism

The four-bar linkage shown in Diagram 2.8 is called the crank-and-rocker mechanism.

Thus link 2, which is the crank, can rotate in a full circle; but the rocker, link 4, can only

oscillate. We shall generally follow the accepted practice of designating the frame or

fixed link as link 1. Link 3 in Diagram 2.8 is called the coupler or connecting rod. With

the four-bar linkage the position problem generally consists of finding the positions of

the coupler and output link or rocker when the dimensions of all the members are given

together with the crank position.

Tangent line





Diagram 2.8

To obtain the analytical solution we designate s as the distance AO4 in Diagram 2.8. The

cosine law can then be written twice for each of the two triangles O4O2A and ABO4. In

terms of the angles and link lengths shown in the figure we then have

There will generally be two values of λ corresponding to each value of θ2. If θ2 is in the

range 0 ≤ θ2 ≤ , the unknown directions are taken as

θ3 = φ β; θ4 = λ β

However, if θ2 is in the range ≤ θ2 ≤ 2, then

θ3 = φ + β; θ4 = λ + β

Finally, the transmission angle is given by the

2.3 CONCEPT OF TRANSMISSION ANGLE

However, even at the stage of kinematic design, we should ensure that the output

member receives, along its direction of movement, a large component of the force (or

torque) from the member driving it. Assuming all binary links as two-force members

(i.e., neglecting gravity, inertia, and frictional effects), we can express the free-running

quality of simple mechanisms (like 4R or slider-crank mechanisms) through an index

termed as the transmission angle.

Diagram 2.9

For a 4R linkage, the transmission angle (μ) is defined as the acute angle between the

coupler (AB) and the follower (O4B), as shown in Diagram 2.9. If ABO4 is acute, then μ

= ABO4. On the other hand, if ABO4 is obtuse, then μ = ‒ ABO4. As explained in

this diagram, if μ = /2, then the entire coupler force is utilized to drive the follower.

For good transmission quality, the minimum value of μ (μmin) > 30°. For a crank-rocker

Component producing

(driving) torque

Coupler force

B

A

2

4

3

B

A





mechanism, the minimum value of μ, occurs when the crank becomes collinear with the

frame, i.e.,

or

It is not at all difficult to prove the last assertion. If the swing angle of the rocker is

increased, maintaining the same quick-return ratio, then the maximum possible value of

μmin decreases. If the forward and return strokes of the rocker take equal time, then

(μmin)max is restricted to . Therefore, such a crank rocker will have a poor

transmission quality if > 120°.

Application 2.2: A crank-rocker 4R linkage without the quick-return effect has to have

a swing angle and a minimum transmission angle μmin. Determine the link-length

ratios l2/ l1, l3/ l1, and l4/ l1.

Solution: Using Eqs. (2.6) and (2.7), for such a linkage, we have

Considering the configuration θ2 = 0 in Diagram 2.9, when ABO4 = μmin, we can write

Equations (a), (b), (c) are rewritten, respectively, as

Using (d) in (f), we obtain

Using (e) in (g), we obtain

Using (d), (g), and (h), it is easy to show that

It is obvious from (i) that l3/l1 ≤ 1 and then, from (h), we get

For a slider-crank mechanism, the transmission angle is defined as the acute angle

between the connecting rod and the normal to the slider movement as indicated in

Diagram 2.10. The minimum transmission angle in this case is given by





Diagram 2.10

2.4 GENERAL PLANE MOTION

Before going into the details of various methods of velocity and acceleration analysis of

plane mechanisms (i.e., a series of interconnected rigid bodies), let us briefly

recapitulate the fundamentals of the velocity and acceleration of a particle and a rigid

body in plane motion.

Motion of a Link

Let us suppose a rigid link OA, has uniform angular velocity ω rad/s in the counter-

clockwise direction with radius r, rotate about a fixed point O as shown in Diagram

2.11(a). After a small time interval δt link OA turns through with a small angle δθ and

point A reach at point A' as shown in Diagram 2.11(b).

Diagram 2.11

Now, velocity of A with respect to O is

The direction of VAO is along the displacement of A is perpendicular to OA as shown in

Diagram 2.11(c). The fact that the direction of velocity vector is perpendicular to the link

also emerges from the fact that A can neither approach nor recede from O and thus, the

only possible motion of A relative to O is in a direction perpendicular to OA.

Consider a point B on the link OA. It is also moves with link OA and reaches at point B'

in time interval δt. Again it is seen that the direction of VBO is also perpendicular to OA.

Velocity of B = ω. OB perpendicular to OB.

This means that all the particles (or points) present on OA and moves with link OA,

then direction of velocity of the particles (or points) is perpendicular to link OA.

At other instants, when the link OA assumes other positions, the velocity vectors will

have their directions changed accordingly. Also, the magnitude of the instantaneous

linear velocity of a point on a rotating body is proportional to its distance from the axis

of rotation.

Hence velocity of any point on a link with respect to another point on the same link is

always perpendicular to the line joining these points on the configuration (or space)

diagram.

O

B

A

O

B'

A' B

A

A





Now

Rubbing Velocity

The rubbing velocity at a pin joint is defined as the algebraic sum between the angular

velocities of the two links which are connected by pin joints, multiplied by the radius of

the pin.

Consider two links A and B which are connected by a pin joint at O as shown in

Diagram 2.12.

Diagram 2.12

Consider ω1 = Angular velocity of link A; ω2 = Angular velocity of link B; r = Radius of

the pin at the joint.

According to definition,

Rubbing velocity at the pin joint O

= (ω1 ‒ ω2) r, if the links move in the same direction

= (ω1 + ω2) r, if the links move in the opposite direction

At gudgeon pin of the connecting rod one member is turning and the other is sliding, the

rubbing velocity is ω. r because the angular velocity of the sliding member is zero.

Plane Motion of a Particle

The path of a particle moving in a plane is shown in Diagram 2.13a. The positions and

velocities of the particle at the instants t and (t + δt) have been indicated. Also, the

radius and centre of curvature corresponding to these instants have been shown. As

depicted in Diagram 2.13b, the change in velocity δV can be resolved into two mutually

perpendicular components δVn and δVt. When δt → 0, the direction of the tangential

component δVt coincides with that of V, and the normal component δVn will be directed

towards the centre of curvature.

Diagram 2.13

Obviously, the magnitude of the velocity of the particle at the time t can be expressed as

Path of particle

Reference line

(a)

O

P' (t + δt ) V + δV

V

P(t) δs

δθ

θ

(b)

δθ

V + δV

δ

V

V

VA

A

AB

B

VB

VA

VBA (⊥r to AB in

sense of ω) VB

B A

O





where θ is the inclination [with respect to a reference line (see Diagram 2.13a)] and ω is

the magnitude of the angular velocity of the radius of curvature. The velocity vector V is

always tangential to tin path and its direction is obtained by rotating ρ through 90° in

the sense of ω. In vector notation,

V = ω × ρ

The normal and tangential components of acceleration can be derived as follows. The

normal component of acceleration is

From Diagram 2.13b, the magnitude of δVn is

δVn = V δθ

So,

Using (2.13) in this relation, we obtain

The normal component of acceleration is always directed towards the centre of

curvature. In vector notation,

an = ω × (ω × ρ)

The magnitude of the tangential component of acceleration is

Using (2.13), we find the final expression for at becomes

where α is the magnitude of the angular acceleration of the radius vector.

Plane Motion of a Rigid Body

The position of a rigid body in plane motion is completely defined by specifying either

the positions of any two points on the body or the position of a point and the orientation

of a line fixed on the rigid body.

Similarly, the velocity (of all points) of a rigid body in plane motion is completely defined

by specifying the velocity of any point on the body along with the angular velocity ω of

the body. If the rigid body is in pure translation (without rotation), then the motion of

all points on it is identical. The difference in velocity of two points A and B (Diagram

2.13c) is entirely due to ω and this difference is expressed as (Diagram 2.13d)

VBA = VB VA = ω × AB

It is easy to identify that the difference in the motions of B and A, due to ω, is a circular

motion of B about A since AB remains constant, i.e., ρ in (2.13)-(2.15) is constant and

replaced by AB.

For the velocity analysis of a mechanism, the following form of the foregoing equation is

used:

VB = VA + VBA = VA + ω × AB [2.16]

It should be noted that for this equation to be valid, A and B must be on the same rigid

body and ω must be the angular velocity of the body.

The acceleration of a rigid body is defined by specifying either the accelerations of any

two points on it or the acceleration of any point on it and its angular acceleration





(assuming that the angular velocity of the body is already known). The acceleration of a

point B on a rigid body can be expressed in terms of the acceleration of a point A and the

angular velocity and acceleration of the body. The difference in acceleration is given by

From (2.14) and (2.15), with ρ = AB = constant,

written in vector notation as

The normal component is always along BA and the tangential component is

perpendicular to AB in the sense of . The commonly-used form of the acceleration

relation is

Motion Difference between Two Instantaneously Coincident Points

In a mechanism, a link is quite often guided along a prescribed path in another moving

link. For the velocity and acceleration analyses of such a mechanism, the differences in

the velocities and accelerations of two instantaneously coincident points belonging to the

two links have to be determined. In this section, we shall derive the expressions for

these quantities.

Diagram 2.14

Diagram 2.14a shows a rotating rigid link (labeled 2) on which link 3 is moving along a

straight line. The configurations at the instants t and (t + δt) are, respectively, shown by

the symbols without and with a prime. Further, P2 and P3 represent two points on links

2 and 3, respectively, coincident at the instant t. The displacement of P3 can be written

as

where represents the displacement of P2 and

represents the

displacement of P3 with respect to link 2. Dividing both sides of the foregoing equation

by δt and taking the limit δt → 0, we get

where Vp3/2 is the velocity of P3 as seen by an observer attached to link 2. The direction

of Vp3/2 is tangential to the path of P3 in link 2.

From (2.18), it may appear that the absolute acceleration of P3 can also be obtained from

an equation similar to it, i.e., ap3 = ap2 + ap3/2, where ap3/2 is the acceleration of P3 as seen

Link 2

Link 3

A

C

2

3





by an observer on link 2. However, this is incorrect since an extra term has to be added

to the right-hand side of this equation, as now explained.

From Diagram 2.14b, we see that the block has moved through an additional transverse

distance because of the rotation of link 2 and the radial motion of link 3 with respect

to link 2. When δt→ 0,

From this equation, we observe that the additional displacement term is proportional to

the square of the time elapsed. Therefore, this displacement must be due to an

additional acceleration of P3 in the transverse direction. If the magnitude of this

additional acceleration is ac, then

or

In vector notation, (2.19) is written as

This extra transverse component of acceleration is known as the Coriolis component.

The final expression for ap3 will then be

It should be noted that the direction ac is obtained by rotating Vp3/2 through 90° in the

sense of ω2. For a straight-line path of P3 on link 2, the direction of ap3/2 is along the

straight line.

When link 3 moves along a curvilinear path on the rotating link 2 (see Diagram 2.14c),

equation (2.20) can be written in terms of the components of ap3/2 as

It may be noted that the magnitude of is equal to

, where ρ is the radius of

curvature of the path of P3 on link 2. The direction of Vp3/2 is obviously tangential to this

path. For a straight-line path of P3 on link 2, ρ becomes infinite and = 0.

It may be pointed out that for (2.18) and (2.21) to be valid, P2 and P3 need not belong to

two adjacent links. We can use these relations for two coincident points, Pi and Pj,

belonging to the i-th and j-th links, respectively. The important point is that the path of

Pj in the i-th link should be known (generally, the path of Pi in the j-th. link is not the

same as this one) and we can write

or

with

To identify the path of Pj in the i-th link, it is advisable to consider a kinematic

inversion of the mechanism keeping the i-th link fixed.

Brief Solution of Coriolis Component of Acceleration

Generally we know that the acceleration of fixed point with respect to another fixed

point on moving link has two components. These are normal (radial or centripetal) and

other is tangential component.





Now suppose a case in which acceleration of moving point relative to fixed point on the

moving link. This will have three acceleration components, these are

(a) Normal, (radial, centripetal) component an;

(b) Tangential component at; and

(c) Coriolis component ac, ac.

Now, consider a link OA rotates with an angular velocity ω about a fixed point O. Point

B is slider which moves radially along OA as shown in Diagram 2.15.

Consider at time interval t at any given instant,

ω = Angular velocity of the link;

= Angular acceleration of the link

V = Linear velocity of the slider on the link

a = Linear acceleration of the slider on the link

r = Radial distance of point B on the slider.

After a short interval of time dt, the link has been taken angular displacement dθ, and

dr is the radial displacement of the slider in the outward direction.

Diagram 2.15

Now acceleration in vertical (y) direction (which is parallel to OA).

From diagram initial velocity in y-direction = V

Final velocity in y-direction = (V + a.dt) cos dθ ‒ (ω + .dt)(r + dr) sin dθ.

If angle is very small, then cos dθ ≈1, and sin dθ ≈ dθ, then acceleration will be

After neglecting the small terms, then we get

Now acceleration in horizontal (x) direction (which is ⊥r to OA).

From diagram initial velocity in x-direction = rω

Final velocity in x-direction = (V + a.dt) sin dθ ‒ (ω + .dt)(r + dr) cos dθ.

If angle is very small, then cos dθ ≈1, and sin dθ ≈ dθ, then acceleration will be

where

A





After neglecting the small terms, then we get

ax = 2Vω + r [b]

First term in Eq. (b) is termed as Coriolis component of acceleration

ac = 2Vω

Radial component of acceleration of point B with respect to O, is

Tangential component of acceleration of B with respect to O, is

Radial component of the slider B is

Tangential component of the slider is

Direction between VOB, ω, and ac:

Diagram 2.16

2.5 INSTANTANEOUS CENTRE OF VELOCITY

In Diagram 2.17, a rigid body 2 is shown to be in plane motion with respect to the fixed

link 1. The velocities of two points A and B of the rigid link 2 are shown by VA and VB,

respectively. Two lines drawn through A and B in directions perpendicular to VA and VB

meet at P. Let PA = r1 and PB = r2. The velocity of the point B in the direction of AB is

VB cos , and that of the point A in the same direction is VA cos θ. As the length of AB is

fixed, the component of VBA in the direction of AB is zero. Thus, VB cos = VA cos θ.

Diagram 2.17

From the triangle PAB, we have

Thus, the velocities of the points A and B are proportional and perpendicular to PA and

PB, respectively. So, instantaneously, the rigid body can be thought of as being

1

2 B A

C VC

VB

VA

P





momentarily in pure rotation about the point P. The velocity of any point C on the body

at this instant is given by VC = PC.VB /r2 in a direction perpendicular to PC. This point P

is called the instantaneous centre of velocity, and its instantaneous velocity is zero. So,

alternatively, the instantaneous centre of velocity can be defined as a point which has no

velocity with respect to the fixed link.

If both links 1 and 2 are in motion, in a similar manner, we can define a relative

instantaneous centre P12 (sometimes called centro) to be a point on 2 having zero relative

velocity (i.e., the same absolute velocity) with respect to a coincident point on 1.

Consequently, the relative motion of 2 with respect to 1 appears to be pure rotation (for

the instant) about P12. It is evident that P12 and P21 are identical. Thus, if a mechanism

has N links, the number of relative instantaneous centres is N(N ‒ l)/2. The absolute

instantaneous centre of velocity of a link is the relative instantaneous centre of velocity

with respect to the fixed link. Note carefully that the instantaneous centres can lie

outside the physical boundary of the links and anywhere in the plane of motion.

However, they are considered to be integral points of the concerned two links (imagined

to be extended).

How to Find Out the Location of Instantaneous Centres in a Mechanism:

In certain cases, the relative instantaneous centres are easily identified by inspection of

the geometry of motion. Examples of such situations are as follows:

1. If two links have a hinged joint, the location of the hinge is the relative

instantaneous centre because one link is in pure rotation with respect to the other

about that hinge. Such a instantaneous centre is of permanent nature, but if one

of the links is fixed, the instantaneous centre will be of fixed type. (Diagram

2.18a).

2. If the relative motion between two links is pure sliding, the relative instantaneous

centre lies at infinity on a line perpendicular to the direction of sliding. (Diagram

2.18b)

3. If one link is rolling (without slipping) over another, the point of contact is the

relative instantaneous centre. (Diagram 2.18c)

4. If a link is sliding over a curved element, the centre of curvature is the relative

instantaneous centre. (Remember the curved slider and the equivalent hinge)

(Diagram 2.18d)

5. If the relative motion between two links is both rolling and sliding, the relative

instantaneous centre lies on the common normal to the surfaces of these links

passing through the point of contact. (Diagram 2.18e)

The determination of its exact location requires some more information.

Diagram 2.18

Fixed link 1

(b)

B A

VA

VB

Link 2

(Slider)

P12 at ∞

Link 1

(Slider)

Fixed link 2

(c)

P12 VB

VA

B A

Link 1

(Slider)

Fixed link 2

(e)

B A

VB

VA

P12

Link 1

(f)

B A

VB

VA

Link 2

(Slider)

P12

(a) P12

Link 2

Link 1





The instantaneous centre of acceleration is defined as a point on a link having zero

relative acceleration with respect to a coincident point on the other link, and in general

is different from the instantaneous centre of velocity.

2.6 ARDNHOLD-KENNEDY THEOREM OF THREE CENTRES

The Aronhold-Kennedy theorem states that if three bodies are in relative motion with

respect to one another, the three relative instantaneous centres of velocity are collinear.

Diagram 2.19

Proof: Diagram 2.19 shows three links 1, 2, and 3 in relative motion with respect to one

another. Since we are interested only in the relative motion, without any loss of

generalities we can assume one of the three links, say, link 1, is fixed. P12(O2) and

P13(O3) are the points about which links 2 and 3 are rotating. If P23 is not on the line

joining P12P13, let it be somewhere else as shown in the diagram. Considering P23 as a

point on link 2, its velocity must be in a direction perpendicular to P12P23. If P23 is taken

as a point on link 3, its velocity must be in a direction perpendicular to P13P23. By

definition, P23 must have the same velocity whether it is considered to be on link 2 or 3.

This cannot be so unless P23 is on the line P12P13, otherwise the directions will be

different as has been shown in the diagram.

This theorem will be used very often for determining the relative instantaneous centres

of a mechanism.

Types of Instantaneous Centres

The instantaneous centres for a mechanism are of the following three types:

1. Fixed instantaneous centres, does not change at their position in space with time.

2. Permanent instantaneous centres, for two links at joint which in space when

mechanism moves, and

3. Neither fixed nor permanent instantaneous centres, also known as movable

instantaneous centres and finding by using Kannedy’s theorem.

The first two types i.e. fixed and permanent instantaneous centres are together known

as primary instantaneous centres and the third type is known as secondary

instantaneous centres.

Consider a four bar mechanism O2ABO4 as shown in Diagram 2.20. The number of

instantaneous centres (N) in a four bar mechanism is given by

The instantaneous centres P12 and P14 are called the fixed instantaneous centres as they

remain in the same place for all configurations of the mechanism. The instantaneous

centres P23 and P34 are the permanent instantaneous centres as they move when the

mechanism moves, but the joints are of permanent nature. The instantaneous centres

2 1

3 2





P13 and P24 are neither fixed nor permanent instantaneous centres as they vary with the

configuration of the mechanism.

Diagram 2.20

Angular Velocity Ratio Theorem

It’s state that velocity of the link in a given mechanism equals the product of the angular

velocity and radius of I-centre at the instant.

When the angular velocity of a link is known and it is required to find the angular

velocity of another link, locate their common I-centre. The velocity of this I-centre

relative to a fixed third link is the same whether the I-centre is considered on the first or

the second link. First consider the I-centre to be on the first link and obtain the velocity

of the I-centre. Then consider the I-centre to be on the second link and find its angular

velocity.

Proof: Consider an example of four bar link as shown in Diagram (2.20).

Now according to theorem

VBA = ω2 AB

V2 = ω2 (P12 P23) (a)

Let ω3 be the angular velocity of link 3, then

V2 = ω3 (P13 P23) (b)

From Eqs. (a) and (b), then we have

In general

If the velocity of any link x to be determine, when angular velocity of link y is known,

then

2.7 VELOCITY AND ACCELERATION IMAGES

The concepts of velocity and acceleration images are used extensively in the kinematic

analysis of mechanisms having ternary, quaternary, and higher-order links. If the

velocities and accelerations of any two points on a link are known, then, with the help of

images, the velocity and acceleration of any other point on the link can be easily

determined. An example to illustrate this concept follows. The method used will be

better appreciated when the kinematic analysis of complex mechanisms is considered

later in this chapter.

2

A

B

4

1

3





Diagram 2.21

A rigid link BCDE having four hinges is shown in Diagram 2.21a. Let the angular

velocity and acceleration of this link be ω and (CCW). The absolute velocity vectors of

the points E, B, C, and D are shown in Diagram 2.21b as VE, VB, VC, and VD,

respectively. The velocity difference vectors are

eb = VBE; bc = VCB; ec = VCE; bd = VDB

and their magnitudes are, respectively,

EB; BC; EC; BD

So,

Hence, the velocity diagram bcde is a scale drawing of the link BCDE. The diagram bcde

(formed by the tip of the absolute velocity vectors) it called the velocity image of the link

BCDE. The velocity image is rotated through 90° in the direction ω, as all the velocity

difference vectors are perpendicular to the corresponding lines. The scale of the image is

determined by ω, and therefore the scale will be different for each link of a mechanism.

The letters identifying the end points of the image are in the same sequence as that in

the link diagram BCDE. The absolute velocity of any point X on the link is obtained by

joining the image of X(x) with the pole of the velocity diagram o.

The absolute acceleration vectors of the points E, B, C, D are shown in Diagram 2.21c as

aE, aB, aC, aD respectively. The acceleration difference vectors are

Thus,

Similarly,

So, the diagram ebcd (formed by the tip of the absolute acceleration vectors) is a scale

drawing of the link EBCD and is called the acceleration image. The scale will be

different for each link of a mechanism. The letters identifying the end points of the

image are in the same sequence as that in the link diagram BCDE. The absolute

acceleration of any point X on the link can be obtained by joining the image of X(x) with

(c) Acceleration diagram

(a) Space diagram

E

X B

C

D

(b) Velocity diagram





the pole of the acceleration diagram o. The orientation of the acceleration image from

the link diagram is (180°‒ 0) in the counter-clockwise direction, where θ = tan-1

[bb1/(b1e)] = tan-1(/ω2), with a positive in the CCW sense (as explained in Diagram

2.21c). It should be noted that once the absolute velocities and accelerations of any two

points (e.g., E and C) of a rigid link are known, those of any other point on the link (such

as X, B, and D) can be determined just by drawing the respective images.

2.8 VELOCITY AND ACCELERATION ANALYSIS (GRAPHICAL)

The general principal for carrying out the kinematic analysis of most problems is to

construct the velocity and acceleration diagrams starting from input link. In these

diagrams, the fixed link is representing by a point. Such a point is termed as the pole of

velocity diagram or the acceleration diagram.

Slider-Crank Mechanism

Consider a slider-crank mechanism OAB as shown in Diagram 2.22(a). Crank OA

rotates clockwise with angular velocity ω rad/sec. It is required to draw the velocity and

acceleration diagrams.

Diagram 2.22

Velocity Diagram

Velocity of point A with respect to fixed point O is VAO = ω.OA in the clockwise sense.

First of all, velocity polygon is drawn as follows:

1. Take any point o and from it draw oa = VAO = ω.OA with some suitable scale as

shown in Diagram 2.22 (b). Now, VBO and VBA were known in direction only. VAO

was known in direction and magnitude both.

2. The velocity of point B with respect to point A. (on connecting rod) is

perpendicular to line AB. So from point ‘a’ draw a line perpendicular to AB

representing VAB.

3. The velocity of slider B relative to O (VBO) is parallel to the line OB. From o draw

a line parallel to OB to intersect VAB point b.

Measure VBA = ab and VBO = ob from the velocity diagram.

Angular velocity of coupler AB is

Acceleration Diagram

We know that radial/normal acceleration of A with respect to O is given as

(a) Single slider

mechanism

B

O

A


⊥

⊥


⊥

⊥





The tangential acceleration of point A with respect to O is zero because crank OA rotates

with constant angular velocity.

Thus = 0. The acceleration diagram is drawn as:

1. Draw vector oa' = parallel to AO with some suitable scale as shown in

Diagram 2.22 (c).

2. From point a' draw vector a'x = which is the radial component of acceleration

of B with respect to A. It is parallel to BA. VBA can be taken from velocity

diagram.

3. From x draw vector xb' perpendicular to a'x or AB. The vector xb' represents the

tangential component of acceleration of B with respect to A i.e. . It is known in

direction only and contains b' .

4. The acceleration of point B with respect to O is parallel to the line of motion of

slider B i.e. along OB. It is not radial acceleration as the slider has reciprocating

motion. So from point o draw a line parallel to OB representing aB to intersect

vector xb' at b'.

5. Join a' to b'. Vector a'b' represents acceleration of B with respect to A i.e. aBA.

aB, , aBA., can be found by measurement with the scale.

S.No. Vector Magnitude Direction Sense

1 OA → O

2 AB → A

3 ⊥ AB

4 to line of motion B

Four-bar Chain Mechanism

Consider a four-bar chain ABCD is shown in Diagram 2.23(a). AB is the crank and

rotates at ω rad/sec in the clockwise sense. It is required to draw the velocity and

acceleration diagrams.

Diagram 2.23

Velocity diagram

The velocity of point B with respect to point A (VBA) can be written as

VBA = ω.AB in the clockwise sense

1. Take points a, d as fixed centres as shown in Diagram 2.23(b).

2. The velocity of point B with respect to A i.e. VBA is perpendicular to AB. VBA is

given. Draw vector ab = VBA perpendicular to AB with some suitable scale.

3. The velocity of point C with respect to point B i.e. VCB is perpendicular to line BC.

Only the direction of VCB is known. So from point b draw a vector bc

perpendicular to BC to represent VCB.

(a) Four-bar mechanism

D

B

A ω

C


⊥

⊥

⊥

⊥


⊥

⊥





4. The velocity of point C with respect to D i.e. VCD is known in direction only. From

a (or d) draw a vector ac perpendicular to CD to represent VCD.

5. The vectors ac and be intersect at c. The vector ac = VCD.

By measurement of vectors ac and bc the velocities VCD and VCB can be found. The

angular velocity of link BC can be found as

ωCB = VCB/BC = rad/sec

Similarly,

ωCD = VCD/CD = rad/sec

Acceleration diagram

The angular acceleration of link AB is not given, so it may be assumed that = 0 or

tangential component of acceleration of B with respect to A, will be zero. Radial

acceleration = ω2. BA is known in direction (parallel to BA) also. Diagram 2.23(c).

1. Draw vector a'b' = = ω2. AB parallel to BA. a' and d' are fixed points and

=

0.

2. From b' draw vector b'x = =

. BC = /BC. VCB is taken from velocity

diagram, b'x is parallel to CB. From x draw vector xc' perpendicular to b'x. Its

magnitude is unknown. It contains point c'.

3. From point a' draw vector a'y parallel to CD as it represents radial acceleration

of point C with respect to D i.e. . The magnitude of a'y =

= /CD and

tangential component of acceleration is perpendicular to a'y.

Draw vector yc' representing , from point y.

and intersect at point c'.

4. Join c' to y and b' to c'. Also join a' to c'.

Here from diagram the values of various components of acceleration can be measured.

Angular accelerations of links BC and CD can be determined as:

Four-bar Chain Mechanism (with off-set)

Consider a four-bar chain ABCD with off-set BFC as shown in Diagram 2.24(a). AB is

the crank and rotates with ω rad/sec in the clockwise sense and angular acceleration

rad/sec in the clockwise sense. It is required to draw the velocity and acceleration

diagrams.

Velocity Diagram:

To draw the velocity diagram same as pervious here we discuss only the construction of

velocity due to off-set BFC.

Intermediate point: The velocity of an intermediate point on any of the links can be

found easily by dividing the corresponding velocity vector in the same ratio as the point

divides the link. For point E on the link BC,

be/bc = BE/BC

ae represents the absolute velocity of E.

Offset Point: Write the vector equation for point F,

The vectors VBA and VCD are already there on the velocity diagram.

VFB is ⊥r to BF, draw a line ⊥r to BF through b;





VFC is ⊥r to CF, draw a line ⊥r to CF through c;

The intersection of the two lines locates the point f.

af or df indicates the velocity of F with respect to A (or D) or the absolute velocity of F.

Diagram 2.24



1 AB → A

2 ⊥ AB or ab → b

3 BC → B

4 ⊥ BC

5 DC → D

6 ⊥ DC

Construct the acceleration diagram as follows:

1. Select the pole point a' or d'.

2. Take the 1st vector from the above table, i.e. take a'x to a convenient scale in the

proper direction and sense.

3. Add the second vector to the first and then the third vector to the second.

4. For the addition of the fourth vector, draw a line perpendicular to BC through

the head y of the third vector. The magnitude of the fourth vector is unknown

and c' can lie on either side of y.

5. Take the fifth vector from d'.

6. For the addition of the sixth vector to the fifth, draw a line perpendicular to DC

through head z of the fifth vector.

The intersection of this line with the line drawn in step (4) locates the point c'.

Total acceleration of B = a'b'

Total acceleration of C with respect to B = b'c'

Total acceleration of C = d'c'.

Intermediate Point: The acceleration of intermediate points on the links can be obtained

by dividing the acceleration vectors in the same ratio as the points divide the links. For

point E on the link BC (Diagram 2.24d),

(a) Four-bar mechanism

E

F

D

B

A

C


⊥

⊥

⊥


(d) Acceleration diagram of off-set

⊥

⊥

⊥

⊥





BE/BC = b' e'/b' c'

a'e' gives the total acceleration of point E.

Offset Points: The acceleration of an offset point on a link, such as F on BC (Diagram

2.24d), can be determined by applying any of the following methods:

BA already exists on the acceleration diagram. Now,

to BF, direction towards B

⊥

Shaper (Crank and Slotted Lever) Mechanism

A shaper mechanism is shown in Diagram 2.25(a). Crank O2A rotates in anticlockwise

direction with angular velocity ω2. It is required to draw the velocity diagram of the

configuration. The velocity of point A which is located at one end of the crank O2A can be

written as VAO2 = ω2.O2A. It is perpendicular to O2A. The point B is located on link 4

(O1C) such that it coincides with point A. The velocity of point B (VBO1) is perpendicular

to O1B. The points O1 and O2 are fixed, so they may be treated as one point in the

velocity diagram, as shown in [Diagram 2.25(b)].

Diagram 2.25

Velocity diagram:

1. First of all, take point o2. From o2 draw vector o2a to some suitable scale such

that VA = VAO2 = ω2.O2A. It is perpendicular to O2A.

2. The velocity of point B with respect to point A is along line O1B. Thus, VBA is

parallel to O1B. From a draw a vector ab representing VBA.

3. The velocity of point B with respect to O1 i.e. VBO1 is perpendicular to O1B. From

o1 draw a vector o1b (i.e. VBO1) to intersect ab at point b. Extend o1b to o1c such

that

Now, point c is located.

4. The velocity of point D with respect to C i.e. VDC is perpendicular to DC. From c

draw a vector cd perpendicular to DC, it represents VDC.

(a) Shaper mechanism

Point A on crank

O2A and B on link

O1C

2

6 D

5

C

3

B

A

4


⊥

⊥

⊥


⊥

⊥

⊥





5. The velocity of point D with respect to point O1 i.e. VDO1 is along the line of stroke

of the slider link 6. So from o1 draw a vector o1d to intersect cd at point d. Vector

o1d represents the velocity of point D on the slider. It can be written as

Angular Velocity of O1C (link 4) can be determined as



1 O2A → O2

2

O1B → O1

3 ⊥ O1B

4 O1C

5 ⊥ O1C

where ω1 is the angular velocity of O1C.

2.9 GOODMAN'S INDIRECT METHOD

Goodman's indirect approach to the acceleration analysis of a complex mechanism is

based on the following two properties of a constrained mechanism:

(i) The angular velocities and accelerations of the links are linear functions of the

respective input quantities.

(ii) The relative angular velocities and accelerations between different links of a

linkage remain unaffected by a kinematic inversion.

Basic Relations

Let i denote the input link and l denote any other link. The angular velocity of l at any

instant can be expressed in terms of that of i. Thus,

Cl is a geometrical property depending only on the configuration of the mechanism

(except at dead-centre locations, when two links are collinear) and is independent of

velocities and accelerations. Equation (2.25) is the mathematical statement of the

obvious fact that the velocity polygons of a mechanism at any instant with different

input velocities are scale drawings of (i.e., they are similar to) one another. The angular

acceleration of the link l is

is also a geometrical property. The first term on the right-hand side of (2.26)

represents the acceleration of the link l, the acceleration of the input link being zero and

the actual velocity being ωi. Thus, (2,26) can be written as





where 0l denotes the angular acceleration of the link l obtained from an auxiliary

acceleration diagram drawn with true velocities but with zero input acceleration.

Diagram 2.26

For a sliding input link, we have

where Vi and ai denote the velocity and acceleration of the input link.

Diagram 2.26a shows two consecutive links of some mechanism moving with angular

velocities and accelerations as indicated. The corresponding velocity diagram is shown

in Diagram 2.26b. The auxiliary acceleration diagram with i = 0 is shown in Diagram

2.26c and the true acceleration diagram with αi is shown in Diagram 2.26d. Note that

the normal components of accelerations (⊥r to the corresponding velocities) remain the

same in Diagrams 2.26c and 2.26d.

Let us suppose the motion of the point C. Without losing any generality, for simplicity of

analysis, the point O is taken as fixed. ωr and ωs are the angular velocities and r and s

are the angular accelerations of the two links. Since the normal components remain the

same in Diagrams 2.26c and 2.26d, the difference between and

is entirely due to

the difference between the tangential components ( l to the corresponding velocities). So,

Using Eq. (2.27), then, we have

or

Similarly, for a sliding input link, we have

⊥

⊥

⊥

⊥

⊥

⊥

O

C

B

⊥





In sliders, the total acceleration is in the direction of the velocity (i.e., the tangential

component is the total acceleration). So, for a rotating input link, we can write

and for the sliding input link, we have

The absolute velocities and accelerations mean those with respect to the fixed link f (i.e.,

the frame) Thus, (2.27) and (2.28) can be written as

Applications

The form of equations (2.27a) and (2.28a) is applicable to any inversion of the

mechanism where the use of f is no longer restricted to the frame. The subscript i

denotes any alternative input link (not necessarily the actual input link) with assumed

zero acceleration, on the basis of which the auxiliary acceleration diagram should be

drawn.

The indirect approach can be applied to a mechanism with a low degree of complexity in

the following manner:

(i) Choose an alternative input link to transform the mechanism to a simple one.

The auxiliary analysis is carried out with zero acceleration of this alternative

input link.

(ii) Find the actual values by using (2.27) to (2.32).

For mechanisms with a high degree of complexity, a direct kinematic inversion is made

to transform the mechanism to a simple one. The auxiliary velocities and accelerations

are obtained first. Thereafter, using (2.27a) and (2.28a), the actual values can be

determined.

2.10 VELOCITY AND ACCELERATION ANALYSIS (ANALYTICAL)

The analytical method of velocity and acceleration analysis starts from the loop-closure

equations which were discussed during displacement analysis. These equations are valid

at all times, and therefore successive differentiations of these equations with respect to

time establish the relationships between the velocity and acceleration quantities of

various links of a mechanism. The most important point to note is that, once the

configuration of the mechanism is known (i.e., the displacement analysis is complete),

the velocity and acceleration equations are linear in the unknown quantities and,

therefore, are very easy to solve. Consequently, when the velocity and acceleration

analysis has to be carried out for a large number of configurations, the analytical

method turns out to be more advantageous than the graphical method.

We shall now derive in detail the angular velocity and acceleration of the coupler and

the follower of a 4R linkage when the configuration and the crank motions are

prescribed. Referring back to Diagram 2.2, assume that the configuration of the





mechanism has already been determined, i.e., l1, l2, l3, l4 and θ2 are prescribed and θ3

and θ4 have been solved. Now we determine the angular velocity and acceleration of the

coupler and the follower if those of the crank are given.

Towards this end, differentiate (2.2a) and (2.2b) with respect to time and obtain

We should note that (2.33a) and (2.33b) are two simultaneous linear equations in the

two unknowns, viz., and , which can be easily solved to yield

Differentiating (2.33a) and (2.33b) once more with respect to time, we get

Once the velocity analysis is complete, (2.34a) and (2.34b) again provide two linear

equations in which are obtained as

The same methodology can be extended to a mechanism having prismatic pairs. The

only difference will be that all vectors appearing in the loop-closure equation will not be

constant.

2.11 KLEIN'S CONSTRUCTION

A graphical method to find the velocity and acceleration of mechanism was given by

Professor Klein.

Single Slider Crank-Mechanism

A slider crank-mechanism OCP is shown in Diagram 2.27 for which velocity and

acceleration diagrams are shown in diagram.

Diagram 2.27

O

L

K

X

Q

M C

N P





Klein's construction is shown in Diagram 2.27. The construction of Diagram 2.27 is as

follows:

1. The line PC to cut a line through O perpendicular to the line of action at M.

2. Draw a circle with diameter CP at centre X.

3. Draw again circle with radius CM at centre C.

4. KL, the chord common to these two circles, with cut CP at Q and QP at N.

5. Quadrilateral OCQN is acceleration diagram.

6. OCN is velocity diagram.

Four Bar Mechanism

A four-bar mechanism as shown in Diagram 2.28.

Diagram 2.28

Klein's construction for velocity and acceleration diagram is made as under:

1. Draw a circle with centre B with radius BM.

2. Draw a second circle on the link BC as diameter.

3. The chord common to these two circles intersect BC at X and is produce to Q.

4. Two further circle draw, one with centre C and radius AM, and other on CD as

diameter T.

5. The cord of common to these two circles cut CD at S along AM mark of AT equal to

CS.

6. Through T draw TQ at right angles to M.

With this construction

Q

X B

D

S

C

M

T

A b

t

c, m, q s

o

θ

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