Lect Position Analysis (2)

ME 3507: Theory of Machines

Position Analysis

Dr. Faraz Junejo

Introduction A principal goal of kinematic analysis is to determine

the accelerations of all the moving parts in the

assembly. Why ?

The design engineer must ensure that the proposed mechanism or machine will not fail under its operating conditions. Thus the stresses in the materials must be kept well below allowable levels.

From Newton's second law, F = ma, one typically needs to know the accelerations (a) in order to compute the dynamic forces (F) due to the motion of the system's mass (m).

Introduction (contd.)

Kinematic Analysis• We determine relative motion characteristic of

a given mechanism.

• Can be classified into:- Position analysis

- Velocity analysis

- Acceleration analysis

• For all these three type of problems, we can use either:

- Graphical Method or- Analytical Method

Position Analysis

• Given the kinematic dimensions and

position or movement of the input

link determine the position or

movement of all other links

Objective

• Determine the positions of links

and points on mechanisms.

Graphical ApproachIn the graphical method, the kinematic

diagram of the mechanism is drawn to a

suitable scale, and

The desired unknown quantities are

determined through suitable geometrical

constructions and calculations.

Graphical approach

• We will have to do an independent graphical

solution for each of the positions of interest

• None of the information obtained graphically

for the first position will be applicable to the

second position.

• It is useful for checking the analytical results.

Analytical approach

• Derive the general equations of motion

– Solve analytical expressions

– Once the analytical solution is derived for a

particular mechanism, it can be quickly solved

(with a computer) for all positions.

• Graphical Position Analysis – Is more simple then the algebraic approach• Graphical Velocity and Acceleration analysis – Becomes quite complex and difficult then the

algebraic approach

• Graphical analysis is a tedious exercise and

was the only practical method available in the

day B.C.(Before Computer) , not so long ago.

Graphical vs. Analytical approach

Graphical vs. Analytical approach (contd.)

Coordinate System• Global or Absolute: Master frame reference

fixed in space.

• Local: Typically attached to a link at some point of interest.

- This might be a pin joint, a center of gravity, or a line of centers of a link.

- These local coordinate system may be either rotating or non-rotating as we desire.

Position & Displacement (Point motion)

• The position of a point in the plane can be defined by the use of a position vector.

• Polar coordinate / Cartesian coordinate

• A position vector can be expressed in:

– Polar form : a magnitude and angle of vector

– Cartesian form : X and Y components of the vector

Position Vector in Cartesian and Polar Form

Coordinate Transformation

• The system’s origins are coincident and the

required transformation is a rotation.

Coordinate Transformation

Displacement of a point

Is the change in its position and can be

defined as the straight line between the initial

and final position of a point which has moved in

the reference frame.

• Note that displacement is not necessarily the same as the path length which the point may have traveled to get from its initial to final position.

Displacement (contd.)

• Figure a shows a point in two positions, A and B. The

curved line depicts the path along which the point

traveled.

The position vector RBA defines the displacement ofthe point B with respect to point A .

• Figure b defines this situation with respect to a global reference frame XY.


The vectors RA and RB define, respectively, the absolute positions of points A and B with respect to this global XY reference frame.


The vector RBA denotes the difference in position, or the displacement, between A and B. This can be expressed as the position difference eq:

RBA= RB – RA or RBA=RBO-RAO

The position of B with respect to A is equal to the (absolute) position of B minis the (absolute) position of A, where absolute means with respect to the origin ofthe global reference frame.

Case 1 – One body in two successive position• position difference

Case 2 – Two bodies simultaneous in separate position• relative or apparent position


Summary• Cartesian (Rx, Ry)

• Polar (RA, q)• Converting between the two

• Position Difference, Relative position– Difference (one point, two times)– relative (two points, same time)

RBA=RB-RA

xy

yxA

RR

RRR

arctan

22

sin

cos

Ay

Ax

RR

RR

X

Y

RB

RA

ABRBA

Translation

All points on the body have the same displacement, as

No change in angular orientation

Can be curvilinear or rectilinear translation

Rotation

• Different points in the body

undergo different displacements

and thus there is a displacement

difference between any two

points chosen

• The link now changes its angular

orientation in the reference frame

Complex Motion The sum of the translation and rotation components.

total complex displacement =translation component + rotation component The total complex displacement of point B can be defined as:

Whereas, the new absolute position of point B w.r.t origin at A is:

Theorems

Euler’s theorem The general displacement of a rigid body with

one point fixed is a rotation about some axis.

This applies to pure rotation as mentioned earlier.

Chasles’ theorem describes complex motion Any displacement of a rigid body is equivalent

to the sum of a translation of any one point on that body and a rotation of the body about an axis through that point.

Summary: Translation, Rotation, and Complex motion

• Translation: keeps the same angle

• Rotation: one point does not move, such as A

in preceding examples

• Complex motion: a combination of rotation

and translation

Example: 1

• The path of a moving point is defined by the

equation y = 2x2 – 28. Find the position

difference from point P to point Q, when

3 and 4 x

Q

x

P RR

Example: 1 (contd.)• The y-components of two vectors can be written as

• Therefore, the two vectors can be written as

• Thus, position difference from point P to Q is

102832 and 428-42 22 y

Q

y

P RR

j10 ˆ3 and j4 ˆ4 iRiR QP

4.24318043.637

14tan

and 65.15)14((-7)As,

243.415.65j14 ˆ7

1

22

iRRR PQQP

Remember:Angles will always be measured ccw from +ve x-axis.

Example: 2

2link w.r.t 3link of , 2/3 ntdisplacemeRWhere P

Example: 2 (contd.)

Ans 7.118421.4ˆ879.3ˆ121.2

)1()2(

.ˆ6ˆ90sin6ˆ90cos6)2(

906)22()2(

.ˆ121.2ˆ121.2ˆ45sin3ˆ45cos3)1(

453)21()2(

3

333

3

3

3

3

42

2

41

242

jiR

RRR

jjiR

eR

jijiR

eetreR

P

PPP

P

j

P

P

jtjj

P

Example: 2 (contd.)

Ans ˆ3)1()2(

ˆ6ˆ)22()2(

ˆ3ˆ)21()1(

ˆ)2(0)2(

0)2(

2

22

2

22

2

2

22

02

2/32/32/3

2/3

2/3

2/3

2/3

iRRR

iiR

iiR

ittR

etreR

PPP

P

P

P

jj

P

Objective of Position Analysis• The main task in position analysis is the find the

output variables knowing: – The input variable – The length of all the links

Objective of Position Analysis (contd.)• As discussed earlier, there are 2 ways of doing this: – Graphical method (use drawing tools) – Analytical method (use equations)• Reminder: All angles are measured counter clockwise

from the positive x-axis, as shown below

Graphical Position Analysis• For any one-DOF linkage, such a four-bar, only one

parameter is needed to completely define the positions of all the links. The parameter usually chosen is the angle of the input link; i.e.

Construction of the graphical solution

1. The ground link (1) and the input link (2) are drawn to a convenient scale such that they intersect at the origin O2 of the global XY coordinate system with link 2 placed at the input angle θ2.

2. Link 1 is drawn along the X axis for convenience.

Construction of the graphical solution (contd.)

3. The compass is set to the scaled length of link 3 (i.e. length b), and an arc of that radius swung about the end of link 2 (point A) i.e. draw an arc centered at end of Link 2 (point A)

Construction of the graphical solution (contd.)

4. Set the compass to the scaled length of link 4 (i.e. length c), and draw another arc centered at end of Link 1 (point O4). Label the intersection of both arcs B and B’

Note that intersection of both arcs B and B’ define the two solution to the position problem for a four-bar linkage which can be assembled in two configurations, called circuits, labeled open and crossed.

O2-A-B-O4 is first config.O2-A-B’-O4 is second config.

First Config. (Open Config.)

• Measure θ3 and θ4 with protractor• Called ‘Open’ configuration because both links

adjacent to the shortest link (Links 1 and 3) do NOT cross each other

Second Config. (Cross Config.)

• Measure θ3’ and θ4’ with protractor (CCW from positive x-axis)

• Called ‘Cross’ configuration because both links adjacent to the shortest link (Links 1 and 3) cross each other

Summary: Example 1Given the length of the links (a,b,c,d), the ground position, and q2. Find q3 and q4

a

d

b

cq3

q4

A

B

O2 O4

b

c

q2

Example 1: Graphical Linkage Analysis

• Draw an arc of radius b, centered at A

• Draw an arc of radius c, centered at O4

• The intersections are the two possible positions for the linkage, open and crossed

adq2

b

cq3

q4

A

O2 O4

B1

B2

Summary: Graphical Position Analysis

Shaping machine• A photographic view of general configuration of shaping

machine is shown in Figure. The main functions of shaping machines are to produce flat surfaces.

Example: 2• Model of Slotted quick return mechanism used in

Shaping machines

Cutting toolBull gear rotated at constant speed

Tool holder moves in a slot, in the frame of machine

Block Hinged to the bull gear, and moves up & down in the slotted lever

Slotted lever hinged to frame

Link connecting slotted lever with tool holder

• So, we have a six link mechanism, where continuous uniform rotation of the bull gear is converted into to and fro motion of the cutting tool.

• It can be seen that cutting tool is doing useful work during forward motion/stroke, so we have to maintain a proper cutting speed. However, during return stroke it is not doing any useful work, so we would like to make return stroke faster, hence it is referred as quick return mechanism.

Example: 2 (contd.)

Link 2, O2A Bull gear

Link 3, block that is hinged to bull

gear and goes up & down in the

slotted lever, which is link 4

Link 5 connects slotted lever with

tool holder, which is represented

by link 6. So, we have a 6 link

mechanism.

Here, we have got 5 revolute pairs at 02, O4, A, B and D respectively.

There are two prismatic pair, one between link 1 & 6 in the horizontal direction, second one is between link 3 & 4 along the slotted lever.

Example 2: Statement • Determination of quick return ratio (ratio of the durations

of the forward stroke and the return stroke) & stroke length

of a slotted lever mechanism used in shapers, with constant

angular speed ω2 of input link 2 i.e. bull gear

toolcutting theofmotion return during 2 ofrotation

toolcutting theofmotion forward during 2 ofrotation

;

..

link

link

where

rrq

r

f

r

f

Example 2: Solution

1. Note that Point A moves along the circle drawn whose centre is at O2 with radius O2A. Therefore, this circle represent path of point A i.e. KA.

2. To determine the extreme positions of the link 4 (i.e. slotted lever), we draw two tangents to the circle (representing path of point A) from point O4.

3. Consequently, tangent drawn on right hand side (R.H.S) represent right most position of slotted lever (i.e. link 4), indicated by AR, whereas tangent on L.H.S. represent right most position of slotted lever (i.e. link 4), indicated by AL.

Example 2: Procedure

4. Since the distance O4B does not change, so we can also locate rightmost position of revolute pair at B (indicated by BR), by drawing a circular arc with O4 as centre and radius O4B. In similar manner, on L.H.S. we can locate BL.

5. It should be noted that the distance BD does not changes, as D (i.e. tool holder) moves horizontally. Hence, BR location can be used to locate rightmost position of tool holder (indicated by DR) by drawing a circular arc with BR as centre and radius BD. In similar manner, DL i.e. leftmost position of tool holder can be obtained.

6. Distance between DR and DL represent the stoke length of the cutting tool as per scale of the figure.

Determination of Q.R.R It can be seen that input link O2A rotates from O2AR to O2AL for

forward motion (i.e. right to left), hence the angle between O2AR and O2AL represent qf i.e. rotation of input link (i.e. link 2) during forward motion.

Similarly, it can be seen that for return motion (i.e. left to right) input link travels from O2AL to O2AR now indicating this angle with qr i.e. rotation of input link (i.e. link 2) during return motion.

It can be seen that qf is larger then qr resulting in quick return motion of tool holder.

Example 2: Procedure (contd.)

Example 2: Discussion• It should be noted that if stroke length needs to be

decreased, we need to:

decrease the length of input link O2A, because as a result, tangent from O4 to

circle KA i.e. AR and AL points representing rightmost and leftmost position of

slotted lever will move up, resulting in qf qr . (i.e. qf approaches qr) implying

a decrease in quick return ratio.

It can be concluded, that this mechanism is OK for

producing quick return effect so long the stroke length

is sufficiently large, and quick return effect decreases

as stroke length decreases.

Slotting machine• Slotting machines can simply be considered as vertical shaping

machine where the tool reciprocates vertically.• Unlike shaping machines, slotting machines are generally used to

machine internal surfaces, implying smaller stroke length.

Example: 3

• Determination of quick return ratio of

Whitworth quick return mechanism used in

slotting machines.

• Here, the quick return ratio is independent of

the stroke-length.

• Model of Whitworth quick return mechanism used in Slotting machines

Cutting tool

Kinematic diagram

It is a 6 link mechanism, with five revolute pairs at O2, O4, A, C and D, and two prismatic pairs between link 3 & 4, and link 6 & 1 respectively.

• Link 2 (O2A) is input link that rotates at constant angular speed, and is hinged to fixed link at O2.

• Link 3 is the block that moves along link 4 via a prismatic pair.

• Link 4 is hinged to fixed link at O4.

• Link 4 & Link 5 are connected by a revolute pair at point C.

• Link 5 & Link 6 are connected by a revolute pair as well.

• Link 6 has prismatic pair with fixed link 1 for horizontal motion.

Kinematic diagram description

Example 3: Solution

Example: 4• For a six link mechanism shown below, determine

stroke-length of the output link i.e. the slider 6. Also determine the quick return ratio assuming constant angular speed of link 2.

Example 4: Solution

Exercise: 1

Figure shows a kinematic diagram of a mechanism that is driven by moving link 2. Graphically reposition the links of the mechanism as link 2 is displaced 30° counterclockwise. Determine the resulting angular displacement of link 4 and the linear displacement of point E . Use suitable scale.

dq4= 26o, CCW

DR E = 0.95 in.

• Link 2 is graphically rotated 30° counterclockwise, locating the position of point B’.

• Being rigid, the shape of link 3 cannot change, and the distance between points B and C remains constant. Because point B has been moved to B’, an arc can be drawn of length rBC , centered at B’. This arc represents the feasible path of point C’. The intersection of this arc with the constrained path of C (obtained by Drawing an arc of radius 3.3 in, centered at D) yields the position of C’.

Exercise: 1 (sol)

• This same logic can be used to locate the position of point E’. The shape of link 5 cannot change, and the distance between points C and E i.e. rCE remains constant.

• Because point C has been moved to C’, an arc can be drawn of length rCE, centered at C’. This arc represents the feasible path of point E’. The intersection of this arc with the constrained path of E yields the position of E’.

• Finally, with the position of C’ and E’ determined, links 3 through 6 can be drawn. The displacement of link 4 is the angular distance between the new (C’D) and original position (CD), approx.: 26 Degrees counterclockwise

• The displacement of point E is the linear distance between the new (E’) and original position (E)of point . approx.: 0.95 in

Exercise: 2

• A point Q moves from A to B along link 3while

link 2 rotates from . Find the

absolute displacement of Q.

120 to30 '

22

Algebraic Position Analysis• The same procedure that was used earlier to solve geometrically

for the intersections B and B’ and angles of links 3 & 4 can be encoded into an algebraic algorithm.

• Coordinates of point A

• Coordinates of point Bare found using equationsof circles about A & O4

Eq: 1

• Coordinates of point B are found using equations of circles about A & O4 respectively

• Above equation provides a pair of simultaneous equations in Bx and By

• Now subtracting eq. 3 from eq: 2, and solving yields:

Eq: 2

Eq: 3

Eq: 4

• Now substituting eq:4 in to eq:3 gives a quadratic equation in By , which has two solutions corresponding to Figure 4.5 (i.e. graphical solution shown earlier)

Eq: 5

Algebraic Position Analysis (contd.)

• Now eq: 5 can be solved using quadratic equation

• Where,

• Solution to Eq:6;- Imaginary implies links can’t connect at given input angle- Real both values of By can be substituted into eq:4 to find

their corresponding x component

Eq: 5

Eq: 6

Link angles for the given position can be found from

Equations 1 - 7 can be encoded in any computer language or equation solver, such as MATLAB, and the value of θ2 varied over the linkage's usable range to find all corresponding values of the other two link angles.

Algebraic Position Analysis (contd.)

Eq: 7

Vector Loop Representation of Linkage

• An alternate approach to linkage

position analysis creates a vector

loop (or loops) around the linkage.

Vector Loops of a Mechanism

• The main difference between freely moving bodies and

the moving links in a mechanism is that they have a

constrained motion due to the joints in between the

links.

• The links connected by joints form closed polygons

(flat shape consisting of straight lines that are joined to

form a closed chain) that we shall call a loop.

• The motion analysis of mechanisms is based on

expressing these loops mathematically.

Loop closure equation

• Let us consider a four-bar mechanism as shown above as a simple example (figure A).

• In this mechanism A0 ,is a permanently coincident point between links 1 and 2, A is permanently coincident point between links 2 and 3, B between 3 and 4 and B0 between 1 and 4.

• Let us disconnect joint B.

• In such a case we will obtain two open kinematic chains A0 AB

(links 1,2,3) with two degrees of freedom and A0 B0B (links

1,4) with one degree of freedom

Loop closure equation (contd.)

Note that Except A0B0 , the other three vectors will be a function of time (since the distances between the two points on the same link are fixed, the magnitudes will remain constant but the directions of these vectors will change in time).

For the open kinematic chain, the position of point B may be defined in two different forms as:

A0A + AB = A0B3 (1,2,3 open loop)

A0B0 + B0B = A0B4 (1,4 open loop)

• However, when considering a mechanism, at every instant the revolute joint between links 3 and 4 must exist and point B must remain a permanently coincident point.

• Therefore the vector A0B3 and A0B4 obtained from the two equations using the two open kinematic chains must be equal and this results with the vector equation:

A0A + AB = A0B0 + B0B


A0A + AB = A0B0 + B0B

This vector equation must be valid for all positions

due to the permanently coincident points. If this

vector equation can not be satisfied for a given input

angle, then that position cannot exist (mechanism

cannot be assembled at that position).

In a four-bar mechanism there is a single loop formed

and the vector equation describes the closure of this

loop mathematically.


Vector Loop Representation of Linkage• The links are represented as position vectors which

form a vector loop. • This loop closes on itself making the sum of the

vectors around the loop zero.

• The directions of the position vectors in Figure 4-6 are chosen so as to define their angles where we desire them to be measured. By definition, the angle of a vector is always measured at its root, not at its head.

• We would like angle θ4 to be measured at the fixed pivot 04, so vector R4 is arranged to have its root at that point.

• On the same note, we would like to measure angle θ3 at the point where links 2 and 3 join, so vector R3 is rooted there.

Vector Loop Representation of Linkage (contd.)

• A similar logic dictates the arrangement of vectors R1 and R2.

• Note that the X (real) axis is taken for convenience along link 1 and the origin of the global coordinate system is taken at point 02, the root of the input link vector, R2.

• These choices of vector directions and senses, as indicated by their arrowheads, lead to this vector loop equation:

Vector Loop Representation of Linkage (contd.)

Multi-loops Mechanisms

Complex Numbers as Vectors

• Vectors may be defined in polar coordinates by their magnitude and angle, or in Cartesian coordinates as x and y components.

• As mentioned earlier, they are easily convertible from one to other using following notations.

• The position vector in Figure 4-6, can be

represented by any of the following expressions.

Complex Numbers as Vectors (contd.)

Vectors: Regular Notation

• Above expression uses unit vector to represent x and y component directions in Cartesian form

Vectors: Complex Number Notation• A vector can be represented

by a complex number

• Real part is x-axis

• Imaginary part is y-axis

• j is imaginary number, =

• Please note this imaginary number is used in a complex number as operator, not as a value.

RealAxis

ImaginaryAxis

Point A

RA

q

R cos q

jR sin q1

Reason for using complex number notation will be apparent in subsequent discussion

• This complex number notation to represent planar vectors comes from the Euler identity

• Advantage: It is easier to differentiate, because

Complex Number Notation (contd.)

Derivatives, Vector Rotations in the Complex Plane

• Taking a derivative of a complex number will result in multiplication by j

• Each multiplication by j rotates a vector 90° CCW in the complex plane

Real

Imaginary

RA

RB = j R

RC = j2 R = -R

RD = j3 R = - j R

A

B

C

D

It can be seen that each multiplication of j operator result in a 90 Degree counterclockwise (ccw) rotation

Position Analysis of Pin-Jointed Fourbar Linkage using Vector Loop Method

Vector Loop Method (contd.)1. Constants, Input & Output Variables - Constants: a, b, c, d - Variables: θ2 (input), θ3 & θ4 (output)

2. Draw Vector Loop Diagram

Vector Loop Method (contd.)3. Write associated vector loop equation

d

A

B

a

b

cq2

q3

q4

R1

R2

R3R4

O2 O4

Alternative notation: RAO2 + RBA - RBO4 - RO4O2 = 0nomenclature - tip then tail

4. Represent Each Vector using Complex Number Notation

• These equations can be solved for two unknowns.

• There are four variables, namely the four link angles, in above equation and since θ1 = 0, because it is a ground link, thereby only one independent variable θ2.

• We therefore need to find the algebraic expressions which define θ3 and θ4 as functions of the constant link lengths and the one input angle θ2.

Vector Loop Method (contd.)

5. Substitute Euler Identity into above expression

6. Above equation can now be separate into real and imaginary parts, and each set to zero.


Real:a cos q2 + b cos q3 - c cos q4 - d cos q1 = 0

a cos q2 + b cos q3 - c cos q4 - d = 0,

since q1 = 0 cos q1 = 1

Imaginary: ja sin q2 + jb sin q3 - jc sin q4 - jd sin q1 = 0

a sin q2 + b sin q3 - c sin q4 = 0,

since q1 = 0 sin q1 = 0 and the j divides out


a cos q2 + b cos q3 - c cos q4 - d = 0

a sin q2 + b sin q3 - c sin q4 = 0

• a,b,c,d are known• One of the three angles is given

• 2 unknown angles remain

• 2 equations given above

• Solve simultaneously for remaining angles


Solving these two equation simultaneously is straightforward, but a tedious job. First step is to eliminate q3 and solve for q4

a cos q2 + b cos q3 - c cos q4 - d = 0a sin q2 + b sin q3 - c sin q4 = 0


• Using half angle identities, which will convert the sinq4 and cosq4 terms to tanq4

• This result in following simplified from, where the link lengths and known input value (q2) terms have been collected as constants, A, B and C

• Note that above equation is in quadratic form, and the solution is


Summary: Position Analysis of Pin-Jointed Fourbar Linkage using Vector Loop Method

Example: 1

Calculate the values of θ4 when the input

angle, θ2 = 30°

Example: 1 (contd.)

Example: 1 (contd.)

Example: 1 (contd.)

Plot of Output Variable Versus InputVariable

• Using a Matlab program, we can plot the output variables for all values of the input variable (0° ≤ θ2 ≤ 360°)

So Why Should we Choose Analyticalover Graphical Methods?

A plot showing all output values for any input value (0° ≤ θ2 ≤ 360°) can be obtained by using

the analytical method together with a math program (as here in Matlab)

To obtain the same results using the graphical method, 360 diagrams have to be made, which is cumbersome and not practical

Has Two Solutions, solution may be:

• Real and Equal one value of q4 corresponding to any one

value of q2

• Real and Unequal i.e. two values of q4 corresponding to

any one value of q2 (Crossed or Open)

• Complex Conjugates Link lengths are incapable of

connection for the value of q2 under consideration.


Vector Loop Method: Summary

1. Draw and label vector loop for mechanism2. Write vector equations3. Substitute Euler identity4. Separate into real and imaginary parts5. 2 equations, 2 unknown angles6. Solve for 2 unknown angles

Note: there will be two solutions since mechanism can be open or crossed

Solution for θ3

Solving these two equation simultaneously is straightforward, but a tedious job. First step is to eliminate q4 and solve for q3

a cos q2 + b cos q3 - c cos q4 - d = 0a sin q2 + b sin q3 - c sin q4 = 0

Solution for θ3

Example 1:Analytic Position Analysis

• Input position q2 given

• Solve for q3 & q4

q2=51.3°

a=1.6

b=2.14

c=2.06

d=3.5

q3=?°

q4=?°

Example 1:Vector Loop Equation

R2 + R3 - R4 - R1 = 0

aejq2 + bejq3 - cejq4 - dejq1 = 01.6ej51.3Þ + 2.14ejq3 - 2.06ejq4 - 3.5ej0° = 0

q2=51.3°

a=1.6

b=2.14

c=2.06

d=3.5

q3=?°

q4=?°

R2

R3

R4

R1

Example 1:Analytic Position Analysis (contd.)

aejq2 + bejq3 - cejq4 - dejq1 = 0a(cosq2+jsinq2) + b(cosq3+jsinq3) - c(cosq4+jsinq4)

- d(cosq1+jsinq1)=0

Real part:a cos q2 + b cos q3 - c cos q4 - d = 0

1.6 cos 51.3 + 2.14 cos q3 - 2.06 cos q4 - 3.5 = 0

Imaginary part:a sin q2 + b sin q3 - c sin q4 = 0

1.6 sin 51.3 + 2.14 sin q3 - 2.06 sin q4 = 0

q2=51.3°

a=1.6

b=2.14

c=2.06

d=3.5

q3=?°

Example 1: Solution - Open Linkage2 equations from real & imaginary equations

1.6 cos 51.3 + 2.14 cos q3 - 2.06 cos q4 - 3.5 = 0

1.6 sin 51.3 + 2.14 sin q3 - 2.06 sin q4 = 0

2 unknowns: q3 & q4

Solve simultaneously to yield2 solutions.

Open solution: q3 = 21°, q4 = 104°

q2=51.3°

a=1.6

b=2.14

c=2.06

d=3.5

q3=21°

q4=104°

RecapIn case of a four-bar, the vectors in the loop closure

equation have fixed magnitudes.

However, the angular inclinations of the three vectors representing the moving links will change. Hence, there are three position variables (q12, q13 and q14). Note q11 for fixed link is 0.

If one of these variables is defined, the remaining two variables can be solved from the vector equation; using scalar and imaginary parts respectively.

Recap (contd.)If we refer to the definition of the degree-of-freedom

of a mechanism, the variable that must be defined is

the input variable; and, for a constrained motion the

number of input variables must be equal to the

degree-of-freedom of the joints involved.

In case of a four-bar, since all the connections are

revolute joints, the variables are all rotation

variables.

Loop closure Eq. of Slider-crank mechanism

In case of a prismatic joint, the variable will be the magnitude of a vector or a vector component.

Consider a slider-crank mechanism as shown in Figure A. Let us disconnect the revolute joint at B (Figure B)

In order to determine the positions of links 2 and 3 we must define q12and q13.

To locate the position of link 4 its displacement along the slider axis must be known and the position variable s14 must be defined. The resulting loop closure equation is:

AoA + AB = AoB

Again there are three variables (q12, q13 and s14) one of which must be specified as the input. In this case the vectors AoA and AB have fixed magnitudes and varying directions. The vector AoB has a fixed y component (length c) and a changing x component (s14).

Depending on the applications either q12(i.e. in pumps) or s14 (i.e. internal combustion engines) is the input.

In complex numbers the vector loop equation will be:

Loop closure Eq. of Slider-crank mechanism (contd.)

lyrespective 3 & 2link of lengths:,

;

32

14321312

aa

where

jcseaea jj

Important Note!!The vectors defined and the variables used in the loop

closure equations are not unique. For example, for the slider crank mechanism, rather than

disconnecting the revolute joint at B, one can as well disconnect the revolute joint at A between links 2 and 3, as shown in figure below.

We must now define the angle q13‘ i.e. ,<BA instead of the angle q13 to determine the position of link 3.

Note that the angles q13 and q13‘ differ by a constant angle (In this case by 180o). Thus, the resulting loop equation is:

AoA = AoB + AB

In complex numbers the vector loop equation will be:

Important Note!! (contd.)

'1312

3142

jj eajcsea

Example: 2

• The linkage is driven by moving the sliding block 2. Write the loop-closure equation. Solve analytically for the position of sliding block 4 i.e. RA.

45;15,200, mmRwhere AB

Example: 2 (contd.)At first, drawing kinematic diagram with XY coordinate system

Now, the loop closure equation by disconnecting at A is

RA = RB + RAB

B

A

RAB

Example: 2 (contd.)Now writing the loop closure equation in complex form

.54615sin

135sin200

15sin

135sin

135sin15sin

get weequation, above of

componentsimaginary theequating Now

135.15.

)45(12/

mmRR

RR

eRReR

eRReR

ABA

ABA

j

ABB

j

A

j

ABB

j

A

The fourbar Slider-crank Position solution• Following figure shows an offset fourbar slider-crank

linkage.• The term off set‑ means that the slider axis extended

does not pass through the crank pivot.

This linkage could be represented by only three position vectors, R2, R3, and Rs, but one of them (Rs) will be a vector of varying magnitude and angle.

It will be easier to use four vectors, R1, R2, R3, and R4 with R1 arranged parallel to the axis of sliding and R4 perpendicular.

It can be noted that the pair of vectors R1 and R4 are orthogonal components of the position vector Rs from the origin to the slider.

The variable-length, constant-direction vector R1 then represents the slider position with magnitude d. The vector R4 is orthogonal to R1 and defines the constant magnitude offset of the linkage.

The coupler's position vector R3 is placed with its root at the slider which then defines its angle q3 at point B.

This particular arrangement of position vectors leads to a vector loop equation similar to the pin jointed fourbar example:

R2 - R3 - R4 - R1 = 0

Letting the vector magnitudes (link lengths) be represented by a, b, c, d as shown, we can substitute the complex number equivalents for the position vectors.

Now separating the real and imaginary components

Real:a cos 2 - b cos 3 - c cos 4 - d cos 1 = 0

a cos 2 - b cos 3 - c cos 4 - d = 0,

since q1 = 0 cos q1 = 1

Imaginary: ja sin 2 + jb sin 3 - jc sin 4 - jd sin 1 = 0

a sin 2 - b sin 3 - c sin 4 = 0,

since q1 = 0 sin q1 = 0 and the j divides out

a cos q2 - b cos q3 - c cos q4 - d = 0 (Eq:A)

a sin q2 - b sin q3 - c sin q4 = 0 (Eq:B)

• We want to solve above equations simultaneously for the two unknowns, link length d and link angle q3.

• The independent variable is crank angle q2.

• Link lengths a and b, the offset c, and angle q4 are known.

• But note that since we have set up the coordinate system to be parallel and perpendicular to the axis of the slider block, thus the angle q1 is zero and q4 is 90°.

• Equation B can be solved for q3 and the result substituted into equation A to solve for d. The solution is:

• Arcsine function is multivalued. For instance:sin-1(0) = 0° and sin-1(0) = 180°

• Calculator will only give the arcsine value between ±90° representing only one configuration of the linkage

The fourbar Slider-crank Position solution (contd.)

To get the other value of θ3 for the second

configuration:

The fourbar Slider-crank Position solution (contd.)

Example: 1

Determine the values of θ3 and d when the input angle, θ2 = 80°

Example: 1 (contd.)

Example: 1 (contd.)

Position of Any Point on a Linkage

• Determine position of points on the mechanism,

instead of finding the output variables.

Can only be done once all the output variables are

found

This is the ULTIMATE goal of this chapter

Position of points on the links

Consider finding the position of Point S (relative to O2)

• Draw a position vector from the fixed pivot 02 to point S.

• This vector RS02 makes an angle with the vector RA02.

• This angle is completely defined by the geometry

of link 2 and is constant. • The position vector for point S is then:

2

2

Consider finding the position of Point P (relative to O2)

• Similarly, the position of point U on the link 4

can be found using the angle which has a

constant angular offset with the link 4.

Consider finding the position of Point U (relative to O4)

4

Example: Determine position of point P

RPO2= 3.99 @26.5o

Review - Law of Cosines

AB

C

q

cosq=A2 + B2 - C 2

2 AB

q=arccosA2 + B2 - C 2

2 AB

é

ë ê

ù

û ú

Transmission Angles• The transmission angle μ is defined as the angle

between the output link and the coupler.

• For our four bar linkage example, it would be the

difference between q3 and q4 .

• It is usually taken as the absolute value of the acute

angle.

Extreme Values of the Transmission Angle• For a Grashof crank-rocker fourbar linkage the

extreme values of the transmission angle will occur when the crank is collinear with the ground link as shown in Figure.

The values of the transmission angle in these positions

are easily calculated from the law of cosines since the

linkage is then in a triangular configuration.

The sides of the two triangles are link 3, link 4, and

either the sum (when extended) or difference (when

overlapped) of links 1 and 2.

Measure of quality of force transmission. How? Ideally, as close to 90° as possible. Why?

Extreme Values of the Transmission Angle (contd.)

Because…..• If the measured angle between the coupler and output link is

greater than 90°, the transmission angle is calculated as 180° minus the measured angle.

• Thus, the maximum possible, and optimal, transmission angle is 90°.

• At this angle, all of the force (generated from the torque) is transferred to the output link.

• As the transmission angle deviates from 90°, some component of the force is not acting on the output link.

• At 45°, only about 70% of the force is producing desirable work. As a rule of thumb, machine designers try to keep the minimum transmission angle above ~40°.

• Transmission of motion is impossible when transmission angle is either 0 or 180 degrees, as in such case no load can be realized on output link.

• This figure shows that a small input transmission angle tends to maximize the axial force on the coupler, given an input -motor torque T that produces a force F = T/s (s= AD) normal to the crank at D

• Whereas, a large output transmission angle between coupler and rocker BC tends to: Maximize the torque, produced by the axial force on the coupler at point C, about the R-joint at B

• One extreme value of the transmission angle occurs

when links 1 and 2 are collinear and non-overlapping

(i.e. Extended) as shown in Figure.


as μ1 > 90°,

qtrans1 = 180° - μ1

• The other extreme transmission angle occurs when links 1 and 2 are collinear and overlapping as shown in Figure.


Transmission Angles

Transmission Angles: Summary

• Angle between coupler and output link should be 40º≤γ≤140º

Zero torque at output link if γ=0º or γ=180º

• A – ground link• B – input link• C – coupler• D – output link

A

BD

Cγ

Example

• Figure illustrates a crank-and-rocker four-bar linkage in the first of its two limit (toggle) positions. In a limit position, points 02,A, and B lie on a straight line; that is, links 2 and 3 form a straight line. The two limit positions of a crank-rocker describe the extreme positions of the rocking angle. Suppose that such a linkage has r1 = 400 mm, r2 = 200 mm, r3 = 500 mm, and r4 = 400 mm.

Toggle angles are a measure of when the crank torque will create a maximum force

a. Find θ2 (i.e. toggle angles of the crank link) and θ4

corresponding to each limit position.b. What is total rocking angle of link 4?c. What are the transmission angles at the extremes?

Example (contd.)

• From isosceles triangle (as r1 = r4 = 400mm) O4O2B we can calculate or measure

From triangle O2BO4 using law of cosines

29;58;29 42

From triangle O2B’ O4 using law of cosines

68

;136

;24868180

'

'4

'2

b. Since two limit positions of a crank-rocker describe the extreme positions of the rocker, therefore rocking angle is given by

c. The transmission angle μ is defined as the angle between the output link i.e. rocker in this case and the coupler, therefore the transmission angles at the extremes are given by

Example (contd.)

7858136' 444

68' & 29

Exercise: 1

• For a four-bar mechanism shown below, find extreme transmission angles. Find the two toggle angles of the crank AB as well.

6.228;1.40: '22 anglesToggle

)(1.53

);(9.81:

overlapped

extendedanglesonTransmissi

Transmission Angle: Slider Crank

• In the slider-crank, the transmission angle is measured between the coupler and a line normal to the sliding direction.

• The values for the minimum (@ 270 degrees) and maximum (@ 90 degrees) transmission angles can be determined by geometrically constructing the configurations as shown in Figure

• Alternatively, the minimum and maximum transmission angles for a slider-crank can be calculated from

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Lect Position Analysis (2)