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3. Velocity Analysis of Linkages

• Velocities and accelerations in mechanisms are determined by different methods.

i. Velocity and acceleration analysis using vector mathematics

• velocity and acceleration of a point are expressed relative to fixed or moving coordinates.

ii. Velocity and acceleration analysis using equations of relative motion

• Can be solved graphically by velocity and acceleration polygons or by using trigonometric relations.

iii. Velocity and acceleration analysis by using complex numbers.

iv. Vectors velocity analysis using the instant center method.

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3.1. Velocity Analysis by Vector Mathematics• Consider the motion of point P moving with respect to the x- y- z

coordinate system, which in turn, moves relative to the X-Y-Z coordinate system as shown.

- is the position vector of P relative to the X-Y-Z system. - is the position vector of P relative to the x-y-z system - is the position vector of the origin of the moving coordinate

system x-y-z relative to the fixed coordinate system X-Y-Z.

pR

R

oR

3

The position vector of P relative to the X-Y-Z system Rp is expressed as:

(3.1)

Introducing unit vectors i, j, and k along the x, y, and z axes respectively,

= xi + yj + zk (3.2)

Velocity of P relative to the X-Y-Z coordinate system is

(3.3)

- is the velocity of the origin of x-y-z system relative to the fixed system

(3.4)

RRRV opp

oo VR

)()( kzjyixkzjyix

zkyjxidt

dR

RRR op

R

4

Let

(3.5)

note that

(3.6)

Where is the angular velocity vector of x-y-z system relative to X-Y-Z.

(3.7)

- Thus the velocity of P relative to the moving coordinate system is

(3.8)

Vkzjyix

kk

jj

ii

R

zkyjxi

kzjyixkzjyix

)(

)()()(

RVR

5

the velocity of point P relative to the fixed system is:

= velocity of the origin of the x-y-z system relative to the X-Y- Z system

= velocity of point p relative to x-y-z system

= angular velocity of the x-y-z system relative to X-Y-Z system

= position vector of P with respect to the origin of the x-y-z system

)9.3(RVVV op

oV

V

R

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3.2. Velocity analysis by using equation of relative motion

3.2.1. Velocity of points on a common link• A and B are two points on a common rigid link AB as shown.• The points are moving with velocities VA and VB respectively. • Using the equation of relative motion, velocity of one point can be

determined relative to the other.

VA = VB + VA/B (3.10)

Where VA/B = velocity of A relative to B• All absolute velocity vectors originate from the same point O2. • Note that the velocity of A relative to B and the velocity of B

relative to A are equal in magnitude, collinear and opposite in direction, i.e.

• VA/B = -VB/A (3.11)

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3.2.2. Velocity of a block sliding on a rotating link

- As shown in the figure block A slides on the rotating link O2B.

- The angular velocity of the link and the velocity of the block are assumed to be known.

- Let A’ be a point on the link coincident with the block A for the instant represented.

- The velocity of A’ relative to O2 is perpendicular to O2B at A’.

- The velocity of A relative to A’ is along the link parallel to O2A’

- VA = VA’ + VA/A’ (3.12)

- Relative velocity of coincident particles on separate links is effected by physical constraints such as guides.

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3.2.3. Relative velocity of coincident particles at the point of rolling element

- Rolling contact exists when there is no sliding at the contact point b/n two links.

The velocity component along the tangential direction must be zero.

- For pure rolling contact of links 2 and 3, the point P2 on link 2 and P3 on link 3 have the same velocities i.e.

VP2 = VP3

Let a point P be common to the links 2 & 3,which have relative motion to each other.The relative velocity equation may be written:

VP3 = VP2 + VP3/P2 (3.13)

- The condition for pure rolling is that:VP3/P2 = 0 (3.14)

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• This condition is met when the point of contact lies on the line of center O2O3 .

• If VP3/P2 ≠ 0, its direction would be along the tangent t - t.

link 3 would slide relative to link 2 along the t – t direction.

• Rolling of circles or cylinders is a special case of rolling motion.

For pure rolling:)15.3(

2

3

3

2

r

r

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3.2.4 . Relative velocity of crank and connecting rod

• Let 2 be the angular velocity of the crank O2A.

• Velocity of B can be determined using the velocity of point A as the reference which can easily be determined.

VB = VA + VB/A (3.16)

Where VA is known both in magnitude and direction;

VB is known in direction, magnitude is unknown;

VB/A is known in direction, magnitude is unknown.

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3.2.5. Algebraic solution of the slider-Crank mechanism

• Taking the origin of the coordinate system at the crank center, the position of the slider is defined by x.

(3.17)

and

(3.18)

Where r is the crank radiusl is length of connecting rod

From the position of the mechanism it can be noted that

sinsin lr

coscos lrx

)19.3()sin(1)sin(

cos 222

l

r

l

rl

12

• Substituting for cos in equation (3.18), the position of the slider is obtained to be:

(3.20)

• Alternatively, equation (3.20) can be obtained from the law of cosines given by

and by solving quadratic equation for x.• Differentiating equation (3.17)

(3.21)

2)sin(1cos l

rlrx

cos2222 rxxrl

dt

d

l

r

dt

ddt

d

l

r

dt

d

cos

cos

coscos

13

• Differentiating equation (3.18) with respect to time:

(3.22)

Substituting for d/dt from equation (3.21), the velocity of the slider is:

(3.23)

• For small values of r/l, which usually is the case in slider crank mechanism,

and the velocity is (3.24)

dt

dl

dt

drx

sinsin

2)sin(1

cossinsin

)cos

cossin(sin

lr

l

rrV

l

r

llrxV

0sin l

r

)2

2sin(sin

l

rrV

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3.3. Velocity Analysis by Complex Numbers

• Most of the systems of analysis using complex polar notation are based on the following fundamental law:– If the elements of a mechanism are replaced by position

vectors such that their sum is zero, then their time derivatives are also equal to zero.

• This law means that if one takes any linkage or mechanism and replaces the members of the mechanism by vectors such that their sum is zero, then the sum of the velocity vectors is zero, so also the sum of the acceleration vectors.

Considering the slider crank shown:• Link 2 is the driver (crank) and has a constant angular velocity 2 & for

the instant under consideration an angular position of 2.

• Dimensions of linkages are assumed to be known, so the angular position of the follower, link 4, can be obtained.

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• Replacing each link by a vector such that the position polygon closes as shown in figure (b), a mathematical expression for the summation law can be written as:

Where R1 = vector for the ground link

R2 = vector for the crank

R4 = vector to determine the position of link 3. Note that the magnitude of R4 is variable.

• The position of a particle on a link represented by a vector Rp as shown below may be expressed in any of the following equivalent forms:

• Using this complex representation, equation (3.25) is transformed into

)25.3(0421 RRR

2

)26.3(sincos 22

i

pp

pp

p

erR

irR

ibaR

)27.3(0421421 iii ererer

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• Differentiating the above equation we obtain

Note that r1 and 1 are constants and r4 is variable.

• Let

• Separating equation (3.29) into real and imaginary terms:

• The unknown quantities in the above pair of equations are 4 and

solving for these unknown variables:

)28.3(044244422 iii ereireir

ii

)29.3(044244422 iii ereireir

)30.3(0sincossin

0cossincos

44444222

44444222

rrr

rrr

4r

)31.3()cos(

)sin(

244

224

24224

r

r

rr

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3.4. Analysis of Velocity Vectors by Instant-Center Method.

Consider bodies 2 and 3 which move relative to each other.

• Point A on a body 2 has a velocity VA2/A3 relative to point A on body 3 and point B on 3 has velocity VB3/B2 relative to point B on body 2.

• Perpendiculars to both velocities intersect at point P which is the instantaneous center of rotation of body 2 relative to 3 or vise versa.

• Point P may be considered as a point on body 2 about which body 3 is instantaneously rotating or vise versa.

VP2/P3 = VP3/P2 = 0 An instantaneous center of rotation is

defined as a point common to two links

which has the same velocity in each link; It is a point at which the two bodies

have no relative velocity. It is also a point on one link about

which another link is instantaneously

rotating.

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3.4.1 Types of Instantaneous centersThe instantaneous centers of a mechanism are of three types:1. Fixed instantaneous centers: these instantaneous centers (I.C.)

remain fixed for all configurations of the mechanism.2. Permanent instantaneous centers: these instantaneous centers

move with the mechanism but joints are of permanent nature.3. Neither fixed nor permanent instantaneous centers: these

instantaneous centers vary with the configuration of the mechanism.

• In the four bar linkage shown, points O2, O4, A and B are the obvious I.C.

– O2 and O4 are fixed I. C;

– A and B are Permanent I. C.• In the four bar mechanism there

is one I.C. for each pair of links. • In general for n links in a mechanism:

Number of I. C. =2

)1( nn

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3.4.2 The Arnhold- Kennedy Theorem of Three Centers

• Since an instantaneous center is a point common to two links, it is usually denoted by the number of links.

• For the four-bar linkage shown centers 12, 23, 34, and 14 are located by inspection.

• The instantaneous center of the perpendiculars to the velocities of points A and B yield the instantaneous center 13.

• To locate the center 24, the Arnhold–Kennedy theorem is applied. The theorem states that:

When three bodies move to one another they have three instantaneous centers, all of which lie on the same straight line.

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• For links 1, 2, and 4 the instantaneous centers 12, 24, and 14 should lie on a straight line. Similarly, for links 2, 3, and 4 centers 23, 34, and 24 should lie on another straight line, where 24, common to both lines, is located at the intersection of these lines.

• In general, to determine the instantaneous centers for the four bar linkage we could proceed as follows.

– For each combination of 3 links the known and unknown instantaneous centers should lie on a straight line.

– Intersection of such line that contain instantaneous centers give the unknown instantaneous center.

Table. 3.1Link

Instantaneous Centers

Known Unknown

1, 2, 3 12, 23 13

1, 2, 4 12, 14 24

1, 3, 4 34, 14 13

2, 3, 4 23, 34 24

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3.4.3 Determination of Velocity using the line-of-centers method.

• The line of centers method can be summarized as follows

Step 1- Identify the link containing a point whose velocity is known, the link containing the point whose velocity is to be determined and a reference link, usually the ground or the frame.

Step 2- Locate the three instantaneous centers defined by the three links and draw the line of centers.

Step 3- Consider the common instantaneous center as a point on the link which contains the point whose velocity is known and by using similar triangles find the velocity of the common instantaneous center.

Step 4- next consider the common instantaneous center as a point on the link which contains the point whose velocity is unknown. From similar triangles determine the velocity of the point.

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