Acceleration And Velocity

1.1 Acceleration and velocity Kinematic analysis of a mechanism consists of calculating position, velocity and acceleration of any of its points or links. Such an analysis requires knowledge of linkage dimensions, its position, velocity and acceleration of as many number of links as the degrees of freedom the linkage has. Velocity analysis is usually performed following a position analysis; i.e., the position and orientation of all the links in a mechanism are assumed known. We will use two different methods to find the velocity of a point or link in a mechanism. 1. The relative velocity method. 2. The instant center of rotation method.
1.1.1 Relative Velocity Method of determining velocity of a point or link in a mechanism Motion of a link OA is a rigid link, rotating about O with uniform velocity ω rad/s in counter clockwise direction. Let OA turn a small angle δθ in time δt. The point A in this time will travel from A to A’, hence, Motion of a link (Fig.- 6.1) Motion of a link (Fig.- 6.2) Now, since AO is a rigid body the point A with respect to O can neither approach O nor recess O. This means that the velocity of A can only be perpendicular to the line joining A and O. Let B be any other point on AO. Hence, velocity of B with respect to O will be Velocity of B = ω , Ob perpendicular to OB. If ob represents the velocity of B, It can be observed that $\frac {ob}{oa} = \frac { \omega OB}{\omega OA} = \frac {OB}{OA} \qquad --- \hbox{ (Eqn.-6.1) }$ i.e. ob represents the velocity in the same ratio as B divides the link.
1.1.2 Polygon Method Velocity polygon is a graphical pencil-and-paper approach for determining unknown velocities of a single degree-of-freedom mechanism. The method requires constructing a velocity loop equation (a polygon) graphically. A polygon may have three or more edges depending on the number of velocity vectors in the equation. For a vector loop equation, the polygon method is the graphical procedure of solving two algebraic equations in two unknowns. The velocity polygon method is demonstrated for several commonly used mechanisms. Four Bar Mechanism In a given four bar mechanism (Length of the links known) and the crank speed ω is known let us try to find the angular velocity of other two links. Velocity Polygon 4Bar Secondary Equation To determine the velocity of a secondary point, such as a coupler point, we refer to the position expression and the corresponding velocity expression : $R_{PO_2}= R_{AO_2}+R_{PA} \qquad --- \hbox{ (Eqn.-6.2) }$ $V_P =V_A+V_{PA} = \omega_2 R_{AO_2} + \omega_3 R_{PA} \qquad --- \hbox{ (Eqn.-6.3) }$ since the angular velocities are already known, VA and VPA are constructed. We add these two vectors graphically to determine VP. Velocity Secondary
1.1.3 Slider Crank mechanism (inversion 1) This slider-crank mechanism in the given configuration has a known angular velocity of the crank, ω2. We want to determine ω3 and the velocity of the slider block. In this example we assume ω 2 is CCW. Velocity Slider Crank Mechanism Slider Crank mechanism (Fig.- 6.5) The position vector loop equation is: $R_{AO_2} + R_{BA} - R_{BO_2} = 0 \qquad --- \hbox{ (Eqn.-6.4) }$ The velocity(loop) equation is expressed as $V_A + V_{BA} - V_B = 0 \qquad --- \hbox{ (Eqn.-6.5) }$ We note that VA and VBA are tangential and VB is of slip type ( along the axis of RBO2 ). Therefore the velocity equation can be expressed as $\omega_2 R_{AO_2} + \omega_3 R_{BA} - V_B = 0 \qquad --- \hbox{ (Eqn.-6.6) }$ The instantaneous Center of rotation method Instant center of velocities is a simple graphical method for performing velocity analysis on mechanisms. The method provides visual understanding on how velocity vectors are related.
1.1.4 Instant center :- Instant center of velocities between 2 links is the location at which two coinciding points, one on each links, have velocities.
1.1.5 Number of instant Centers:- In a mechanism with ‘n’ links (count ground as one of the links), the number of instant center is determined as. $C = \frac {(n (n-1))}{2} \qquad --- \hbox{ (Eqn.-6.7) }$
1.2 Type of instantaneous center :- Instantaneous center can be fixed , permanent , neither fixed nor permanent . Fixed and permanent instantaneous center are primary instantaneous center and third type is called as secondary instantaneous centers .
1.2.1 Kennedy theorem :- The three instant centers between three planer links must lie on a straight line.
1.2.2 Location of instantaneous center :- When 2 links are connected by a pin joint , the instantaneous center lies on the center of the pin. The three instant centers between three planer links must lie on a straight line. 2 Bar When the two links have a pure rolling contact , the instantaneous center lies on their point of contact. Circle Case 2 (Fig.- 6.7) When the two links have a sliding contact , the instantaneous center lies on the common normal at the point of contact. a) When the link2 (slider) moves on fixed link 1 having straight surface. 3 Case 3a (Fig.- 6.8) b) When the link 2 (slider) moves on fixed link 1 having curved surface , the instantaneous center lies on the center of curvature of the curvilinear path in the configuration at that instant. Down Curv case 3b (Fig.- 6.9) c) When the link2 (slider) moves on a fixed link 1 having a constant radius of curvature , the instantaneous center lies at the center of curvature , is the center of circle for all configuration of the links. Up Curv Case 3c (Fig.- 6.10) Method of locating instantaneous center in a mechanism :- 1. First determine the number of instantaneous center by below formula $N = (n-1)/2 \quad \hbox {( Where n = Number of links)} \qquad --- \hbox{ (Eqn.-6.8) }$ $N = Nr \quad of \quad IC$ 2. Make a list of the instant centers. 3. Locate the fixed and permanent IC by inspection. 4. Locate the remaining neither fixed nor permanent IC by Kennedy’s theorem. This is done by the use of circle diagram. Mark points on a circle equal to the number of links in a mechanism. 5. Join the point by solid lines for all the IC that have been found. 6. For other instant centers , join two such points that line joining them forms two adjacent triangle in the circle diagram. The line which is responsible for completing two triangles, should be a common side to the two triangles.
1.2.3 Instant center for four bar mechanism A four-bar mechanism has six instant centers ((n-1)/2 = 6)regardless of the dimensions or orientation of the links. Instant center for four bar mechanism Since pin joints are instant centers, for a four-bar with four pin joints, four IC's are immediately identified. Each found IC is marked on the circle as a line drawn between the two corresponding link indices. These four IC's are actual (not imaginary) pin joints. In order to find the other IC's ,we apply Kennedy's rule. The IC’s between links 2, 3 and 4 must lie on a straight line. These are I 2,3 , I3,4 , and I2,4 . Since we already have I2,3 and I3,4 , we draw a line through them; I 2,4 must also be on this line. The IC’s between links 1, 2 and 4 must lie on a straight line. These are I1,2 , I1,4 , and I2,4 . Since we already have I1,2 and I1,4 , we draw a line through them; I2,4 must also be on this line. The intersection of these two lines is I2,4. In the circle diagram the red line between links 2 and 4 indicates the center we are after. This line is shared between two triangles with known IC’s. The triangles tell us to draw a line between I1,2 and I1,4 , then draw another line between I2,3 and I3,4 . The intersection is I2,4 . According to the circle, the last center to find is between links 1 and 3. The two triangles that share this new red line tell us to draw a line between I1,2 and I2,3 , and a second line between I 1,4 and I3,4. The intersection of these two lines is I1,3 .
1.2.4 Instant centre of a slider crank mechanism A slider-crank mechanism has six instant centers regardless of which inversion it is. Again, for book keeping purposes, we draw a circle with link indices. Instant centre of a slider crank mechanism
1.2.5 Instant centres of a six bar mechanism A slider-crank mechanism has six instant centers regardless of which inversion it is. Again, for bookkeeping purposes, we draw a circle with link indices. Instant centres of a six bar mechanism
1.2.6 Acceleration Analysis In kinematic analysis of mechanisms, acceleration analysis is usually performed following a velocity analysis; i.e., the positions and orientations, and the velocities of all the links in a mechanism are assumed known. Acceleration of a point - The acceleration of a point is the relationship between the change of its velocity vector and time. Relative Acceleration of Two Points in the Same Rigid Body As the distance between two points of a rigid body cannot change, relative motion between them is a rotation of one point about the other. Let point B rotates about point A, both being part of a link that moves with angular velocity ω and angular acceleration α. The relative acceleration vector of point B with respect to point A can be broken into two components: Acceleration Analysis (Fig.- 6.14) The normal component, anBA, is always perpendicular to the relative velocity vector and it points towards the center of curvature of the trajectory. In this case, it points towards point A. The tangential component, atBA, has the same direction as the relative velocity vector. In the example above as the direction of angular acceleration α opposes the direction of angular velocity ω, the tangential component points in the opposite direction to relative velocity vBA, which means that the magnitude of this velocity is decreasing. These normal and tangential components of the relative acceleration of point B with respect to point A can be obtained by the below equations $a_{BA}^n= ω^2.AB \qquad --- \hbox{ (Eqn.-6.9) }$ $a_{BA}^t= AB.\alpha \qquad --- \hbox{ (Eqn.-6.10) }$ Where, ω is the angular velocity of the link. α is the angular acceleration of the link. AB is the length of the link. The magnitude of the acceleration can be determined by the magnitudes of its normal and tangential components. $a_A = \sqrt { {a^n_A}^2 + (a^t_A)^2 } \qquad --- \hbox{ (Eqn.-6.11) }$ The angle formed by the acceleration vector and the normal direction to the trajectory is given by $\phi = tan^{-1} \frac {a^t_A}{a^n_A} = tan^{-1} \frac { \alpha }{\omega^2} \qquad --- \hbox{ (Eqn.-6.12) }$ The above equations are valid only when the radius is constant. Acceleration in four bar mechanism -Polygon method Acceleration 4Bar For a known four-bar mechanism, in a given configuration and known velocities, and a given angular acceleration of the crank, α2 (say CCW), construct the acceleration polygon. Determine α3 and α4 . Secondary equation : We can use the polygon method to determine the acceleration of a coupler point, such as P. It is assumed that all the angular velocities and acceleration have already been determined. For the position vector $R_{PO_2} = R_{AO_2}+ R_{PA}\qquad --- \hbox{ (Eqn.-6.13) }$ acceleration expression becomes $A_P = A_A + A_{PA} = A_A^n +A_A^t +A_{PA}^n + A_{PA}^t \qquad --- \hbox{ (Eqn.-6.14) }$ $-\omega^2_2 R_{AO_2} + \alpha_2 R_{AO_2} - \omega^2_3 R_{PA} + \alpha_3 R_{PA} \qquad --- \hbox{ (Eqn.-6.15) }$ All four vector can be constructed graphically. The vector sum is the acceleration of P. Acceleration Secondary
1.3 Slider-crank(inversion 1) For a known slider-crank mechanism(inversion 1) in a given configuration and for known velocities, the acceleration of the crank , α2. is given. Construct the acceleration polygons, then determine α3 and the slider block. Assume α2 is given to be CCW. Acceleration Slider Crank The position and velocity vector loop equations are: $R_{AO_2} + R_{BA} - R_{BO_2} = 0 \qquad --- \hbox{ (Eqn.-6.16) }$ $V_A + V_{BA} - V_B = 0 \qquad --- \hbox{ (Eqn.-6.17) }$ Assume that all the velocities have already been obtained. The acceleration equation is obtained from the time derivative of the velocity equation: $A_A+ A_{BA} - A_B = 0 \qquad --- \hbox{ (Eqn.-6.18) }$
1.4 Klein's construction for slider crank mechanism The Klein’s construction is graphical method to obtain the velocity, acceleration of links or important points on the links. This construction is drawn directly on the configuration diagram. hence it does not involve two or three separate figures. However, it is applicable only for the slider-crank mechanism. Steps to find the velocity and acceleration of various links Draw the basic configuration diagram by measuring the angle made by crank and also other dimension of crank and connecting rod. Klein Klein's construction for slider crank mechanism (Fig.- 6.24) Extend the connecting rod up to the vertical center line of the crank circle and mark point M , the triangle created ΔOAM is the velocity triangle. Locate the midpoint of the connecting rod as point G. With Centre as “A” and radius equal to AM draw the circle. With Centre as “G” and radius equal to GA or GB draw the circle. Both circles will intersect each other at two points, join these two points This line will intersect the connecting rod at point “Q” and line of stroke at point “N”. Name these two points. Now OAM is the velocity triangle and the OAQN is the acceleration diagram. Which can be used to find the required velocity of acceleration of the links of various points on the links.

1.1 Acceleration and velocity

Kinematic analysis of a mechanism consists of calculating position, velocity and acceleration of any of its points or links. Such an analysis requires knowledge of linkage dimensions, its position, velocity and acceleration of as many number of links as the degrees of freedom the linkage has. Velocity analysis is usually performed following a position analysis; i.e., the position and orientation of all the links in a mechanism are assumed known. We will use two different methods to find the velocity of a point or link in a mechanism.

1. The relative velocity method.
2. The instant center of rotation method.

1.1.1 Relative Velocity Method of determining velocity of a point or link in a mechanism

Motion of a link

OA is a rigid link, rotating about O with uniform velocity ω rad/s in counter clockwise direction. Let OA turn a small angle δθ in time δt. The point A in this time will travel from A to A’, hence,

Motion of a link (Fig.- 6.1)

Motion of a link (Fig.- 6.2)

Now, since AO is a rigid body the point A with respect to O can neither approach O nor recess O. This means that the velocity of A can only be perpendicular to the line joining A and O.

Let B be any other point on AO. Hence, velocity of B with respect to O will be

Velocity of B = ω , Ob perpendicular to OB. If ob represents the velocity of B, It can be observed that

$\frac {ob}{oa} = \frac { \omega OB}{\omega OA} = \frac {OB}{OA} \qquad --- \hbox{ (Eqn.-6.1) }$

i.e. ob represents the velocity in the same ratio as B divides the link.

1.1.2 Polygon Method

Velocity polygon is a graphical pencil-and-paper approach for determining unknown velocities of a single degree-of-freedom mechanism. The method requires constructing a velocity loop equation (a polygon) graphically. A polygon may have three or more edges depending on the number of velocity vectors in the equation. For a vector loop equation, the polygon method is the graphical procedure of solving two algebraic equations in two unknowns. The velocity polygon method is demonstrated for several commonly used mechanisms.

Four Bar Mechanism

In a given four bar mechanism (Length of the links known) and the crank speed ω is known let us try to find the angular velocity of other two links.

Velocity Polygon 4Bar

Secondary Equation

To determine the velocity of a secondary point, such as a coupler point, we refer to the position expression and the corresponding velocity expression :

$R_{PO_2}= R_{AO_2}+R_{PA} \qquad --- \hbox{ (Eqn.-6.2) }$

$V_P =V_A+V_{PA} = \omega_2 R_{AO_2} + \omega_3 R_{PA} \qquad --- \hbox{ (Eqn.-6.3) }$

since the angular velocities are already known, VA and VPA are constructed. We add these two vectors graphically to determine VP.

Velocity Secondary

1.1.3 Slider Crank mechanism (inversion 1)

This slider-crank mechanism in the given configuration has a known angular velocity of the crank, ω2. We want to determine ω3 and the velocity of the slider block. In this example we assume ω 2 is CCW.

Velocity Slider Crank Mechanism

Slider Crank mechanism (Fig.- 6.5)

The position vector loop equation is:

$R_{AO_2} + R_{BA} - R_{BO_2} = 0 \qquad --- \hbox{ (Eqn.-6.4) }$

The velocity(loop) equation is expressed as

$V_A + V_{BA} - V_B = 0 \qquad --- \hbox{ (Eqn.-6.5) }$

We note that VA and VBA are tangential and VB is of slip type ( along the axis of RBO2 ). Therefore the velocity equation can be expressed as

$\omega_2 R_{AO_2} + \omega_3 R_{BA} - V_B = 0 \qquad --- \hbox{ (Eqn.-6.6) }$

The instantaneous Center of rotation method

Instant center of velocities is a simple graphical method for performing velocity analysis on mechanisms. The method provides visual understanding on how velocity vectors are related.

1.1.4 Instant center :-

Instant center of velocities between 2 links is the location at which two coinciding points, one on each links, have velocities.

1.1.5 Number of instant Centers:-

In a mechanism with ‘n’ links (count ground as one of the links), the number of instant center is determined as.

$C = \frac {(n (n-1))}{2} \qquad --- \hbox{ (Eqn.-6.7) }$

1.2 Type of instantaneous center :-

Instantaneous center can be fixed , permanent , neither fixed nor permanent . Fixed and permanent instantaneous center are primary instantaneous center and third type is called as secondary instantaneous centers .

1.2.1 Kennedy theorem :-

The three instant centers between three planer links must lie on a straight line.

1.2.2 Location of instantaneous center :-

When 2 links are connected by a pin joint , the instantaneous center lies on the center of the pin. The three instant centers between three planer links must lie on a straight line.

2 Bar

When the two links have a pure rolling contact , the instantaneous center lies on their point of contact.

Circle

Case 2 (Fig.- 6.7)

When the two links have a sliding contact , the instantaneous center lies on the common normal at the point of contact.

a) When the link2 (slider) moves on fixed link 1 having straight surface.

3

Case 3a (Fig.- 6.8)

b) When the link 2 (slider) moves on fixed link 1 having curved surface , the instantaneous center lies on the center of curvature of the curvilinear path in the configuration at that instant.

Down Curv

case 3b (Fig.- 6.9)

c) When the link2 (slider) moves on a fixed link 1 having a constant radius of curvature , the instantaneous center lies at the center of curvature , is the center of circle for all configuration of the links.

Up Curv

Case 3c (Fig.- 6.10)

Method of locating instantaneous center in a mechanism :-

1. First determine the number of instantaneous center by below formula

$N = (n-1)/2 \quad \hbox {( Where n = Number of links)} \qquad --- \hbox{ (Eqn.-6.8) }$

$N = Nr \quad of \quad IC$

2. Make a list of the instant centers.

3. Locate the fixed and permanent IC by inspection.

4. Locate the remaining neither fixed nor permanent IC by Kennedy’s theorem. This is done by the use of circle diagram. Mark points on a circle equal to the number of links in a mechanism.

5. Join the point by solid lines for all the IC that have been found.

6. For other instant centers , join two such points that line joining them forms two adjacent triangle in the circle diagram. The line which is responsible for completing two triangles, should be a common side to the two triangles.

1.2.3 Instant center for four bar mechanism

A four-bar mechanism has six instant centers ((n-1)/2 = 6)regardless of the dimensions or orientation of the links.

Instant center for four bar mechanism

Since pin joints are instant centers, for a four-bar with four pin joints, four IC's are immediately identified. Each found IC is marked on the circle as a line drawn between the two corresponding link indices. These four IC's are actual (not imaginary) pin joints. In order to find the other IC's ,we apply Kennedy's rule.

The IC’s between links 2, 3 and 4 must lie on a straight line. These are I 2,3 , I3,4 , and I2,4 . Since we already have I2,3 and I3,4 , we draw a line through them; I 2,4 must also be on this line. The IC’s between links 1, 2 and 4 must lie on a straight line. These are I1,2 , I1,4 , and I2,4 . Since we already have I1,2 and I1,4 , we draw a line through them; I2,4 must also be on this line. The intersection of these two lines is I2,4.

In the circle diagram the red line between links 2 and 4 indicates the center we are after. This line is shared between two triangles with known IC’s. The triangles tell us to draw a line between I1,2 and I1,4 , then draw another line between I2,3 and I3,4 . The intersection is I2,4 .

According to the circle, the last center to find is between links 1 and 3. The two triangles that share this new red line tell us to draw a line between I1,2 and I2,3 , and a second line between I 1,4 and I3,4. The intersection of these two lines is I1,3 .

1.2.4 Instant centre of a slider crank mechanism

A slider-crank mechanism has six instant centers regardless of which inversion it is. Again, for book keeping purposes, we draw a circle with link indices.

Instant centre of a slider crank mechanism

1.2.5 Instant centres of a six bar mechanism

A slider-crank mechanism has six instant centers regardless of which inversion it is. Again, for bookkeeping purposes, we draw a circle with link indices.

Instant centres of a six bar mechanism

1.2.6 Acceleration Analysis

In kinematic analysis of mechanisms, acceleration analysis is usually performed following a velocity analysis; i.e., the positions and orientations, and the velocities of all the links in a mechanism are assumed known.

Acceleration of a point - The acceleration of a point is the relationship between the change of its velocity vector and time.

Relative Acceleration of Two Points in the Same Rigid Body

As the distance between two points of a rigid body cannot change, relative motion between them is a rotation of one point about the other. Let point B rotates about point A, both being part of a link that moves with angular velocity ω and angular acceleration α. The relative acceleration vector of point B with respect to point A can be broken into two components:

Acceleration Analysis (Fig.- 6.14)

The normal component, anBA, is always perpendicular to the relative velocity vector and it points towards the center of curvature of the trajectory. In this case, it points towards point A.

The tangential component, atBA, has the same direction as the relative velocity vector. In the example above as the direction of angular acceleration α opposes the direction of angular velocity ω, the tangential component points in the opposite direction to relative velocity vBA, which means that the magnitude of this velocity is decreasing. These normal and tangential components of the relative acceleration of point B with respect to point A can be obtained by the below equations

$a_{BA}^n= ω^2.AB \qquad --- \hbox{ (Eqn.-6.9) }$

$a_{BA}^t= AB.\alpha \qquad --- \hbox{ (Eqn.-6.10) }$

Where,

ω is the angular velocity of the link.

α is the angular acceleration of the link.

AB is the length of the link.

The magnitude of the acceleration can be determined by the magnitudes of its normal and tangential components.

$a_A = \sqrt { {a^n_A}^2 + (a^t_A)^2 } \qquad --- \hbox{ (Eqn.-6.11) }$

The angle formed by the acceleration vector and the normal direction to the trajectory is given by

$\phi = tan^{-1} \frac {a^t_A}{a^n_A} = tan^{-1} \frac { \alpha }{\omega^2} \qquad --- \hbox{ (Eqn.-6.12) }$

The above equations are valid only when the radius is constant.

Acceleration in four bar mechanism -Polygon method

Acceleration 4Bar

For a known four-bar mechanism, in a given configuration and known velocities, and a given angular acceleration of the crank, α2 (say CCW), construct the acceleration polygon. Determine α3 and α4 .

Secondary equation :

We can use the polygon method to determine the acceleration of a coupler point, such as P. It is assumed that all the angular velocities and acceleration have already been determined.

For the position vector

$R_{PO_2} = R_{AO_2}+ R_{PA}\qquad --- \hbox{ (Eqn.-6.13) }$

acceleration expression becomes

$A_P = A_A + A_{PA} = A_A^n +A_A^t +A_{PA}^n + A_{PA}^t \qquad --- \hbox{ (Eqn.-6.14) }$

$-\omega^2_2 R_{AO_2} + \alpha_2 R_{AO_2} - \omega^2_3 R_{PA} + \alpha_3 R_{PA} \qquad --- \hbox{ (Eqn.-6.15) }$

All four vector can be constructed graphically. The vector sum is the acceleration of P.

Acceleration Secondary

1.3 Slider-crank(inversion 1)

For a known slider-crank mechanism(inversion 1) in a given configuration and for known velocities, the acceleration of the crank , α2. is given. Construct the acceleration polygons, then determine α3 and the slider block. Assume α2 is given to be CCW.

Acceleration Slider Crank

The position and velocity vector loop equations are:

$R_{AO_2} + R_{BA} - R_{BO_2} = 0 \qquad --- \hbox{ (Eqn.-6.16) }$

$V_A + V_{BA} - V_B = 0 \qquad --- \hbox{ (Eqn.-6.17) }$

Assume that all the velocities have already been obtained.

The acceleration equation is obtained from the time derivative of the velocity equation:

$A_A+ A_{BA} - A_B = 0 \qquad --- \hbox{ (Eqn.-6.18) }$

1.4 Klein's construction for slider crank mechanism

The Klein’s construction is graphical method to obtain the velocity, acceleration of links or important points on the links. This construction is drawn directly on the configuration diagram. hence it does not involve two or three separate figures. However, it is applicable only for the slider-crank mechanism.

Steps to find the velocity and acceleration of various links

Draw the basic configuration diagram by measuring the angle made by crank and also other dimension of crank and connecting rod.

Klein

Klein's construction for slider crank mechanism (Fig.- 6.24)

Extend the connecting rod up to the vertical center line of the crank circle and mark point M , the triangle created ΔOAM is the velocity triangle.

Locate the midpoint of the connecting rod as point G.

With Centre as “A” and radius equal to AM draw the circle.

With Centre as “G” and radius equal to GA or GB draw the circle.

Both circles will intersect each other at two points, join these two points

This line will intersect the connecting rod at point “Q” and line of stroke at point “N”. Name these two points.

Now OAM is the velocity triangle and the OAQN is the acceleration diagram. Which can be used to find the required velocity of acceleration of the links of various points on the links.

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