Velocity Analysis

Velocity Analysis

Chapter 6

Definition • Rate of change of position with respect to time – Angular

d  dt

– Linear

dR V dt

– Position Vector R PA  pe j – Velocity

R PA j d Vpa   pje  pe j dt dt

Definition – Velocity (absolute) R PA j d VPA   pje  pe j dt dt •

The velocity is always in a direction perpendicular to the radius of rotation and is tangent to the path of motion

VPA  pj cos   j sin    p sin   j cos   VPA  VP " Absolute"

Definition – Velocity (difference)

VPA  VP  VA VP  VA  VPA " on the same body "

– Relative Velocity

VPA  VP  VA

Graphical Analysis • Graphical Velocity Analysis VP  VA  VPA V  v  r

– Solve for angular ve locities ; 3 , 4 linear vel ocities; A, B, C

Graphical Analysis • Graphical Velocity Analysis VP  VA  VPA

V  v  r

Graphical Analysis • Example 6-1 – Given θ2, θ3, θ4, ω2 find ω3, ω4, VA, VB and VC – Position analysis already performed – 1. Start at the end of the linkage about which you have the most information. Calculate the magnitude of the velocity of point A,

vA   AO2 2

Graphical Analysis • Example 6-1 – 2. Draw the velocity VA

– 3. Move next to a point which you have some information, point B. Draw the construction line pp through B perpendicular to BO4 – 4. Write the velocity difference equation for point B vs. A

VB  VA  VBA

Graphical Analysis • Example 6-1 – 5. Draw construction line qq through point B and perpendicular to BA to represent the direction of VBA – 6. The vector equation can be solve graphically by drawing the following vector diagram

VB  VA  VBA

Graphical Analysis • Example 6-1 – 7. The angular velocities of link 3 and 4 can be calculated, 4 

vB BO4

3 

vBA BA

– 8. Solve for VC

VC  VA  VCA vCA  c3

Instant Center of Velocity • An instant center of velocity is a point, common to two bodies in plane motion, which point has the same instantaneous velocity in each body • The numbers of IC is calculated with; nn  1 C 2

• Linear graph is a useful way to keep track of which IC have been found

Instant Center of Velocity • Kennedy’s Rule – Any three bodies in plane motion will have exactly three instant centers, and they will lie an the same straight line

Instant Center of Velocity

Instant Center of Velocity • Slider-Crank Linkage

Instant Center of Velocity • Slider-Crank Linkage

Instant Center of Velocity • Slider-Crank Linkage • Check Example 6-4: IC for a CamFollower Mechanism

Velocity Analysis with IC • Once the ICs have been found, they can be used to do a very rapid graphical velocity analysis of the linkage v A   AO2 2 vA 3  AI1,3 

vB 4  BO4 

vB  BI1,3 3

vC  CI1,3 3

Velocity Analysis with IC • A rapid graphical solution for the magnitudes at B and C are found from vectors drawn perpendicular to that line at the intersection of the arcs and line AI1,3 (VB’, VC’)

• Angular Velocity Ratio – Output angular velocity divided by the input angular velocity 4 mV  2

– Can be derived by constructing a pair of effective links

– Effective link pairs is two lines, mutually parallel, drawn through the fixed pivot and intersecting the coupler extended O2 A  O2 Asin O4 B  O2 Bsin 

VA  O2 A2

VA  VB

4 O2 A O2 A sin mV    2 O4 B O4 B sin 

– Now the effective links are colinear and intersect the coupler at the same point, I2,4 4 O2 I 2, 4 mV   2 O4 I 2, 4

• Mechanical Advantage – Power in a mechanical system,

  P  F V  FxVx  FyVy

– For rotating system,

P  T

Tout in  Tin out

Pout – Mechanical efficiency,   Pin

– Mechanical Advantage, Fout  Tout  rin   in  rin   O4 B sin   rin            mA    Fin  Tin  rout   out  rout   O2 A sin  rout 

Centrodes • The path, or locus, created by a IC at successive positions

Centrodes

Velocity of Slip • Used when there is a sliding joint between two links and neither one is the ground – Example 6-5, 6-6

Vslip42  VA4 slip  VA2 slip

4  3 

VA4 AO4

Velocity of Slip

4  3 

VA3 AO3

Analysis Solution • Position Analysis (revisited)   

 R2  R3  R4  R1  0

ae

j 2

 be

j3

 ce

j 4

 de

j1

0

3

  E  E 2  4 DF    2 arctan    2D  

4

  B  B 2  4 AC    2 arctan    2 A  

1, 2

1, 2

Analysis Solution • Velocity Analysis ae j2  be j3  ce j4  de j1  0





d ae j 2  be j3  ce j 4  de j1  0 dt d d d jae j 2 2  jbe j3 3  jce j 4 4  0 dt dt dt

ja2e

j 2

 jb3e

j3

 jc4e

j 4

0

Analysis Solution • Velocity Analysis VA  VBA  VB  0 VA  ja2 e j 2 VBA  jb3e j3 VB  jc4 e j 4

Euler identity real part imaginary part

Analysis Solution • Velocity Analysis a2 sin  4   2  3  b sin 3   4  a2 sin  2  3  4  c sin  4  3  VA  ja2 cos  2  j sin  2   a2  sin  2  j cos  2 

VBA  jb3 cos  3  j sin  3   b3  sin  3  j cos  3  VB  jc4 cos  4  j sin  4   c4  sin  4  j cos  4 

Analysis Solution – Slider-Crank

    R2  R3  R4  R1  0

ae j2  be j3  ce j4  de j1  0

ja2e j2  jb3e j3  d  0 VA  VAB  VB  0

VAB  VBA

VB  VA  VBA

Analysis Solution – Slider - Crank 3 

a sin  2 2 b sin  3

d  a2 sin  2  b3 sin 3

VB  VA  VBA

VA  a2  sin  2  j cos  2 

VAB  b3  sin  3  j cos  3  VBA  VAB Review - Inverted Slider - Crank

Geared Fivebar      R2  R3  R4  R5  R1  0

ae j2  be j3  ce j4  de j5  fe j1  0

ja2e

j 2

 jb3e

5   2  

j3

 jc4e

5   2

j 4

 jd5e

j5

0

Geared Fivebar

3  

2 sin  4 a2 sin  2   4   d5 sin  4  5  bcos3  2 4   cos 3 

a2 sin  2  b3 sin 3  d5 sin 5 4  c sin  4 VA  a2  sin  2  j cos  2 

VBA  b3  sin  3  j cos  3  VC  d5  sin  5  j cos  5  VB  VA  VBA

Velocity of Any Point • Once the angular velocities of all the links are found it is easy to define and calculate the velocity of any point on any link for any input position of the linkage

Velocity of Any Point • To find the velocity of points S & U R S 02  R S  se j 2  2   scos 2   2   j sin 2   2 

VS  jse j 2  2 2  s2  sin2   2   j cos2   2  RU 04  ue j 4  4   ucos 4   4   j sin 4   4 

VU  jse j 4  4 4  u4  sin4   4   j cos 4   4 

Velocity of Any Point • To find the velocity of point P R PA  pe j 3  3   pcos3   3   j sin 3   3  R P  R A  R PA

VPA  jpe j 3  3 3  p3  sin 3   3   j cos3   3  VP  VA  VPA