Velocity & Acceleration Analysis - ariel.ac.il · •Note that: –The direction and magnitude of V A is a known function of the input angular velocity, ω 2 –The mechanism’s

• Time derivatives of the loop-closure expressions allow the analysis of velocities & accelerations, i.e.:

Velocity & Acceleration Analysis

MECH 335 Lecture Notes © R.Podhorodeski, 2009

VELOCITY CLOSURE

ACCELERATION CLOSURE

• Review of time derivatives of displacement



• Example: Velocity analysis of the offset slider-crank



2

3

1

0

Displacement closure:

• Taking the time derivative:

• Rearranging and substituting:















• Splitting into real and imaginary eqns

• The solution for is obtained by solving the imaginary equation as:



REAL

IMAGINARY

• Substituting this solution back into the real equation gives the other unknown:



REAL EQUATION

• The velocity of one point can be expressed as the velocity of another point, plus the relative velocity of the two points

Relative Velocity Analysis


• Example



RA

RB/A

RB

A

B

O4O2

θ2

θ3

θ4

ω2

• Example



RA

RB/A

RB

A

B

O4O2

θ2

θ3

θ4

ω2

VB

Absolute velocity of point B

• Example



RA

RB/A

RB

A

B

O4O2

θ2

θ3

θ4

ω2

VA

Absolute velocity of point A

• Example



RA

RB/A

RB

A

B

O4O2

θ2

θ3

θ4

ω2

VB/A

Relative velocity of point B w.r.t. point A

• Example



RA

RB/A

RB

A

B

O4O2

θ2

θ3

θ4

ω2

VA

VB/A

VB

• Note that:

– The direction and magnitude of VA is a known function of the input angular velocity, ω2

– The mechanism’s joints define the direction of many of the remaining relative and absolute velocities

– This information can be manipulated to find the velocity (direction and magnitude) of points not on the input link



• There are 4 distinct cases where relative velocity analysis is applied (though only 3 are non-trivial)



Same Point Different Points

SameLink

Different Links

Case 1TRIVIAL CASE

Case 2DIFFERENCE

MOTION

Case 3RELATIVE MOTION

Case 4DIFFERENCE & RELATIVE

MOTION

• Case 2: Different points on the same link

– Want to find velocity of point B w.r.t. point A (VB|A)

– Take the derivative of the rel. position vector (RB|A)



RA

RB/A

RB

A

B

O4O2

θ2

θ3

θ4

ω2

VA

VB/A

VB


– Examining this result gives simple method for calculation:



RA

RB/A

RB

A

B

O4O2

θ2

θ3

θ4

ω2

VA

VB/A

VB

Always = 0 for a rigid link(no length change)

Equivalent to rotation through 90° in the sense (CW or CCW) of ωB|A (i.e. ω3)

So, for a rigid link :

VB|A = (rB|A)(ω3), ┴ RB|A


– Recalling that VB = VA + VB|A we can set up a system of equations to solve for VB:



RA

RB/A

RB

A

B

O4O2

θ2

θ3

θ4

ω2

VA

VB/A

VB

Tricky to approach analytically, but graphical methods can be used, and can be much more intuitive

• Case 2 Example (Supp Ex V1)



C

B

AO2 O4

3 4

2

ω2

• Solve using a graphical method



C

B

AO2 O4

3 4

2

ω2

• Find VB by relative velocity analysis:



C

B

AO2 O4

3 4

2

ω2

0V

1 mm = 5 mm/s

• VA: direction, magnitude both known



C

B

AO2 O4

3 4

2

ω2

0V

1 mm = 5 mm/s




0V

1 mm = 5 mm/sC

B

AO2 O4

3 4

2

ω2

0V

VA

1 mm = 5 mm/s

A




C

B

AO2 O4

3 4

2

ω2

0V

VA

1 mm = 5 mm/s

A

• VB|A: direction known, magnitude unknown



C

B

AO2 O4

3 4

2

ω2




C

B

AO2 O4

3 4

2

0V

VA

1 mm = 5 mm/s

A




C

B

AO2 O4

3 4

2

0V

VA

dir(VB|A )

1 mm = 5 mm/s

A

• VB: direction known, magnitude unknown



0V

VA

dir(VB|A )

1 mm = 5 mm/s

A

C

B

AO2 O4

3 4

2

ω2




C

B

AO2 O4

3 4

2

0V

VA

dir(VB|A )

1 mm = 5 mm/s

A




C

B

AO2 O4

3 4

2

0V

VA

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

A

• Solution is obtained by intersection & measurement



0V

VA

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

A

C

B

AO2 O4

3 4

2

ω2




0V

VA

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

A

BVB

C

B

AO2 O4

3 4

2

ω2




0V

VA

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

A

BVB

X

C

B

AO2 O4

3 4

2

ω2

• Noticing that , we measure:



C

B

AO2 O4

3 4

2

ω2

X

0V

VA

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

A

BVB




C

B

AO2 O4

3 4

2

ω2

0V

VA

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

VB|A

A

BVB

X




C

B

AO2 O4

3 4

2

ω2

0V

VA

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

VB|A

A

BVB

X

• And compute:

• Where the direction of rotation is inferred from the direction of



C

B

AO2 O4

3 4

2

ω2

0V

VA

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

VB|A

A

BVB

X X

• Similarly:



C

B

AO2 O4

3 4

2

ω2

0V

VA

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

VB|A

A

BVB

X X X

0V

VA

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

VB|A

A

BVB

• Now, VC can be found in a variety of ways:

– Intersect relative velocity directions w.r.t. A & B

– Compute directly, e.g. VC = VA+(ω3 X AC)



C

B

AO2 O4

3 4

2

ω2

X X X

• Using the first method, note that:



X X X

C

B

AO2 O4

3 4

2

ω2

0V

VA

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

VB|A

A

BVB




C

B

AO2 O4

3 4

2

ω2

X X

and ,

X

0V

VA

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

VB|A

A

BVB




C

B

AO2 O4

3 4

2

ω2

X X

and ,

0V

VA

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

dir(VC|A)

VB|A

A

BVB

X




C

B

AO2 O4

3 4

2

ω2

X X

and ,

X

0V

VA

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

dir(VC|A)

VB|A

A

BVB




C

B

AO2 O4

3 4

2

ω2

X X

and ,

0V

VA

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

dir(VC|A)

VB|A

dir(V

C|B )

A

BVB

X

• Intersection gives the solution for VC



C

B

AO2 O4

3 4

2

ω2

X X X

0V

VA

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

dir(VC|A)

VB|A

dir(V

C|B )

A

BVB




C

B

AO2 O4

3 4

2

ω2

X X

0V

VA

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

dir(VC|A)

VB|A

VC

dir(V

C|B )

A C

BVB

X




C

B

AO2 O4

3 4

2

ω2

X X

0V

VA

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

dir(VC|A)

VB|A

VC

dir(V

C|B )

A C

BVB

X X

• Another Case 2 Example (Supp Ex V2)



C

B

A

O23

2

ω2

4

• Solve using the same graphical method:



C

B

A

O23

2

ω2

40V

1 mm = 10 mm/s

• Find VB by case 2 analysis:



C

B

A

O23

2

ω2

40V

1 mm = 10 mm/s

• VA : magnitude, direction both known



C

B

A

O23

2

ω2

40V

1 mm = 10 mm/s




C

B

A

O23

2

ω2

40V

1 mm = 10 mm/s




C

B

A

O23

2

ω2

40V

AV

A

1 mm = 10 mm/s

• VB|A : direction known, magnitude unknown



C

B

A

O23

2

ω2

40V

AV

A

1 mm = 10 mm/s




C

B

A

O23

2

ω2

40V

AV

A

1 mm = 10 mm/s




C

B

A

O23

2

ω2

40V

dir

(VB

|A)

AV

A

1 mm = 10 mm/s

• VB : direction known, magnitude unknown



0Vdir

(VB

|A)

AV

A

1 mm = 10 mm/s

C

B

A

O23

2

ω2

4

• VB : direction known, magnitude unknown



0Vdir

(VB

|A)

AV

A

dir(VB)

1 mm = 10 mm/s

C

B

A

O23

2

ω2

4

• Intersection & measurement give:



C

B

A

O23

2

ω2

40V

dir

(VB

|A)

AV

A

dir(VB) VBV

B|A

B

1 mm = 10 mm/s

• Intersection & measurement give:



C

B

A

O23

2

ω2

40V

dir

(VB

|A)

AV

A

dir(VB) VBV

B|A

B

1 mm = 10 mm/s

X

• And ω3 is found from:



C

B

A

O23

2

ω2

4

X0V

dir

(VB

|A)

AV

A

dir(VB) VBV

B|A

B

1 mm = 10 mm/s

X

• Solve for VC by intersection:



C

B

A

O23

2

ω2

40V

dir

(VB

|A)

AV

A

dir(VB) VBV

B|A

B

1 mm = 10 mm/s

X X




C

B

A

O23

2

ω2

40V

dir

(VB

|A)

AV

A

dir(VB) VBV

B|A

B

dir(V

C|A )

1 mm = 10 mm/s

X X




C

B

A

O23

2

ω2

40V

dir

(VB

|A)

AV

A

dir(VB) VBV

B|A

dir(VC|B

)

B

dir(V

C|A )

1 mm = 10 mm/s

X X

• Measuring then gives:



C

B

A

O23

2

ω2

40V

dir

(VB

|A)

AV

A

dir(VB) VBV

B|A

dir(VC|B

)

B

C VC

dir(V

C|A )

1 mm = 10 mm/s

X XX

• Case 3: Coincident points on different links

– Occurs for slides & pistons, cams & followers:

• Two points on different links momentarily occupy the same point in the plane

• Each has a different absolute velocity, therefore a relative velocity exists




– Calculate the slide (relative) velocity, VB3|B4



B2, B3, B4

42

3

• So far, we have taken the derivative of the relative position vector, RB3|B4

• But how can we express this vector for two coincident points?

• Intuitively, we can imagine displacing the slide by some small distance along the slide, then drawing RB3|B4

• Taking the limit as the displacement approaches zero, we can see that RB3|B4

has zero length, and is directed along the tangent to the slide (link 4) at point B





θslide

B2, B3, B4

42

3

• So:

• Taking the derivative:





θslide

B2, B3, B4

42

3

• Simplifying gives the final expression:

• Note that this deceptively simple expression hides the potentially difficult task of finding the slide tangent angle

• In the following examples, straight slides are used to avoid this hassle (tangent angle = link angle for a straight slide)

• Case 3 Example (Supp Ex V3)



C

B

O2

2

ω2

4

O4

3

C

B

O2

2

ω2

4

O4

3

• Solve graphically



0V

1 mm = 10 mm/s

C

B

O2

2

ω2

4

O4

3

• Use case 3 analysis to find VB4:



0V

1 mm = 10 mm/s

C

B

O2

2

ω2

4

O4

3

0V

1 mm = 10 mm/sC

B

O2

2

ω2

4

O4

3

• VB2: direction, magnitude both known:



0V

B2

VB2

1 mm = 10 mm/s

0V

B2

VB2

1 mm = 10 mm/s

0V

B2

VB2

dir(V

B2|B

4 )

1 mm = 10 mm/sC

B

O2

2

ω2

4

O4

3

• VB2|B4: only direction is known



C

B

O2

2

ω2

4

O4

3

C

B

O2

2

ω2

4

O4

3

C

B

O2

2

ω2

4

O4

3

0V

B2

VB2

dir(V

B2|B

4 )

1 mm = 10 mm/s

0V

dir(VB4)

B2

VB2

dir(V

B2|B

4 )

1 mm = 10 mm/s

• VB4: direction known, magnitude unknown



• Obtain VB4 by intersection



C

B

O2

2

ω2

4

O4

3

0V

dir(VB4)

B2

VB2

dir(V

B2|B

4 )

VB4

B4

VB

2|B

4

1 mm = 10 mm/s

X

• ω4 follows immediately from VB4:



C

B

O2

2

ω2

4

O4

3

0V

dir(VB4)

B2

VB2

dir(V

B2|B

4 )

VB4

B4

VB

2|B

4

1 mm = 10 mm/s

X X

• VC follows immediately from ω4, or by:

•



C

B

O2

2

ω2

4

O4

3

0V

dir(VB4)

B2

VB2

dir(V

B2|B

4 )

VB4

B4

VB

2|B

4

1 mm = 10 mm/s

X X


• ,



C

B

O2

2

ω2

4

O4

3

X X0V

dir(VB4)

B2

VB2

dir(V

B2|B

4 )

VB4

B4

VB

2|B

4

dir(VC|B4)

1 mm = 10 mm/s


• , ,



C

B

O2

2

ω2

4

O4

3

X X0V

dir(VB4)

B2

VB2

dir(V

B2|B

4 )

VB4

B4

VB

2|B

4

dir

(VC)

dir(VC|B4)

1 mm = 10 mm/s

C

B

O2

2

ω2

4

O4

3

• Measuring:



0V

dir(VB4)

B2

VB2

dir(V

B2|B

4 )

VB4

B4

VB

2|B

4

dir

(VC)

dir(VC|B4)

C

VC

1 mm = 10 mm/s

X X X

• Another Case 3 Example (Supp Ex V4)



C

B

AO2 O4

3 4

2

ω2

D5, D6

5

6

• Note: The 4-Bar was solved in Ex V1



C

B

AO2 O4

3 4

2

ω2

D5, D6

5

6

C

B

AO2 O4

3 4

2

ω2

D5, D6

5

6

• VC is found by scaling arguments:



VA

VB

VB|A

A

B

0V

1 mm = 5 mm/s

C

B

AO2 O4

3 4

2

ω2

D5, D6

5

6

• VC is found by scaling arguments:



VA

VB

VB|A

A

B

VC|A

0V

1 mm = 5 mm/s

C

B

AO2 O4

3 4

2

ω2

D5, D6

5

6

• Measuring:



VA

VB

VB|A

A

B

VC|A

0V

C

VC

1 mm = 5 mm/s

X

C

B

AO2 O4

3 4

2

ω2

D5, D6

5

6

• VD5 is found by Case 3 analysis:



X

VA

VB

VB|A

A

B

VC|A

0V

C

VC

1 mm = 5 mm/s

C

B

AO2 O4

3 4

2

ω2

D5, D6

5

6

• VD5 : only the direction is known



X

VA

VB

VB|A

A

B

VC|A

dir(VD5 )

0V

C

VC

1 mm = 5 mm/s

C

B

AO2 O4

3 4

2

ω2

D5, D6

5

6

• VD5|C : only the direction is known



X

VA

VB

VB|A

A

B

VC|A

dir(VD5 )

0V

dir(VD5|C

)

C

VC

1 mm = 5 mm/s

C

B

AO2 O4

3 4

2

ω2

D5, D6

5

6

• Measuring:



X

VA

VB

VB|A

A

B

VD5

VC|A

dir(VD5 )

0V

dir(VD5|C

)

VD5|C

C

D5

VC

1 mm = 5 mm/s

X

C

B

AO2 O4

3 4

2

ω2

D5, D6

5

6

• ω5 is found immediately from:



VA

VB

VB|A

A

B

VD5

VC|A

dir(VD5 )

0V

dir(VD5|C

)

VD5|C

C

D5

VC

1 mm = 5 mm/s

X X X

END OF LECTURE PACK 3

Velocity & Acceleration Analysis - ariel.ac.il · •Note that: –The direction and magnitude of V A is a known function of the input angular velocity, ω 2 –The mechanism’s

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