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

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:

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

Absolute velocity of point B

• Example

Absolute velocity of point A

• Example

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

• Example

• 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)

– Examining this result gives simple method for calculation:

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:

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

• Case 2 Example (Supp Ex V1)

AO2 O4

• Solve using a graphical method

AO2 O4

• Find VB by relative velocity analysis:

AO2 O4

1 mm = 5 mm/s

• VA: direction, magnitude both known

AO2 O4

1 mm = 5 mm/s

1 mm = 5 mm/sC

AO2 O4

1 mm = 5 mm/s

AO2 O4

1 mm = 5 mm/s

• VB|A: direction known, magnitude unknown

AO2 O4

1 mm = 5 mm/s

AO2 O4

dir(VB|A )

1 mm = 5 mm/s

• VB: direction known, magnitude unknown

dir(VB|A )

1 mm = 5 mm/s

AO2 O4

dir(VB|A )

1 mm = 5 mm/s

AO2 O4

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

• Solution is obtained by intersection & measurement

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

AO2 O4

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

AO2 O4

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

AO2 O4

• Noticing that , we measure:

AO2 O4

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

AO2 O4

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

AO2 O4

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

• And compute:

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

AO2 O4

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

• Similarly:

AO2 O4

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

• 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)

AO2 O4

• Using the first method, note that:

AO2 O4

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

AO2 O4

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

AO2 O4

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

dir(VC|A)

AO2 O4

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

dir(VC|A)

AO2 O4

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

dir(VC|A)

• Intersection gives the solution for VC

AO2 O4

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

dir(VC|A)

AO2 O4

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

dir(VC|A)

AO2 O4

dir(VB)

dir(VB|A )

1 mm = 5 mm/s

dir(VC|A)

• Another Case 2 Example (Supp Ex V2)

• Solve using the same graphical method:

1 mm = 10 mm/s

• Find VB by case 2 analysis:

1 mm = 10 mm/s

• VA : magnitude, direction both known

1 mm = 10 mm/s

• VB|A : direction known, magnitude unknown

1 mm = 10 mm/s

• VB : direction known, magnitude unknown

1 mm = 10 mm/s

• VB : direction known, magnitude unknown

dir(VB)

1 mm = 10 mm/s

• Intersection & measurement give:

dir(VB) VBV

1 mm = 10 mm/s

• Intersection & measurement give:

dir(VB) VBV

1 mm = 10 mm/s

• And ω3 is found from:

dir(VB) VBV

1 mm = 10 mm/s

• Solve for VC by intersection:

dir(VB) VBV

1 mm = 10 mm/s

dir(VB) VBV

1 mm = 10 mm/s

dir(VB) VBV

dir(VC|B

1 mm = 10 mm/s

• Measuring then gives:

dir(VB) VBV

dir(VC|B

1 mm = 10 mm/s

• 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

• 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

• So:

• Taking the derivative:

θslide

B2, B3, B4

• 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)

• Solve graphically

1 mm = 10 mm/s

• Use case 3 analysis to find VB4:

1 mm = 10 mm/s

1 mm = 10 mm/sC

• VB2: direction, magnitude both known:

1 mm = 10 mm/s

1 mm = 10 mm/sC

• VB2|B4: only direction is known

1 mm = 10 mm/s

dir(VB4)

1 mm = 10 mm/s

• VB4: direction known, magnitude unknown

• Obtain VB4 by intersection

dir(VB4)

1 mm = 10 mm/s

• ω4 follows immediately from VB4:

dir(VB4)

1 mm = 10 mm/s

• VC follows immediately from ω4, or by:

dir(VB4)

1 mm = 10 mm/s

dir(VB4)

dir(VC|B4)

1 mm = 10 mm/s

• , ,

dir(VB4)

dir(VC|B4)

1 mm = 10 mm/s

• Measuring:

dir(VB4)

dir(VC|B4)

1 mm = 10 mm/s

• Another Case 3 Example (Supp Ex V4)

AO2 O4

D5, D6

• Note: The 4-Bar was solved in Ex V1

AO2 O4

D5, D6

AO2 O4

D5, D6

• VC is found by scaling arguments:

1 mm = 5 mm/s

AO2 O4

D5, D6

• VC is found by scaling arguments:

1 mm = 5 mm/s

AO2 O4

D5, D6

• Measuring:

1 mm = 5 mm/s

AO2 O4

D5, D6

• VD5 is found by Case 3 analysis:

1 mm = 5 mm/s

AO2 O4

D5, D6

• VD5 : only the direction is known

dir(VD5 )

1 mm = 5 mm/s

AO2 O4

D5, D6

• VD5|C : only the direction is known

dir(VD5 )

dir(VD5|C

1 mm = 5 mm/s

AO2 O4

D5, D6

• Measuring:

dir(VD5 )

dir(VD5|C

1 mm = 5 mm/s

AO2 O4

D5, D6

• ω5 is found immediately from:

dir(VD5 )

dir(VD5|C

1 mm = 5 mm/s

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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