1. Instant Velocity Centers – A ‘Fast Track’ to Linkage Control
R. Lindeke, Ph. D.
ME 3230
ME 3230
9/22/2018

2. Topics:
“At every instant during motion of a rigid body in a plane there exists one point that is instantaneously at rest”
This is called the Center of Velocity
Methods for Finding and Using IC’s
From 2 velocities
For Various Joints
Kennedy-Aronholdt Theorem
Using Circle Diagrams and the co-linear proof
Rotating Radius Approach
ME 3230
9/22/2018

3. Mathematically:
We define the Instant Center as existing between two arbitrary links (or reference frames). It is the one existing point (P) in the plane of motion of any 2 links that have the exact same velocity at the same instant
If we define a point B on one Link and a Point C on a second, both can be measured in the reference (null?) frame R. If they are attached to their own link and are at the velocity center between the links then they can be called IBC and ICB and instantly have zero velocity wrt each other!
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9/22/2018

4. Finding the Location of IC (the instant center of velocity)
Starting with two points (P, Q) in a rigid body – body C
If we “know” the velocity of these two points from a (more fundamental) body B
And we wish to search for some point on body C that has zero velocity relative to Body B – this we will designate as: IBC (the instant center of velocity linking the two bodies)
This ‘connecting’ velocity point is the single point of intersection between the normal's from the two velocity vectors (as measured in Body B space) of the points of interest on Body C
This is a general observation!
ME 3230
9/22/2018

5. Mathematically:
Both of these radii vectors are normal to the general angular rotation of C from B so, for both equations to be satisfied, the single “center of Velocity” must lie at their intersection!!!
If they don’t intersect in a finite sense, then IBC must lie “at infinity”
ME 3230
9/22/2018

6. Application to real Joint Cases:
Revolute Joints (coupling Link 1 & 2)
The joint itself is a permanent velocity center labeled both I12 & I21
Curved Slider
The Instant center is at the center of curvature of the curve or center of an Osculating Circle over each portion of the sliders curve
Prismatic Joint
The Instant Center of Velocity is at Infinity normal to the direction of sliding
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9/22/2018

7. Application to real Joint cases:
Rolling Contact Pair
The instant center is at the point of contact of the two bodies
General CAM-Pair Contact
The instant center lies along the common normal to the line of tangency of the two components at the point of contact
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8. General CAM-Pair IC’s In the upper figure we see the general case
In the lower figure we see the case of frame-mounted cams (more realistic!)
The IC satisfies two conditions:
It is normal to the line of tangency between the 2 members
It is on a line that connects the IC’s of each cam member to the ground
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9. Finding IC’s – a general model
If 3 bodies are in relative planar motion with each other (or 2 bodies and a reference frame) there are 3 instant centers pertaining to the relative motion of the bodies and these must lie along the same line (be co-linear)
The Kennedy-Aronholdt Theorem
Furthermore for a linkage having n-links, the total number of instant centers is:
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10. Using “Circle Diagrams” to find IC’s
Useful when the number of links is large
The method:
Draw a kinematic diagram (to scale)
Draw a circle and add tick marks to represent each link
Determine (by inspection) as many IC’s as possible – see earlier rules
If a line (co-line from KATH) such that the line is the only unknown side in 2 triangles in the ‘circle,’ the physical IC represented by that line in the circle must lie at the intersection of the two lines that include the IC’s represented by the two known sides of each triangle
Repeat this last step as needed …
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9/22/2018

11. Using the Circle diagram for a 4-bar Linkage
Here we expect: {4*(4-1)}/2 = 6 IC’s
Using ‘revolute joint rule’ we identify: IC12, IC14, IC23 & IC34 readily
In the “Circle” we need to find 1-3 (yellow line) and 4-2 (red line)
1-3 lies at the intersection of the 2 lines: 1-2 & 2-3, with 1-4 & 3-4
2-4 similarly: 1-2 & 1-4 with 2-3 & 3-4
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12. Using the Circle diagram for a Slider Crank Linkage
Uses the same ideas (closing triangles) to select intersecting lines
Add one more thought: If an instant center is at infinity (normal to a prismatic joint or slider) it can be drawn as any convenient line using ‘parallaxing principles’
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13. Trying one: Pr 4.1 (pg 165)
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14. All by the rules (but with a circle twist)
Solution:
IC12 & IC23 were obvious
IC14 normal to slider (at infinity)
IC 13 along Line IC12 – IC23 (at infinity) as a ‘semi-prismatic joint’
IC34 – Normal to Block C at Infinity
IC 24 where vert. line IC12 – IC14 strikes line IC34 – IC23
All by the rules (but with a circle twist)
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9/22/2018

15. This applies to any arbitrary point on the link or all points!
Making Use of the IC – Using Rotating Radius Method for link and point velocities
This is a graphical method – very simple
Based on this simple idea:
This applies to any arbitrary point on the link or all points!
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9/22/2018

16. This last point leads to:
This states that the magnitude of the velocity of any point on a link is known if any other(s) is known
Direction of the velocity will change, however it is normal to the line from the point to the appropriate IC!
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17. This is useful for finding relevant velocity between links as well
Here we are working to Find the velocity of Pt B (on link 5) knowing the velocity of Pt A on Link 2 (and 3)
To work this problem – while using rotating radius – we need IC of the link where ‘info’ is know, IC of link where ‘info’ is needed and finally the IC that connects these two links
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18. Continuing: We know IC12 – revolute joint
We need to find IC15 (+++) and IC25 (x) – see above thus follow >, >>, and >>> indicated lines
To find vIC25 we rotate vA2 about IC12 ‘til it struck the line containing IC25 and note it is proportional to vA2
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19. Finally: We rotate IC25 about IC15 to a line from IC15 thru pt. B
Then we recognize that vB5 is proportional to the velocity of IC25 by rotating radii!
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20. Let’s Try one: 4.10 vA2 = 2 x ra/c = 1*1.4 = 1.4”/s
Direction is: 35 + 90 = 125
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21. Starts with Finding IC’s
Use Circle Method 
Knowns: I12, I23; I16; I14; I56; I34; I35
Consider Co-linear effects to find others
We need information for IC12; IC16 and IC26 to use Rotating Radii approach
Start with Scale Model of the linkage
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22. Making Scale Model: Put in Pt F (coor. given)
Draw CA: 1.4” at 35; Draw hor. sketch line from C to right
Draw Line from A (3.15” long) to intersect Hor. Line at E
Make construction Circle 1.6” rad. from F were it intersects AE is Pt. D
Make 2nd construction circle from F (1.25” radius)
Make 3rd construction Circle from D (0.8” rad.)
Were 2nd & 3rd circles intersect defines B (draw lines BD and BF)
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23. Working from Known IC’s we build I13 and I36
Since I14 is “at infinity” where a line I12, I23 intersects I34, I14 we find I13
Then since I35 is normal to AE “at infinity” where a line from I56 (in this dir.) intersects a line from I13 thru I16 we find I36
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24. Building Location of I26
Extend a line from I36 to I23 (Pt A2) it defines the position of I26 where it intersects the line from I12 to I16
We now have a ‘intellectual linkage’ between Pt A2 and Point B6!
We will perform radii rotation to find velocity of I26 and then rotate it to find VB6
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25. Getting vI26 value – rotating vA2 to I12-I26-I16 line
Construct a line from C to A
Measure its angle w/line containing I26
Rotate the construction line (uses CATIA Tool on Transformation Toolbar) by this angle
Strike a line from the end of the rotated line the same length as vA2 – this is vA2’
The line from C to the end of vA2’ sets the value of vI26 (parallel to vA2’)
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26. Uses the Rotate Transformation tool to rotate the radii:
Highlighted Line FB, clicked Pt. F then entered  as the rotate value.
This rotated FB to the line containing I26 about I16
The end of the rotated line Defines Pt B’ which is used to determine vB6
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27. The line vB’ is the rotated radii value of vB6 – it gives us magnitude (parallel to vI26). Real vB6 is this magnitude w/ direction normal to line FB starting at the original Pt. B
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28. What we learned (from this):
vI26 is 1.204”/s in the direction 
vB6 is 0.562”/s in the direction 
While we didn’t ask for it:
We also know that the angular velocity of link 6 is:
6 = (vB6/rB6/F6) = (.562/1.25) = rad/s CW
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29. Using Instant Centers
A “fast track” to velocity issues as they move around a linkage
With a single input, we can move to other links and points velocities very easily
So why not Acceleration Centers – computationally they are (at best) a trade off versus graphical method we discussed earlier!
ME 3230
9/22/2018

30. Homework: Do these problems for practice: Do to “Turn in”: 4.24
4.7; 4.8; 4.13; 4.25; 4.28
Key will be posted Monday Mar 3
Do to “Turn in”: 4.24
This real homework will be due at class time Mar 3
ME 3230
9/22/2018