Instantaneous center of velocity for a inverted slider crank

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SUMMARY

The discussion focuses on determining the instantaneous centers (ICs) of an inverted slider crank mechanism with four links. The formula for calculating the number of ICs, n(n-1)/2, indicates there are six ICs for this linkage. Participants express confusion regarding the classification of links, particularly the welding of links 2 and 3, and the nature of their connections to the other components. Clarification is sought on the rigid connections and the sliding mechanism of the black hoop and bar.

PREREQUISITES
  • Understanding of kinematic chains and linkages
  • Familiarity with the concept of instantaneous centers in mechanical systems
  • Knowledge of rigid body dynamics
  • Basic proficiency in interpreting mechanical diagrams
NEXT STEPS
  • Study the principles of kinematic analysis in mechanical linkages
  • Learn about the application of the instantaneous center method in mechanism design
  • Explore the dynamics of slider-crank mechanisms
  • Investigate the effects of welding on the classification of links in mechanical systems
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Mechanical engineering students, educators, and professionals involved in the design and analysis of mechanical linkages and mechanisms.

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Homework Statement


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Find the instantaneous centers of the linkage below.

upload_2015-11-1_18-14-26.png

Homework Equations



# of ICs = n(n-1) / 2

The Attempt at a Solution


[/B]
From the equation above, there are 4(4-1)/2 = 6 ICs in this linkage. But one thing I don't understand is how there are 4 links when 2 and 3 are welded together to make one piece? And how is the IC of point 2,3 at the center of the circle?
 
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I'm not sure I understand the diagram. I'm guessing 2 represents a solid disk which pivots eccentrically about a point near its base, while the black hoop is free to slide around the disk. But as you say, the black hoop is rigidly connected to the black bar. How is the black bar connected to the block numbered 4? That looks rigid too, so I only see three pivot points, so only two links. Or can it slide through the block?
 

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