4 bar linkage and instantaneous centers help

[Luchekv]

We are to pick any 4 bar mechanism, fix one point and plot its movement/the movement of the instantaneous centre.

So far I've been trying to do this by using Freudensteins equations, but I'm not sure if I'm on the right track. The pdf document that I've been using can be found here:[/PLAIN]
http://www.upet.ro/annals/mechanical/pdf/2010/Annals-Mechanical-Engineering-2010-a27.pdf[/URL] The configuration of the 4 bars is the same as Figure 1 in the pdf.
My dimensions are as follow:
Left bar: 2
Top bar: 4
Right bar: 3
Bottom bar: 4(Fixed)

For ϕ: 20 degrees
As a result of following the equations in that document I got for ψ: 65.81 and -57.89 degrees.

To be honest. Not really sure what to do with those numbers..What I'm guessing is, if the left hand bar is calculated at 20 degrees the new angle of the right hand bar is 65.81 degrees?

Thanks in advance.

 

[BvU]

I see two solutions, a $\psi^+$ and a $\psi^-$, reasonably interpreted in Figure 2.

 

[Luchekv]

"These two values correspond to the two ways in which a four-bar linkage may be closed" is what confuses me. Am I meant to use both values? to plot the new position of the bar? or only one depending on which 'way' I want my linkage to be 'closed'. The way I'm thinking is if I redo the calculation at increments of 5 degrees for example. I'll have a decent path of motion. Again..that is, if in fact that those equations are for plotting motion..

I hope you understand what I mean.

 

[BvU]

"We are to pick any 4 bar mechanism, fix one point and plot its movement/the movement of the instantaneous centre" is the exact wording of the exercise?

instantaneous centre is what according to the relevant equations under 2. in the template ?

A and B move over the circles, so perhaps the halfway point is meant ? You may know, but I can't deduce that from your post.

 

[Luchekv]

We are to base it off a real four bar mechanism and use its dimensions. Fortunately, I have one in the same configuration as the one seen in figure 1 in the pdf. At the moment, I'm simply trying to plot the motion of the four bar mechanism I'm not worrying about the instantaneous center.

I know how it moves, I've done this using simulators. I just need to prove this mathematically. I've included 2 pictures. One, is of what I mean by instantaneous center and the second one is what I'm trying to explain. IF I calculate the mechanism at 20 degrees and my ouput from the equations in pdf yields a 65.81. Is that the new position of that bar?

 

[BvU]

I see. Picture 1 provides a clear idea of what is meant with instantaneous center. Picture 2 is one of the two solutions with $\phi=20^\circ$. For every $\phi$ there will be two points B at distance a₂ from A. Once that ambiguity is resolved, the path of the instantaneous center is determined.

You ask 'Is that the new position of that bar ?' but
1 you haven't given the initial postion
2 the $65.8^\circ$ is an angle, not a position.

 

[Luchekv]

Initial position is shown in the picture.. The mechanism stands at 90 degree angles. I know that 65.8 is an angle, but if I plot the bar at THAT angle..that gives me the new position of the bar in relation to what it was at 90.. I just wanted to confirm that if I drew the bar at that new angle. Will that be its motion/displacement/new position in relation to the ϕ=20 degrees.

I just need to know. If I start calculating ϕ at increments of 10 from the initial position of 90 degrees...and using the new ψ+ value to redraw the the right bar. Will this give me an accurate path of motion. I just need that confirmed.

 

[BvU]

PF is meant to help you with your homework, not to provide approval stamps. If you want confirmation, hand in your work and teacher will tell you. What I am trying to help you with is that you can convince yourself that it's correct.

I hope the ambiguity issue is cleared up by now. If not, one more shot: think of the thing as two triangles, O_B A B on top of O_B O_A A, with O_B A in common. B can be 'above' O_B A or below, but once that's established, it doesn't flip over any more by cranking O_A A: continuity, also in the equations, takes care of that.

Your description of the initial configuration is in conflict with the given information in post #1: with sides 2 4 3 4 you can't have found right angles. But my estimate is that you started with with $\phi = 90^\circ$ and B 'above' O_B A. So that takes care of resolving the ambiguity.

The path of motion asked for in the original post is the path of motion of the instantaneous center, so you still have some work to do ! Having $\psi$ written out as a function of $\phi$ is a good intermediate step for that.

 

[Luchekv]

Apologies for the delayed response. I just thought it would be more beneficial to ask if I'm on the right track rather than what the answer is, but I see the merit in your approach. Thank you :)

One more question if I were to choose a scissor mechanism instead. Would I be correct to assume the instantaneous center would remain in the joint?

 

[BvU]

I'm not sure what you mean by "scissor mechanism" ? Is that the other solution with B^- or do you mean something else ?


(If you do mean the B^- solution, then figure 2 already shows it does not "remain in the joint" )