- #1
Luchekv
- 66
- 1
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.
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.
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