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Kinematic equation for 4-bar linkage
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Posted by: gpriyavct ®

01/13/2010, 03:09:48

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Looking For Help to derive the kinematic equation for the 4-bar linkage shown; (to find the angular position of the cylinder at different values of theta )

 

1_1.JPG (94.7 KB)  






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: Kinematic equation for 4-bar linkage
: Kinematic equation for 4-bar linkage -- gpriyavct Post Reply Top of thread Engineering Forum
Posted by: zekeman ®

01/14/2010, 22:01:31

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Trying to post my response, but having problems. Just testing to see if sketch will show up and then I'll post equations

 

quad-cyl.bmp73.4 KB  






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: : Kinematic equation for 4-bar linkage -- zekeman Post Reply Top of thread Engineering Forum
Posted by: zekeman ®

01/14/2010, 22:21:57

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Well, that wasn't too good. I'll make it bigger and write the equations relating theta to psi=#:
Note that r1 and x are variable and # is the angle the cylinder makes with the horizontal

w-r1*cost(@)+X*cos(#)+r2*sin(#)=0

h-r1*sin(@)-X*sin(#)+r2*cos(#)=0

solution
X=sqrt(w^2+h^2-2r1*(w*cos(@)+h*sin(@)+r1^2-r2^2)
#=INVERSE COS((w-r1COS(@)/SQRT(x^2+r2^2)+INVERSE TAN(r2/X)

 

1_quad-cyl.bmp116.3 KB  






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Posted by: gpriyavct ®

01/25/2010, 02:39:12

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we cant able to understand that drawing which you attached, so please clarify the drawing once again







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: Kinematic equation for 4-bar linkage
: Kinematic equation for 4-bar linkage -- gpriyavct Post Reply Top of thread Engineering Forum
Posted by: jboggs ®

01/13/2010, 11:17:32

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If this attachment works, just apply the centerlines shown below to your layout, identify knowns and unknowns, apply a few trigonometry equations, and you should have what you need.

 

Drawing2-Model.jpg (45.0 KB)  






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Posted by: gpriyavct ®

01/25/2010, 02:45:29

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1. please can you make an equation for this drawing which you given.

2. And we attached one more drawing here for your reference.


 

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Posted by: jboggs ®

01/25/2010, 09:47:01

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Homework for school?







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Posted by: gpriyavct ®

01/27/2010, 01:36:04

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no, its not an homework for school, we are designing a welding fixture for that purpose we need the equation.







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Posted by: zekeman ®

02/15/2010, 10:45:43

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Can't seem to get a better dwg but, I'll try to clarify:

The angles psi=#
theta=@


w,h are the rectangular coordinates of the upper pivot with respect to the cylinder pivot (0,0).

r2 is perpendicular distance from cylinder pivot to the cylinder rod line extended

x is the variable length of the cylinder rod

r1 is the the variable length of the distance between the rod end and the upper pivot.

The 2 equations I provide should give the answer you need .









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Posted by: zekeman ®

02/15/2010, 10:54:18

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Taking another look






Modified by zekeman at Mon, Feb 15, 2010, 12:50:30


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Posted by: zekeman ®

02/15/2010, 10:50:05

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Noticed error in stating that r1 is variable, In fact all links except x are fixed in length.

Edited:
However, the equations ly posted have a sign error.

They should be

x=sqrt(w^2+h^2-2r1*(w*cos(@)+h*sin(@)+r1^2-r2^2)
#=INVERSECOS((+r1COS(@-w))/SQRT(x^2+r2^2)+INVERSETAN(r2/X)


If there are any further questions I will be glad to answer them.

Good luck.







Modified by zekeman at Mon, Feb 15, 2010, 15:12:47


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