Posted by: gpriyavct ® 01/13/2010, 03:09:48 Author Profile eMail author Edit

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 )

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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 Author Profile eMail author Edit

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

Posted by: zekeman ® 01/14/2010, 22:21:57 Author Profile eMail author Edit

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

Posted by: gpriyavct ® 01/25/2010, 02:39:12 Author Profile eMail author Edit

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 Author Profile eMail author Edit

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.

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

Posted by: gpriyavct ® 01/25/2010, 02:45:29 Author Profile eMail author Edit

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

Posted by: jboggs ® 01/25/2010, 09:47:01 Author Profile eMail author Edit

Homework for school?

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

Posted by: gpriyavct ® 01/27/2010, 01:36:04 Author Profile eMail author Edit

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

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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 ® 02/15/2010, 10:45:43 Author Profile eMail author Edit

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

Posted by: zekeman ® 02/15/2010, 10:54:18 Author Profile eMail author Edit

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

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

Posted by: zekeman ® 02/15/2010, 10:50:05 Author Profile eMail author Edit

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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