DOF in kinematics

[sudhakarn]

Hi,
I have a question here in Kinematics.After defining the joints and a fixed part and the commands,CATIA says that the mechanism can be simulated with the condition of DOF = 0. But actually speaking for a revolute joint there is one rotational dof that is free.If this is the case then why it shows DOF = 0.

or in general what is actually meant by the condition DOF=0 in a kinematics mechanism in CATIAV5.Awaiting your replies.

 

[tbuelna]

I believe the kinematic/constraint analysis result showing DoF=0 simply means the components in the model are fully constrained with respect to each other for the given set of commands. Obviously, in kinematic terms each body potentially has 3 rotational and 3 translational degrees of freedom. Depending on how you applied the commands to your kinematic model and how you applied the constraints, it is possible for some components in the model to not be fully constrained and still obtain a determinant result. For example, if a shaft in the kinematic model is only rotationally constrained on one axis and is free to translate along that same axis, then it has 2 degrees of freedom. But if the command simply forces the shaft to rotate about the bushing axis, the kinematic model should still interpret the case as being determinant.

 

[itsmyjob]

the revolute does have 1 DOF but you control it with a command so at the end DOF=0

Eric N.
indocti discant et ament meminisse periti ​

 

[sudhakarn]

Thank you so much for your quick replies.So i understand that if the mechanism satisfies all the conditions to be defined for the joint(s)then it is taken as DOF = 0.