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action for a  through the point n 4 (n 4 is end of vector cn 4). This
DC
line should be perpendicular to the link CD or to the vector cn 4.
Next one draw vertical line of action for acceleration of the point
D through the pole р a of the accelerations polygon. Two
mentioned straight lines will be crossed in the point d. Determine
accelerations by measuring of the line segments n 4d and р d
2
a   n 4 d  a  39 0, 059  2, 301 m/s ,
DC
a  p a d   92 0, 059  5, 428 m/s .
2
D
a
Angular accelerations of the links:
a  4, 012
-2
  BА   10, 03 s ,
2
l AB 0, 4
a  3, 422
-2
  BO 1   13, 688 s ,
3
l O 1 B 0, 25
a  2, 301
-2
  DC   7, 67 s .
4
l CD 0, 3
Directions of the angular accelerations match with directions
of tangential components of mentioned linear accelerations.

1.6 Research of mechanism’s output link velocity by analytical
method

Research of kinematics of the output link will be done by
method of vector closer loop. Determine number of loops which
are created by the vectors that represent mechanism’s links
к  р  n  7  5  2,
where р – is number of kinematic pairs; n – is number of mobile
links in a mechanism.
Calculated diagram of the mechanism is drawn on the
fig. 1.4.

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