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4) find extreme position of the point D, in which links AB
and OA will lie on one line. For it draw arc with center in the point
l  l 0, 4  0, 12
O with radius equals R 4  AB   OA  0 005  104 mm. This
l
arc should cross trajectory of the point B. We got the point B 0 as
cross point. Let’s draw a straight line from the point B 0 till point O.
This line crosses circle in the point A 0;
5) draw vertical line with left side position relative to the
b 0, 3
point O 1 on distance   60 mm. This line is the slider’s
 l 0, 005
(link 5) direction of motion.
6) draw straight line through the point B 0 from the point O 1
till intersection with trajectory of motion for the point C. We got
the point C 0;
7) than, draw arc with centre in the point C 0 and with radius
l 0, 3
CD
equals R 5    0, 005  60 mm on the slider’s (link 5)
1
trajectory of motion. The point D 0 is gotten;
8) draw mechanism’s scale diagram in one of extreme
position by the connection of all points by the straight lines;
9) further, divide circle with radius R 1=OA on twelve equals
parts. Draw straight lines from centre of this circle till crossing
with this circle. We got cross points. Designate these points as А О,
А 1, А 2…А 12 in ω 1 direction.
Drawings of another 11 positions are similarly.

1.5 Drawing of velocities and accelerations polygons for
mechanism’s control position

Drawing of velocities polygon. The link 1 moves rotational
(see fig. 1.3), therefore
V A  l  7 0, 12  0, 84 m/s.
1 OA
Choose free position of the velocities polygon’s pole. Draw
line perpendicular to OA – direction of point A velocity. Put the
vector р va with length 70 mm on this line. Scale factor of the
velocities polygon will be next:
V 0, 84
  A   0, 012m/s∙mm.
v
p v a 70
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