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Analytical Acceleration Analysis Page 1 of 3 Dr. Keith
Hekman
The vector loop equation for a fourbar linkage is
01432 =+ decebeae jjj (1)
Assuming the initial vector loop equation has been solved using
methods from chapter 4, we
can take the derivative with respect to time to get
( ) ( ) ( ) 0432
432 =+ &&& jcejbejae jjj (2)
Taking the derivative with respect to time to again results
in
( ) ( ) ( ) ( ) ( ) ( ) 042
43
2
32
2
2443322 =+++ &&&&&&&&&
jcejcejbejbejaejae jjjjjj (3)
Setting &= and &&= we have
04433224
2
43
2
32
2
2=+++ jjjjjj cejcebejbeaejae (4)
This can be written as
0=+++ttNNtt BBBAA
aaaarrrr
( )
4
3
2
4
2
4
3
2
3
2
2
2
jj
jj
jj
aejaaea
aejaaeaaejaaea
T
T
T
==
====
r
r
r
6
ur goal is to find the unknown velocities
based on the known parameters, i.e. to find
( )
( )43434
23232
,,,,,,,,,,
,,,,,,,,,,
cbg
dba
=
=(7
where the angular acceleration of the input link is given, and
the position and velocity
med.
Arranging (4) so that the known values are on one side, and the
unknowns are on the
other and dividing by j, we have
)
Zceaeae
bejjj
j 432
4 43
2
43
(8
Taking the conjugate to get the second equation leads to
cebejj = 43 4
9
[ ] [ ]BA =
4
3
( )
= Z
ZB
cebe
cebeA
j
jj
3
4
. 11)
Thus
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Analytical Acceleration Analysis Page 2 of 3 Dr. Keith
Hekman
[ ] [ ]B1
4
3
(12)
Note that the A matrix is the same as in the velocity analysis,
only the B matrix changes.
The vector loop equation for a fourbar slider-crank linkage
is
01432 = decebeae jjj (13)
Taking the derivative with respect to time, we have
( ) ( ) 03232 = djbejae jj & (14)
Taking the derivative with respect to time again, we have
03322 32
32
2
2 =++ dbejbeaejaejjjj && (15)
This also can be written as
0=+++ BBABAAA aaaaa ttNttNrrrrr
(16)
where
da
aejaaea
aejaaea
B
j
BA
j
BA
j
A
j
A
TN
TN
&&r
rr
rr
=
==
==33
22
3
2
3
2
2
2
(17)
The unknown variables in this equation are 3 and d&& .
Gathering the known values all on one
side, we have
Zbeaejaedbejjjjj =++=+ 3223
2
32
2
23
&& (18)
Taking the conjugate to get the second equation leads to( )
Zdjbe j =+ &&3
3
(19)
These two equations can be written in matrix form of
[ ] [ ]Bd
A =
&&
3(20)
where
=
=
Z
ZB
jbe
jbeA
j
j
and1
1
3
3
. (21)
Thus
[ ] [ ]BAd
13 =
&&
. (22)
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7/29/2019 Acc El Analysis
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Page of 3 Dr. Keith Hekman
The vector loop equation for a fourbar inverted slider crank
linkage is
01432 = decebeae jjj ( )
3jbe
term, both b and 3 change so you have to differentiate by parts.
Thus the derivative is
04332432
= jcejbeebjae jjjj & . (24)
Again, taking the derivative results in
02 443333224
2
43
2
32
2
2 =+++ jcecejbebejebebjaeaejjjjjjjj &&& (25)
One of the jebj 3 & terms comes from taking the derivative
of 3
jeb& and the other from
32
3
j
be . Recall that the relationship between 3 and 4 remains fixed
at
+= 43 (26)
Taking the derivatives of this, we have
43
43
=
=(27)
Thus our unknown variables are 3=4 and b&& , so we need
to group our equation accordingly.
This leads to
( ) Zcebejebjaeaecebejeb jjjjjjjj =+++=++ 43322433 24
2
32
2
232
&&& . (28)
Taking the conjugate to get the second equation gives
( ) Zcebejeb jjj =+ 433 3 && . (29)
These two equations can be written in matrix form of
[ ] [ ]Bb
A =
3
&&
(30)
where
( )( )
=
+
+=
Z
ZB
cebeje
cebejeA
jjj
jjj
and433
433
. (31)
Thus
[ ] [ ]BAb 1
3
=
&&
. (32)
