Components of acceleration vector.
One can rewrite the above equation in such a way that the directions of all vectors are
referenced with respect to the direction of vector r. Then Equation 2.112 takes the form
ṙ˙ = ṙ˙ cos
– r cos
T
sin
+
T
+ 2ṙ cos
sin
T
+
+ --2
2
sin
(2.113)
+ --2
T
+ r cos
+ --2
sin
+ --2
The directions of four vectors and their magnitudes are shown in Figure 2.40,
where one can see that the direction of first vector coincides with the direction of
vector r; the direction of third vector is opposite to that of vector r, and the two
other vectors are perpendicular to vector r whereas their directions are found by
rotating vector r counterclockwise by /2. The first vector in Equation 2.113 is called
the translational component of acceleration, the third is called the centripetal component of acceleration, the fourth is called the angular component of acceleration,
and the second is called the coriolis component of acceleration.
Note that centripetal acceleration is caused by the rotation of the vector (irrespective of whether this rotation is time dependent or time independent), whereas
coriolis acceleration is caused by the rotation of a translationary moving vector. Both
of these components are functions of velocities only, and thus can be found based
on the velocity analysis. The other two components of acceleration, translational
and angular, are found as a result of acceleration analysis.
The equations for accelerations follow from the loop-closure equation for positions,
Equation 2.15, if the equation is differentiated twice with respect to time. As a result,
the loop-closure equation for accelerations is obtained. Note that it is assumed that
all parameters are time–dependent variables.
N
ṙ˙i cos
i
sin
+ 2r˙i – sin
T
i
T
i
cos
i
i
cos
i
(2.114)
i
i=1
–r i cos
i
sin
T
i
2
i
+ r i – sin
T
i
= 0
In Equation 2.114 the unknowns are ṙ˙i and i, and the system defining them is
linear. As before, the loop-closure equation for accelerations can have only two
unknowns. This again entails five possible combinations of these unknowns. In this
case the solutions for each can be found from the solutions for velocities in a
straightforward manner.
The unknowns, ˙˙ j and ṙ˙j , are found by differentiating Equations 2.77 and 2.78 with
respect to time assuming that all the variables are time dependent. The results, taking
into account Equations 2.77 and 2.78, are
1
= ---- – ḃ˙x sin
rj
j
j
+ b˙˙y cos
j
– 2ṙ j
(2.115)
j
and
ṙ˙j = ḃ˙x cos
j
+ b˙˙y sin
2
+ rj
j
(2.116)
j
In this case the two unknowns, ˙˙ j and ṙ˙i , are found by differentiating Equations
2.80 and 2.81, respectively.
j
[
1
= --------------------------------r j cos j – i
j
–
rj
i
j
– ṙ˙j sin
–
˙
i ḃ y + ḃ x sin
–2
i
j
i + –
–
i
– 2
–
j
ṙ j cos
i
˙
i ḃ x + ḃ y cos
i –
– ri
i
j
–
i
i ṙ i – r i i ]
(2.117)
and
[ ṙ i
1
ṙ˙i = ----------------------------cos j – i
j
–r i
i
sin
+ ḃ˙x + ḃ y
j
j
–
i
cos
j
i
j
–
+ ḃ˙y – ḃ x
j
cos
j
–
– ṙ˙j
i
sin j ]
(2.118)
The unknowns in this case are the translational accelerations, ṙ˙i and ṙ˙j . The expressions for them are found by differentiating Equations 2.82 and 2.83.
– ṙ i 2
1
ṙ˙i = ---------------------------sin i – j
–ṙ j
i
j
–
– rj
j
+r i
i
cos
˙
j – ḃ x + ḃ y
–
i
+r i
j
i
i
–
j
sin
j
sin j + ḃ˙y – ḃ x
i
cos
j
–
j
(2.119)
j
and the equation for ṙ˙j is obtained by interchanging indices i and j in the above.
– ṙ j 2 j – i + r j j cos
1
ṙ˙j = ---------------------------sin j – i
–ṙ i i – r i i – ḃ˙x + ḃ y
j
i
–
i
+r j
j
j
–
sin i + ḃ˙y – ḃ x
i
i
sin
cos
j
i
–
i
(2.120)
The angular accelerations are found by differentiating Equations 2.84 and 2.85.
–
1
= -------------------------------r i sin j – i
i
j
–
i
–ṙ˙i cos
–ṙ˙j + ḃ˙x + ḃ y
and the equation for
j
iri
1
= -------------------------------r j sin i – j
j
–
j
j
–
+ṙ i
i
cos
j
j
–2
+ ḃ˙y – ḃ x
i
sin
sin
j
j
–
i
(2.121)
j
is obtained by interchanging indices i and j in the above.
jr j
i
–
j
–ṙ˙j cos
–ṙ˙i + ḃ˙x + ḃ y
j
i
–
j
+ṙ j
i
–2
˙
i + ḃ y – ḃ x
cos
i
j
sin
sin
i
i
–
j
(2.122)
In this case the accelerations, ˙˙i and ṙ˙j , are found by differentiating Equations 2.90
and 2.91 taking into account that and are constants.
i
A i cos i – – Bi sin i – – ṙ˙i sin – ṙ˙k sin –
= ------------------------------------------------------------------------------------------------------------------------------– d x sin i – + d y cos i –
(2.123)
and
1 – A j cos i – – B j sin i – + C j cos i + D j sin
ṙ˙j = ----Tj
+ K j cos i – + L j sin i – + Q j
i
(2.124)
where it is denoted
2
A i = – ḋ y
i
+ dx
i
Bi = – d˙x
i
+ dy
i
2
– ḃ x
i
+ ḃ˙y
(2.125)
– ḃ y
i
+ ḃ˙x
(2.126)
A j = – d˙y + d x
B j = d˙x + d y
i
i
ṙ j
(2.127)
ṙ j
(2.128)
C j = ṙ˙i d y + ṙ i ḋ i – ṙ i d x
i
(2.129)
D j = – ṙ˙i d x – ṙ i d˙x – ṙ i d y
i
(2.130)
K j = ṙ˙k d y + ṙ k ḋ y – ṙ k d x
i
(2.131)
L j = – ṙ˙k d x – ṙ k ḋ x – ṙ k d y
i
(2.132)
Q j = – ḋ y ḃ x – d y ḃ˙x + d˙x ḃ y + d x ḃ˙y
(2.133)
Angular acceleration of the connecting rod vs. crank angle.
T j = – d y cos
i
–
+ d y sin
i
(2.134)
–
where bx, by are given by Equation 2.45, and dx, dy are defined by Equations 2.88
and 2.89, respectively.
•
Figure 1.14a with the driving crank
This mechanism falls into the second case category. The solutions are given by
Equations 2.117 and 2.118. In this case i = 1, j = 3, bx = –r2 cos 2, by = –r2 sin 2,
ḃ x = r2 ˙ 2 sin 2 , ḃ y = – r 2 ˙ 2 cos 2 , r2 = const., r3 = const., and 1
. Assume also
that the crank rotates with constant angular velocity, i.e., 2 = const. Then,
2
2
ḃ˙x = r 2 2 cos 2 , and ḃ˙y = r 2 2 sin 2 . Taking all this into account, the general formulas, Equations 2.117 and 2.118, are reduced to
2
3
2
r 2 2 sin 2 + r 3 3 sin 3
= ------------------------------------------------------r 3 cos 3
(2.135)
and
1
ṙ˙1 = -------------- ṙ 1
cos 3
3 sin
3–r2
2
2
–
3
cos
2
–
3
(2.136)
The above equations can also be obtained by differentiating the expressions for the
corresponding velocities (Equations 2.93 and 2.94).
A plot of the change of the angular acceleration of the connecting rod with the
crank angle is shown in Figure 2.41. A change of the slider acceleration with the
crank angle is shown in Figure 2.42.