Graphics Programs Reference
In-Depth Information
fromEq. (5.9)
2
2
g
(0
.
1)
−
g
(0
.
2)
4(0
.
9675)
−
0
.
8918
f
(0)
≈
G
=
=
=
0
.
9927
2
2
−
1
3
which is a finite difference approximation of
O
(
h
4
).
EXAMPLE 5.3
C
b
β
B
c
a
α
A
D
d
The linkage shown has the dimensions
a
=
100 mm,
b
=
120 mm,
c
=
150 mm
and
d
=
180 mm. It can be shown by geometry that the relationship between the
α
β
angles
and
is
)
2
)
2
c
2
(
d
−
a
cos
α
−
b
cos
β
+
(
a
sin
α
+
b
sin
β
−
=
0
For a givenvalueof
α
, wecan solvethistranscendentalequation for
β
by one of the
0
◦
,
5
◦
,
10
◦
,...,
30
◦
, the
root-finding methods in Chapter 4. This was done with
α
=
results being
α
(deg)
0
5
10
15
20
25
30
β
(rad)
1
.
6595
1
.
5434
1
.
4186
1
.
2925
1
.
1712
1
.
0585
0
.
9561
If link
AB
rotates with the constant angular velocity of 25 rad/s, use finite difference
approximationsof
(
h
2
)totabulate the angular velocity
d
O
β/
dt
of link
BC
against
α
.
Solution
The angular speed of
BC
is
d
dt
d
d
dt
25
d
d
=
=
rad/s
α
α
β/
α
where
d
iscomputed fromfinite difference approximations using the datainthe
table.Forward and backward differences of
d
(
h
2
) are usedat the endpoints, central
differences elsewhere. Note that the incrementof
O
α
is
=
5deg
180
rad/deg
h
=
0
.
087266 rad
The computations yield
(0
◦
)
(5
◦
)
(10
◦
)
25
−
3
β
+
4
β
−
β
25
−
3(1
.
6595)
+
4(1
.
5434)
−
1
.
4186
˙
(0
◦
)
β
=
=
2
h
2 (0
.
087266)
=−
32
.
01rad/s
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