1
RCRRRR Spatial Mechanism – Summary of How to Obtain
θ
1
Group 2 Spatial Mechanism
given:
α
12
,
α
23
,
α
34
,
α
45
,
α
56
,
α
61
a
12
, a
23
, a
34
, a
45
, a
56
, a
61
S
1
, S
2
, S
3
, S
4
, S
6
θ
6
(input angle)
find:
θ
1
,
θ
2
,
θ
3
,
θ
4
,
θ
5
, S
5
.
Want to get four equations of the form:
(a
i
x
m
2
+ b
i
x
m
+ d
i
) x
n
+ (e
i
x
m
2
+ f
i
x
m
+ g
i
) = 0
i = 1..4
(9.2)
where the coefficients a
1
through g
4
are quadratic in x
1
.
The first pair of equations is derived from the following subsidiary tanhalfangle laws
for a spherical hexagon:
X
¯
4
 X
612
= (Y
¯
4
 Y
612
)x
3
,
(9.19)
(X
¯
4
+ X
612
)x
3
= (Y
¯
4
+ Y
612
) .
(9.20)
which are converted to
s
34
(X
¯
4
 X
612
) = (c
34
Z
612
 s
34
Y
612
 c
45
)x
3
,
(9.23)
s
34
(X
¯
4
+ X
612
)x
3
= (c
34
Z
612
+ s
34
Y
612
 c
45
) .
(9.24)
Secondary tanhalf angle laws are now created realizing that
.
2
x
+
1
=
2
tan
+
1
2
1
=
d
x
d
2
3
3
2
3
3
θ
θ
6
R
R
R
R
R
C
5
4
3
2
1
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2
After much algebraic manipulation the result is
{[a
34
(Z
612
c
34
c
45
)  s
34
c
34
Z
0612
+ s
34
2
Y
0612
+ S
3
s
34
2
X
612
] s
45
 a
45
s
34
+
a
45
s
34
c
45
(c
34
Z
612
 s
34
Y
612
)} x
3
+
(9.44)
a
45
c
45
s
34
2
X
612
+ S
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 Spring '10
 Chancey
 Trigonometry, Pythagorean Theorem, E3 binding protein

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