1
2
Suppose we have 2 equations in the format below
where A1J2 are known constants.
cj(A1ci+B1si+D1) + sj(E1ci+F1si+G1) + (H1ci+I1si+J1) = 0
cj(A2ci+B2si+D1) + sj(E2ci+F2si+G2) + (H2ci+I2si+J2) = 0
How do we find the values for i and j which
simultaneously satisfy the two equations?
3
cj(A1ci+B1si+D1) + sj(E1ci+F1si+G1) + (H1ci+I1si+J1) = 0
cj(A2ci+B2si+D1) + sj(E2ci+F2si+G2) + (H2ci+I2si+J2) = 0
Substitute the tan-half angle expressions and multiply
throughout by (1+xi2)(1+xj2).
k
2 xk
1  xk 2
xk  tan , sk 
, ck 
2
2
1  xk
1  xk 2
(1–xj2) [Aq(1–xi2)+Bq(2xi)+Dq(1+xi2)]
+ 2xj [Eq(1–xi2)+Fq(2xi)+Gq(1+xi2)]
+ (1–xj2) [Hq(1–xi2)+Iq(2xi)+Jq(1+xi2)] = 0, q = 1,2
4
regroup as follows:
xj2(a1xi2+b1xi+d1) + xj(e1xi2+f1xi+g1) + (h1xi2+i1xi+j1) = 0
xj2(a2xi2+b2xi+d2) + xj(e2xi2+f2xi+g2) + (h2xi2+i2xi+j2) = 0
where
aq = Aq – Dq – Hq + Jq,
eq = 2(Gq – Eq),
hq = -Aq + Dq – Hq + Jq,
q = 1,2
bq = 2(Iq – Bq),
fq = 4 Fq,
iq = 2(Iq + Bq),
dq = -Aq – Dq + Hq + Jq
gQ = 2(Gq + Eq
jq = Aq + Dq + H q + Jq
5
How do we solve 2 bi-quadratic equations for the two
unknowns xi and xj?
How many solution sets will exist?
xj2(a1xi2+b1xi+d1) + xj(e1xi2+f1xi+g1) + (h1xi2+i1xi+j1) = 0
xj2(a2xi2+b2xi+d2) + xj(e2xi2+f2xi+g2) + (h2xi2+i2xi+j2) = 0
6
Write the equations in the following form:
L1 xj2 + M1 xj + N1 = 0
L2 xj2 + M2 xj + N2 = 0
where
L1 = a1xi2+b1xi+d1
M1 = e1xi2+f1xi+g1
N1 = h1xi2+i1xi+j1
L2 = a2xi2+b2xi+d2
M2 = e2xi2+f2xi+g2
N2 = h2xi2+i2xi+j2
We will present 2 solution methods:
Sylvester’s Dialytic Method
Bezout’s Method
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Sylvester’s solution method:
L1 xj2 + M1 xj + N1 = 0
L2 xj2 + M2 xj + N2 = 0
Let t = xj2, u = xj, v = 1. The equations can be written as
L1 t + M1 u + N1 v = 0
L2 t + M2 u + N2 v = 0
Now multiply the original equations by xj and let s = xj3.
This gives
L1 s + M1 t + N1 u = 0
L2 s + M2 t + N2 u = 0
8
We now have four equations:
L1 xj2 + M1 xj + N1 = 0
L2 xj2 + M2 xj + N2 = 0
L1 t + M1 u + N1 v = 0
L2 t + M2 u + N2 v = 0
Write the four equations in matrix format as:
0
0

 L1

 L2
L1
L2
M1
M2
M1
M2
N1
N2
N1   s  0 
N 2   t  0 
  
0  u   0 
   
0   v  0
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0
0

 L1

 L2
L1
L2
M1
M2
M1
M2
N1
N2
N1   s  0 
N 2   t  0 
  
0  u   0 
   
0   v  0
In order for there to be a solution to these ‘homogeneous’
equations, either [s, t, u, v]T = 0 or the equations are
linearly dependent.
It must be the case that the equations are linearly
dependent.
0
0
L1
L2
L1
L2
M1
M2
M1
M2
N1
N2
N1
N2
0
0
0
10
0
0
L1
L2
L1
L2
M1
M2
M1
M2
N1
N2
N1
N2
0
0
0
Since the terms L1 through N2 are quadratic in the variable
xi, expansion of the determinant will yield an 8th degree
polynomial in xi.
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Corresponding values for xj can be obtained by taking any
three of the equations such as the first three as:
0
0

 L1
L1
L2
M1
M1
M2
N1
 s  0
N1     
t  0


N2 

u   0 
0     
 v  0
Substituting back for s, t, u, and v and rearranging gives
0
0

 L1
L1
L2
M1
M 1   x j 3    N1 
 2 

M 2   x j     N 2 
N1   x j   0 
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The vector [xj3, xj2, xj]T can be solved as
 x j3   0
 2 
xj    0
 x j   L1
 
L1
L2
M1
M1 
M 2 
N1 
1
  N1 
 N 
 2
 0 
The third component of this vector gives the xj value
corresponding to each xi value.
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Bezout’s solution method:
Start with
L1 xj2 + M1 xj + N1 = 0
L2 xj2 + M2 xj + N2 = 0
where
L1 = a1xi2+b1xi+d1
M1 = e1xi2+f1xi+g1
N1 = h1xi2+i1xi+j1
L2 = a2xi2+b2xi+d2
M2 = e2xi2+f2xi+g2
N2 = h2xi2+i2xi+j2
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Factor the equations into the form:
xj(L1xj + M1) + N1 = 0
xj(L2xj + M2) + N2 = 0
These two new ‘linear’ equations must be linearly
dependent if there is to be a common solution. Thus
N2 (L1xj + M1) – N1 (L2xj + M2) = 0
Write as
L1
xj
L2
N1 M 1

N2 M 2
N1
0
N2
(1)
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Now factor the equations into the form:
xj2 (L1) + (M1 xj + N1) = 0
xj2 (L2) + (M2 xj + N2) = 0
These two new ‘linear’ equations must be linearly
dependent if there is to be a common solution. Thus
L1 (M2xj + N2) – L2 (M1xj + N1) = 0
Write as
L1
xj
L2
M 1 L1

M 2 L2
N1
0
N2
(2)
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We now have 2 equations:
L1
xj
L2
N1 M 1

N2 M 2
N1
0
N2
(1)
L1
xj
L2
M 1 L1

M 2 L2
N1
0
N2
(2)
These two linear equations will have a solution only if they
are linearly independent. Thus
L1
L2
M1 M1
M2 M2
N1 L1

N 2 L2
2
N1
0
N2
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L1
L2
M1 M1
M2 M2
N1 L1

N 2 L2
2
N1
0
N2
Since the terms L1 through N2 are quadratic in the variable
xi, expansion of the determinant will yield an 8th degree
polynomial in xi.
Corresponding values of xj can be obtained from either (1)
or (2).
L1
xj
L2
N1 M 1

N2 M 2
N1
0
N2
(1)
L1
xj
L2
M 1 L1

M 2 L2
N1
0
N2
(2)
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Since this is a group 2 mechanism, we will have to generate
additional equations that relate the twist angles, joint angles,
link lengths, and joint offsets.
We will show three techniques:
• projection of vector loop equation
• self-scalar product of vector loop equation
• secondary cosine laws
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22
given: a12, a23. a34. a45. a51
12, 23, 34, 45, 51
S1,, S3, S5
5 (input angle)
find:
1, 2, 3, 4, S2, S4
Obtain an equation
that does not have S2
or S4.
23
24
Obtain an equation
that does not have S2
or S3.
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26
Set 2
S2 (
S3 (
S4 (
S5 (
S1 (
0,
0,
X¯3,
X43,
X543,
0,
-s23,
Y¯3,
Y43,
Y543,
1
c23
Z¯3
Z43
Z543
)
)
)
)
)
a23
a34
a45
a51
a12
(
(
(
(
(
1,
c3,
W43,
W543,
c2,
0,
s3c23,
-U*432,
-U*5432,
-s2,
0
U32
U432
U5432
0
)
)
)
)
)
Set 9
S3 (
S2 (
S1 (
S5 (
S4 (
0,
0,
X2 ,
X12,
X512,
0,
s23,
-Y2,
-Y12,
-Y512,
1
c23
Z2
Z12
Z512
)
)
)
)
)
a23
a12
a51
a45
a34
(
(
(
(
(
1,
c2,
W12,
W512,
c3,
0,
-s2c23,
U*123,
U*5123,
s3,
0
U23
U123
U5123
0
)
)
)
)
27
)
28
the equation contains 1 and 2 as the only unknowns
it can be paired with
Z512 = c34
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Factor the spherical equation, Z512=c34 into the format:
c2[A1c1 + B1s1 + D1] + s2[E1c1+F1s1+G1] + [H1c1+I1s1+J1] = 0
30
Factor the vector loop projection equation into the format:
c2[A2c1 + B2s1 + D2] + s2[E2c1+F2s1+G2] + [H2c1+I2s1+J2] = 0
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32
33
34
35
36
37
38
39
40
2nd equation
A spherical equation that also has 1 and 2 as its only unknowns is:
1st equation
41
c2
s2
42
c2
s2
43
44
45
insert dual angles into
Z671 = Z43
46
insert dual angles into
Z671 = Z43
47
insert dual angles into
Z671 = Z43
48
49
50
51
This equation has 1 and 6 as its only unknowns.
A second equation in these unknowns is:
Z56712 = c34
52
53
…
54
Obtain remaining variable joint
parameters.
6
3
4
S2
S5
55
56
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