75
".
28
'
42
83
ο
=
∠
AOB
weight 3
26
".
43
'
15
102
ο
=
∠
BOC
weight 2
22
".
27
'
38
94
ο
=
∠
COD
weight 4
77
".
23
'
23
79
ο
=
∠
DOA
weight 2
Adjust the angles by method of Correlates.
Solution:
75
".
28
'
42
83
ο
=
∠
AOB
Weight 3
26
".
43
'
15
102
ο
=
∠
BOC
Weight 2
22
".
27
'
38
94
ο
=
∠
COD
Weight 4
77
".
23
'
23
79
ο
=
∠
DOA
Weight 2
____________________________________
Sum
=
00
."
03
'
00
360
ο
Hence, the total correction E = 360º  (360º0’3”)
= 3”
Let e
1
,e
2,
e
3
and e
4
be the individual corrections to the four angles
respectively. Then by the condition equation, we get
e
1
+ e
2
+e
3
+ e
4
= 3”
 (1)
Also, from the least square principle, Σ(we
2
) = a minimum
3e
1
2
+ 2e
2
2
+4e
3
2
+ 2e
4
2
= a minimum  (2)
Differentiating (1) and (2), we get
δe
1
+ δe
2
+ δe
3
+ δe
4
= 0
 (3)
0
2
4
2
3
4
4
3
3
2
2
1
1
=
+
+
+
e
e
e
e
e
e
e
e
δ
δ
δ
δ
 (4)
Multiplying equation (3) by –λ and adding it to (4), we get
δe
1
(3e
1
– λ) + δe
2
(2e
2
λ) + δe
3
(4e
3
λ) + δe
4
(2e
4
λ) = 0  (5)
Since the coefficients of δe
1
,δe
2
,δe
3
,δe
4
must vanish independently, we have
3e
1
– λ = 0 or e
1
=
3
λ
2e
1
– λ = 0 or e
2
=
2
λ
 (6)
`4e
1
– λ = 0 or e
3
=
4
λ
2e
1
– λ = 0 or e
4
=
2
λ
Substituting these values in (1), we get
3
2
4
2
3

=
+
+
+
λ
λ
λ
λ
3
)
12
19
(

=
λ
19
12
*
3

=
λ
Hence
"
63
.
0
19
12
19
12
*
3
*
3
1
1

=

=
=
e
"
95
.
0
19
18
19
12
*
3
*
2
1
2

=

=
=
e
"
47
.
0
19
9
19
12
*
3
*
4
1
3

=

=
=
e
"
95
.
0
19
18
19
12
*
3
*
2
1
4

=

=
=
e
_______________
Sum =
3.0”
_______________
Hence the corrected angles
AOB
= 83º42’28”.75 – 0”.63
= 83º42’28”.12
BOC
= 102º15’43”.26 – 0”.95
= 102º15’42”.31
COD
= 94º38’27”.22 – 0”.47
=
94º38’26”.75
DOA
= 79º23’23”.77 – 0”.95
= 79º23’22”.82
_______________
360º00’00”.00
8. The following round of angles was observed from central station to the
surrounding station of a triangulation survey.
A = 93º43’22”
weight 3
B = 74º32’39”
weight 2
C = 101º13’44”
weight 2
D = 90º29’50”
weight 3
In addition, one angle
(
)
B
A
+
was measured separately as combined
angle with a mean value of 168º16’06” (wt 2).
Determine the most probable values of the angles A, B, C and D.
Solution:
A + B+C+D = 359º59’35”.
Total correction E = 360º  (359º 59’ 35”)
= + 25º
Similarly,
(
)
B
A
+
=
(A+B)
Hence correction E’
= A + B 
(
)
B
A
+
= 168º16’01” – 168º16’06”
= 5”
Let e
1
,e
2,
e
3
,e
4
and e
5
be the individual corrections to A, B, C, D and
(
)
B
A
+
respectively. Then by the condition equation, we get
e
1
+ e
2
+e
3
+ e
4
= 25”
 (1(a))
e
5
– e
1
– e
2
= 5”
 (1(b))
Also, from the least square principle, Σ(we
2
) = a minimum
3e
1
2
+ 2e
2
2
+2e
3
2
+ 3e
4
2
+ 2e
5
2
= a minimum  (2)
Differentiating (1a) (1b) and (2), we get
δe
1
+ δe
2
+ δe
3
+ δe
4
= 0
 (3a)
δe
5
 δe
1
 δe
2
= 0
 (3b)
0
2
3
2
2
3
5
5
4
4
3
3
2
2
1
1
=
+
+
+
+
e
e
e
e
e
e
e
e
e
e
δ
δ
δ
δ
δ
 (4)
Multiplying equation (3a) by –λ
1
, (3b) by λ
2
and adding it to (3), we get
δe
1
(3e
1
–λ
1
+λ
2
) +δe
2
(2e
2
λ
1
+ λ
2
) + δe
3
(2e
3
λ
1
) + δe
4
(3e
4
λ
1
)
+δe
5
(λ
2
+2e
5
) = 0  (5)
Since the coefficients of δe
1
,δe
2
,δe
3
,δe
4
etc. must vanish independently, we
have
0
3
1
2
1
=
+
+

e
λ
λ
or
3
3
2
1
1
λ
λ

=
e
0
2
2
2
1
=
+
+

e
λ
λ
or
2
2
2
1
2
λ
λ

=
e
0
2
3
2
=
+

e
λ
or
2
1
3
λ
=
e
 (6)
0
3
4
1
=
+

e
λ
or
3
1
4
λ
=
e
0
2
5
2
=
+

e
λ
or
2
2
5
λ

=
e
Substituting these values of e
1
,e
2,
e
3
,e
4
and e
5
in
Equations (1a) and (1b)
)
1
(
25
3
2
2
2
3
3
1
1
2
1
2
1
a
from
=
+
+

+

λ
λ
λ
λ
λ
λ
or
25
6
5
3
5
2
1
=

λ
λ
5
6
1
3
2
1
=

λ
λ
 (I)
)
1
(
5
32
2
3
3
2
2
1
2
1
2
b
from

=
+

+

λ
λ
λ
λ
λ
5
6
5
3
4
1
2

=

λ
λ
 (II)
Solving (I) and (II) simultaneously, we get
11
210
1
+
=
λ
11
90
2
+
=
λ
Hence
64
".
3
11
"
40
11
90
.
3
1
11
210
.
3
1
1
+
=
+
=

=
e
`
45
".
5
11
"
60
11
90
.
2
1
11
210
.
2
1
2
+
=
+
=

=
e
55
".
9
11
"
105
11
210
.
2
1
3
+
=
+
=
=
e
36
".
6
11
"
70
11
210
.
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 Spring '20
 Celestial coordinate system, great circle, base line