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CBLAMSIS UNIVERSITY OF THE CORDILLERAS MISSING DATA
OMITTED MEASUREMENTS/MISSING DATA
When for any reason it is impossible or impractical to determine by field observations the
length and bearing of every side of a closed traverse, the missing data may generally be
calculated, provided not more than two quantities (lengths and/or bearings) are omitted.
It must be assumed that the observed values are without error, and hence all errors of
measurements are thrown into the computed lengths or bearings.
Omitted measurements in the field may be cause by several condition or problems
encountered like the presence of obstacle in the area that can not be resolve, rugged
terrain, and sometimes unfriendly or hostile landowners. It is therefore evident that if the
field measurements for any lengths and directions of a closed traverse are to be omitted,
it is always essentially important to employ checks on the computed values which will be
done later in the office.
The principle of omitted measurements is advantageous in land partitions. For example,
a large tract of closed traverse is to be subdivided into several smaller tracts of closed
traverse, the dividing line may be considered as an omitted measurement.
The common types of omitted measurements are:
- Length and bearing of one side unknown.
- Length of one side and bearing of another side unknown.
- Lengths of two sides unknown for which the bearings have been observed.
- Bearings of two sides unknown for which the lengths have been observed.
A. Length and bearing of one side unknown
The problem of determining the length and bearing of one side of a closed traverse is
exactly the same as that of computing the length and bearing of the linear error of
closure in any closed traverse for which field measurements are complete. The
latitudes and departures of the known sides are computed. Get the algebraic sum of
the latitudes and departures. If the algebraic sum of the latitudes and the algebraic
sum of the departures of the known sides are designated by ΣL and ΣD, respectively,
then the length S of the unknown side is
2
2
meaning the algebraic sum of latitudes ΣL is the latitude of the unknown side but with
opposite sign and the algebraic sum of departures ΣD is the departure of the unknown
side but with opposite sign also. And the tangent of the bearing angle θ is
tan 𝜃 =
SAMPLE PROBLEM NO. 1
A closed traverse has the following data as given in the table below. Determine the
length and bearing of course 3 – 4.
Course Bearing Distance (m)
1 – 2 N 09
0
16’ E 58.
2 – 3 S 88
0
26’ E 27.
4 – 5 S 05
0
18’ E 35.
5 – 1 S 72
0
02’ W 78.
CBLAMSIS UNIVERSITY OF THE CORDILLERAS MISSING DATA
Determine the latitudes and departures of the given sides assuming that all
measurements are correct.
Course Bearing Distance (m) Latitudes (m) Departures (m)
1 – 2 N 09
0
16’ E 58.7 + 57.934 + 9.
2 – 3 S 88
0
26’ E 27.3 – 0.746 + 27.
4 – 5 S 05
0
18’ E 35.0 – 34.850 + 3.
5 – 1 S 72
0
02’ W 78.96 – 24.356 – 75.
ΣL = – 2.018 ΣD = – 35.
Thus, the latitude of course 3 – 4 is – (ΣL) = - (- 2.018) = 2.018 m and the departure
of the same course is – (ΣD) = - (- 35.135) = 35.135 m. Its distance then,
3 − 4
2
2
3 − 4
2
2
And its bearing angle is
tan 𝜃 =
tan 𝜃 =
θ = 86
0
′
"
0
The bearing of course 3 – 4 is N 86
0
43’ E, north- east since its latitude and departure
are both positive in sign.
Final Tabulation
Course Bearing Distance (m) Latitudes (m) Departures (m)
1 – 2 N 09
0
16’ E 58.7 + 57.934 + 9.
2 – 3 S 88
0
26’ E 27.3 – 0.746 + 27.
3 – 4 N 86.
0
E 35.193 + 2.018 + 35.
4 – 5 S 05
0
18’ E 35.0 – 34.850 + 3.
5 – 1 S 72
0
02’ W 78.96 – 24.356 – 75.
EL = 0 ED = 0
The linear error of closure is equal to zero since EL and ED are both equal to zero.
2
3
4
5
N
CBLAMSIS UNIVERSITY OF THE CORDILLERAS MISSING DATA
And its bearing angle is
tan 𝜃 =
tan 𝜃 =
θ = 49. 446
0
0
′
"
The bearing of course 3 – 5 is S 49.
0
E, south- east since its latitude is negative
and departure is positive in sign.
Final tabulation of traverse 1- 2 - 3 - 5 - 1
Course Distance Bearing Latitude (m) Departure (m)
1 – 2 58.7 N 09
0
16’ E + 57.934 + 9.
2 – 3 27.3 S 88
0
26’ E – 0.746 + 27.
3 – 5 50.498 S 49.
0
E – 32.832 38.
5 – 1 78.96 S 72
0
02’ W – 24.356 – 75.
EL = 0 ED = 0
Consider traverse 3- 4 - 5 - 3. Determine length of course 3 – 4 and bearing of
course 4 – 5.
The traverse 3- 4 - 5 - 3 is a triangle with two sides known, lengths of course 4 – 5 and
course 5 – 3, and one interior angle θ. Using sine law, the other three elements of the
triangle can be determined.
0
0
0
Using sine law
3 − 5
sin 𝜌
4 − 5
sin 𝜃
sin 𝜌
sin 43. 841
0
sin 𝜌 =
sin 43. 841
0
0
0
3
4
5
θ
β
N
ρ
α
CBLAMSIS UNIVERSITY OF THE CORDILLERAS MISSING DATA
Taking the sum of the interior angles of the triangle
0
0
0
The bearing angle of course 4 – 5 , α , can now be determined,
0
0
0
0
0
′
And the bearing of course 4 – 5 is S 5
0
19 ’ E (a minute difference with original
given bearing of course 4 – 5 in sample problem no. 1)
Using sine law, solve for the length of course 3 – 4,
3 − 4
sin 𝛽
4 − 5
sin 𝜃
3 − 4
4 − 5
(sin 𝛽)
sin 𝜃
0
sin 43. 841
0
The distance of course 3 – 4 is 35.180 m (0.013 m difference with the computed
distance of course 3 – 4 in sample problem no. 1)
C. Length of Two Sides Unknown
SAMPLE PROBLEM NO.
A closed traverse has the following data as given in the table below. Determine the
lengths of course 1 – 2 and course 5 – 1.
Course Bearing Distance (m)
1 – 2 N 09
0
16’ E ----
2 – 3 S 88
0
26’ E 27.
3 – 4 N 86.
0
E 35.
4 – 5 S 05
0
18’ E 35.
5 – 1 S 72
0
02’ W ----
2
3
4
5
N
Figure B
CBLAMSIS UNIVERSITY OF THE CORDILLERAS MISSING DATA
ø = 180 − 72
0
′
0
0
′
"
check
𝜃 + ø + 𝛽 = ( 3 − 2 ) 180 = 180
0
′
0
′
"
0
′
"
0
0
Sine Law
1 − 2
sin ø
5 − 1
sin 𝛽
2 − 5
sin 𝜃
sin 62
0
′
1 − 2
- 746 𝑠𝑖𝑛ø
sin 62
0
′
0
′
"
sin 62
0
′
5 − 1
sin 62
0
′
0
′
"
sin 62
0
′
Course Bearing Distance (m)
1 – 2 N 09
0
16’ E 58.
2 – 3 S 88
0
26’ E 27.
3 – 4 N 86.
0
E 35.
4 – 5 S 05
0
18’ E 35.
5 – 1 S 72
0
02’ W 78.
D. Bearings of two sides unknown
SAMPLE PROBLEM NO. 4
A closed traverse has the following data as given in the table below. Determine the
bearings of course 4 – 5 and course 5 – 1.
Course Bearing Distance (m)
1 – 2 N 09
0
16’ E 58.
2 – 3 S 88
0
26’ E 27.
3 – 4 N 86.
0
E 35.
2
3
4
5
N
Figure C
CBLAMSIS UNIVERSITY OF THE CORDILLERAS MISSING DATA
In figure C above, the bearing of course 4 – 5 and bearing of course 5 – 1 are unknown.
The analysis will be to isolate the two sides of the traverse with unknown data from
the sides of the traverse of given technical descriptions. For example, in figure C draw
a line connecting corners 1 and 4 forming two traverses (traverse 1- 2 - 3 - 4 - 1 and
traverse 1- 4 - 5 - 1). Traverse 1 - 2 - 3 - 4 - 1 will be a case of omitted measurement with the
distance and bearing of one side unknown while the other traverse, traverse 1- 4 - 5 - 1,
is a triangle that can be solve by the solution of oblique triangles.
Consider traverse 1 - 2 - 3 - 4 - 1. Determine the length and bearing of course 4 – 1.
Course Distance Bearing Latitude (m) Departure (m)
1 – 2 N 09
0
16’ E 58.70 + 57.934 + 9.
2 – 3 S 88
0
26’ E 27.3 – 0.746 + 27.
3 – 4 N 86.
0
E 35.193 + 2.018 + 35.
4 – 1 ---- ---- Lat 4 - 1 Dep 4 - 1
ΣL = + 59.206 ΣD = + 71.
Thus, the latitude of course 4 – 1 is – (ΣL) = - (59.206) = – 59.206 m and the departure
of the same course is – (ΣD) = - (71.877) = – 71.877 m. Its distance then,
4 − 1
2
2
4 − 1
2
2
And its bearing angle is
tan 𝜃 =
tan 𝜃 =
θ = 50. 521
0
0
′
"
The bearing of course 4 – 1 is S 50.
0
W, south-west since its latitude is negative
and departure is negative in signs.
Consider traverse 1 - 4 - 5 - 1. Determine bearings of course 4 – 5 and course 5 – 1.
The traverse 1- 4 - 5 - 1 is a triangle with three sides known and three interior angles
unknown (θ, β, ρ). Using cosine law, solve for the three interior angles.
N
1
4
5
θ
β
ρ
Figure D
