GEFS - Omitted Measurements or Missing Data PDF

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CBLAMSIS UNIVERSITY OF THE CORDILLERAS MISSING DATAOMITTED MEASUREMENTS/MISSING DATAWhen 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 t...

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GEFS (FUNDAMENTALS OF SURVEYING) 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: 1. Length and bearing of one side unknown. 2. Length of one side and bearing of another side unknown. 3. Lengths of two sides unknown for which the bearings have been observed. 4. 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 𝑆 = √(𝛴𝐿)2 + (𝛴𝐷)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 −𝛴𝐷 tan 𝜃 = −𝛴𝐿 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 1–2 2–3 3–4 4–5 5–1

Bearing N 090 16’ E S 880 26’ E -------------S 050 18’ E S 720 02’ W

Distance (m) 58.7 27.3 ------------35.0 78.96

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GEFS (FUNDAMENTALS OF SURVEYING) CBLAMSIS UNIVERSITY OF THE CORDILLERAS

MISSING DATA

N

4

2

3

5

1

Determine the latitudes and departures of the given sides assuming that all measurements are correct. Course 1–2 2–3 3–4 4–5 5–1

Bearing N 090 16’ E S 880 26’ E -------S 050 18’ E S 720 02’ W

Distance (m) 58.7 27.3 ------35.0 78.96

Latitudes (m) Departures (m) + 57.934 + 9.452 – 0.746 + 27.290 --------– 34.850 + 3.233 – 24.356 – 75.110 ΣL = – 2.018 ΣD = – 35.135

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, 𝑆3−4 = √(𝛴𝐿)2 + (𝛴𝐷)2 𝑆3−4 = √(−2.018)2 + (−35.135)2 = 35.193 𝑚 And its bearing angle is −𝛴𝐷 −𝛴𝐿 35.135 tan 𝜃 = 2.018 θ = 860 42′ 43" 𝑠𝑎𝑦 86.7130 tan 𝜃 =

The bearing of course 3 – 4 is N 860 43’ E, north- east since its latitude and departure are both positive in sign. Final Tabulation Course 1–2 2–3 3–4 4–5 5–1

Bearing N 090 16’ E S 880 26’ E N 86.7130 E S 050 18’ E S 720 02’ W

Distance (m) 58.7 27.3 35.193 35.0 78.96

Latitudes (m) Departures (m) + 57.934 + 9.452 – 0.746 + 27.290 + 2.018 + 35.135 – 34.850 + 3.233 – 24.356 – 75.110 EL = 0 ED = 0

The linear error of closure is equal to zero since EL and ED are both equal to zero.

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GEFS (FUNDAMENTALS OF SURVEYING) CBLAMSIS UNIVERSITY OF THE CORDILLERAS

MISSING DATA

B. Length of one side and bearing of another side unknown SAMPLE PROBLEM NO. 2 A closed traverse has the following data as given in the table below. Determine the length of course 3 – 4 and the bearing of course 4 – 5. Course 1–2 2–3 3–4 4–5 5–1

Bearing N 090 16’ E S 880 26’ E S 86.7130 E ---S 720 02’ W

Distance (m) 58.7 27.3 ---35.0 78.96

N

2

4 3

5

1 Figure A

In figure A above, the distance of course 3 – 4 and bearing of course 4 – 5 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 A draw a line connecting corners 3 and 5 forming two traverses (traverse 1-23-5-1 and traverse 3-4-5-3). Traverse 1-2-3-5-1 will be a case of omitted measurement with the distance and bearing of one side unknown while the other traverse, traverse 3-4-5-3, is a triangle that can be solve by the solution of oblique triangles. Consider traverse 1-2-3-5-1. Determine length and bearing of course 3 – 5. Course 1–2 2–3 3–5 5–1

Distance 58.7 27.3 --78.96

Bearing N 090 16’ E S 880 26’ E --S 720 02’ W

Latitude (m) + 57.934 – 0.746 --– 24.356 ΣL = + 32.832

Departure (m) + 9.452 + 27.290 --– 75.110 ΣD = – 38.368

Thus, the latitude of course 3 – 5 is – (ΣL) = - (32.832) = – 32.832 m and the departure of the same course is – (ΣD) = - (- 38.368) = 38.368 m. Its distance then, 𝑆3−5 = √(𝛴𝐿)2 + (𝛴𝐷)2 𝑆3−5 = √(32.832)2 + (−38.368)2 = 50.498 𝑚

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GEFS (FUNDAMENTALS OF SURVEYING) CBLAMSIS UNIVERSITY OF THE CORDILLERAS

MISSING DATA

And its bearing angle is tan 𝜃 = tan 𝜃 =

−𝛴𝐷 −𝛴𝐿

38.368 −32.832

θ = 49.4460 𝑠𝑎𝑦 490 26′ 45" The bearing of course 3 – 5 is S 49.4460 E, south- east since its latitude is negative and departure is positive in sign. Final tabulation of traverse 1-2-3-5-1 Course 1–2 2–3 3–5 5–1

Distance 58.7 27.3 50.498 78.96

Bearing N 090 16’ E S 880 26’ E S 49.4460 E S 720 02’ W

Latitude (m) + 57.934 – 0.746 – 32.832 – 24.356 EL = 0

Departure (m) + 9.452 + 27.290 38.368 – 75.110 ED = 0

Consider traverse 3-4-5-3. Determine length of course 3 – 4 and bearing of course 4 – 5. N 4 ρ

3 θ

β

α

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. 𝜃 = 180 − (86.7130 + 49.4460 ) = 43.8410 Using sine law

𝑆3−5 𝑆4−5 = sin 𝜌 sin 𝜃

50.498 35.00 = sin 𝜌 sin 43.8410

sin 𝜌 =

50.498(sin 43.8410 ) = 𝑠𝑖𝑛(180 − 𝜌) 35.00

𝜌 = 180 − 87.9650 = 92.0350 4 |9

GEFS (FUNDAMENTALS OF SURVEYING) CBLAMSIS UNIVERSITY OF THE CORDILLERAS

MISSING DATA

Taking the sum of the interior angles of the triangle 𝛴𝑖𝑛𝑡𝑒𝑟𝑖𝑜𝑟 𝑎𝑛𝑔𝑙𝑒𝑠 = (𝑛 − 2)180 𝜃 + 𝜌 + 𝛽 = (3 − 2)180 = 180

𝛽 = 180 − 43.8410 − 92.0350 = 44.1240 The bearing angle of course 4 – 5, α, can now be determined, 𝛼 = 49.4460 − 𝛽 = 49.4460 − 44.1240 = 5.3220 ≈ 50 19′ And the bearing of course 4 – 5 is S 50 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, 𝑆4−5 𝑆3−4 = sin 𝛽 sin 𝜃 𝑆4−5 (sin 𝛽) 35 𝑠𝑖𝑛 44.1240 = = 35.180 𝑚 𝑆3−4 = sin 𝜃 sin 43.8410 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.3 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 1–2 2–3 3–4 4–5 5–1

Bearing N 090 16’ E S 880 26’ E N 86.7130 E S 050 18’ E S 720 02’ W

Distance (m) ---27.3 35.193 35.0 ----

N

4

2

5

1 Figure B

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