ERRORS IN MESUREMENT AND HORIZONTAL MEASUREMENTS PDF

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Surveying or land surveying is the technique, profession, art, and science of determining the terrestrial or three-dimensional positions of points and the distances and angles between them.Surveying has been an element in the development of the human environment since the beginning of recorded histo...

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Technological University of the Philippines Ayala Blvd. cor. San Marcelino St., Ermita, Manila

College of Engineering Civil Engineering Department

CE 25 – 2B Elementary and Higher Surveying

Assignment No. 2 Theory of Errors and Measurements

Rivera, Ericka Shane L. 14-205-007 Date of Submission: Dec. 16, 2015

Engr. Jesus Ray M. Mansayon Instructor

PROBLEM SET 1 1. LINEAR MEASUREMENTS. The measured length of airport runways in five major cities in the Philippines are: 1250.00, 1375.50, 1410.75, 1550.25, and 1750.00 meters. Determine the equivalent length of each runways in kilometers, decimeters, and centimeters. Tabulate values accordingly. Given: Runway 1 2 3 4 5 Required:

m 1250.00 1375.50 1410.75 1550.25 1750.00

Length in kilometres, decimetres and centimeters

Illustration: -

----- 1 --------

-

----- 4

Solution: To km

To dm

--------

-------- 2 -------

------- 5 --------

------- -

3 ------

To cm

Runway 1 2 3 4 5

m 1250.00 1375.50 1410.75 1550.25 1750.00

km 1.25 1.3755 1.41075 1.55025 1.750

dm 12,500 13,755 14,107.5 15,502.5 17,500

cm 125,000 137,500 141,075 155,025 175,000

2. AREA MEASUREMENTS. Given the dimensions of the following tracts of land: a. 108.75 m by 76.82 m b. 940.08 m by 1296.73 m c. 13.36 m by 50.08 m d. 1258.30 m by 624.03 m e. 8476.55 m by 121.79 m Determine the area of each tract in square meters, square kilometres, ares and hectares. Given: Tracts of Land 1 2 3 4 5 Required:

Dimension 108.75m by 76.82 m 940.08m by 1296.73m 13.36m by 50.08m 1258.30m by 624.03m 8476.55m by 121.79m

Area in m2, km2, ares and hectares

Illustration: 1

Solution:

2

3

4

5

Area in m2

Area in km2

Area in ares

` Area in hectares

3. AREA MEASUREMENTS. Given the area and width of the following rectangularshaped pieces of property: a. 2.575 ha and 195.42 m b. 125.42 sq m and 545.0 cm c. 0.85 sq km and 925.09 m d. 50.0 ares and 100.0 m e. 42545.19 sq m and 346.72 m Determine the length of each property in meters. Given: Property 1 2 3 4 5 Required:

Area 2.575 ha 125.42 sq m 0.85 sq km 50.0 ares 42545.19 sq m

Width 195.42 m 545.0cm 925.09 m 100.0 m 346.72 m

Length of each property in meters

Illustration: 1

Solution: Property 1

Property 2

2

3

4

5

Property 3

Property 4

Property 5

4. VOLUME MEASUREMENTS. Following are the dimensions for length, width and depth of five excavated borrow pits for a highway project: a. 113.26 m, 35.48 m, and 18.60 m b. 50.08 m, 39.25m, and 7.14 m c. 243.55 m, 76.19 m, and 24.66 m d. 42.055 m, 8.605 m, and 12.332 m e. 9.5 m, 6.3 m, and 4.9 m Determine the volume of each pit in cubic meters. Given: Pit A B C D E

L 113.26 m 50.08 m 243.55 m 42.055 m 9.5 m

Required:

W 35.48 m 39.25 m 76.48 m 8.605 m 6.3 m

D 18.60m 7.14 m 24.66 m 12.332 m 4.9 m

Volume of each pit in cubic meters

Illustration: A

Solution:

B

C

D

E

Pit A (

)(

)

(

)( )

(

)(

(

)(

Pit B

Pit C

)

Pit D )

Pit E ( )( ) 5. VOLUME MEASUREMENTS. Given the approximate flat area and depth of excavation of the following borrow pits: a. 3750.0 sq m and 758.0 cm b. 0.035 sq km and 180.0 m c. 15.6 ares and 495.0 m d. 9.250 ha and 250.0 m e. 46750 sq m and 195.0 m Determine the volume of earth removed from each pit in cubic meters. Given: Pits A B C D E

Area 3750sq m 0.035 sq km 15.6 ares .250 ha 46750 sq m

Required:

Depth 758.0 cm 180.0 m 495.0 m 250.0 m 195.0 m

Volume of earth removed from each pit in cubic meters.

Illustration: A

Solution: Pit A

B

C

D

E

()

( )( )

Pit B

(

()

)( )

Pit C

(

)(

()

)

Pit D

(

()

)( )

Pit E

( )

6. SIGNIFICANT FIGURES. Given the following numbers: 45.63, 5.700, 4010, 0.00037, 0.000940, 6.0090, 7.00, 9.5x108, 4.00x107, 2.604x1018, and 3.00x10-6. For each number, identify the significant figures and state the number of significant figures. Tabulate values accordingly. Given: Number 45.63 5.700 4010 0.00037 0.000940 6.0090 7.00 9.5X108 4.00x107 2.604x1018

3.00x10-6 Required: Number of significant figures of each number Solution: Number Significant Figures 45.63 4 5.700 4 4010 3 0.00037 2 0.000940 3 6.0090 5 7.00 3 9.5X108 2 4.00x107 3 18 2.604x10 4 3 3.00x10-6 7. ROUNDING OFF NUMBERS. Given the following numbers: 0.333333, 0.412342, 50.2155, 48.5455, and 16.3545. Round off each number to three, four and five figures. Tabulate values accordingly. Given: Numbers 0.333333 0.412342 50.2155 48.5455 16.3545 Required: Round off each number to three, four and five figures Solution: Numbers 0.333333 0.412342 50.2155 48.5455 16.3545

3 Figures 0.333 0.412 50.2 48.5 16.4

4 Figures 0.3333 0.4123 50.22 48.54 16.35

5 Figures 0.33333 0.41234 50.216 48.546 16.354

8. MOST PROBABLE VALUE. The three angles of a triangle were measured with the following results: A=42°05’, B=115°38’, and C=22°08’. Determine the most probable value of each angle. Given: A = 42°05’

B = 115°38’

C = 22°08’

Required: Most probable value of each angle Illustration:

A

C

B

Solution: A+B+C = 179°51’

Error = 180-179°51’

Error = 0°9’

Correction: ±0°3’

A’ = A ± 0°3’

B’ = B ± 0°3’

C’ = C ± 0°3’

A’ = 42°5’ +0°3’

B’ = 115°38’ + 0°3’

C’ = 22°08’ + 0°3’

A’ = 42°8’0”

B’ = 115°41’0”

C’ = 22°11’0”

9. MOST PROBABLE VALUE. The angles about a point have the following observed values: 87°07’50”, 125°17’20”, and 147°35’20”. Determine the most probable value of the three angles. Given:

A = 87°07’50”

B = 125°17’20”

C = 147°35’20”

Required: Most probable value of each angle Illustration: C A B

Solution: A + B + C = 360°0’30”

Error = 360° - 360°0’30”

Correction = ±0°0’10”

Error = 0°0’30”

A’ = A ± 0°0’10”

A’ = 87°07’50” - 0°0’10”

A’ = 87°7’40”

B’ = B ± 0°0’10”

B’ = 125°17’20” - 0°0’10”

B’ = 125°17’10”

C’ = C ± 0°0’10”

C’ = 147°35’20” - 0°0’10”

C’ = 147°35’10”

10. MOST PROBABLE VALUE. The interior angles of a quadrilateral were observed to be: A=100°35’40”, B=118°44’15”, C=80°54’35”, and D= 59°45’50”. Determine the most probable value of each of these angles. Given:

A=100°35’40”

B=118°44’15”

C=80°54’35”

D= 59°45’50”

Required: Most probable value of each of these angles Illustration: B

A

D C

Solution: A + B + C = 360°0’20” Error = (n-2)180 - 360°0’20”

Correction = 5”

Error = 360 - 360°0’20” Error = 20” A’ = A ± 5”

A’ = 100°35’40” – 5”

A’ = 100°35’35”

B’ = B ± 5”

B’ = 118°44’15” – 5”

B’ = 118°3510”

C’ = C± 5”

C’ = 80°54’35” – 5”

C’ = 80°54’30”

D’ = D± 5”

D’ = 59°45’50” – 5”

D’ = 59°45’45”

11. PROBABLE ERROR. A quantity was measured ten times with the following results: 34.630, 34.626, 34.634, 34.628, 34.629, 34.626, 34.627, 34.633, 34.625, and 34.624 meters. Determine the probable error of the mean and the relative precision of the mean. Given: Trial

X

1 2 3 4 5 6 7 8 9 10 Required:

34.630 34.626 34.634 34.628 34.629 34.626 34.627 34.633 34.625 34.624

Probable error of the mean and the relative precision of the mean

Illustration: 10x

Solution: ∑

= ±1.052 X10-3

= ± 3.327X10-3 PEm = ± 0.6745(±1.052 X10-3)

PEm = 7.096 X10-4

RP =2.049

12. PROBABLE ERROR. A surveying instructor sent all the 40 students in his class out to measure a distance between two points marked on a runway. The students working in groups of four came up with 10 different measurements as follows: 920.45, 921.05, 921.65, 920.25, 920.15, 921.85, 921.95, 920.45, 921.15, and 921.35 meters. Assuming these values are equally reliable and that variations result only from accidental errors, determine the relative precision of a single measurement and the relative precision of the mean. Given: Groups 1 2 3 4 5 6

Measurement 920.45 921.05 921.65 920.25 920.14 921.85

7 8 9 10 Required:

921.95 920.45 921.15 921.35

RP of single measurement and RP of mean measurement

Illustration: 10x

Solution: ∑

= ± 0.673 PEs =± 0.6745( ) PEs =± 0.454

= ± 0.213 PEm = ± 0.6745 ( ) PEm =± 0.144

13. PROBABLE ERROR. The following values were determine in a series of rod readings made under identical conditions: 3.312, 3.307, 3.304, 3.306, 3.309, 3.301, 3.311, 3.308, 3.312, 3.306, and 3.313 meters. Determine the following: most probable value of the observed rod readings, probable error of a single measurement and of the mean, and the relative precision of a single measurement and of the mean. Given: Rod reading 1 2 3 4 5 6 7 8 9

Value 3.312 3.307 3.304 3.306 3.309 3.301 3.311 3.308 3.312

10 11 Required:

3.306 3.313

most probable value of the observed rod readings, probable error of a single measurement and of the mean, and the relative precision of a single measurement and of the mean

Illustration: 10x

Solution: ∑

= ± 3.754 X10-3

= ± 1.132 X 10-3

PEs = ± 0.6745( )

PEm = ±0.6745( )

PEs = ± 2.532 X10-3

PEm = ± 7.634 X10-4

14. WEIGHTED MEASUREMENTS. A line is measured on a windy day as 338.65 m. The same line measured 338.37 m on a calm day. If the latter measurement is given four times the reliability of the first, determine the most probable value of the measured line. Given: x 338.65 338.37 Required:

weather windy Calm

Weigth 1 4

most probable value of the measured line

Illustration:

Windy

calm

Solution: ∑( )

(

∑

)(

)

15. WEIGHTED MEASUREMENTS. A distance AB is measured five times as 610.03, 610.01, 610.05, 610.04, and 610.02 meters. The three measurements were given weights of 3, .2, 1, 2, and 3 respectively, by the head tapeman. Determine the weighted mean for distance AB. Also, determine what difference results if later judgement revises the weights to 2, 3, 1, 3 and 2. Given: Trial 1 2 3 4 5

x 610.03 610.01 610.05 610.04 610.02

W 3 .2 1 2 3

Required: Weighted mean and difference if weights are 2,3,1,3,2 Illustration: 5x

Solution: ∑( )

∑

Weighted mean = 610.0306522 Weighted Mean if Reverse = 610.0290909 Difference = 610.0306522 - 610.0290909 Difference = 1.561 X10-3 16. WEIGHTED MEASUREMENTS. An angle ABC is measured at different times using various instruments and procedures. The results, which are assigned certain weights, are as follows: 75°09’26”, weight of 4; 75°09’25”, weight of 3; and 75°09’27”, weight of 1. Determine the most probable value of the angle measured. Given: x 75°09’26”

W 4

W*X 300°37’44”

75°09’25” 75°25’27” Required:

3 1

225°28’15” 75°09;27”

Most probable value of angle measured

Illustration:

A

B

C

Solution: ∑( )

∑

17. WEIGHTED MEASUREMENTS. In this problem, the weight of an angle is assumed to be proportional to the number of times it has been measured by repetition. Five angles in a pentagon were measured with the following results: 134°44’35”, 167°02’05”, 86°15’20”, 75°48’50” and 76°08’50”. If the number of repetitions for each measurement were 2, 6, 6, 8, and 4 respectively, determine the adjusted values of the angles. Given: Angle 1 2 3 4 5 Required:

x 134°44’35” 167°02’05” 86°15’20” 75°48’50” 76°08’50”

Adjusted Values or most probable value of angles

Illustration: 3

4

1

2

5

w 2 6 6 8 4

Solution: 1 + 2 + 3 + 4 + 5 = 539°59’40” Error = (n-2)180 - 539°59’40” Error = 540 -539°59’40” Error = 0°0’20” Adjested Values ( 

Correction 0°0’20” X

= 0°0’8.28”

134°44’43.28”

0°0’20” X

= 0°0’2.76”

167°2’7.76”

0°0’20” X

= 0°0’2.76”

86°15’22.76”

0°0’20” X

= 0°0’2.07”

75°48’52.07”

0°0’20” X

= 0°0’4.14”

76°8’54.14”

)

18. WEIGHTED MEASUREMENTS. Two sides and the included angle of a triangle were measured and the probable error of each value were computed as follows: a=267.55 ± 0.05 m, b=564.75 ± 0.06 m, and angle C=57°15’45”. Determine the area of the triangle and the probable error of the area. Given:

a = 267.55m ±0.05 m

Required:

Area and PE of Area

Illustration:

b=564.75m±0.06m

C a

b

Solution: ( A= 63,548.9341m

)

)

C=57°15’45”

) (

√(

)

PEA = ±13.66m2

19. WEIGHTED MEASUREMENTS. Three sides of a triangle were measured with the following results: a=1431.20 m ± 0.02 m, b=570.77 ± 0.03 m, and c= 1767.15 ± 0.04 m. Determine the angles in the triangle together with the probable errors of the angles. Given: Sides a b c Required:

x 1431.20m 570.77m 1767.15m

PE ±0.02m ±0.03m ±0.04m

Angles A,B and C with the probable errors of the angles.

Illustration:

C b

a

A c

B

Solution: c2 = a2 +b2 -2ab cos C (

)

Sin A = 0.72

C = 117°16’31.71”

A = 46°2’29.35”

A + B + C = 180 B = 180 – (A + C) B = 16°40’58.94” ( (

) )

( (

) )

(

)

)

(

20. WEIGHTED MEASUREMENTS. A line AE is divided into segments for measurement with a tape. The results were AB=134.10m ± 0.040m, BC=320.63m ± 0.055m, CD=173.73m ± 0.056m, and DE=160.85m ± 0.050m. Determine the length of the line and the probable error of the measured length. Given: Line AB BC CD DE Required:

x 134.10m 320.6m 173.73m 160.85m

PE ±0.040m ±0.055m ±0.056m ±0.050m

Length of line AE and its PE

Illustration: A

B

C

D

E

Solution: L = AB + BC + CD +DE L = 143.10 + 320.6 + 173.73 + 160.85 L = 789.31 m √( )

√( )

( ) (

)

( ) (

( )

) (

)

21. WEIGHTED MEASUREMENTS. The difference in elevation between two ground points was measured by each of three field parties using different kinds of levelling instruments. The results are as follows: 1st party, DE=18.45m ± 0.05m; 2nd party, DE=18.40m ± 0.04m; 3rd party, DE= 18.48m ± 0.05m. Determine the most probable difference in elevation between two ground points. Given: Party

x

PE

1st 2nd 3rd

18.45m 18.40m 18.48m

Required:

±0.05m ±0.04m ±0.05m

Most probable difference in elevation between two ground points

Illustration: E

D

Solution: W = 1/PE2 400 625 400 

∑

RW=W/Smallest Weight 1 1.5625 1 

∑

RW*X 18.45 28.75 18.48



22. SUMMATION OF ERRORS. The four approximately equal sides of attract of land were measured and the measurements included the following errors: ± 0.05 m, ± 0.14 m, ± 0.175 m, and ± 0.205 m, respectively. Determine the probable error for the total length of the tract. Given: Sides 1 2 3 4 Required:

PE ± 0.085 ± 0.014 ± 0.0175 ± 0.205

PE for perimeter

Illustration: 2 1

4 3

Solution:

√( )

√(

)

( ) ( ) (

( )

) (

)

(

)

23. SUMMATION OF ERRORS. The dimensions of a five-sided tract of land are given by the following measurements and corresponding probable errors: 221.63 m. ± 0.04m, 235.70 m ± 0.002 , 196.05 m ± 0.05m, 296.13 m ± 0.012m, and 303.18 m ±0.015 m. determine the probable error of the sum of the five measurements and the most probable value of the perimeter. Given: Sides 1 2 3 4 5

x 221.63m 235.70m 196.05m 296.13m 303.18m

Required:

PE ± 0.004m ± 0.002m ± 0.005m ± 0.012m ± 0.015m

PEsum and Most Probable Value

Illustration:

2

1

3

5

4

Solution: √( ) √(

)

( ) ( ) (

) (

( ) )

(

( ) )

(

)

24. PRODUCT OF ERRORS. Two sides of a rectangle were measured as being 226.25 ± 0.03 m and 307.28 ± 0.04 m. Determine the area of the figure and the probable error of the area. Given: Side 1 2

x 226.25m 307.28m

Required:

PE ± 0.03m ± 0.04m

Area of the figure and the probable error of the area

Illustration:

1

2

Solution: Area = L*W Area = 226.25*307.28 Area = 69522.1m2 √(

√(

)

(

)

(

)

)

25. PRODUCT OF ERRORS. The base and altitude of triangular-shaped figure were measured with certain estimated probable errors as follows: b=425.67 ± 0.07 m and h=138.63 ± 0.06 m. determine the area of the figure and the probable error in the resulting calculations. Given:

b=425.67m ± 0.07m

h=138.63m±0.06m

Required: area of the figure and the probable error in the resulting calculations Illustration:

h

b

Solution:

(

)(

)

Area = 29,505.316m2 √(

√(

)

PEA = ±13.66m

(

)

(

)

)

UNIT EXAM No. 1

B 1. Surveying is defined as the art and science of determining angular and linear measurements to establish the form, extent, and relative position of points, lines, and areas on or near the surface of the earth or other extraterrestrial bodies through a. b. c. d. ...