elementary surveying lecture part 1

7
ELEMENTARY SURVEYING 1 SURVEYING - art of determining and measuring distance, direction, and elevation. Types of error 1. Systematic error – (cumulative error) effects can be eliminated by applying corrections. 2. Accidental errors - error w/c remain after mistakes and systematic errors have been eliminated. SYSTEMATIC ERRORS 1. Correction due to tape too long or too short General Rule: Tape is Too Short : Laying out distances : Add the correction Measuring distances : Subtract the correction Tape is Too Long : Laying out distances : Subtract the correction Measuring distances : Add the correction MEASURING DISTANCES : LAYING OUT DISTANCES: 2. Correction due to change in Temperature αΔTL C T to be subtracted or added where C .. ree deg .. per .. m 10 x 6 . 11 6 F .. ree deg .. per .. ft 10 x 45 . 6 6 s a T T T change in temperature L length of the tape at standard temperature A B Standard Tape 100.2 m A B Tape too short by 0.2 m 100 m B’ 99.8 m B Tape too long by 0.2 m B’ 100 m 100 m A 99.8 m A B Tape too long by 0.2 m A B Standard Tape 100 m A B Tape too short by 0.2 m B’ B’ 100.2 m 100 m

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Page 1: Elementary Surveying Lecture Part 1

ELEMENTARY SURVEYING

1

SURVEYING - art of determining and measuring distance, direction, and elevation.

Types of error 1. Systematic error – (cumulative error) effects can be eliminated by applying corrections.

2. Accidental errors - error w/c remain after mistakes and systematic errors have been eliminated.

SYSTEMATIC ERRORS

1. Correction due to tape too long or too short General Rule: Tape is Too Short : Laying out distances : Add the correction

Measuring distances : Subtract the correction Tape is Too Long : Laying out distances : Subtract the correction

Measuring distances : Add the correction

MEASURING DISTANCES :

LAYING OUT DISTANCES:

2. Correction due to change in Temperature

αΔTLCT to be subtracted or added

where C..reedeg..per..m10x6.11 6 F..reedeg..per..ft10x45.6 6

sa TTT change in temperature

L length of the tape at standard temperature

A B

Standard Tape

100.2 m

A B

Tape too short by 0.2 m

100 m

B’ 99.8 m

B

Tape too long by 0.2 m

B’

100 m

100 m

A

99.8 m

A B

Tape too long by 0.2 m

A B

Standard Tape

100 m

A B

Tape too short by 0.2 m

B’

B’

100.2 m

100 m

Page 2: Elementary Surveying Lecture Part 1

ELEMENTARY SURVEYING

2

3. Correction due to change in Pull

AEPLCp to be added or subtracted

where sa PPP = change in pull

A = cross sectional area of tape

E = modulus of elasticity of the tape

L = length of the tape at standard pull

4. Correction due to sag

___P24LwC 2

32

sg to be subtracted only

where w = weight per linear meter

L = unsupported length of the tape

P = applied pull

5. Correction due to Slope

___S2

hC2

sp to subtracted only

where h = difference in elevation ; S = slope distance

6. Correction due to reduction to sea level

)hR('S

RS

where S = level distance at sea level S’ = level distance above sea level R = earth’s radius (R =6400 km ) h = vertical distance at sea level

h

R

Mean sea level

S’

S

Page 3: Elementary Surveying Lecture Part 1

ELEMENTARY SURVEYING

3

Problems

1. Using a 100 m tape that is 0.02 m too short, the measured distance from A to B is 160.42m, what is the correct distance of line AB? Given: L = 100 m * measure *

C = 0.02 m too short ML = 160.42 m

Req’d: T.L.

Sol’n: m 160.388

TL100

42.16002.042.160CMLTL

2. 30-m steel tape, known to be 30.006 (under standard conditions) was used to record a measurement of 119.898m. What is the correct distance for erroneous tape length? Given: Ltape = 30 m

True length (tape) = 30.006 m

898.119MLlong too m 006.0C

Sol’n:

m 119.922

TL

30898.119006.0898.119CMLTL

3. The correct distance between two points is 220.45m. Using a 100m tape that is “x” m too long, the length to be laid on the ground should be 220.406 m. Find the value of “x”? Sol’n: T.L. = 220.45 L.L. = 220.406

C = “x” m too long

m 0.02

x100

45.220x45.220406.220

C.L.T.L.L

4. A tape has a standard length at 20o C. a line was measured at a temperature of 3oC. If the coefficient of thermal expansions is 0.0000116m/oC and its true horizontal length is 865.30.What is the measured length in meters? Given: Ts = 20oC Ta = 3oC

T.L. = 865.30 Cm/ 106.11 6

Rea’d: M.L. Sol’n:

m 865.47

MLL

MLL10972.1-ML865.30

CMLTLshort tooL10972.1

L203106.11TL

4

4

6

5. A steel tape is 100 m long at a standard pull of 65 N. Compute the pull correction in mm if during measurement the applied pull is 40N. The tape has a cross sectional of 3.18 mm2 and E=200GPa. If the measured length of the line is 865.30, what is the corrected distance? Given: L 100 m

Ps = 65 N ML = 865.30

Pa = 40 N A = 318 mm2 E = 200GPa

Req’d : TL

Page 4: Elementary Surveying Lecture Part 1

ELEMENTARY SURVEYING

4

Sol’n:

865.266

.L.T100

30.86510003.93-865.30T.L

short too mm 93.3000,200318

10001006540AEPLCp

6. A 50m steel tape weighing 1.75kg is constantly supported at mid- length and at its end points, and is used to measure a line AB with a steady pull of 6.5kg. If the measured length of AB is 1349.60m, determine the correct length of line AB. Given: m = 1.75 kg L = 50 m

mkg035.0

5075.1

Pa = 6.5 kg

ML (AB) = 1,349.60 m Supports at end point and midpoint, unsupported L = 25 m

For 25 m:

0.125

132501888.0C

5.62425035.0

P24LC

sg

2

32

2

32

sg

For 24.6 m

m 1,348.582

018.160.1349TL

018.025

132501888.0C

5.6246.24035.0C

sg

2

32

sg

7. Find the correction for the horizontal distance of 20,000 m 10 km above sea level Given:

Sol’n:

m 31.20

8.968,19000,20C8.968,19S

6400S

10400.6000,20

RS

hR'S

sl

8. Slope distances AB and BC measures 450.60m and 1005.81m, respectively. The difference in elevation are 5.3m for points A and B and 3.6m for points B and C. Line AB has a rising slope and BC has a falling slope. Determine the horizontal distance from pt A to pt C Given:

Req’d: AC

h = 10 km

R = 6400 km

S’ = 20,000

S = ?

Mean sea level

S1 = 450.6

5.3 m

S2 = 1,005.81

3.6 m

A

B

B’ C

Page 5: Elementary Surveying Lecture Part 1

ELEMENTARY SURVEYING

5

Sol’n:

m 1,456.372

C'B'ABAC803.10056.381.1005C'B

569.4503.56.450'AB22

22

9. A line was measured with a 50m tape. There were 2 tallies, 8 pins, and the distance from the last pin to the end was 2.25m. Find the length of the line in meters.

Note: 1 tally = 1 pin

1 pin = 1 full tape

Given: F.T.L. = 50 m

No. of tallies = 2

No. of excess pins = 8

Partial tape length = 2.25 m

m 1,402.25

LL

25.2508102line of Length

10. A line 100m long was paced by a surveyor for four times with the following data 142, 145.5, 145 and 146. Then another line was paced for four times again with 893,893.5,891 and 895 paces. Determine the length of the line.

m 617.149

125.893691.0LLpaces 125.893Pace .Ave

48958915.893893Pace .Ave

:line of length unknown For

pace/m 691.0625.144

100PF

paces of no. ave.dist. tape(PF) Factor Pace

11. Two points A and B are established along the same direction from a theodolite station. If the subtended angle read on a subtense bar held at A and B are 0˚55’20” and 0˚23’44” respectively. Find the distance between two points.

m 165.444BCCABA

m 289.697

223'44"0tan

1BC

m 124.253

255'20"0tan

1CA

A B

AC

C

AB SUBTENSE BAR

Page 6: Elementary Surveying Lecture Part 1

ELEMENTARY SURVEYING

6

ACCIDENTAL ERRORS

Probable Error of Single Observation (PEs)

1n)xx(

6745.0PEs2

Probable Error of the Mean (PEm)

)1n(n)xx(

6745.0PEm2

Most Probable Value (MPV)

nx

MPV (mean) w

)wx(MPV

(weighted mean)

where

1. KNw , (weight of observation is directly proportional to no. of observation)

2. )K(1/ew 2 (weight of observation is inversely proportional to the square of the probable error)

3. 1/dw (weight of observation is inversely proportional to the distance where measurement is taken)

SAMPLE PROBLEMS 12. Number of measurements as tabulated in the table as shown.

DISTANCE 612.12

612.14 612.16 612.18 612.20 Determine the most probable error of the mean

x x 2xx

612.12 612.16 204.0

612.14 612.16 202.0

612.16 612.16 0

612.18 612.16 202.0

612.20 612.16 204.0

2xx 0.004

0.01612.16

01.0155

004.06745.0PE

02.0PE15

004.01nxx6745.0PE

m

s

2

s

Page 7: Elementary Surveying Lecture Part 1

ELEMENTARY SURVEYING

7

13. ELEVATION DISTANCE (km)

62.12 m 2.8 62.85 m 3.8 63.16 m 3.0 Determine the most probable value in the difference in elevation

Elev D D1w x*W

62.12 2.8 0.3571 62.12(0.3571) 62.85 3.8 0.2632 62.85(0.2632)

63.16 3.0 0.3333 63.16(0.3333)

0.9536 59.7764

62.684

9536.07764.59

WW*xMPVx

14. Number of measurements as tabulated in the table as shown. DISTANCE MEASUREMENTS 612.12 2 612.14 4

612.16 3 612.18 5 612.20 6 Determine the most probable value of the measurements having different values.

x N W=N x*W

612.12 2 2 612.12(2)

612.14 4 4 612.14(4)

612.16 3 3 612.16(3)

612.18 5 5 612.18(5) 612.20 6 6 612.20(6)

20 12,243.38

612.169

20

38.243,12WxWMPVx

15. Number of measurements as tabulated in the table as shown. DISTANCE PROBABLE ERROR 612.12 0.2 612.14 0.4

612.16 0.3 612.18 0.5 612.20 0.6 Determine the most probable value of the measurements having different values.

x e 2e1w x*W

612.12 0.2 25 612.12(25)

612.14 0.4 6.25 612.14(6.25)

612.16 0.3 11.11 612.16(11.11)

612.18 0.5 4 612.18(4)

612.20 0.6 2.78 612.20(2.78) 49.14 30,080.6086

612.14

WW*xMPV