elementary surveying lecture part 1
DESCRIPTION
CE120-0TRANSCRIPT
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
ELEMENTARY SURVEYING
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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
ELEMENTARY SURVEYING
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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
ELEMENTARY SURVEYING
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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
ELEMENTARY SURVEYING
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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
ELEMENTARY SURVEYING
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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
ELEMENTARY SURVEYING
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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