surveying presentation
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surveying discussionTRANSCRIPT
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Surveying Set 1Corrections in Measuring Distances
Pace Factor
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Correction for Measuring DistancesGeneral Formula
CD = RD + Cwhere:
CD = Correct DistanceRD = Recorded DistanceC = all applicable Corrections
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Tape Correction
πΆπ·=π π·Β±πΆπ‘ππππ π·πΏπ‘πππ
+ too long-too short
π π·=πΆπ· Β±πΆπ‘ππππ π·πΏπ‘πππ
+ too long-too short
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Correction Due to Temperature (Ct)
Ct = k( Tm β Ts ) LWhere:
k β coefficient of linear expansion 0.0000116/0C 0.00000645/0FTm β Temperature during measurementTs β Standard Temperature
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Problem 1:When the temperature was 480C, the
measured distance from B to C was 318m. The steel tape used has a standard length at 200C, with a coefficient of thermal expansion of 0.0000116/0C. Find the correct distance BC in m.Solβn:
Ct = k( Tm β Ts ) LCt = 0.0000116/0C(480C- 200C)(318m)Ct = +0.103m too longCD = CD Β± CtCD = 318 + 0.103CD = 318.103m
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Problem 2:When the temperature was 50C, the
measured distance from D to E was 674.25m. The steel tape used has a standard length at 200C, with a coefficient of thermal expansion of 0.0000116/0C. What is the correct distance from D to E in m?Solβn:
Ct = k( Tm β Ts ) LCt = 0.0000116/0C(50C- 200C)(674.25m)Ct = -0.117m too shortCD = CD Β± CtCD = 674.25 β 0.117CD = 674.133m
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Correction due to Pull (Cp)
Where: Pm = Pull during measurement
Ps = Standard Pull A = Cross Sectional Area of Tape E = Modulus of Elasticity of Tape
Cp= (Pm - Ps )LAE
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Problem 3:A line was determined to be 2395.25m
when measured with a 30m steel tape supported throughout its length under a pull of 4 kg. Tape used is of standard length at a pull of 5kg. Cross sectional area of tape is 0.03 sq.cm. Modulus of elasticity of tape is 2x106 kg/cm2.Solβn:
AELPsPmCp )(
)102)(03.0()25.2395)(54(
6xCp
mCp 0399.0
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Correction due to Sag (Csag)
Where:w β weight of tape per linear
meterWtape / Ltape
Pm β Pull during measurement
Csag =w2Lsag3
24 Pm2
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Problem 4:A 100m tape weighs 0.0508 kg/m.
During field measurements, the tape was subjected to a tension of 45N and was supported at the end points, midpoints and quarter points. Find the correction per tape length due to sag.Solβn:
2
32
24PmLsagwCsag
2
32
)45(24)25()81.90508.0( xCsag
mCsag 0798.0
cmormxCsagTotal 3232.040798.0
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Pace Factor
ππΉ=ππ£ππππππππ π‘ππππ
ππ£πππππππ.ππ πππππ
πΏ=ππ£πππππππ .ππ πππππ π₯ ππΉ
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A line 100m long was paced by a surveyor for four times with the following data. 142, 145, 145.5, and 146. Then another line eas paced for four times again with the following results, 893, 893.5, 891 and 895.5.
1. Determine the pace factor.2. Determine number of paces for the new line.3. Determine the distance of the new line.
Problem 5: CE Board Nov. 1998
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Solution: 1. Pace factor:
No. of paces = (142+145+145.5+146)/5No. of paces = 144.625Pace factor = 100/144.625 = 0.691
2. Number of paces for the new line:No. of paces = (893+893.5+891+895.5)/5No. of paces = 893.25
3. Distance of the new line:Distance of new line = 893.25(0.691)Distance of new line = 617.236m.
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ChainingProb1: A line was measured to have 5 tallies, 6 marking pins and 63.5 links. How long is the
line in feet?Prob 2:Determine the length of the line, in m, it there were 3 tallies, 8pins and the last pin
was 9m from the end of the line. The tape used was 50m.Conversion:(For a 100ft Tape) (For a 50m tape)1 link = 1 ft 1 link = 0.5m1 chain = 100 ft 1 chain = 50m1 pin = 1 chain 1 pin = 1 chain1 pin = 100 ft 1 pin = 50m1 tally = 10 pins 1 tally = 10 pinsSolβn:(1) Dist = 5 tallies(10pins/tally)(100ft/1pin)+(6pins)(100ft/1pin)+(63.5links)(1ft/1link) Dist = 5663.5ft(2) Dist = 3 tallies(10pins/tally)(50m/1pin)+(8pins)(50m/1pin)+9m Dist = 1909m
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Probable Value and Error
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Probable Error (PE)
Where:E = Probable error
= sum of the squares of the residualsn = number of observation
ππΈ=0.6745β β π£2
π (πβ1)
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Standard Error (SE)
ππΈ=β β π£2
π(πβ1)
Where:SE = Standard error
= sum of the squares of the residualsn = number of observation
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Standard Deviation
ππ·=β β π£2
(πβ1)Where:SD = Standard Deviation
= sum of the squares of the residualsn = number of observation
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Precision of MeasurementPrecision of Measurement = PE/PVWeight
Probable ValueW = k x M (no. of measurement)W = k x 1/D (distance)W = k x 1/PE (probable error)
Correct ValueW = k x M (no. of measurement)W = k x 1/D (distance)W = k x 1/PE (probable error)
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The following interior angles of a triangle traverse were measured with the same precision.
1. Determine the most probable value of angle A.2. Determine the most probable value of angle B.3. Determine the most probable value of angle C.
Angle Value (Degrees) No. of Measurements
A 41 5B 77 6C 63 2
Problem 5: CE Board May 2003
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1. Probable value of angle A.A + B + C = 41 + 77 + 63 = 1810 Error = 181-180 = 10 Error = 60 mins.LCD of 5, 6, and 2 is 30
Sta. WeightCorrection
A
B
C
Solution:
30 = 6 5
6 (60) = 13.84β 26
30 = 5 6
5 (60) = 11.54β 26
30 = 15 2 26
15 (60) = 34.62β 26 60β
Corrected value of A = 410 β 13.84β = 40046.16β
3. Probable value of angle B: Corrected value of B = 770 β 11.54β = 76048.46β
4. Probable value of angle C: Corrected value of C = 630 β 34.62β = 62025.38β
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Leveling
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