Surveying Presentation

Surveying Set 1Corrections in Measuring Distances

Pace Factor

Correction for Measuring DistancesGeneral Formula

CD = RD + Cwhere:

CD = Correct DistanceRD = Recorded DistanceC = all applicable Corrections

Tape Correction

𝐶𝐷=𝑅𝐷±𝐶𝑡𝑎𝑝𝑒𝑅𝐷𝐿𝑡𝑎𝑝𝑒

+ too long-too short

𝑅𝐷=𝐶𝐷 ±𝐶𝑡𝑎𝑝𝑒𝑅𝐷𝐿𝑡𝑎𝑝𝑒

+ too long-too short

Correction Due to Temperature (Ct)

Ct = k( Tm – Ts ) LWhere:

k – coefficient of linear expansion 0.0000116/0C 0.00000645/0FTm – Temperature during measurementTs – Standard Temperature

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.Sol’n:

Ct = k( Tm – Ts ) LCt = 0.0000116/0C(480C- 200C)(318m)Ct = +0.103m too longCD = CD ± CtCD = 318 + 0.103CD = 318.103m

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?Sol’n:

Ct = k( Tm – Ts ) LCt = 0.0000116/0C(50C- 200C)(674.25m)Ct = -0.117m too shortCD = CD ± CtCD = 674.25 – 0.117CD = 674.133m

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

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.Sol’n:

AELPsPmCp )(

)102)(03.0()25.2395)(54(

6xCp

mCp 0399.0

Correction due to Sag (Csag)

Where:w – weight of tape per linear

meterWtape / Ltape

Pm – Pull during measurement

Csag =w2Lsag3

24 Pm2

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.Sol’n:

2

32

24PmLsagwCsag

2

32

)45(24)25()81.90508.0( xCsag

mCsag 0798.0

cmormxCsagTotal 3232.040798.0

Pace Factor

𝑃𝐹=𝑎𝑣𝑒𝑟𝑎𝑔𝑒𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑎𝑣𝑒𝑟𝑎𝑔𝑒𝑛𝑜.𝑜𝑓 𝑝𝑎𝑐𝑒𝑠

𝐿=𝑎𝑣𝑒𝑟𝑎𝑔𝑒𝑛𝑜 .𝑜𝑓 𝑝𝑎𝑐𝑒𝑠 𝑥 𝑃𝐹

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

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.

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 pinsSol’n:(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

Probable Value and Error

Probable Error (PE)

Where:E = Probable error

= sum of the squares of the residualsn = number of observation

𝑃𝐸=0.6745√ ∑ 𝑣2

𝑛 (𝑛−1)

Standard Error (SE)

𝑆𝐸=√ ∑ 𝑣2

𝑛(𝑛−1)

Where:SE = Standard error

= sum of the squares of the residualsn = number of observation

Standard Deviation

𝑆𝐷=√ ∑ 𝑣2

(𝑛−1)Where:SD = Standard Deviation

= sum of the squares of the residualsn = number of observation

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)

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

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) = 13.84’ 26

30 = 5 6

5 (60) = 11.54’ 26

30 = 15 2 26

15 (60) = 34.62’ 26 60’

Corrected value of A = 410 – 13.84’ = 40046.16’

3. Probable value of angle B: Corrected value of B = 770 – 11.54’ = 76048.46’

4. Probable value of angle C: Corrected value of C = 630 – 34.62’ = 62025.38’

Leveling