Ge 105 lecture 2 (TAPING CORRECTION) by: Broddett B. Abatayo

ENGR. BRODDETT B. ABATAYO, GE, REA

Part-time Lecturer – GE division, CEIT, CSU, Ampayon, Butuan CityResearch Assistant – Phil-LiDAR 2 Project, CSU, Ampayon, Butuan City

Proprietor – BPA ABATAYO Land Surveying Services1

with CASIO fx-991 es plus Calculator Technique

Lecture 2

Caraga State University

College of Engineering and Information Technology

Ampayon, Butuan City 8600

TAPING CORRECTIONS

GE 105 – Theory of Errors and Adjustments

Rules for Applying Tape Corrections

Measured distance:

1. Add correction - tape too long

2. Subtract correction - tape too short

Laying out distance:

1. Subtract correction - tape too long

2. Add correction - tape too short

A B

MEASURE

Standard

Too short

Too long

True Distance AB is equal to 8cm

8cm

LAYING OUT

Standard

Too short

Too long

A

Taping CorrectionsA. Correction due to Temperature:

(to be added or subtracted)

Where;

α = coefficient of thermal expansion (0.0000116/°C)

To = observed temperature during measurement

Ts = standard temperature

L = Nominal length of tape or total measured distance

Example:

The measured distance from B to C was 318m. The steel tape used has a standard length at 20°C with a coefficient of thermal expansion of 0.0000116/°C. The corrected distance B to C is 318.103m. Find the temperature during measurement.

LTsToC t )(

shift CALC =

Ans. 47.92°C

Calculator technique:

B. Correction due to Pull:

(To be added or subtracted)

Where;

Po = applied pull during measurement

Ps = standard pull

L = Nominal length of tape or total measured distance

A = cross-sectional area of tape

E = modulus of elasticity of tape

Taping Corrections

AE

LPsPoCp

)(

Example:

The measured distance from A to B was 318musing tape having a cross-sectional area of0.05cm2 has been standardized at a tension of5.5kg. If the modulus of elasticity E = 2.10x106

kg/cm2, determine the pull applied if thecorrected distance A to B 1s 318.012m.

shift CALC =

Ans. 9.46kg


Taping CorrectionsC. Correction Due to Sag:

(to be subtracted only)

Where:

ω = weight of tape per unit length

W = total mass or weight of tape

L = unsupported length of tape

Po = applied pull during measurement

2

2

2

32

2424 Po

LW

Po

LCsag

Ans. 1. 0.0162m2. 29.984m

Example

A 30m tape is supported only at itsends and under a steady pull of 8kg.If the tape weighs 0.91kg. Determinethe following:

1.Sag correction2.Correct distance between the ends of the tape

A line was determined to be 2395.25m when measured with a 30m steel tape supported throughout its length under a pull of 4kg. Determine the temperature during measurement if the tape used is of standard length at 20°C under a pull of 5kg. The cross-sectional area of the tape is 0.03sq.cm, its coefficient of linear expansion is 0.0000116/°C, the corrected distance is 2395.63m and the modulus of elasticity of steel is 2.10x106

kg/cm2.

Combined Corrections

shift CALC =


Ans. 35.04°C

D. Correction due to slope:

(to be subtracted only)

Where;

S = inclined/slope distance

H = correct horizontal distance

h = vertical distance at ends of tape during measurement

Taping Corrections

S

hCs

2

2

CsSH

Example

Slope distances AB and BC measure 330.49mand 660.97m, respectively. The difference inelevation is 10.85m for B and C. Using theslope correction formula, determine thedifference in elevation for A and B. If thehorizontal length of line ABC is 991.145m.Assume the line AB has a rising slope and BCa falling slope.

Ans. 12.22m

shift CALC =


Taping Corrections

CsagCp

2

2

24

)(

Pn

LW

AE

LPsPn

)(204.0

PsPn

AEWPn


shift CALC =

1. Determine the most probable value of the angles about a given point. (10 points)

Angle Value repetition

A 130°15‘03" 5

B 142°37‘21" 6

C 87°08‘17" 2

Determine the most prob. value of A, B and C.

2. A base line measured with an invar tape, and with a steel tape as follows: (20 points)

Set I (Invar tape) Set II (Steel tape)

571.185 571.193

571.186 571.190

571.179 571.185

571.170 571.179

571.193 571.192

Determine the following:1. Probable error in set II.2. Standard error in set I.3. Most probable value of the two

sets.4. Probable error of the general

mean.

QUIZ 2 ½ cross wise

Reduction to Sea-Level

Taping CorrectionsF. Reduction to Sea-Level

Where;

D = measured distance bet. two points

D’ = corresponding sea-level distance of these points

R = average radius of curvature

(1-h/r) = sea-level reduction factor

h = average elevation above sea level

R

hDD 1'

shift CALC =


Ans. 6844.35m

Prob 1• When the temperature was 48°C, the

measured distance from B to C was 318 m. The steel tape used has a standard length at 20°C with a coefficient of thermal expansion of 0.0000116/°C. Find the correct distance BC in meters.

Ans. 318.103 m

Prob 2• When the temperature was 3°, the

distance from E to F was measured using a steel tape that has a standard length at 20 °C with a coefficient of thermal expansion of 0.0000116/ °C. If the correct distance from E to F is 836.5m, what was the measured distance in meters?

Ans. 836.665 m

Prob 3• A 50 m tape was standardized and

was found to be 0.0042m too long than the standard length at an observed temperature of 58 °C and a pull of 15kg. If the same tape was used to measure a certain distance and was recorded to be 673.92m long at an observed temperature of 68 °C and a pull of 15kg, and the coefficient of thermal expansion is 0.0000116/ °C, determine the following:

1. Standard Temperature2. Total correction3. True length of the line

Ans. 1. 50.76 °C2. 0.1348m3. 674.05 m

Prob 4• A 50 m tape having a cross-sectional

area of 0.05cm2 has been standardized at a tension of 5.5kg. If the modulus of elasticity E = 2.10x106

kg/cm2, determine the elongation of the tape if a pull of 12 kg. is applied.

Ans. 0.003m

Prob 5• It takes 20 kg of normal tension to

make the elongation of a steel tape offset the effect of sag when supported at the end points. The tape has a cross-sectional area of 0.05cm2 and E = 2x106 kg/cm2. If the tape is 50m long and has a standard pull of 8kg. What is its unit weight in kg/m?

Ans. 0.0215 kg/m

Prob 6• A 30m tape is supported only at its

ends and under a steady pull of 8kg. If the tape weighs 0.91kg. Determine the following:

1. Sag correction2. Correct distance between the ends of

the tape

Ans. 1. 0.0162m2. 29.984m

Prob 7• A 100m tape weighs 0.0508 kg/m.

During field measurements, the tape was subjected to a tension of 45 N, and was supported at the end points, midpoint, and quarter points, find the correction per tape length due to sag.

Ans. 0.319 m

Prob 8• A line 100 m long was measured with

a 50m tape. It was discovered that the first pin was stuck 30cm to the left of the line and the second pin 30cm to the right. Find the error in the measurement in cm.

Ans. 0.45cm

Prob 9• A line was determined to be

2395.25m when measured with a 30m steel tape supported throughout its length under a pull of 4kg at a mean temperature of 35°C. The tape used is of standard length at 20°C under a pull of 5kg. If the cross-sectional area of the tape is 0.03cm2, coefficient of thermal expansion is 0.0000116/°C, and E = 2x106 kg/cm2, determine the following:

1. Temperature correction2. Pull correction3. Correct length of the line

Ans. 1. +0.4168m2. -0.0399m 3. 2395.6269m

Solution: MODE 1

458.65 + A + B - C = 456.8209015

9 pin + 8.65 = 458.65

Example: A civil engineer used a 100 m tape which is of standard length at 32°C in measuring a certain distance and found out that the length of tape have different lengths at different tensions were applied as shown: K = 0.0000116 m/°C

Length of tape @ 32°C Tension applied

99.986 m 10 kg

99.992 m 14 kg

100.003 m 20 kg

1. What tension must be applied to the tape at a temp. of 32°C so that it would be of standard length?

2. What tension must be applied to the tape at a temp. of 40.6°C so that it would be of standard length?


A civil engineer used a 100 m tape which is of standard length at 32°C in measuring a certaindistance and found out that the length of tape have different lengths at different tensions were applied as shown: K = 0.0000116 m/°C

Length of tape @ 32°C Tension applied

99.986 m 10 kg

99.992 m 14 kg

100.003 m 20 kg




Solution: MODE 199.986 →A 10 →D99.992 →B 14 →E100.003 →C 20 →F

MODE 3 2

X Y

D

E

F

AC shift 1 5 5 ← 100 = (18.35874439)

18.35874439 kg

C

B

A





X Y

D

E

F

0.0000116(40.6-32)(100)= →X

Length of tape @ 40.6°C Tension applied

99.995976 10 kg

100.001976 14 kg

100.012976 20 kgC + X

B + X

A + X

AC shift 1 5 5 ← 100 = (12.54313901)

12.54313901 kg

shift 1 2 (table)





X Y

D

E

FC + Y

B + Y

A + YLength of tape @ 30°C Tension applied

99.98368 10 kg

99.98968 14 kg

100.00068 20 kg

0.0000116(30-32)(100)= →Y

shift 1 2 (table)

AC shift 1 5 5 ← 100 = (19.71121076)

19.71121076 kg




18.35874439 kg

12.54313901 kg

19.71121076 kg

35⁰ 43’ 53.2" →A29⁰ 37‘ 05.8" →B23⁰ 29‘ 36.7" →C65⁰ 20‘ 58.2" →D53⁰ 06‘ 43.1" →E

88⁰ 50‘ 36.2" →F

Solution: MODE 1

a b c d

1

2

3

MODE 5 2

C + E + F321

B+D+E+F242

A + D + F123

Press =

X = 35⁰ 43’ 52.98"

Y = 29⁰ 37‘ 5.75"

Z = 23⁰ 29‘ 37.18"

Press =

Press =

X = 35⁰ 43’ 52.98"

Y = 29⁰ 37‘ 5.75"

Z = 23⁰ 29‘ 37.18"

5.369

SY 2014-15 PRELIM EXAM: Measured from point A, angles BAC, CAD, and BAD were recorded as follows:

C

DA

BAngle Value # of repetitions

BAC 28⁰24‘00" 2

CAD 61⁰15‘00" 2

BAD 89⁰29‘40" 4

a. Most Probable Value of angle BAC. b. Most Probable Value of angle BAD. c. Most Probable Value of angle CAD.

Determine the following:

Solution:

Angle Value # of repetitions

BAC 28⁰24‘00" 2

CAD 61⁰15‘00" 2

BAD 89⁰29‘40" 4

MODE 1

28⁰24‘00" A

61⁰15‘00" B

89⁰29‘40" C

D

C

A

B

DA

B

BAC + CAD > BAD

28⁰24‘00“ + 61⁰15‘00“ > 89⁰29‘40"

89⁰39‘40“ > 89⁰29‘40"

- corr

- corr

+ corr

(A + B) – C =

(A+B)-C0⁰9‘20"

D Math

X

2¯¹ + 2¯¹ + 4¯¹ = D Math

Y

D Math

A

D Math

B

D Math

C

Check: A + B = C

SY 2015-16 PRELIM EXAM: An angle was carefully measured 5 times with an optical theodolite by observers A and B on two separate days. The calculated results are as fallows:

Observer A Observer B

Mean = 42°16‘25" Mean = 42°16‘20"

Em = ± 03.02" Em = ± 01.06"

Compute the most probable angle between observers A and B.

Ans. 42°16’20.58"

SY 2015-16 PRELIM EXAM: If the astronomical azimuths at P-100 to P-101 are as follows:

93⁰ 28‘ 16“ 93⁰ 28‘ 20“93⁰ 28‘ 10“ 93⁰ 28‘ 13“

Find the probable error of the mean of observation.

Ans. ± 1.44“

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Contact No. 09468504583

Broddett B. Abatayo, GE, REA

Lecturer

THANK YOU !!!