In the preparation for the Geodetic Engineering Licensure Examination, the BSGE students must memorized the fastest possible solution for the TAPING CORRECTION using casio fx-991 es plus calculator technique in order to save time during the said examination. note: lec 2 and above wala akong nilagay na solution para hindi makupya techniques ko. just add me on fb para ituro ko sa inyo solution. Kasi itong solution ko wala sa google, youtube, calc tech books at hindi rin itinuro sa review center.

1. ENGR. BRODDETT B. ABATAYO, GE, REA
Part-time Lecturer – GE division, CEIT, CSU, Ampayon, Butuan City
Research Assistant – Phil-LiDAR 2 Project, CSU, Ampayon, Butuan City
Proprietor – BPA ABATAYO Land Surveying Services
1
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

2. 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

3. A B
MEASURE
Standard
Too short
Too long
True Distance AB is equal to 8cm
8cm

4. LAYING OUT
Standard
Too short
Too long
A

5. Taping Corrections
A. 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.
LTsToCt )( 
shift CALC =
Ans. 47.92°C
Calculator technique:

6. 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
LPsPo
Cp
)( 

Example:
The measured distance from A to B was 318m
using 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 pull applied if the
corrected distance A to B 1s 318.012m.
shift CALC =
Ans. 9.46kg
Calculator technique:

7. Taping Corrections
C. 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
L
Csag 

Ans. 1. 0.0162m
2. 29.984m
Example
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 correction
2.Correct distance between the ends
of the tape

8. 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 =
Calculator technique:
Ans. 35.04°C

9. 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
h
Cs
2
2
 CsSH 
Example
Slope distances AB and BC measure 330.49m
and 660.97m, respectively. The difference in
elevation is 10.85m for B and C. Using the
slope correction formula, determine the
difference in elevation for A and B. If the
horizontal length of line ABC is 991.145m.
Assume the line AB has a rising slope and BC
a falling slope.
Ans. 12.22m
shift CALC =
Calculator technique:

10. Taping Corrections
CsagCp 
2
2
24
)(
Pn
LW
AE
LPsPn


)(
204.0
PsPn
AE
WPn


Calculator technique:
shift CALC =

11. 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

12. Reduction to Sea-Level

13. Taping Corrections
F. 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
h
DD 1'
shift CALC =
Calculator technique:
Ans. 6844.35m

14. 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 Temperature
2. Total correction
3. True length of the line
Ans. 1. 50.76 °C
2. 0.1348m
3. 674.05 m

15. 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 correction
2. Correct distance between the ends of
the tape
Ans. 1. 0.0162m
2. 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

16. 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 correction
2. Pull correction
3. Correct length of the line
Ans. 1. +0.4168m
2. -0.0399m
3. 2395.6269m

18. Solution: MODE 1
458.65 + A + B - C = 456.8209015
9 pin + 8.65 = 458.65

19. 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?
3. What tension must be applied to the tape at a temp. of 30°C
so that it would be of standard length?

22. 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?
3. What tension must be applied to the tape
at a temp. of 30°C so that it would be of
standard length?
Solution: MODE 1
99.986 →A 10 →D
99.992 →B 14 →E
100.003 →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

23. 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
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?
3. What tension must be applied to the tape
at a temp. of 30°C so that it would be of
standard length?
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 kg
C + X
B + X
A + X
AC shift 1 5 5 ← 100 = (12.54313901)
12.54313901 kg
shift 1 2 (table)

24. 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
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?
3. What tension must be applied to the tape
at a temp. of 30°C so that it would be of
standard length?
X Y
D
E
FC + Y
B + Y
A + Y
Length 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

25. 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?
3. What tension must be applied to the tape
at a temp. of 30°C so that it would be of
standard length?
18.35874439 kg
12.54313901 kg
19.71121076 kg

27. 35⁰ 43’ 53.2" →A
29⁰ 37‘ 05.8" →B
23⁰ 29‘ 36.7" →C
65⁰ 20‘ 58.2" →D
53⁰ 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 =

28. X = 35⁰ 43’ 52.98"
Y = 29⁰ 37‘ 5.75"
Z = 23⁰ 29‘ 37.18"

30. 5.369

31. SY 2014-15 PRELIM EXAM: Measured from point A, angles BAC,
CAD, and BAD were recorded as follows:
C
DA
B
Angle 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:

32. 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)-C
0⁰9‘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

33. 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"

34. 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“

35. https://www.sites.google.com/site/bbabatayo/lecturer/ge-105
Email Add: broddett_25@yahoo.com
Contact No. 09468504583
Broddett B. Abatayo, GE, REA
Lecturer
THANK YOU !!!