GE 105 Theory of errors and adjustments with fx 991-es plus calculator techniques by Broddett Bello Abatayo, Geodetic Engineer 3rd Placer 2014.

1. Theory of Errors and Adjustments
Lecture 1
Caraga State University
College of Engineering and Information Technology
Broddett B. Abatayo, GE
Lecturer

2. Grading System
• Major Exams--------50%
• Quizzes--------------20%
• Assignments--------10%
• Participation---------10%
• Attendance----------10%
100%
Passing rate------------60%

3. Conversion
• 1 tally = 10 pins
Metric System:
• 1 pin = 1 tapelength
English System
• 1 pin = 100 links
• 1 link = 1 ft
• 1 perch = 1 rod= 16.5 ft
• 1 vara = 33 inches
• Pace – Length of a step
in walking; maybe
measured from toe to toe,
or heel to heel
• Stride – Equivalent to two
paces or a double step
Pace Factor = _distance_
ave. # of paces
Pace Factor

4. Prob 1
1. A line was measured to have
5 tallies, 6 pins, and 63.5
links. How long is the line in
feet?
2. A line was measured with a
50 m tape. There were 5
tallies, 8 pins, and the
distance from the last pin to
the end line was 2.25 m. Find
the length of the line in
meters.
3. A distance was measured
and was recorded to have a
value equivalent to 8
perches, 6 rods, and 45
varas. Compute the total
distance in meters.
Ans. 1.) 5,663.5 ft
2.) 2,902.25 m
3.) 108.12 m
Prob 2
• A line 100 m long was paced
by a surveyor for four times
with the following data: 142,
145, 145.5 and 146. Another
line was paced four times
again with the following
results: 893, 893.5, 891, and
895.5.
1. Determine the pace factor
2. Determine the average
number of paces for the new
line
3. Determine the distance of the
new line
Ans. 1.) 0.691 m/pace
2.) 893.25 paces
3.) 617.236 m

5. Errors and Mistakes
• Error – the difference between the true
and measured value of a quantity.
• Mistakes – inaccuracies in measurements
which occur because some aspect of
surveying works were done with
carelessness, poor judgment, improper
execution

6. Statistical Formula’s
A. Probable Error of Single
Observations, E
B. Probable Error of the Mean,
Em
C, Standard Deviation, S.D.
D. Standard Error, S.E.
E. Precision
Where;
V = x – x
x = observed/measured value of a
quantity
x = mean value
n = number of measurements
1
6745.0
2
−
Σ
±=
n
V
E
)1(
6745.0
2
−
Σ
±=
nn
V
Em
1
..
2
−
Σ
±=
n
V
DS
)1(
..
2
−
Σ
±=
nn
V
ES
__
x
E
P m
±=

7. Prob. 3 The following data shows the
difference in elevation between
A and B.
1. Determine the most probable
difference in elevation?
2.The standard deviation?
3. The probable error of the mean?
4. The standard error?
5. And the precision?
Ans. 1.) 520.19
2.) ±0.04
3.) ±0.014
4.) ±0.02
5.) ±1/37,048
Trial No. Diff. in Elevation
1 520.14
2 520.20
3 520.18
4 520.24

8. Using fx-991 es plus
1.) Press ON MODE 3 1
3 Stat mode
1 Single variable x
Input 520.14 = 520.20 = 520.18 =
520.24 =
Press AC SHIFT 1 4 2 =
Ans. X = 520.19
2.) Press SHIFT 1 4 4 =
Ans. sx = 0.04163331999
3.) Press 0.6745 SHIFT 1 4 4 ÷ √
SHIFT 1 4 1 ) =
Ans. 0.6745sx√(n) = 0.01404083717
4.) Press SHIFT144 ÷ √ SHIFT 141) =
Ans. sx√(n) = 0.02081665999
5.) Press 0.6745 SHIFT 1 4 4 ÷ √
SHIFT 1 4 1 ) SHIFT 1 4 2 =
Ans. 0.6745sx√(n)x= 2.6991x10^-5
Press Xˉ¹=
Ansˉ¹ = 37048 (denominator)

9. Rules for Weighted
Measurements
 The weight is directly
proportional to the number of
observations or
measurements.
 The weight is inversely
proportional to the square of
the probable errors.
 The weight is inversely
proportional to the distance.
Prob 4
• The following data shows the
difference in elevation between
A and B.
1. Determine the most probable
difference in elevation?
2. The standard deviation?
3. The probable error of the mean?
4. The standard error?
5. And the precision?
Ans. 1.) 520.208
2.) ±0.03
3.) ±0.005
4.) ±0.0076
Trial No.
Diff. in
Elevation
No. of
Measurements
1 520.14 1
2 520.20 3
3 520.18 6
4 520.24 8

10. Using fx-991 es plus
1.) Press ON
Press MODE 3 1
3 Statistic mode
1 Single variable x
To change set up:
Press SHIFT MODE DOWN 4 1
4 Stat mode set up
1 Frequency(weight) turn on
Input 520.14 = 520.20 = 520.18 =
520.24 = RIGHT DOWN
1= 3= 6= 8=
Press AC
Press SHIFT 1 4 2 =
Ans. X = 520.2077778

11. Using fx-991 es plus
2.) Press SHIFT 1 4 4 =
Ans. sx = 0.03227739248
3.) Press 0.6745 SHIFT 1 4 4 ÷ √
SHIFT 1 4 1 ) =
Ans. 0.6745sx√(n) = 0.005131497
4.) Press SHIFT144 ÷ √ SHIFT 141) =
Ans. sx√(n) = 0.007607854
5.) Press 0.6745 SHIFT 1 4 4 ÷ √
SHIFT 1 4 1 ) SHIFT 1 4 2 =
Ans. 0.6745sx√(n)x= 9.8643x10^-6
Press Xˉ¹=
Ansˉ¹ = 101375.427 (denominator)

12. QUIZ 1 ½ cross wise
1.) Lines of levels between B and C
are run over four different routes.
B is at elevation 825m, and is
higher than C.
a. Determine the most probable
difference in elevation.
b. Most probable elevation of C.
Ans. a.) 0.826m
b.) 824.174m
2.) From the measured values of
distance AB, the following trials were
recorded.
a.Determine the Most Probable Dist.?
b.Probable Error of the Mean ?
c.Standard Deviation?
d.Standard Error?
e.Precision?
Ans. a.) 120.73m
b.) ± 0.0299m
c.) ±0.0877m
d.) ±0.0443m
e.) ±1/4036
Route
No.
Distance
(km)
Diff. in
Elevation(m)
1 2 0.86
2 6 0.69
3 4 0.75
4 8 1.02
Trial
No.
Distance (m)
1 120.68
2 120.84
3 120.76
4 120.64

13. QUIZ 1 solution(BESAVILLA)
W = k/d
Wd = k
W1d1=k W2d2 = k W3d3 = k W4d4 = k
W1d1 = W2d2 = W3d3 = W4d4
W1d1= W4d4
W1(2)= W4(8)
Let W4 = 1
W1(2)= (1)(8)
W1 = 4
W2d2 = W4d4
W2(6) = (1)(8)
W2 = 4/3
W3d3 = W4d4
W3(4) = (1)(8)
W3 = 2
Route Dist. Weights Wt. x Diff. in Elev.
1 2 4 4(0.86) = 3.44
2 6 4/3 4/3(0.69) = 0.92
3 4 2 2(0.75) = 1.5
4 8 1 1(1.02) = 1.02
Route Dist. (km) Diff. in Elev (m)
1 2 0.86
2 6 0.69
3 4 0.75
4 8 1.02
8.33 6.88
most probable difference in elevation:
6.88 / 8.33 = 0.8256m.
most probable elevation of C :
825 - 0.8256 = 824.174m.

14. X FREQ
0.86 2ˉ¹
0.69 6ˉ¹
0.75 4ˉ¹
1.02 8ˉ¹
QUIZ 1 solution
ON MODE 3 1 (stat mode)
SHIFT MODE DOWN 4 1 (freq on)
To find the most probable
difference in elevation or x:
AC SHIFT 1 4 2
To find the most probable elevation
of C:
185 - x

15. Ex. From the measured values of
distance AB, the following trials were
recorded.
Trial Dist. (m) Probable Error
1 120.68 ± 0.002
2 120.84 ± 0.030
3 120.76 ± 0.100
4 120.64 ± 0.050
Determine the :
a. Most Probable Distance?
b. Probable Error of the Mean ?
c. Standard Deviation?
d. Standard Error?
e. Precision?
Ans. a.120.68m
b. ±0.0000146m
c. ±0.0108481m
d. ±0.0000216m
e. ±1/8273064m

16. Solution:
ON MODE 3 1 (stat mode)
SHIFT MODE DOWN 4 1 (freq on)
X FREQ
120.68 1/(0.002)²
120.84 1/(0.030)²
120.76 1/(0.100)²
120.64 1/(0.050)²
a. AC SHIFT 1 4 2 =
b.0.6745 SHIFT 1 4 4÷ √ SHIFT 1 4 1 )=
c. SHIFT 1 4 4 =
d. SHIFT 1 4 4 ÷ √ SHIFT 1 4 1) =
e. 0.6745 SHIFT 1 4 4 ÷ √ SHIFT 1 4 1
) SHIFT 1 4 2 =

17. Ex. The following interior angles of a
triangle traverse were measured with
the same precision:
To determine the total weight:
5ˉ¹ + 6ˉ¹ + 2ˉ¹ = ans. 13/15
SHIFT RCL Y
The corrected angle A is
A + (5ˉ¹)X/Y =
ans. 40º46‘9.23"
The corrected angle B is
B + (6ˉ¹)X/Y =
ans. 76º48‘27.69"
The corrected angle A is
C + (2ˉ¹)X/Y =
ans. 62º25‘23.08"
40º46‘9.23“+ 76º48‘27.69“+
62º25‘23.08“=180º
Angle
Value
(degrees)
No. of
measurements
A 41 5
B 77 6
C 63 2
Determine the most prob. value of :
a. angle A.
b. angle B.
c. angle C.
Solution:
41 SHIFT RCL A
77 SHIFT RCL B
63 SHIFT RCL C
To determine the error:
180-(A+B+C)= ans. -1
SHIFT RCL X
181

18. Ex. Determine the most probable
value of the angles about a given
point.
To determine the error:
360-(A+B+C)= Ans. 18" → X
To determine the total weight:
5ˉ¹ + 6ˉ¹ + 2ˉ¹ = Ans. 13/15 →
Y
The corrected angle A is
A + (5ˉ¹)X/Y =
ans. 130º15‘7.15"
The corrected angle B is
B + (6ˉ¹)X/Y =
ans. 142º37’24.46"
The corrected angle A is
C + (2ˉ¹)X/Y =
ans. 87º07’28.38"
Angle
Value
(degrees)
No. of
measurements
A 130º15‘03" 5
B 142º37‘21" 6
C 87º07‘18" 2
C
BA
•
Determine the most prob. value of :
a. angle A.
b. angle B.
c. angle C.
Solution:
130º15‘03" → A
142º37‘21“ → B
87º07‘18" → C

19. QUIZ 2 ¼ sheet
(50 pts) The following measured interior
angles of a five sided figure, compute the
following:
Station Angles # of measure
A 110º 2
B 98º 3
C 108º 4
D 120º 6
E 105º 4
1. Probable value of angle A?
a. 109º48‘00“ c.109º40‘00"
b. 109º53‘41.05“ d. 109º56’33.33“
2. Probable value of angle B?
a. 97º48‘00“ c. 97º50‘31.58"
b. 97º46‘40“ d. 97º52‘33.33"
3. Probable value of angle C?
a. 107º50‘00“ c. 107º42‘33.33"
b. 107º48‘00“ d. 107º47‘22.11“
4. Probable value of angle D?
a. 119º53‘20“ c. 119º48‘00"
b. 119º41‘3.16“ d. 119º43‘33.33“
5. Probable value of angle D?
a. 104º48‘00“ c. 104º47‘22.11"
b. 104º50‘00“ d. 104º44‘46.68“
Ans. 1. C
2. B
3. A
4. A
5. B

20. QUIZ 2 solution
110 → A
98 → B
108 → C
120 → D
105 → E
A+B+C+D+E = 541 > 540
540-(A+B+C+D+E)=-1 → X
To determine the total weight:
2ˉ¹+3 ˉ¹+4 ˉ¹+6 ˉ¹+4 ˉ¹=3/2 → Y
A + 2ˉ¹X/Y = 109º40‘00"
B + 3ˉ¹X/Y = 97º46‘40“
C + 4ˉ¹X/Y = 107º50‘00“
D + 6ˉ¹X/Y = 119º53‘20“
E + 4ˉ¹X/Y = 104º50‘00“
Station Angles # of measure
A 110º 2
B 98º 3
C 108º 4
D 120º 6
E 105º 4

21. Ex. A base line measured with an invar
tape, and with a steel tape as follows:
Set I (Invar tape) Set II (Steel tape)
571.185 571.193
571.186 571.190
571.179 571.185
571.180 571.189
571.183 571.182
1. What is the S.D. for set II.
2. What is the MPV in set II.
3. What is the S.E. in set II.
4. What is the most probable value
of the two sets.
5. And what is the P.E. of the
general mean.

22. MODE 3 2 (input all data)
Press AC
1.SHIFT 1 4 7 =
Ans. sy=0.004324...
2. SHIFT 1 4 5 =
Ans. y = 571.1878…
3. SHIFT 1 4 7 ÷ √ SHIFT 1 4 1) =
Ans. sy÷√(n) = 0.001933…
4. (x + y)÷2 = 571.1852
5. ±
X Y
571.185 571.193
571.186 571.190
571.179 571.185
571.180 571.189
571.183 571.182