ge 105 lecture 1 (least squares adjustment) by: broddett b. abatayo
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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 1
Caraga State University
College of Engineering and Information Technology
Ampayon, Butuan City 8600
LEAST SQUARES
ADJUSTMENT
GE 105 – Theory of Errors and Adjustments
GRADING SYSTEM
Major Exam 50%
Quiz 20%
Participation 20%
Attendance 10%
Total 100%
Passing rate 60%
Land Measurement Conversion Table1 vara 33 inches
1 pin 1 tape length
1 pin 100 links
1 tally 10 pins
1 rod = 1 pole = 1 perch 25 links
1 link 0.66 ft.
1 yard 3 ft.
1 chain 4 rods
1 mile 8 furlongs
1 furlong 40 rods
1 acre
1 chain 4 rods
1 knot 6080.2 ft.
1 township 36 sections
1 section 640 acres
Pacing
heel
toe
1 pace
1 pace
(Toe to toe)
(heel to heel)
1 stride
(Toe to toe)
1 stride
(heel to heel)
Example:
20 m
Given: distance AB is 20 meters.
number of paces is 23.
Determine the pace factor?
A B
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
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
Statistical Formula’sA. 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
16745.0
2
n
VE
)1(6745.0
2
nn
VEm
1..
2
n
VDS
)1(..
2
nn
VES
__
x
EP m
The following data shows the difference in elevation between A and B.
Determine:
1. the most probable diff. in elevation?
2.The standard deviation?
3. The probable error of the mean?
4. The standard error?
5. And the precision?
Trial No. Diff. in Elevation
1 520.14
2 520.20
3 520.18
4 520.24
Using fx-991 es plus
1. Most probable diff. in elevation?
w – mode type
3 – statistics mode
1 – single variable (x)
Using fx-991 es plus
2. Standard deviation ?
3. Probable error of the mean ?
4. Standard error ?
Using fx-991 es plus
5. Precision ?
The following data shows the difference in elevation between A and B.
1. Determine the most probable diff. in elevation?
2.The standard deviation?
3. The probable error of the mean?
4. The standard error?
5. And the precision?
Trial No. Diff. in Elevation
1 520.14
2 520.20
3 520.18
4 520.24
520.19
±0.04
±0.014
±0.02
Rules for Weighted Measurements
The weight (FREQ) is directly proportional to the number of observations or measurements.
The weight (FREQ) is inversely proportional to the square of the probable errors.
The weight (FREQ) is inversely proportional to the distance.
The weight (FREQ) is inversely proportional to the number of set ups.
Ex. The following data shows the diff. in elevation between A and B.
Determine the :
1. most probable difference in elevation?
2. standard deviation?
3. probable error of the mean?
4. standard error?
5. precision?
Trial No. Diff. in Elevation No. of Measurements
1 520.14 1
2 520.20 3
3 520.18 6
4 520.24 8
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
w
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)
w
Ex. The following data shows the diff. in elevation between A and B.
Determine the :
1. most probable difference in elevation?
2. standard deviation?
3. probable error of the mean?
4. standard error?
5. precision?
Trial No. Diff. in Elevation No. of Measurements
1 520.14 1
2 520.20 3
3 520.18 6
4 520.24 8
520.208
±1/101,375
±0.0076
±0.005
±0.03
QUIZ 1 ½ cross wise
1. From the measured values of
distance AB, the following trials were
recorded. (10pts)
Determine the:
a. Determine the Most Probable Dist.?
b. Probable Error of the Mean ?
c. Standard Deviation?
d. Standard Error?
e. Precision?
Trial No. Distance (m)
1 120.68
2 120.84
3 120.76
4 120.64
2. The following data shows the measured
distance between A and B. (20pts)
Determine the:
a. Most probable dist. Bet. A and B.
b. Standard deviation.
c. Probable error of the mean.
d. Standard error.
e. Precision.
Trial Distance (m) Probable Error
1 100.860 ± 0.02
2 100.690 ± 0.04
3 100.750 ± 0.06
4 101.020 ± 0.08
REVIEW :Rules for Weighted Measurements
The weight (FREQ) is directly proportional to the number of observations or measurements.
The weight (FREQ) is inversely proportional to the square of the probable errors.
The weight (FREQ) is inversely proportional to the distance.
The weight (FREQ) is inversely proportional to the number of set ups.
Ex. The following interior angles of a
triangle traverse were measured with
the same precision:
AngleValue
(Degrees)
No. of
Measurements
A 39 3
B 65 4
C 75 2
Determine the most prob. value of :
a. angle A.
b. angle B.
c. angle C.
Ex. The following interior angles of a
triangle traverse were measured with
the same precision:
AngleValue
(degrees)
No. of
measurements
A 39 3
B 65 4
C 75 2
Determine the most prob. value of :
a. angle A.
b. angle B.
c. angle C.
Solution:
39
65
75
A+B+C =179º
179
AB
C
3 9 shift RCL (-)
7 5 shift RCL hyp
w
A
B
C
X
Y
To determine the error:
180-(A+B+C)=
To check:
A+B+C =180º
To determine the total weight:
The corrected angle A is
The corrected angle B is
The corrected angle C is
w
Ex. 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?
2. Probable value of angle C?
3. Probable value of angle D?
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?
2. Probable value of angle C?
3. Probable value of angle D?
Solution:
110 → A
98 → B
108 → C
120 → D
105 → E
A+B+C+D+E= 541º
541
1 0 shift RCL (-)1
13.33333333
6.666666667
The following interior angles of a
seven sided figure: Determine the following :
1. MPV of angle A. ________
2. MPV of angle B. ________
3. MPV of angle C. ________
4. MPV of angle D. ________
5. MPV of angle E. ________
6. MPV of angle F. ________
7. MPV of angle G. ________
Angle Value measurements
A 138º 2
B 140º 4
C 121º 7
D 119º 3
E 137º 8
F 126º 6
G 120º 5
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 are the most probable value under each set.
2. What are the probable errors under each set.
3. What is the most probable value of the two sets.
4. What is the probable error of the general mean.
MODE 3 2 (input all data and FREQ turn off)
Press AC
1. SHIFT 1 4 2 =
Ans. x=571.1826
SHIFT 1 4 5 =
Ans. y=571.1878
X Y
571.185 571.193
571.186 571.190
571.179 571.185
571.180 571.189
571.183 571.182
A
B
2. 0.6745 SHIFT 1 4 4 ÷ √ SHIFT 1 4 1 ) =
Ans. 0.000919895
0.6745 SHIFT 1 4 7 ÷ √ SHIFT 1 4 1 ) =
Ans. 0.00130442
C
D
571.1826
571.1878
X FREQ
A
B
SHIFT 1 4 2 =
Ans. x = 571.1843271
3. MODE 3 1 (FREQ turn on)
Press AC
4. MODE 1
MODE 3 2 (input all data and FREQ turn off)
Press AC
1. SHIFT 1 4 2 =
Ans. x=571.1826
SHIFT 1 4 5 =
Ans. y=571.1878
2. 0.6745 SHIFT 1 4 4 ÷ √ SHIFT 1 4 1 ) =
Ans. 0.000919895
0.6745 SHIFT 1 4 7 ÷ √ SHIFT 1 4 1 ) =
Ans. 0.00130442
X Y
571.185 571.193
571.186 571.190
571.179 571.185
571.180 571.189
571.183 571.182
X FREQ
A
B
A
B
C
D
3. SHIFT 1 4 2 =
Ans. x = 571.1843271
MODE 3 1 (FREQ turn on)
Press AC
MODE 1
FULL SOLUTION
1. The following data observed are
the difference in between BM1 and
BM2 running a line levels over four
different routes.
Route Diff. in Elev. (m) Probable Error
1 340.22 ± 02
2 340.30 ± 04
3 340.26 ± 06
4 340.32 ± 08
a. What is the weight of route 2
assuming of route 1 is equal to 1?
b. Determine the most probable
value of difference in elevation ?
c. If the elevation of BM1 is
650.42m. What is the elevation of
BM2 assuming it is higher than
BM1?
2. Determine the most probable value
of the angles about a given point.
AngleValue
(degrees)
No. of
measurements
A 130º15‘03" 5
B 142º37‘21" 6
C 87º07‘18" 2
Determine the most prob. value of :
a. angle A.
b. angle B.
c. angle C.
Solution:
ON MODE 3 1 (stat mode)
SHIFT MODE DOWN 4 1 (freq on)
a. weight of route 2
b. AC SHIFT 1 4 2 =
c. AC 650.42 + SHIFT 1 4 2 =
X FREQ
1 340.22
2 340.30
3 340.26
4 340.32
2
8
0.0625 / 0.25 = 0.25
0.25 / 0.25 = 1
Weight of route 2 if
weight of route 1 is
equal to 1
2. Determine the most probable value
of the angles about a given point.
To determine the error:
360-(A+B+C)= → X
To determine the total weight:
5ˉ¹ + 6ˉ¹ + 2ˉ¹ = → Y
The corrected angle A is
A + (5ˉ¹)X/Y = → A
ans. 130º15‘7.15"
The corrected angle B is
B + (6ˉ¹)X/Y = → B
ans. 142º37’24.46"
The corrected angle A is
C + (2ˉ¹)X/Y = → C
ans. 87º07’28.38"
Angle Value measurements
A 130º15‘03" 5
B 142º37‘21" 6
C 87º07‘18" 2
C
B
A•
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
To check:
A+B+C =360º
1. The following data shows the measured
distance between A and B. (30pts)
Determine the:
a. Most probable distance between
A and B.
b. Standard deviation.
c. Probable error of the mean.
d. Standard error.
e. Precision.
Trial# of
set-ups
Distance
(m)
Probable
Error
1 3 100.860 ± 02
2 2 100.690 ± 04
3 4 100.750 ± 06
4 1 101.020 ± 08
QUIZ 2 ½ cross wise
2. Determine the most probable value
of the angles about a given point.
(20pts)
AngleValue
(degrees)
No. of
measurements
A 130º15‘03" 5
B 142º37‘21" 6
C 87º07‘18" 2
Determine the most prob. value of :
a. angle A.
b. angle B.
c. angle C.
Trial# of set-
ups
Distance
(m)
Probable
Error
1 3 100.860 ± 02
2 2 100.690 ± 04
3 4 100.750 ± 06
4 1 101.020 ± 08
2. Determine the most probable value
of the angles about a given point.
To determine the error:
360-(A+B+C)= → X
To determine the total weight:
5ˉ¹ + 6ˉ¹ + 2ˉ¹ = → Y
The corrected angle A is
A + (5ˉ¹)X/Y = → A
ans. 130º15‘7.15"
The corrected angle B is
B + (6ˉ¹)X/Y = → B
ans. 142º37’24.46"
The corrected angle A is
C + (2ˉ¹)X/Y = → C
ans. 87º07’28.38"
Angle Value measurements
A 130º15‘03" 5
B 142º37‘21" 6
C 87º07‘18" 2
C
B
A•
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
To check:
A+B+C =360º
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
END
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