ge 105 lecture 1 (least squares adjustment) by: broddett b. abatayo

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)


2. Standard deviation ?

3. Probable error of the mean ?

4. Standard error ?


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


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


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


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


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




360-(A+B+C)= → X


5ˉ¹ + 6ˉ¹ + 2ˉ¹ = → Y


A + (5ˉ¹)X/Y = → A

ans. 130º15‘7.15"


B + (6ˉ¹)X/Y = → B

ans. 142º37’24.46"


C + (2ˉ¹)X/Y = → C

ans. 87º07’28.38"


A 130º15‘03" 5

B 142º37‘21" 6

C 87º07‘18" 2

C

B

A•


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



(20pts)

AngleValue

(degrees)

No. of

measurements

A 130º15‘03" 5

B 142º37‘21" 6

C 87º07‘18" 2


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




360-(A+B+C)= → X


5ˉ¹ + 6ˉ¹ + 2ˉ¹ = → Y


A + (5ˉ¹)X/Y = → A

ans. 130º15‘7.15"


B + (6ˉ¹)X/Y = → B

ans. 142º37’24.46"


C + (2ˉ¹)X/Y = → C

ans. 87º07’28.38"


A 130º15‘03" 5

B 142º37‘21" 6

C 87º07‘18" 2

C

B

A•


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: [email protected]

Contact No. 09468504583

Broddett B. Abatayo, GE, REA

Lecturer

END