GEODETIC Least Square with fx 991-es plus

Theory of Errors and AdjustmentsLecture 1

Caraga State UniversityCollege of Engineering and Information Technology

Broddett B. Abatayo, GE Lecturer

Grading System

• Major Exams--------50%• Quizzes--------------20% • Assignments--------10%• Participation---------10%• Attendance----------10% 100%

Passing rate------------60%

Conversion• 1 tally = 10 pinsMetric System:• 1 pin = 1 tapelengthEnglish 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

Prob 11. 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 factor2. Determine the average number

of paces for the new line3. 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 – xx = observed/measured value of a

quantityx = mean valuen = number of measurements

16745.0

2

nVE

)1(6745.0

2

nnVEm

1..

2

nVDS

)1(..

2

nnVES

__

x

EP m

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

Using fx-991 es plus1.) 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 = 520.192.) Press SHIFT 1 4 4 = Ans. sx = 0.041633319993.) Press 0.6745 SHIFT 1 4 4 ÷ √ SHIFT 1 4 1 ) = Ans. 0.6745sx√(n) = 0.014040837174.) Press SHIFT144 ÷ √ SHIFT 141) = Ans. sx√(n) = 0.020816659995.) 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)

Rules for WeightedMeasurements

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 5.) ±1/101,375

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

Using fx-991 es plus

2.) Press SHIFT 1 4 4 = Ans. sx = 0.032277392483.) Press 0.6745 SHIFT 1 4 4 ÷ √ SHIFT 1 4 1 ) = Ans. 0.6745sx√(n) = 0.0051314974.) Press SHIFT144 ÷ √ SHIFT 141) = Ans. sx√(n) = 0.0076078545.) 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)

QUIZ 1 ½ cross wise1.) 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

QUIZ 1 solution(BESAVILLA)

W = k/dWd = k

W1d1=k W2d2 = k W3d3 = k W4d4 = kW1d1 = W2d2 = W3d3 = W4d4

W1d1= W4d4W1(2)= W4(8)

Let W4 = 1W1(2)= (1)(8)

W1 = 4W2d2 = W4d4W2(6) = (1)(8)

W2 = 4/3W3d3 = W4d4W3(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.

X FREQ0.86 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

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

Solution:ON MODE 3 1 (stat mode)SHIFT MODE DOWN 4 1 (freq on)

X FREQ120.68 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 =

Ex. The following interior angles of a triangle traverse were measured with the same precision:

To determine the total weight:5ˉ¹ + 6ˉ¹ + 2ˉ¹ = ans. 13/15SHIFT RCL Y

The corrected angle A isA + (5ˉ¹)X/Y = ans. 40º46‘9.23"

The corrected angle B isB + (6ˉ¹)X/Y = ans. 76º48‘27.69"

The corrected angle A isC + (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 A77 SHIFT RCL B63 SHIFT RCL CTo determine the error:180-(A+B+C)= ans. -1SHIFT RCL X

181

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 isA + (5ˉ¹)X/Y = ans. 130º15‘7.15"

The corrected angle B isB + (6ˉ¹)X/Y = ans. 142º37’24.46"

The corrected angle A isC + (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" → A142º37‘21“ → B 87º07‘18" → C

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

QUIZ 2 solution110 → A 98 → B108 → C120 → D105 → E

A+B+C+D+E = 541 > 540540-(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

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.

MODE 3 2 (input all data)

Press AC1.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