advanced surveying solutions

7/24/2019 Advanced Surveying Solutions

1/17

Advanced Surveying Home Work Solutions

Problem 2.2

Discuss the three types of errors and their influence upon survey work.

Blunders:human errors that tend to be noticeable and result in degraded work.

Systematic:mechanical errors due to equipment configuration or maintenance,these are errors that can be corrected. If uncorrected, they degrade work.

Random:this type of error is inherent to the act of making measurements and is

not preventable, however, it is usually as likely to be on the long side as the short

side and therefore in the overall picture cancels out when multiple measurementsof the same quantity are made.

Problem 2.3

Give examples of systematic and random errors.

Chaining included numerous corrections for systematic error such as for the standard

length of the chain, the tension used, the temperature, the slope and sag all of which

could be modeled mathematically and the measurement made adjusted to compensate forthe error. Random errors can not be modeled physically; they can be modeled

statistically by determining the most probable value and the distribution of measurements

about that value.

Problem 2.5

Discuss compensating for error in measuring distance and angles.

Common methods of measuring horizontal distance are by chaining or EDM. Inprecision chaining all of the adjustments mentioned in Problem 2.3 must be accounted for

(many of these may not be significant to any particular problem, but they should be

considered). When using an EDM, corrections such as offset distances for the EDM andthe prism being used, and corrections for the speed of light due to changes in

temperature, humidity and barometric pressure can be employed. When measuring

angles it is important to insure that the instrument in use is vertical. It is very difficult toinsure this, however, buy measuring the angle first with the instrument in the upright

position and then inverting the scope and repeating the measurement small deviations

from vertical can be cancelled out.


2/17

Problem 2.7

Define: Mean, Standard Deviation, Covariance and Correlation.

Mean, the average of a set of values, the most likely value

Standard Deviation, a measure of the range of the expected difference between

measured value and the expected or most likely value (the mean) of a measurement.Covariance, a measure of variance for measurements involving multiple variables.

Correlation, a measure of the dependency between variables in multivariable situations.

Problem 2.8

Define reference variance.

An arbitrarily selected scale or proportionality factor used to determine measurement

weighting factors.

Problem 2.14

Construct a histogram of the following measurement data using 2mm intervals thendetermine the mean median, mode, mid range, variance, standard deviation and the

standard deviation of the mean.

Note: 206 m subtracted from each data point.


3/17

Problem 2.14

Data x-x' (x-x')^2 Bin Occurs Probabili

0.159 0.018 0.000314 0.120 0.120 1 4%

0.161 0.020 0.000389 0.122 0 0%

0.145 0.004 1.38E-05 0.124 0 0%

0.140 -0.001 1.64E-06 0.126 0 0%0.143 0.002 2.96E-06 0.129 0.128 1 4%

0.139 -0.002 5.2E-06 0.130 0 0%

0.137 -0.004 1.83E-05 0.133 0.132 0.131 0.132 3 12%

0.135 -0.006 3.94E-05 0.135 0.134 1 4%

0.144 0.003 7.4E-06 0.137 0.137 0.136 0.136 3 12%

0.148 0.007 4.52E-05 0.139 0.138 1 4%

0.143 0.002 2.96E-06 0.140 0.140 0.140 0.140 3 12%

0.137 -0.004 1.83E-05 0.143 0.143 0.142 0.142 3 12%

0.136 -0.005 2.79E-05 0.145 0.144 0.144 2 8%

0.147 0.006 3.27E-05 0.147 0.146 1 4%

0.142 0.001 5.18E-07 0.149 0.148 0.148 2 8%

0.131 -0.010 0.000106 0.150 0.150 1 4%0.133 -0.008 6.86E-05 0.152 0.152 1 4%

0.140 -0.001 1.64E-06 0.154 0 0%

0.140 -0.001 1.64E-06 0.156 0 0%

0.132 -0.009 8.61E-05 0.159 0.158 1 4%

0.129 -0.012 0.000151 0.161 0.160 1 4%

0.150 0.009 7.6E-05 0.162 0 0%

0.149 0.008 5.96E-05

0.152 0.011 0.000115 25.00 100%

0.120 -0.021 0.000453

Sum: 3.532 0.000 0.002

Average: 0.14128 8.15E-05(Mean) (Variance)

Count 25

Mean 0.141

Median 0.140

Mode 0.140

Mid Rang 0.141

Variance 8.488E-05

Std Dev 0.009

SD of Mea 1.843E-03 (Std Dev of Mean)

Bin Sorted Data

Problem 2.14 Histogram

0%

2%

4%

6%

8%

10%

12%

14%

1 3 5 7 9 11 13 15 17 19 21Distance

Probability


4/17

Problem 2.20

The following two equations are provided for calculating X and Y distances.

X = Xo + d Sin(A) Y = Yo + d Cos(A)

Xo and Yo are errorless, the Standard Deviation for d is 0.005m, and the StandardDeviation for A is 5. Form the Covariance matrix for (X,Y) coordinates when d is 500m

and A is 42 30 30. Calculate the Correlation Coefficient between calculated X and Y

coordinates.

The covariance matrix = is calculated by J J

2

2

yyx

xyx

SS

SS T where

J =

AY

dY

AX

dX

; J

T

=

AY

AX

dY

dX

; and =

2

2

AdA

dAd

dX

= Sin(A) ;

AX

= d Cos(A) ;

dY

= Cos(A) ;A

Y

= -d Sin(A)

d= 0.005m;A= 5 or 0.0000242 Radians

dA= 0; distance and angle measurements are uncorrelated

2

2

10066.8072.6

072.6125.9

)()(

)()(

0000242.00

0005.0

)()(

)()(

=

=

xAdSinAdCos

ACosASin

AdSinACos

AdCosASin

Sx2= 9.125 x 10

-5, Sx= 0.009550m

Sy2= 8.066 x 10

-5, Sy= 0.008981m

The correlation coefficient xy= xy/(xy) is estimated as Sxy/(SxSy)

xy= 707.0

)008981.0(009550.0

00006072.0=

=

yx

xy

SS

S

Problem 2.28

Explain the difference between accuracy and precision.

Accuracyis how close a measurement is to the actual or real value.

Precisionis how repeatable a measurement is.


5/17

Problem 2.29

When does adjustment become necessary in a survey network?

Adjustment is necessary when the network includes redundant data. Redundant data is

data beyond the minimum number of measurements required for a solution. Closure data

is redundant as a solution could be obtained prior to making the closing shot, however, bycollecting the closure data the surveyor is able to insure that the data will clearly show the

presence of any blunders and the magnitude of precision in the work completed. Because

all measurements are subject to error, one should not anticipate closure without anobservable error.

Problem 2.30

What is the basic objective of a survey adjustment?

The purpose of adjusting a survey is to distribute the observed closure error throughout

the survey in a meaningful manner.

Problem 2.31

What is the Least Squares criterion and why is it needed?

The least squares solution is a method of obtaining a solution by the minimization of

deviations between observations and all possible solutions. Using least squares it is quite

possible to assign different weights to each observation and therefore distributeadjustments anywhere from uniformly using all points to focusing upon specific points

and even a single point. By using mathematically derived weighting functions to controlthe adjustment, complex distributions of error are possible. A method of adjustment is

needed anytime redundant data is collected. The least squares method of adjustment is

not well suited to hand calculations.

END OF HW SET ONE


6/17

Problem 2.17

For the five given angles compute the standard deviation of a single value and the meanangle. Determine the 95% confidence interval for the 33 4210 measurement.

The standard deviation of a single value is the typical standard deviation computed for a

set of observed values (using n-1 to calculate the variance). The standard deviation of themean is the standard deviation of the population divided by the square root of the number

of observations in the population.

Problem 2.17

Angles given:

Deg Min Sec DD

33 42 10 33.7028

33 42 0 33.7000

33 42 15 33.7042

33 42 20 33.7056

33 42 30 33.7083

Average 33 42 15 33.70417

Std Dev 11.18034 0.003106

To calculate the 95% confidence interval for a single value, use the student t distribution

probability tabulation on page 1149 of the text. Since we are looking for the probability

of a single occurrence, use one degree of freedom at t = 0.95. The number of standarddeviations ranging from 0% (the infinite left side) to 95% of the distribution with 1 dof is

6.31. The confidence interval around the 33 4210 measurement then is 33 4210 +/-

6.31(11.18); 6.31(11.18) is 1 and 10.55.

Therefore the single value 95% confidence interval is from 33 4100 to 33 4320.

To calculate the 95% confidence interval aroundthe mean, use n-1 dof (4) and becausethe interval is symmetricabout the mean it runs from 2.5% to 97.5%. The number of

standard deviations between the mean (50% probability) and 97.5% probability to the

right (or 2.5% probability to the left) is 2.78. The confidence interval around the mean is+/- 2.78(11.18) or 31.08. Therefore the 95% confidence interval around the mean of

33 4215 is from 33 4144 to 33 4246.


7/17

Problem 2.19

The estimated standard deviation for measurement of a distance, determined by EDM is0.005m + 10 ppm of the distance measured. Calculate estimated standard deviations for

distances or 300.05, 30100.60 and 2500.45 m measured with the EDM. Assuming a

reference variance of 1.00, form the covariance and weight matrices for the distances,

which are assumed to be un-correlated.

The standard deviation, S = 0.005 + d (10/1000000), and the variance is S2.

Distance Measured Std Dev iat ion Var iance Inv o f Var

300.05 0.00800 6.4008E-05 15623.05

30100.60 0.30601 9.3640E-02 10.68

2500.45 0.03000 9.0027E-04 1110.78

The covariance matrix () is a 3x3 diagonal matrix (all off diagonal terms are 0.0) since

the distances are un-correlated and is composed of the Variance values shown above.When the measurements are uncorrelated, the matrix of weights (W) is the inverse of ().

=

03.000

030601.00

00000064.0

; and

=

78.111000

068.100

0005.156231

Problem 2.21

Two sides and the included angle of a plane triangle are measured to give the following

values: side a is 100.000m with S = 0.030m; side b is 200.00m with S = 0.040m and theincluded angle C is 45 00 00 with S = 30. These measurements are uncorrelated.

Calculate the area of the triangle formed and the standard deviation of the area.

Convert the standard deviation for the included angle to radians: S = 30/(3600(180));Sc=0.0001454 Radians

Derive an equation for the area of the triangle: A = Base x Height; take the base to belength b and the height to be length a Sin(C).

A = abSin(C)

Calculate SA2= J J

Twhere J is the matrix of partial differentials.

J =

)(

2

1)(

2

1)(

2

1CabCosCaSinCbSin and =

2

2

2

0001454.000

004.00

0003.0


8/17

SA2= {b

2Sa

2Sin

2(C) + a

2Sb

2Sin

2(C) + a

2b

2SC

2Cos

2(C)}

SA2= 0.25(200

20.03

2Sin

2(45) + 100

20.04

2Sin

2(45) + 100

2200

20.0001454

2Cos

2(45))

SA2= (18 + 8 + 4.231)

SA2= 7.5577 m

2

SA= 2.749 m

A = 200 100 Sin(45)

A = 7071.068 m2

Problem 2.27

An angle was measured 16 times by repetition and the standard deviation of a singleangle was found to be 4. If the reference variance ois 162, calculate the weight for the

average angle.

The weight for the average angle will be the inverse of the variance of the average angledivided by the reference variance. The variance of the average angle is the variance of a

single angle divided by the number of observations.

Sx2= Sx/n

Sx2= (4)2/16

Sx2= 12

W = o/ Sx2

W = 16/1

W = 16


9/17

Problem 3.2

What factors should be considered by the surveyor when setting the specifications for

accuracy on a given project?

The project tolerances will control the survey. If constructed works need to be located to

+/- 1 foot the survey would not need to supersede that tolerance (common constructionpractices today require tolerances on the order of +/- 0.01 feet). The instruments and

methods employed in the survey are a direct result of the precision that is required. Oneshould not specify precision that is not obtainable with the instruments and methods

employed!

Problem 3.7

A point is to be established on the ground at a distance of 300.00 m from a given point by

means of one linear and one angular measurement. This point is to be established to

within 5 cm of its true location. What angular error could be allowed to achieve thisaccuracy in position?

Locate the point 300 meters away within 5 cm or 0.05 m of arc length (r sin(a) or ra,where a is in radians). a then is 0.05/300 or 1.66x10

-4Radians (or 34).

a = 34

Problem 3.8

A horizontal distance has been measured with a relative precision of 1 part in 25000.This specification for accuracy in a distance of 500.000 m corresponds to what tolerance

in horizontal angular measurement?

S = 500 (1)/25000

S = 0.02 m

S = ra; r = 500 m

a = S/r

a = 0.02/500

a = 4x10-5Radians

a = a180/

a = 0.00004(180)/

a = 2.2918x10-3

or 8.25


10/17

Problem 3.10

An angle of 50 00 00 is measured with a theodolite having a least count of 20.

Assume the maximum error is one-half the least count of the theodolite. What is the

relative precision if the sine, cosine and tangent of the angle are to be used (individually)

in computations?

Use Figures 3.3 and 3.4 on page 83 of text with S = 10

Precision for Sine is less than 1/20,000(off the provided chart).

Precision for Cosine is 1/17,000

Precision for Tangent is 1/10,000

Problem 3.17

Discuss the advantages and disadvantages of data collectors and the electronic notebookconcept for collecting and processing survey data.

Electronic data collectors and field notebooks allow more data to be collected in a givenamount of time since there is no human read/write interface, the data is manipulated

electronically. Many of these devices also provide the surveyor powerful analysis tools

in the field during the data collection process. This results in fewer opportunities totranspose information, better legibility of recorded data (as it will all be machine printed),

and fewer trips to the field as many office calculations (and the decisions dependent upon

those calculations) can now be done during the fieldwork.

Electronic data is subject to corruption and loss. In some cases the data is easilymanipulated and difficult to secure. In attempting to protect the original data from

tampering, it is also difficult to amend an errant condition that has been entered into the

logger or notebook (like the wrong prism height on a single shot).

The legality of electronic data is a complicated question with pros and cons. A

suggestion: download electronic data often and file a printed copy of the original field

work in a job folder.

Hand written field notes and sketches should accompany electronic data and notes.

Digital pictures are nice additions to electronic notes.

END OF HOME WORK ASSIGNMENT 2


11/17

Problem 10.1

Define the term Celestial Sphere as used in astronomy.

A spherical surface (fixed radius of celestial proportion) upon which the position of

celestial bodies are projected.

Problem 10.2

Describe how an observers position on the surface of the earth is defined in the earthsequatorial system.

The earth is modeled as a sphere and the position of any point on the surface of thismodel can be described by the intersection of two planes, one containing the polar axis

and one perpendicular to the polar axis. The orientation of these two planes is described

by angles from fixed reference planes. For the planes containing the polar axis, thereference plane is that plane passing through Greenwich, England and the angle ismeasured to the West of East and is called longitude. The reference plane perpendicular

to the polar axis is located at the midpoint of the axis and its intersection with the surface

of the sphere is called the equator. Planes parallel to this reference form intersectionswith the surface called lines of Latitude and are located by their declination angle from

the equator to the N or S pole.

Problem 10.3

Define a) Vernal equinox, b) Hour angle, and c) Right ascension.

a) Vernal equinox: that point on the celestial sphere where the sun crosses thecelestial equator. This occurs each year around March 21

st.

b) Hour angle: the angular distance measured along the celestial equator between themeridian of a reference point the meridian of an observer. Common referencepoints are Greenwich, England and the Vernal Equinox. Hour angles are

measured positive clockwise (to the West) when observing from the N pole.

c) Right ascension: is the angular distance measured along the celestial equatorbetween the Vernal Equinox and the hour circle (meridian) through a celestial

body; it is measure positive to the East in either degrees of arc or hours of time.


12/17

Problem 10.4

Explain the relationship between the horizon system of spherical coordinates and the

equatorial system.

The equatorial system is defined by extending the polar axis of the earth and the equatorto the celestial sphere. The horizon system is defined by extending an axis in the

direction of gravity from the observers location on earth to the celestial sphere locating aZenith and Nadir, and a plane perpendicular to this axis through the center of the earth

which intersects the celestial sphere in a circle called the horizon. North on the horizon is

the meridian of the North pole and Zenith Any point on the celestial sphere may belocated by an angle of altitude from the horizon or the Zenith angle (90-altitude) and an

angle of Azimuth from North on the horizon to the intersection of a vertical circle

through the celestial body and the horizon.

Problem 10.5

On a given day, 0hGreenwich civil time occurs at 5

h17

m30

sGreenwich sidereal time.

At that instant, what is the local sidereal time at a place whose longitude is 8h10

m40

sW?

LST = GST West

LST = 5h1730 - 8

h1040 = -3

h0650

The LST is 3 hours before the current day or 21h0650 of the previous day.

Problem 10.6

When the local apparent time is 9h1030 at a place where the longitude is 961030W,

what is the Greenwich apparent time?

GAT = LAT + West

West= 961030 / 15 degrees per hour

West= 96.175/15 = 6.4117 hours or 6h2442

GAT = 9h1030 + 6h2442

GAT = 17h3512


13/17

Problem 10.10

From an ephemeris find the equation of time for the instant of 2h

1010 PM Pacific

Standard Time on June 10, 2005. If the longitude of the place is 7h4603W, calculate

the local civil and local apparent times.

GCT = PCT + 8h

GCT = (2

h1010 + 12

h) + 8

h= 22

h1010

EOT June 10, 2005: +00 39.73EOT June 11, 2005: +00 27.62

Change in EOT: 27.62 39.73 = -12.11Change in EOT per hour: -12.12/24 = -0.5046 seconds per hour

EOT @ 22

h

1010: -12.11(22+10/60+10/3600)/24 = -11.19

GAT = GCT + EOT = 22h1010 11.19 = 22

h0958.81

LAT = GAT - West

LAT = 22h0958.81 - 7

h4603

LAT = 14h2355.81 or 2

h2355.81 PM

Problem 10.13

The mean radius of the earth is 3956 miles and the mean distance to the sun is 92,900,000

miles. What is the suns mean horizontal parallax: What is the parallax correction whenthe altitude is 25 degrees?

Alpha = 90-25 = 65


14/17

Find Theta 1: r = R Theta 1 when Theta 1 is in radians; Theta 1 = r/R

Theta 1 = 3956/92,900,000 = 4.26x10-5

Radians

Theta 1 = 4.26x10-5

Radians or 4.26x10-5

(180/)3600 = 8.8

Theta 1 = 8.8

Cp= ChCos(h) = Theta 2

Cp= 8.8 Cos(25)

Cp= 7.96 or 8

Problem 10.14

The observed altitude of a star is 261030. The temperature is 90F and theatmospheric pressure is 28.70 inches Hg. Find the refraction correction and compute the

true altitude of the star.

Cr= 0.27306 P Tan(z)/(460+T) or 0.27305 P Cot(h)/(460+T)

P = atmospheric pressure in inches Hgz = Zenith angle

h = altitude angleT = temperature in degrees Fahrenheit

Cr= 0.27306(28.70)Cot(261030)/(460+90)

Cr= 0.02898 deg or 144.4

h = h - Cr

h = 261030 144.4

h = 260845.6


15/17

Problem 10.21

On October 9, 2005 at a place where the latitude is 375221 and the longitude is

1221533W, the following data were recorded for a set of solar observations by the

hour angle method:

Instrument (Wilde T-2 theodolite No. 68730) at point 24, sight taken on point number 49

Sightings taken to center of sun using a solar circle attachement.

PI BS FS Scope Horiz Circ UT1 Time

24 49 D 00042

Cntr Sun R 2221359

(421359)

23h1533.9

Cntr Sun R 2222402

(422402)

23h1623.5

Cntr Sun D 424424 23h1803.5

Cntr Sun D 425633 23h1903.5

49 R 1800038

Using the average UT1 time and average horizontal angle, compute the azimuth of thesun and the azimuth of line 24 to 49 from North.

Average UT1: 23h1716.1 or 23.2778 hrs; Average Horizontal Circle: 423404.5

From ephemeris: www.cadastral.com/2005oct.htm

Date Dec GHA EOT Semi Diam

Oct 9, 2005 -61306.4 1830926.5 +12 37.77 16 00.7

Oct 10, 2005 -63553.2 1831330.0 +12 54.00 16 01.0

EOT = GAT GMT

Difference in EOT for Oct 9th

: 12 54.00 12 37.77 = 12.9000 12.6295 = 0.2705

EOT correction: 0.2705/24 = +0.01127 min/hr

GMT = UT1


16/17

Correction for time at average observation: +0.01127(23.2878) = 0.2625 or 16.75

EOT @ average time of observation: +12 37.77 + 16.75 = 12 54.52 or 12.91 or

0.2151 hours

GAT = GMT + EOT

GAT = 23.2778 + 0.2151 = 23.4929 hrs or 23h2935

GAT = 23h2935

GHA = GAT - 12h= 11

h2935

Convert GHA to degrees: 15(11.4929) = 172.3942 or 172 2339

LHA = GHA West Longitude

LHA = 172 2339 - 1221533 = 500806

LHA = 500806

Change in Declination of sun over the day: -63553.2 - -61306.4 = +2246.8

Declination at time of average observation: -63553.2 + 2246.8( 23h2935)/24 + B

Dec = -63553.2 + 2246.8(0.9789) + B

Dec = -61335.2 + B

B = 0.0000395 Dec0Sin (7.5 GCTobs) = 0.0000395(-6.2184)Sin(7.5(23.2778))

B = -2.3185x10-5 degrees or -0.08

Dec = -61335.28 or -6.2265

Z = ArcTan(-Sin(LHA)/(Cos(Lat)Tan(Dec) - Sin(Lat)Cos(LHA)))

Z = ArcTan{-Sin(50.1350)/[Cos(37.8725)Tan(-6.2265) Sin(37.8725)Cos(50.1350)]}

Z = ArcTan{-0.7676/[0.7894(-0.1091) 0.6139(0.6410)]} = ArcTan(1.6004)

Z = 58.0011 or 580004

Azimuth sun= 180 + Z = 180 + 580004

Azimuth sun= 238 00 04

Azimuth 24-49= Azimuth sun Horizontal Circle

Azimuth 24-49= 238 00 04- 42 34 04.5


17/17

Azimuth 24-49= 195 25 59.5

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