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## Section 9: Geodetic Surveying

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### Overview

Unless otherwise instructed, latitude and longitude will be presented as degrees, minutes, and seconds. Direction indicators N or W will prefix the value and seconds will be carried out five (5) places right of the decimal when using published National Geodetic Survey (NGS) latitude/longitude (geographic) values.

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### Units

The coordinate system used by TxDOT is the State Plane Coordinate System (SPCS) in NAD 83 datum. However, units of length for northings and eastings will be in U.S. Survey Feet, rather than the defining units of length which is in meters. It should be noted that the U.S. Survey Foot is designated by the Texas Legislature to be the state standard, rather than the International Foot, used in some states (Texas Natural Resources Code, Subchapter D, §21.071 – 79). Horizontal coordinates should be carried out to .001 ft. unless otherwise instructed.

Processing and adjusting GPS data may be done in the metric system, but all project data must be converted and delivered in U.S. Survey Feet. Conversion from meters to U.S. Survey Feet must be made using the following formula:

Meters * 3937/1200 = U.S. Survey Feet

The factor is 3.280833333333 and working with state plane coordinates (SPC’s) in the millions, one must carry the factor out to 12 places to the right of the decimal just as shown here.

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### Datums

All geodetic surveying will be done in the NAD 83 horizontal datum. A horizontal datum is defined as an ellipsoid model with a designated reference point. The most recent datum used in Texas, before NAD 83 was the NAD 27. The latitude and longitude coordinates of a particular point on the ground will differ by about one second of arc between NAD 27 and NAD 83, which equates to about 100 ft.

An adjustment, done in Texas (using GPS) resulted in the 1993 High Accuracy Reference Network (HARN). The network was extended to nearly all old, conventionally surveyed federal monumentation monuments. Projects should be referenced to the published HARN coordinates of NGS monumentation monuments.

Elevations will be referenced to the NAVD 88 vertical datum. An older datum, NGVD 29, has been superseded. Differences between the two range from almost none in Southeast Texas to about a foot in Northwest Texas. NGS data sheets specifically indicate bench mark elevations as 1.) monumented and surveyed in the newer NAVD 88 datum, 2.) re-surveyed to the new datum or 3.) transformed to the new datum by means of an interpolation software called VERTCON.

Where design survey accuracy is required, TxDOT will not accept any datum transformations based upon a single control point. There is no way to transfer NAD 27 coordinates to NAD 83 datum accurately. CORPSWIN and other conversion software are based on NADCON algorithms, which perform a rubber sheeting adjustment, are not at all accurate in most parts of the state.

If a change to the NAD 83 project datum is needed for comparison of old surveys, two control points can be resurveyed (GPS or conventional) from references in the new datum. Then a translation-rotation-scale can be done holding to the two points common to both datum. Alternatively, if original raw GPS data is available, it can be reprocessed holding the new datum coordinates rather that than the original datum coordinates.

There are several programs, such as CORPSWIN, available through the NGS for conversion of geographic coordinates to state plane coordinates. CORPSWIN, one of the best software available is for the mathematical conversion of a.) Metric to U.S. Survey Foot (or visa-versa) b.) SPC zones to adjoining SPC zones c.) UTM to SPC’s and d.) Lat/Lon to SPC’s.

In addition to these strictly mathematical conversions, CORPSWIN provides useable outputs to determine a combined adjustment factor (CAF) at the specific location of a point, if the elevation is included in the input. Users should note that this is computed using the sea level factor of the NAD 27 datum, and does not utilize the height above ellipsoid (HAE) method of NAD 83. The programs published by NGS use a mean radius of the earth (20,906,000 feet) for the continental U.S. Other commercially published programs, such as Trimble Geomatics Office, compute a CAF for each point using a terrestrial radius computed for the latitude at that point.

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### State Plane Coordinates

Background. Because surveys are performed on the curved surface of the earth and at different elevations, measurements from different starting points will not mesh nor will the lines be rectilinear. This may not present a problem in a few miles of work, but in order to seamlessly connect survey work over greater distances, a working plane with a rectangular grid must be used. For engineering applications, the State Plane Coordinate System (SPCS) is used most often.

The SPCS was developed in the 1930’s and was based on NAD 27. With a few changes, it was adapted to the present day NAD 83 datum. Coordinates from NAD 27 and NAD 83 do not at all resemble each other, however. Differing false eastings and northings were assigned to the two datums to prevent confusion between coordinates computed on each datum.

Coordinates are given as X (east) and Y (north) in meters. The projection used in Texas to project from the earth’s curved surface to a developable plane was the Lambert conformal conic projection. Five plane coordinate projections zones — North (TX4201), North-central (TX4202), Central (TX4203), South-central (TX4204) and South (TX4205) are used.

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### The State Plane-to-Surface Transition

One of the most troublesome issues often faced when dealing with state plane coordinate systems (SPCSs) is the fact that, except for a few very small areas, a distance measured on the ground (surface) will differ from the inversed distance between SPCS values. Depending on how far north or south the project falls in the state plane zone and depending on the elevation of the area, coordinates in the SPCS most likely will need to be adjusted so that lengths measured on the surface will coincide with lengths inversed on the state plane grid.

One component of the adjustment involves the projected position from the curved surface of the earth to the plane. It is a ratio, which reflects the difference between a length measurements on the grid versus one lengths made computed on the ellipsoid. A factor suitable for a conversion of any position on the surface to a position on the state plane grid can be found in tables or derived by a number of software programs. This is called the scale factor.

The other component of the conversion is a function of a point’s elevation. It reflects the fact that a sea level (or ellipsoid-level) distance measured between two lat/lon points is shorter than a surface distance. The elevation factor is computed by the formula:

20,906,000 / (20,906,000 + n)

where n is the orthometric height and 20,906,000 is the mean radius of the earth

For most surveys, this calculation is adequate (the sample calculations beginning on page 56). For extremely accurate work, the difference between the orthometric elevation and height above the ellipsoid should be taken into account.

Since a user would not use one of these factors without the other, they may as well be combined into one. To derive a combined adjustment factor (CAF), we multiply the scale factor by the elevation factor. TxDOT has calculated a very general combination factor for each county in the state. It is the reciprocal value of the CAF meaning that it is multiplied times a state plane baseline length (or coordinate) to arrive at a surface value. This multiplier, known as a surface adjustment factor (SAF) and a list can be obtained from the district surveyor if the use of project specific factors is not required.

The only way of maintaining integrity between points on the surface and the state plane grid over long distances is to apply an individual CAF for each leg of a traverse between two stations of known state plane coordinates. With the size and nature of most of TxDOT’s projects, the practice of choosing a single CAF for a project has become common. Just as with the countywide SAF, the factor is applied equally to all project coordinates rather than an individual factor for each baseline length.

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### Choosing an Appropriate Project CAF

A CAF can be calculated for any individual point. There may be quite a change in elevation from one end of the project to the other, and if the corridor runs north and south, there may be a significant scale change. The factor should be chosen to minimize these influences in surface to grid distances. A CAF for use on the entire project is usually feasible for projects of less than three to five miles. For larger projects or smaller surface to grid differences, the project will have to be broken down into segments and a separate CAF be determined for each segment.

There are several ways to select a single factor for a project or segment of a project. The TxDOT surveyor or engineer may take an average of the CAF’s of each control point of a project or simple pick a point central to the project. Using the CAF provided on the NGS datasheet for a station located near the center of the project is an easy way to obtain a useable value.

Another source is one of the software programs that deal with coordinate calculations. CORPSCON, for instance provides the CAF for any latitude, longitude (or state plane coordinate) and elevation given. The value is generally carried out eight decimal places so that others using it will arrive at the same figure to the nearest thousandth of a foot.

An option, if rapidly increasing differences outward of the center of the project is not of much concern, is to use the less precise TxDOT surface adjustment factor assigned for the entire country. It is not important what value is used, but it is absolutely necessary that the value be included in the metadata so that the coordinates can be backed down to state plane coodinates for meshing with adjoining work or for reassigning a preferred CAF at a later date.

Highway projects with several CAF’s pose no problem when all coordinates can be taken back to state plane coordinates so that everything will mesh. However, the juncture of two project segments needs to be well identified to ensure that all measurements stop or begin at a common point of that juncture.

Occasionally, surface coordinates may need to be returned to state plane coordinates (SPC’s) so that a more accurate CAF or one used by another consultant can be applied. Ideally, if enough care is taken to arrive at accurate CAF’s, and if the areas (to which new CAF’s are calculated) are kept small enough, there would be very little accumulation of irreconcilable surface/grid difference.

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### Truncating Coordinates

With NAD 83 SPC’s, coordinates have values well into the millions. If an across-the-board CAF is applied to project points in state plane coordinates, the coordinates change by a huge amount. Two coordinate pairs that both represent the same point, could be over 1,000 feet apart. A shift of that magnitude will occur for every coordinate pair to which the factor is applied. The intent was to change the relationship between points by just a few hundredths or tenths but in the process all the coordinate values were shifted and moved about a thousand feet. Note that the inverse azimuths between the points will not change in this process, only the distances.

When the calculation is done to convert from grid-to-surface or vice versa, the new coordinates can be easily confused with state plane coordinates. In addition, if one truncates the coordinate value – reduce all northings by the same amount and reduce all eastings by another amount – the solution can cause more dilemma than the original inconvenience. Coordinates in this altered state will also not fit the standard TxDOT seed files in MicroStation®. Furthermore, if the truncated amount is not recorded, there will be many more problems later.

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### Identifying Delivered Coordinates

All coordinates delivered, whether hardcopy or in digital medium, must contain metadata indicating the CAF (or SAF), horizontal datum and adjustment, vertical datum (Geoid model if applicable), units of measure and the date of the fieldwork. This would include hard copy drawings, CAD drawings, the data sheets, and each sheet containing coordinates in a report, and ASCII or LandXML files.

If coordinates have been truncated for easier calculations or for identification, they must be returned to their full configuration before delivery. This is not only for standardization but also so that they will work in the seed files for MicroStation®.

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### Calculations

Note: Software is available to perform all of the following calculations; however, “by-hand” solutions are used to illustrate the procedures involved.

This section covers most of the different types of calculations that might be encountered when working on the State Plane Coordinate System. With few exceptions, placing a survey on the system will require only conversion of surface distance to grid. On some surveys, it may be necessary to convert a geodetic (“true”) azimuth to grid azimuth. Fortunately, both of these calculations are very simple.

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### Surface Distance to Geodetic Distance

Before a surface distance can be converted to grid it must first be converted to geodetic (reduced to sea level).

From the figure of surface distance below the following relationship can be derived:  where Rm is the mean radius of the earth or 20,906,000 ft.

Figure 3-15. Surface Distance to Geodetic Distance.

To be more precise, the ellipsoid (GRS80) does not coincide with sea level by an amount equal to the geoid height (approximately 100 ft. for Texas). Neglecting this difference will result in an error of approximately 1/200,000. If this is significant, then the following equation should be used: The term Rm/(Rm + Elevation) is frequently referred to as the sea level factor. Rm and elevation must be in the same units, either both in feet or both in meters. Depending on survey accuracy requirements and differences in elevation within the project area, the average elevation taken from a USGS quadrangle map is sufficient. For high accuracy requirements, the average elevation of each line should be used. The mean radius of the earth (Rm) is assumed to be 20,906,000 feet.

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### Geodetic Distance to State Plane Grid Distance

To convert geodetic distance to grid, the projection scale factor (scale ratio) is used where: or Scale factor is a function of latitude and can be obtained from the appropriate projection table. Depending on accuracy requirements and the extent of the project area in a north-south direction, the average latitude as obtained from a quad sheet is sufficient. For high-precision surveys a weighted average of scale factor for each line can be used. But there are usually more important things to worry about—adjustment of optical plummets, ppm correction, baseline calibration, etc. It should be noted that for all practical purposes, scale factors did not change from SPCS 27 to SPCS 83 in Texas.

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### Surface Distance to State Plane Grid Distance

Converting a surface distance to the state plane grid involves multiplying the distance times both the elevation factor and scale factor.

Grid Distance = Surface distance x Elevation factor

If an average elevation and latitude for the project area is being used, these two factors can be multiplied to obtain a single factor (combined adjustment factor (CAF), which is in turn multiplied times all of the field measured distances, either slope or horizontal.

EXAMPLE:

Given:

Horizontal Distance = 2,640.00 ft.

Average Elevation = 1,400 ft.

Average Latitude = 32o 54'

83 North-central Zone

Solution:

Scale Factor = 0.99987611

(Refer to Appendix B, Second Term Calculations, enter with Latitude = 32o 54') Combined Scale Factor (CSF) = 0.99987611 x 0.99993304 = 0.99980916

Grid Distance = 2,640.00 x 0.99980916 = 2,639.50 ft.

EXAMPLE:

Given:

Horizontal Distance = 2,640.00 ft.

Average Elevation = 350 ft.

Average Latitude = 32o 04'

83 Central Zone

Solution:

Scale Factor = 1.00005440

(Refer to Appendix B, Second Term Calculations, enter with Latitude = 32o 04') Combined Scale Factor = 1.00005440 x 0.99998326 = 1.00003766

Grid Distance = 2,640.00 x 1.00003766 = 2,640.10 ft.

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### Geodetic Azimuth to Grid Azimuth

To convert geodetic azimuth to grid azimuth the following equation is used: The second term can be neglected for most surveys (see Appendix B, for use of second term); therefore, the above equation reduces to: Figure 3-16. Geodetic to Grid Azimuth.

Mapping angle or grid declination is a function of longitude and can be computed from the following equation: = (Longitude of Central Meridian - Longitude of Station) x 1

where Longitude of Central Meridian and l are projection zone constraints. They are listed below as well as in the projection tables found in Appendix B, Second Term Calculations, of this manual.

Anchor: #i1021761Table 3.14 State Plane Zones

Zone

LCM

l

83 North

101o30'00''

0.579535862261

83 North-central

98o30'00''

0.545394412971

83 Central

100o20'00''

0.515058882235

83 South-central

99o00''00''

0.489912625143

83 South

98o30'00''

0.454006848165

For most work, the longitude of the beginning point of the survey must be accurately known to convert geodetic azimuth to grid. Since this is usually a triangulation station or other point of known coordinates, the longitude is given or can be computed from the coordinates. Scaling longitude from a USGS quad sheet may be sufficient; however, for Texas a scaling error of 5'' of longitude will result in approximately 2'' to 3''of error in EXAMPLE:

Given:

Geod. Az. from Tri Station to Pt. A = 241o12'37''

Long. of Tri Station = 105o10'15''2758

83 Central Zone

Solution:

Long. of Central Meridian = 100o20'00'' (Refer to Appendix B, or the above table)

l = 0.515058882235 = -2.49164o

= -2o29'30''

Grid Az. = 241o12'37''- (-2o29'30'') = 243o42'07''

Given:

Geod. Az. To Backsight = 127o48'36''

Long. of Instrument Station = 94o39'28''

83 South-central Zone

Solution:

Long. of Central Meridian = 99o00'00'' (Refer to Appendix B, or the table above)

l = 0.489912625143

= (99o00'00'' - 94o39'28'') x 0.489912625143 = +2o07'38''

Grid Az = 127o48'36'' - (+2°07'38'') = 125o40'58''

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### Astronomic Azimuth to Geodetic Azimuth

Astronomic azimuth is based on the true shape and rotation of the earth; whereas, geodetic azimuth is based on the mathematical approximation of the earth’s shape. Astronomic azimuth is obtained when making a celestial observation (sun or Polaris). Astronomic azimuth can be converted to geodetic from the following equation:

Geodetic Azimuth = Astronomic Azimuth + Laplace Correction

The Laplace correction is less than 3'' for most of Texas and does not exceed 6''. If accuracy requires use of the correction, values can be obtained from NGS.

Geodetic Angle to Grid Angle. If the second term is neglected, grid angle is the same as geodetic (grid) angle. Therefore, field angles can be used when computing grid azimuths.

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### Plane Coordinates to Geographic Coordinates (Lat. and Long.)

To compute latitude and longitude from N and E coordinates, the following equations are used:

N' = N - Nb

E' = E - Eo   Enter the projection tables with R and interpolate to obtain latitude.

Nb, Eo, Rb and l are constants for the zone. When working on the 83 system, coordinates used in the above equations must be in meters. In solving the above equations, the calculator/computer must be capable of computing to at least 10 digits.

Figure 3-17. Computation Sample. Plane Coordinates to Geographic Coordinates.

EXAMPLE:

Given:

N = 2,256,876.543 m

E = 225,025.678 m

83 North-central Zone

Solution:

Nb = 2,000,000.000 m

Eo = 600,000.000 m

Long. Cent. Meridian = 98o 30'00''

Rb = 9,964,225.7538 m

l = 0.545394412971

N' = 2,256,876.543 - 2,000,000.000 = 256,876.543

E' = 225,025.678 - 600,000,000 = - 374,974.322

Rb – N' = 9,964,225.7538 - 256,876.543 = 9,707,349.2108

Using Equations:  R = 9,707,349.2108

cos (-2.2121147004) = 9,714,588.7428

Or using R - P (for HP: E' in y reg. and Rb-N' in x reg)

R = 9,714,588.7428 (display) Long. = 98o 30' 00'' - (- 2.2121147004/0.545394412971)

= 98o30'00'' - (- 4o03'21''.5667)

= 102o33'21''.5667

Enter projection tables with R and interpolate for latitude. Computed R is between 33o55' and 33o56'.

R = 9,714,712.237 & diff. = 30.81105 (at lat. = 33o55', Appendix B, Second Term Calculations)  Lat. = 33o55'04''.0081

Enter projection tables with R and interpolate for latitude. Computed R is between 33o55' and 33o56'.

R is obtained from projection tables by entering with latitude and interpolating.

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### Geographic Coordinates (Lat. and Long.) to Plane Coordinates

To compute N and E coordinates from latitude and longitude, the following equations are used:   Nb, Eo, Rb and l are constants for the zone. When working on the 83 system, coordinates will be in meters. In solving the above equations, the calculator or computer must be capable of computing to at least 10 digits.

Figure 3-18. Computation Sample. Geographical Coordinates Lat./Lon. to Plane Coordinates.

EXAMPLE:

Given:

Latitude = 29o55'41''.2345

Longitude = 94o52'36''.5432

83 South-central Zone

Solution:

Nb = 4,000,000.000 m (Appendix B)

Eo = 600,000.000 m (Appendix B)

Long. Cent. Meridian = 99o00'00 (Appendix B)

Rb = 11,523,512.5584 m (Appendix B)

l = 0.489912625143 (Appendix B)

Enter projection tables with latitude

R = 11,292,620.462 & diff. = 30.78941 (For= 29°55', Appendix B) R = 11,292,620.462 - 1,269.586

=11,291,350.876 Using Equations

N = 11,523,512.5584 - 11,291,350.876 x cos (2.0199991353) + 4,000,000.000

= 4,239,178.303 m

E = 11,291m350.876 x Sin (2.0199991353) + 600,000.000

= 998,001.292 m

Using P - R (For HP: in y reg. and R in x reg.)

(Rb – N') = 11,284,334.255 (display)

N' = 11,523,512.5584 - 11,284,334.255 + 4,000,000.000 = 4,239,178.303 m

E’ = 398,001.292 (x<>y)

E = 600,000.000 - 398,001.292 =998,001.292 m

Conversion from SPCS 27 to SPCS 83: For all practical purposes, there is no accurate means to convert from NAD 27 to NAD 83. NGS has software available (CORPSCON) to make a rough transformation from one system to the other, but its accuracy is limited to approximately one-half foot.

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### Example Problems

(1.) Compute state plane coordinates in area.

Figure 3-19. Computation Sample: Area.

83 Central Zone Grid Az. to Az. Mark = 195o54'16''

Ave. Lat. = 31o21' NT =3,190,394.533 m

Ave. Elev. = 2,900' ET = 420,496.711 m

Convert beginning coordinates to U.S. Survey Feet.

NT = 3,190,394.533x3937/1200 = 10,467,152.73 ft.

ET = 420,496.711x3937/1200 = 1,379,579.63 ft.

Combined Scale Factor = 0.99990018 x 20,906,000 / (20,906,000 + 2,900) = 0.99976150

Angle Rt.: Az. Mk/T/A = 268o31'41' (given)

Grid Az. T - A = 195o54'16' + 464o22'57'' = 104o22'57''

The following is a listing of computation information:

Table 3.15 Computation Information for Problem 1

Pt.

(given)

Grid Az.

Hor. Dist.

(given)

Grid D

Lat. (N)

Dep. (E)

T

104-25-57

4,163.05'

4,162.06'

-1,037.35'

4,030.71'

A

84-27-11

8-54-08

2,769.68

2,769.02

2,735.66

428.50

B

99-37-46

288-31-54

5,307.22

5,305.95

1,686.38

-5,030.82

C

64-53-08

170-25-02

3,433.23

3.432.41

-3.384.52

571.40

T

114-00-55

15,669.44

+0.17

-0.21

Continuation of Table 3.15

Pt.

N

E

T

10,467,152.73'

1,379,579.63'

-1037.39

4,030.76'

A

10,466,115.34

1,383,610.39

2,735.63

428.54

B

10,468,850.97

1,384,038.98

1,686.32

-5,030.75

C

10,470,537.29

1,379,008.18

-3,384.56

571.45

T

10,467,152.73

1,379,579.63

Grid Area = 13,767,229 sq. ft. = 316.0521 acres

Surface Area = Grid Area / (Combined Scale Factor)2 = 316.2029 acres

For the above problem, assume azimuth mark cannot be found.

From Polaris observation: Az. T - A = 102ο 55'11 Nb = 3,000,000.000 m (Appendix B)

Eo = 700,000.000 m (Appendix B)

Rb = 10,770,561.1034 m (Appendix B)

N' = N - Nb = 3,190,394.533 - 3,000,000.000 = 190,394.533

E' = E - Eo = 420,496.711 - 700,000.000 = -279,503.289

Rb – N' = 10,770,561.1034 - 190,394.533 =10,580,166.5704

Using R - P: Grid Az. T - A = 102o55'11'' - (1o30'48'') = 104o25'59''

(2.) Compute state plane coordinates (in meters) for the traverse below.

Figure 3-20. Computation Sample: Traverse.

Given:

N1 = 4,253,992.557 m N4 = 4,253,195.719 m South-Cent. Zone

E1 = 734,141.809 m E4 = 737,212.927 m

Ave. Lat. = 30o07' Ave. Elev. = 700 ft.

From Polaris: Geod. Az. 1 - 2 = 96o45'39'' (neglecting Laplace)

Geod. Az. 4 - 3 = 331o13'49''

Solution:

Convert Geod. Az. To Grid Az.

Nb = 4,000,000.000 m

Eo = 600,000.000 m

Rb = 11,523,512.5584 m

At Sta. 1: N' = N - Nb = 253,992.557

E' = E - Eo = 134,141.809

Rb – N' = 11,269,520.001 Grid Az = 96o4539'' - (0°4055') = 96°04'44''

At Sta. 4: N' = N - Nb = 253,195.719

E'= E - Eo = 137,212.927

Rb = N' = 11,270,316.839 Grid Az. =331o13'49'' - (0o41'51'') = 330o31'58''

CSF = 0.99995611 x (20,906,000 / (20,906,000 + 700)) = 0.99992263

The following is a listing of computation information:

Anchor: #i1074193Table 3.16 Traverse Computation for Problem 1

Pts.

Ang. Rt.

Grid Az.

Grid Dist.

Lat.

Dep.

1

96-44-04

96-44-06

1,219.700

143.043

1,211.283

2

151-38-15

68-22-19

68-22-23

1,309.875

482.770

1,217.664

3

262-09-31

150-31-50

150-31056

1,305.675

-1,136.675

642.256

4

3,835.149

796.948

3,071.203

* Az. Closure =150o31'50'' - (330o31'58'' - 180o) = -08''

Angles measured at four stations: Adj./Sta. = +02''

Computed Closure:

N4 = 4,253,992.557 - 796.948 = 4,253,195.609

E4 = 734,141.809 + 3,071.203 = 737,213.012

Error in N (Lat.) = 4,253,195.609 - 4,253,195.719 = -0.110

Error in E (Dep.) = 737,213.012 - 737,212.927 = +0.085

Adjust Lats. and Deps. by compass rule and compute coordinates.

Anchor: #i1095918Table 3.17 Adjustments for Problem 2

Pt.

N

E

1

4,253,992.557

734,141.809

143.008

1,211.256

2

4,253.849.549

735,353.065

482.808

1,217.635

3

4,254,332.357

436,570.700

-1,136,638

642.227

4

4,253,195.719

737,212.927

(3.) Convert coordinates and grid azimuth from N.- Cent. to Cent. Zone.

Given:

N = 2,041,990.909 m 83 N.-Cent. Zone

E = 203,858.434 m

Grid Azimuth = 207o 40'32''

Compute:

N, E and Grid Azimuth in 83 Central Zone.

Procedure:

Convert N and E to Lat. and Lon., and Grid Az. to Geod. Az. Using constants and table for 83 North-central Zone. Then convert Lat. and Lon. to N and E, and Geod. Az. to Grid Az. using constants and table for 83 Central Zone.

For 83 North-central Zone:

Nb = 2,000,000.000 m (from Appendix B)

Eo = 600,000.000 m (from Appendix B)

Long. Cent. Meridian = 98o30'00'' (from Appendix B)

Rb = 9,964,225.7538 (from Appendix B)

= 0.545394412971 (from Appendix B)

N' = N - Nb = 41,990.909

E' = E - Eo = -396,141.566

Rb – N' = 9,922,234.845

Using R - P

R = 9,930,139.599  = 98o30'00'' - (-2.2862986088/0.545394412971)

= 102.692009589o

Enter tables with R and interpolate for Latitude.

R = 9,930,957.217 & diff = 30.80324 (At = 31o58', from Appendix B)  Lat = 31o58' 26.5432'' = 207o40' 32'' + (-228629861o)

= 205o 23' 21'' 3

For 83 Central Zone:

Nb = 3000,000.000 m (from Appendix B)

Eo = 70,000.000 m (from Appendix B)

Long. Cent. Meridian = 100o 20' 00 (from Appendix B)

Rb = 10,770,561.1034 (from Appendix B)

l = 0.581899128040 (from Appendix B) = (100o 20' 00''- 102.692009589o) x 0.515058882235

= 1.21485715596o

Enter tables with latitude and interpolate for R.

R = 10,515,580.239 & diff. = 30.80257 (For = 31o58', from Appendix B) R = 30.80257 x 26.5432'' = 817.599

R = 10,515,580.239 - 817.599 = 10,514,762.640

Using P - R

Rb – N' =10,512,399.125

N' = 258,161.978

N = N' + Nb = 3,258,161.978 m

E - Eo = 222,930.513

E = 477,069.487 m = 205o 23' 21'' 3 - (-1.2148571560o)

= 206o 36' 15''

Summary:

Anchor: #i1083179Table 3.18 Converted Coordinates for Grid Azimuth for Problem 3

Zone

N

E

Grid Az.

83 N.-cent.

2,041,990.909 m

203,858.434 m

207°40'32''

83 Cent.

3,258,161.978 m

477,069.487 m

206°36'15''

Latitude = 31o 58' 26.5432''

Longitude = 102o 41' 31.2345''