Home

Instructors

Schedule

Grading

Readings

A worked example

A worked example. Here are the field notes. Note that skipping a line makes it easier to discern which measures are taken between which stations.

The first step is to transcribe the field notes into Excel or whatever tabular data manipulation environment you will be using. Note we have used sta.from and sta.to in order to keep things straight.

raw data
  sta.from sta.to  fs  bs slope    sd
1        1      2 130 310    -4  79.7
2        2      3 205  27    -2 202.2
3        3      4 345 165     0  72.8
4        4      1  20 200     2 173.8

Declination correction
  sta.from sta.to  fs  bs slope    sd fs.dec
1        1      2 130 310    -4  79.7    147
2        2      3 205  27    -2 202.2    222
3        3      4 345 165     0  72.8      2
4        4      1  20 200     2 173.8     37

Correct horizontal distance for slope :

HD = SD * cos(slope° * pi / 180)

Note that the differences between slope and horizontal distances are small because there was not much elevation change over the surveyed area.

Horizontal distance
  sta.from sta.to  fs  bs slope    sd fs.dec        hd
1        1      2 130 310    -4  79.7    147  79.50585
2        2      3 205  27    -2 202.2    222 202.07683
3        3      4 345 165     0  72.8      2  72.80000
4        4      1  20 200     2 173.8     37 173.69413

We should also obtain the sum of horizontal distance, which will be needed in the latitudes and departures corrections:

Sum of HD: 528.08 

Calculate interior angles. This is done by using foreshots and backshots. For example, angle 4-1-2 is calculated:

200 - 130 = 70

Interior Angles
  sta.from sta.to  fs  bs slope    sd fs.dec        hd  ia
1        1      2 130 310    -4  79.7    147  79.50585  70
2        2      3 205  27    -2 202.2    222 202.07683 105
3        3      4 345 165     0  72.8      2  72.80000  42
4        4      1  20 200     2 173.8     37 173.69413 145

Sum of interior angles: 362 

Correct the azimuths based on the interior angle error. In this case the error was +2, and the number of angles was 2. To make the corrected interior angles sum to 360, it is necessary to subtract 0.5 from each measured azimuth.

Corrected azimuths
  sta.from sta.to  fs  bs slope    sd fs.dec        hd  ia fs.cor
1        1      2 130 310    -4  79.7    147  79.50585  70  146.5
2        2      3 205  27    -2 202.2    222 202.07683 105  221.5
3        3      4 345 165     0  72.8      2  72.80000  42    1.5
4        4      1  20 200     2 173.8     37 173.69413 145   36.5

To calculate corrected interior angles, add or subtract the correction factor to each interior angle (not shown, because we don't need the corrected interior angles for further calculation).

Calculate latitudes and departures. For example, look at station 2:

latitude = hd * cos(azimuth) = 202.07 * cos(222 * pi / 180) = 150.17

Latitudes and Departures
  sta.from sta.to  fs  bs slope    sd fs.dec        hd  ia fs.cor        lat         dep
1        1      2 130 310    -4  79.7    147  79.50585  70  146.5  -66.29881   43.882222
2        2      3 205  27    -2 202.2    222 202.07683 105  221.5 -151.34659 -133.900156
3        3      4 345 165     0  72.8      2  72.80000  42    1.5   72.77505    1.905682
4        4      1  20 200     2 173.8     37 173.69413 145   36.5  139.62521  103.317224

Calculate the sums of latitudes and departures.

Sums of L&D: -5.25 15.2

Under perfect conditions these should sum to 0. Because they don't, determine the correction. For example, station 1:

-(-5.25 / 528.08 * 79.51) = 0.79

Latitude and Departure correction factors
  sta.from sta.to  fs  bs slope    sd fs.dec        hd  ia fs.cor        lat         dep   lat.cor   dep.cor
1        1      2 130 310    -4  79.7    147  79.50585  70  146.5  -66.29881   43.882222 0.7896932 -2.289221
2        2      3 205  27    -2 202.2    222 202.07683 105  221.5 -151.34659 -133.900156 2.0071314 -5.818420
3        3      4 345 165     0  72.8      2  72.80000  42    1.5   72.77505    1.905682 0.7230872 -2.096138
4        4      1  20 200     2 173.8     37 173.69413 145   36.5  139.62521  103.317224 1.7252197 -5.001194

Balance the latitudes and departures by adding the correction factors to the originally calculated latitudes and departures. For example, station 1:

-66.3 + 0.8 = -65.5

Balanced Latitude and Departure
  sta.from sta.to  fs  bs slope    sd fs.dec        hd  ia fs.cor        lat         dep   lat.cor   dep.cor    lat.bal      dep.bal
1        1      2 130 310    -4  79.7    147  79.50585  70  146.5  -66.29881   43.882222 0.7896932 -2.289221  -65.50911   41.5930012
2        2      3 205  27    -2 202.2    222 202.07683 105  221.5 -151.34659 -133.900156 2.0071314 -5.818420 -149.33946 -139.7185753
3        3      4 345 165     0  72.8      2  72.80000  42    1.5   72.77505    1.905682 0.7230872 -2.096138   73.49814   -0.1904563
4        4      1  20 200     2 173.8     37 173.69413 145   36.5  139.62521  103.317224 1.7252197 -5.001194  141.35043   98.3160304

Sums of balanced lat & dep: 0 0 

Calculate X and Y coordinates by starting at a known point (station1 in this example), and then sequentially adding the latitudes and departures of subsequent points.

X and Y
  sta.from sta.to  fs  bs slope    sd fs.dec        hd  ia fs.cor        lat         dep   lat.cor   dep.cor    lat.bal      dep.bal       x        y
1        1      2 130 310    -4  79.7    147  79.50585  70  146.5  -66.29881   43.882222 0.7896932 -2.289221  -65.50911   41.5930012 1277180 241854.7
2        2      3 205  27    -2 202.2    222 202.07683 105  221.5 -151.34659 -133.900156 2.0071314 -5.818420 -149.33946 -139.7185753 1277221 241789.2
3        3      4 345 165     0  72.8      2  72.80000  42    1.5   72.77505    1.905682 0.7230872 -2.096138   73.49814   -0.1904563 1277082 241639.9
4        4      1  20 200     2 173.8     37 173.69413 145   36.5  139.62521  103.317224 1.7252197 -5.001194  141.35043   98.3160304 1277082 241713.3

It is possible to double-check the calculated X and Y coordinates by adding the latitudes and departures for the last station. They should result in the coordinates for the first station:

1277082 + 98 = 1277180

241713.3 + 141.3 = 241854.7

Another check can be done by calculating the distances between stations (using the Pythagorean theorem) and comparing them with the measured distances. For example, line 3-4 (note the XY coordinates shown in the table are displayed in a truncated form):

sqrt((1277081.77 - 1277081.58)^2 + (241639.85 - 241713.35)^2) = 73.5

Distance
  sta.from sta.to  fs  bs slope    sd fs.dec        hd  ia fs.cor        lat         dep   lat.cor   dep.cor    lat.bal      dep.bal       x        y      dist
1        1      2 130 310    -4  79.7    147  79.50585  70  146.5  -66.29881   43.882222 0.7896932 -2.289221  -65.50911   41.5930012 1277180 241854.7  77.59782
2        2      3 205  27    -2 202.2    222 202.07683 105  221.5 -151.34659 -133.900156 2.0071314 -5.818420 -149.33946 -139.7185753 1277221 241789.2 204.50808
3        3      4 345 165     0  72.8      2  72.80000  42    1.5   72.77505    1.905682 0.7230872 -2.096138   73.49814   -0.1904563 1277082 241639.9  73.49839
4        4      1  20 200     2 173.8     37 173.69413 145   36.5  139.62521  103.317224 1.7252197 -5.001194  141.35043   98.3160304 1277082 241713.3 172.18010

The last calculations are:

• Area, which can be calculated from the XY coordinates.

Area: 11308.6 ft² = 0.26 ac

• Linear error of closure

EOC = sqrt((-5.25 ft)² + (15.2 ft)²) = 16 ft

• Precision

  Precision = EOC / perimeter = 16 / 528 = 1:33

Finally, a few maps can be made using ArcGIS (click for full-sized versions):

source code