CogoKit

Traverse adjustment

Reducing a ground traverse to grid with a combined factor

A link traverse measured on the ground is tied to state plane control. The combined factor is built from the grid scale factor and the elevation factor, and the closure is computed with and without it.

Analyze· about 32 minutes by hand· 6 steps

Given

  • A three-course link traverse GPS-1, 201, 202, GPS-2, tying two state plane control monuments.
  • Published grid coordinates, US survey feet: GPS-1 N 1,842,650.22 E 2,315,880.14; GPS-2 N 1,844,215.55 E 2,317,608.90.
  • Grid azimuths, already adjusted to the control azimuths at both ends: GPS-1 to 201, 45°00′16″; 201 to 202, 50°06′08″; 202 to GPS-2, 48°15′12″.
  • Measured horizontal ground distances: GPS-1 to 201, 749.58 ft; 201 to 202, 795.20 ft; 202 to GPS-2, 789.61 ft.
  • Mean project orthometric height 4250 ft above sea level; geoid height at the project −22.4 m; mean earth radius taken as 6,372,000 m.
  • Grid scale factor for the project area = 0.99995780.
  • Specification 1:15,000.

Required

  • The elevation factor and the combined factor.
  • Grid distances for the three courses.
  • The closure at GPS-2 and the precision, computed on grid.
  • What the closure would have been if the ground distances had been used directly.
  • Adjusted grid coordinates of 201 and 202.

Work it through yourself before reading on — the solution below shows every step, so there is no way to skim it without giving the answer away.

Worked solution

Elevation factor

A distance measured on the ground at project elevation is longer than the corresponding distance on the ellipsoid, by the ratio of the two radii. The elevation, or sea level, factor is the earth radius divided by the earth radius plus the height of the measurement above the ellipsoid.

The height above the ellipsoid is the orthometric height plus the geoid height. Get the sign of the geoid height right: over most of the continental United States it is negative, meaning the geoid lies below the ellipsoid, so the ellipsoidal height is smaller than the orthometric height and the factor is slightly larger than it would otherwise be.

Work in meters for this step, because the earth radius is a metric quantity. Convert the height using the US survey foot, since the coordinates are published in US survey feet.

H = 4250 US survey ft × (1200/3937) m per ft = 1295.403 m h = H + N = 1295.403 + (−22.4) = 1273.003 m elevation factor = R / (R + h) = 6,372,000 / 6,373,273.003 = 0.99980026

Combined factor

The combined factor is the product of the grid scale factor and the elevation factor. Multiply a ground distance by it to get a grid distance; divide a grid distance by it to get back to the ground.

Here the two factors pull the same way, both being less than one, so they reinforce rather than cancel. The combined factor is 242 parts per million below unity.

combined factor = grid scale factor × elevation factor CF = 0.99995780 × 0.99980026 = 0.99975807 1 − CF = 0.00024193, that is 242 parts per million 1 / CF = 1.00024199 (the multiplier for grid to ground)

Reduce the distances to grid

Multiply each measured ground distance by the combined factor. The reduction is 0.18 to 0.19 ft on courses of about 780 ft, which is small on any one course and decidedly not small when three of them accumulate in the same direction.

The azimuths need no reduction here. They are already grid azimuths, having been adjusted onto the two control azimuths at the ends of the traverse. A separate convergence correction would be needed only if the directions had been observed astronomically or by gyro.

Ground to grid reduction, combined factor 0.99975807
CourseGrid azimuthGround distance (ft)Reduction (ft)Grid distance (ft)
GPS-1 to 20145°00′16″749.58−0.18749.40
201 to 20250°06′08″795.20−0.19795.01
202 to GPS-248°15′12″789.61−0.19789.42
Totals2334.39−0.562333.83

Closure on grid

Resolve the grid distances on the grid azimuths and compare the sums with the coordinate difference between the two published control monuments.

The closure is 0.13 ft over 2333.83 ft of traverse, giving 1:17,385, which meets the 1:15,000 specification.

required ΔN = 1,844,215.55 − 1,842,650.22 = 1565.33 ft required ΔE = 2,317,608.90 − 2,315,880.14 = 1728.76 ft e = √(0.096² + 0.094²) = 0.1342 ft precision = 2333.83 / 0.1342 = 17,385, that is 1:17,385, which meets 1:15,000
Latitudes and departures using grid distances
CourseGrid azimuthGrid distance (ft)Latitude (ft)Departure (ft)
GPS-1 to 20145°00′16″749.40529.865529.947
201 to 20250°06′08″795.01509.935609.924
202 to GPS-248°15′12″789.42525.626588.983
Sums2333.831565.4261728.854
Required from control1565.3301728.760
Closure0.0960.094

What happens if the factor is skipped

Run exactly the same computation with the measured ground distances and nothing else changed. The latitude sum becomes 1565.802 and the departure sum 1729.269, giving a closure of 0.472 and 0.509 ft, a linear misclosure of 0.694 ft and a precision of 1:3,365.

That is a failure by a factor of four and a half, and it looks exactly like a blunder. A crew that goes back to remeasure will find nothing wrong, because nothing is wrong with the measurements.

The size of the false misclosure is predictable: it is the total ground length times one minus the combined factor, 2334.39 × 0.00024193 = 0.56 ft, and it points along the general direction of the traverse. A closure that is both large and aligned with the run of the traverse, on work that ties to grid control, should always prompt a check of the combined factor before anyone goes back to the field.

using ground distances: ΣLat = 1565.802, ΣDep = 1729.269 closure = 0.472 and 0.509 ft, e = 0.694 ft precision = 2334.39 / 0.694 = 3,365, that is 1:3,365 predicted systematic error = 2334.39 × (1 − 0.99975807) = 0.56 ft

Adjust and report grid coordinates

With the reduction correctly applied, the residual 0.13 ft is ordinary random error and the compass rule distributes it in proportion to grid length.

Report the coordinates as grid. If a ground distance is wanted later, between two of these points, inverse on grid and divide by the combined factor. Publishing coordinates without stating the zone, the datum, the foot and the combined factor used is the single most common way for a good traverse to become unusable.

Adjusted grid coordinates
CourseLat correction (ft)Dep correction (ft)To stationNorthing (ft)Easting (ft)
GPS-1 to 201−0.031−0.0302011,843,180.052,316,410.06
201 to 202−0.033−0.0322021,843,689.962,317,019.95
202 to GPS-2−0.032−0.032GPS-21,844,215.552,317,608.90

Answer

  • Elevation factor 0.99980026; combined factor 0.99975807, which is 242 parts per million below unity.
  • Grid distances: 749.40, 795.01 and 789.42 ft, a total reduction of 0.56 ft.
  • Closure on grid: 0.096 ft in latitude and 0.094 ft in departure, a linear misclosure of 0.13 ft. Precision 1:17,385, which meets the 1:15,000 specification.
  • Using ground distances directly, the traverse would appear to miss by 0.69 ft, a precision of 1:3,365, and would be wrongly rejected.
  • Adjusted grid coordinates: 201 at N 1,843,180.05 E 2,316,410.06; 202 at N 1,843,689.96 E 2,317,019.95.

Check

The adjusted traverse must land on the published coordinates of GPS-2, and it does: N 1,844,215.55 and E 2,317,608.90 to the hundredth.

Predict the false misclosure before computing it. The systematic error from omitting the combined factor is the total ground length times one minus the factor: 2334.39 × 0.00024193 = 0.56 ft. The computed false misclosure is 0.69 ft, larger because the courses do not run in exactly the same direction and the random error adds to it. The two agreeing in order of magnitude confirms the diagnosis.

Reverse the reduction on one course: 749.40 / 0.99975807 = 749.58 ft, the measured ground distance, recovered to the hundredth.

Straight inverse from GPS-1 to GPS-2 on grid: √(1565.33² + 1728.76²) = 2332.14 ft on azimuth 47°50′25″. The traverse ran 2333.83 ft of grid distance along a slightly bowed path, so it must be longer than the straight line. It is, by 1.69 ft.

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