Surveying basics and adjustment

Basic formulas and concepts for surveying

Bearing and angles:

Geodetic azimuth (HCR, horizontal circular reading) is the true bearing of a line with respect to the true north or the reference ellipsoid. It can be compute by the arctan(Delta E/ Delta N), if Delta N is — , then + 180, if the Delta E is — , then +360, and if the both Delta N and E are — , then +360 degrees.

Zenith angle (VCR, vertical circular reading) is the angle between the vertical line (plumb line) and the line of sight to the point of interest that pointing to the mirror/ targets, if the control station height = target station, then the VCR is about 90 degrees. The VCR is for Trigonometric heighting, so it is determine the vertical part (z value) of the traverse stations.

Distance measurements:

Delta N and Delta E are the difference between two point, the N is respect to the y axis (Northing to the local coordinate system) and E is respect to the x axis (Easting to the local coordinate system).

HS and HI are the instrument signal and heights relatively, surveyors should measure the height after the centering and levelling process of the total station or mirror.

Slope distance (SD) is the straight-line distance between two points on a slope, including the vertical component of the distance. Slope distance is measured only when the target is reflectable (target plant cannot reflect the EDM signal so no SD can be survyed in ROs).

Horizontal distance (HD) is the distance between two points on a horizontal plane, without taking into account any changes in elevation.

Vertical distance (VD) is the difference in elevation between two points. It can be computed by the formula: SD cos(VCR).

Horizontal coordinate computations

Northing of the target = Original Northing + HD cos (azimuth)

Easting of the target = Original Easting +HD sin (azimuth)

Error propagration model for the Horizontal coordinate computations

The error propagration model will do the partial derivatives for each of the elements in the horziontal coordinate computation process, there are three elements need to be partial derive

1. Original coodinate, σE/ σE
2. HD (Horizontal distance) of the measurement, σd
3. α (Geodetic azimuth) of the measurement, σα

σ are the variance of the elements (original coordiantes, horizontal distance and geodetic azimuth)

The final standard devation result of the Error propagration model for the Horizontal coordinate computations

Sample of the horizontal computation

Example 1

Sample question for the Horizontal coordinate computations

Compute the horizontal coordinates Np and Ep:

Original is the point 1

Compute the geodetic azimuth (α) by arctan (Delta E/ Delta N)

α = arctan (E2-E1 / N2-N1) = arctan (-250/250) = 45

α + given angle = 90° + — 45°

Ep = 500 + 500 sin (45°) = 925.451762267 meters

Np = 500 + 500 cos (45°) = 762.660994409 meters

Point P = (925.451762267 m, 762.660994409 m)

For the Standard deviation computation, by the given statistics

σα = 4" (beware that the " should be converted to radians, π/180) and σd = 0.03 m, so the σα = 4/3600 x π/180 = 0.00001939254724

σE = √((sin α² σd²)+ (d² cos α² σα²))

= √((sin(45)²0.03²)+ (500² cos(45)² 0.00001939254724²))

= 0.02122428 meters

σN = √((cos α² σd²)+ (d² sin α² σα²))

= √((cos(45)²0.03 ²)+ (500² sin(45)² 0.00001939254724 ²))

= 0.022293695 meters

Therefore, Point P = (925.451762267 ± 0.02122428 m, 762.660994409 ± 0.022293695 m)

Example 2

The position of a new point P was measured from the point 1 by using the distance and angular measurements; The horizontal position of the points I and 2 is given by the following grid coordinates:
Point I: N1 = 650 m, E1 = 750 m
Point 2: N2 = 750 m, E2 = 550 m
The angle α= 89° 00' 00" was measured with the accuracy characterized by the standard deviation for the angle of 11"
The distance d = 1350 m was measured with the accuracy
characterized by the standard deviation of 10 cm.

Compute the geodetic azimuth (α) by arctan (Delta E/ Delta N)

α = arctan (E2-E1 / N2-N1) = arctan (-200/100) = — 63.43494882°

α + given angle = 89° + — 63.43494882° = 25° 33' 54.18"

Ep = given coodinate 1 + d sin(α)

= 750 + 1350 sin (25° 33' 54.18") = 1332.573002 meters

Np = given coodinate 1 + d cos(α)

= 650 + 1350 cos (25° 33' 54.18") = 1867.829502 meters

For the Standard deviation computation, by the given statistics

σα =11" (beware that the “ should be converted to radians, π/180) and σd = 0.10 m, so the σα = 11/3600 x π/180 = 0.000053329504

σE = √((sin α² σd²)+ (d² cos α² σα²))

= √((sin(25° 33' 54.18")²0.10²)+ (1350² cos(25° 33' 54.18")² 0.000053329504²))

= 0.077975919 meters

σN = √((cos α² σd²)+ (d² sin α² σα²))

= √((cos(25° 33' 54.18")²0.10²)+ (1350² sin(25° 33' 54.18")² 0.000053329504²))

= 0.09540909704 meters

Therefore, Point P = (1332.573002 ± 0.077975919 m, 1867.829502 ± 0.09540909704 m)

Example 3

The horizontal position of the new point P was measured by total station from the given benchmark. The measured azimuth (bearing) was 90° and its standard deviation 2”. The measured (horizontal) distance was 1 km and its standard deviation 2 cm. Compute the standard deviations of the coordinates Np and Ep of the new point P.

For the Standard deviation computation, by the given statistics

σα =2" (beware that the “ should be converted to radians, π/180) and σd = 0.02 m, so the σα = 2/3600 x π/180 = 0.000009696273622

σE = √((sin α² σd²)+ (d² cos α² σα²))

= √((sin(90° )² 0.02 ²)+ (1000² cos(90°)² 0.00000969627362 ²))

= 0.02 meters

σN = √((cos α² σd²)+ (d² sin α² σα²))

= √((cos(90°)² 0.02²)+ (1000² sin(90°)² 0.00000969627362²))

= 0.009696273 meters

Trigonometric Heighting computation

In the trigonometric heighing computation, it includes:

• hA (height of the benchmark) and hP (height of the target)
• Slope distance (SD) or horizontal distance (HD)
• VCR (zenith angle, z) sometime the zenith angle will be α as well as z

To compute the height of the target (hP), we can use the properties of triangle, then hP = hA + HD/ tan(z)

tan(z) = (hP — hA) / d, where d is the horizontal distance of the triangle

After the change of subject respect to hP, then hP = hA + HD* tan(z), If the question is given in SD (slope distance) HD = SD* sin(z)

For example, given that the hA = 55, d = 450 and z = 52°

hP = 55+450/tan(52°) = 406.5785319 meters

Error propagation model for the Trigonometric Heighting computation

σP = √ (σA ² + cos(z)² σd ² + (d /tan(z) ²cos(z) ²)² σz²)

σA is the standard deviation of original height, σd is the standard deviation of slope distance and σz is the standard deviation of the zenith angle

Step by step explanation

1. Do the partial derivative respect to each elements ∂σP / ∂σA = 2σA, ∂σP / ∂σd = 2cos(z)²σd and ∂σP / ∂d = 2(d /tan(z) ²cos(z) ²)σz²

2. Formulate the σP² by the deviated results = (σA ²)(∂σP / ∂σA)² + (cos(z)² σd ²)(∂σP / ∂σd)² + ((d /tan(z) ²cos(z) ²)² σz²) (∂σP / ∂d)² + (σz²) (∂σP / ∂z)²

3. Re-formulate σP² as σP by taking square root, σP = √ (σA ² + cos(z)² σd ² + (d /tan(z) ²cos(z) ²)² σz²)

Remember that the σz should be degree to radians, for example σz =2" = 2/3600 x π/180 = 0.000009696273622

For example, hA = 55, d = 450, z = 52, σA = 0.007, σd = 0.06 and σz = 15"

So, for σz = 15/3600 x π/180 = 0.000072772052

σP = √ (0.007 ² + cos(52)² 0.06 ² + (450/tan(52)² cos(52)²)² 0.00007277205 ²)

= 0.064736993 meters

On the other hand, if the slope distance is not provided, we can compute it using the pyth. th: SD = √(d² + (hP — hA)²)

Substituting this expression for SD in the formula for σP, then the Error propagration model for the Trigonometric Heighting computation in terms of horizontal distance:

σP = √(σA² + (d/(cos(z) tan(z)))² σz² + ((hP — hA)/SD)² σd²)

for the d is the horizontal distance and σd is the standard deviation of the horizontal distance