QUESTION

OCT09Q2

The distance (/) and bearing (β) of a line RS were independently measured and recorded as

/ = 167.85m ± 0.07m

β = 55° 33' 26" ± 05"

The latitude (lat) and departure (dep) of the line can be determined from the following model

lat = l x Cos (β)

dep = l x Sin (β)

You are required to compute the standard deviation for the latitude and departure of line RS based on

LOPOV in matrix form. Determine and clearly show the values (in three decimal places) of: 25M

a) Partial derivative matrix A

b) Variance-covariance matrix of observations

c) Variance-covariance matrix of latitude and departure

d) Standard deviation for latitude and departure

ANSWER

a) Partial Derivative matrix A

Distance, l = 167.85 + 0.07 m

Bearing, β = 55° 33' 26" ± 05"

Latitude = l Cos β

Departure = l Sin β

1. Differentiate,

������ = Cos β

������ = Sin β

������ = - l Sin β

������ = l Cos β

2. Calculate Matrix A

Matrix A = Cos β � l Sin βSin β l Cos β � = Cos 55° 33� 26" � 167.85 Sin 55° 33� 26"!Sin 55° 33� 26" 167.85 Cos 55° 33� 26" � = "0.566 �138.4240.825 94.933 &

AT =

b) Variance – Covariance matrix of observations

Calculate Σy

Σy = '()² ()β(β) + ,

-./-/01 ²2

Σy = '0.07² 00 + 0"-./-/01 ²2

c) Variance – Covariance matrix of latitude and departure

ΣNE = A Σy AT

= "0.566 �138.4240.825 94.933 & 30.07² 00 + 5"2062651 ²4 "0.566 �138.4240.825 94.933 & = "0.002 0.0020.002 0.003&

σ² N = 0.002 σNE = 0.002

σ² E = 0.003 σEN = 0.002

d) Standard Deviation for Latitude and Departure

Latitude = l Cos β

= 167.85 (Cos 55° 33' 26")

= 94.933 + √σN

= 94.933 + 0.045

Departure = l Sin β

= 167.85 (Sin 55° 33' 26")

= -138.424 + √Σe

= -138.424 + 0.055