SURVEYING II (CE 6404) UNIT II SURVEY ADJUSTMENTS By Mr.R.GOPALAKRISHNAN, Asst.Professor in Civil Engineering, Sri Venkateswara College of Engineering

Error sources – precautions and corrections – classification of errors , true and most probable values –weighed observations – method of equal shifts, principle of least squares – normal equation- correlates – level nets – adjustments of simple triangulation networks.

Kinds of errors Mistakes Systematic errors Accidental errors

Different terms of errors Independent Quantity Conditioned Quantity Direct observation Indirect observation Weight of an observation Observed value of a quantity True value of a quantity Most probable value Residual errors

Probability curve

Probable error for a single measurement: Es = ± 0. 6745. Sq Probable error for a single measurement: Es = ± 0.6745 * Sq. root {∑ V² / n-1} Where, Es – Probable error of single observation V – Difference between any single observation and mean n – no. of observations. Probable error of an average: Em = Es/ Sq.root (n) Where, Em – probable error of the mean. Principle of Least Square: In observations of equal precision , the most probable values of the observed quantities are those that render the sum of the squares of the residual errors a minimum

Laws of weights

Determination of Probable Error Direct observation of Equal weight on a Single Unknown Quantity. p.e. Of single observation of unit weight p.e. Of single observation of weight w p.e. Of single arithmetic mean

Cont’d. Direct observation of Unqual weight on a Single Quantity. p.e. Of single observation of unit weight p.e. Of single observation of weight w p.e. Of weighted arithmetic mean

Direct observation of unequal weight

Problem: The following are the observed values of an angle: Angle 40°20’20” wt. 2, 40°20’18” wt. 2, 40°20’19” wt. 3, Find the P.E. of a single observation of unit weight, P.E. of weighted arithmetic mean, P.E. of single observation of weight 3. Solution: Value Weight Value * Weight V v² wv² 20° 2 40° +1 1 18° 36° -1 19° 3 57° ∑w = 7 Weighted mean= 19° ∑w v² = 4

Problem 2. The angles of a triangle ABC recorded were as follows: Inst station Angle Weight A 77° 14' 20" 4 B 49° 40' 35" 3 C 53° 04' 53" 2 Give the corrected values of the angles. Solution: Sum of observed angles = 77° 14' 20" + 49° 40' 35" + 53° 04' 53" = 179O 59’ 48’’ Error = - 12” Total correction = 12” Let C1, C2 & C3 be the corrections to the observed angles A, B and C. The error will be distributed to the angles in an inverse proportion to their weights. A = 77° 14' 20" + C1 B = 49° 40' 35" + C2 C = 53° 04' 53" + C3

Normal Equations Rule 1: To form a Normal equation for each of the unknown quantities multiply each observation equation by the algebric co- efficient of that unknown quantity in that equation and add the results. Rule 2: To form the Normal equation for each of the unknown quantities, multiply each observation equation by the product of the algebric coefficient of that unknown qty. in that equation and weight of that observation and add the results.

Triangulation adjustments

Figure Adjustment The determination of the most probable value of the angles involved in any geometrical figure so as to fulfil geometrical conditions is called the figure adjustment. Adjustments of a Triangle: