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SURVEYING-II

SURVEYING-II. ADJUSTMENT OF ANGLES IN TRIANGULATION. ADJUSTMENT OF ANGLES. After completion of field work of measurements of angles, it is necessary to adjust the angles. Generally the angles of a triangle and chain of triangles are adjusted under two heads. (i). Station Adjustment.

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SURVEYING-II

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  1. SURVEYING-II

  2. ADJUSTMENT OF ANGLES IN TRIANGULATION

  3. ADJUSTMENT OF ANGLES After completion of field work of measurements of angles, it is necessary to adjust the angles. Generally the angles of a triangle and chain of triangles are adjusted under two heads. (i). Station Adjustment. (ii). Figure Adjustment.

  4. STATION ADJUSTMENT Sum of the angles about a station should be 360o. If not, find the difference and adjust the difference equally to all the angles algebraically to make their sum equal to 360o. Suppose; for a station B.

  5. FIGURE ADJUSTMENT The determination of most probable values of angles involved in any geometrical figure so as to fulfill the geometrical conditions is called the figure adjustment. All cases of figure adjustment necessarily involve one or more conditional equations. The geometrical figures used in a triangulation system are: (a). Triangles. (b). Quadrilaterals. (c). Polygons with central stations.

  6. TRIANGLE ADJUSTMENT Triangulation of ordinary precision, the sum of the angles of a triangle is equal to 180o. For large triangles, covering big area, correction for spherical excess is to be applied because sum of angles of a spherical triangle will be more than 180o and the correction to be applied is as an addition for 01'' for every 75 square miles. For triangle ABC

  7. TRIANGLE ADJUSTMENT

  8. ADJUSTMENT OF BRACED QUADRILATERAL Geometric Condition (a). Sum of all the angles should be equal to 360o. (b). Sum of equal pair of angle should be equal. ∟2 + ∟3 = ∟6 + ∟7 ∟1 + ∟8 = ∟4 + ∟3 Suppose L.H.S > R.H.S by 12'' Divide this 12 seconds by 4; correction = 12''/4 = 3'' Add 3'' to the angles of R.H.S and subtract 3'' from the angles of L.H.S.

  9. ADJUSTMENT OF BRACED QUADRILATERAL

  10. Trigonometric Condition: log (sin 1) + log (sin 3) + log (sin 5) + log (sin 7) = log (sin 2) + log (sin 4) + log (sin 6) + log (sin 8) For this adjustment following procedure is adopted :

  11. PROCEDURE • Record ‘log (sin)’ of each angle obtained after geometric adjustment. • For each angle record the ‘log (sin)’ difference for 01'' i.e., the difference between the previous value in step (1) & the value obtained after adding 01'' to actual angle. • Find the average required change (α) in ‘log (sin)’ by dividing the difference of sum of odd and even angle values by 8.

  12. PROCEDURE • Find the average difference “β” for difference for 01''. i.e., total ‘log (sin)’ difference for 01'' of all angles divided by 8. • The ratio α / β gives the no. of seconds to apply a correction. 6. Add the correction to each of four angles where ‘log (sin)’ is smaller & subtract the correction from the other four angles.

  13. SATELLITE STATION

  14. SATELLITE STATION • A satellite station is used when instrument can not be set up at the main station. The distance of the satellite from its station is usually very small as compared to the length of the sides of the triangulation. • In the figure ABC represent part of a triangulation station. Station A could be observed but it was not possible to set up an instrument there. A satellite station ‘S’ was therefore used and angles from S to A, B & C are measured. Suppose distance AS = S'.

  15. In the triangle ACS: sinδ1 / S' = sinθ1 / l1 sinδ1 = S' / l1 x sin θ1 ----------- (1) For very small value of δ1: sinδ1 = δ1 (in radians) = (δ1 x 180) / π = (δ1 x 180 x 60 x 60) / π (seconds) = 206265 x δ1 (seconds)

  16. Equation (1) can be written as: δ1 = (S' x sinθ1) / l1 (in radians) δ1 = (206265 x S' x sinθ1) / l1 (in seconds) Similarly: δ2 = (206265 x S' x sinθ2) / l2 (in seconds) BSC = θ and BAC =? Since; α + β + θ = (α + δ1) + (β + δ2) + BAC. Therefore; BAC = θ - (δ1 + δ2)

  17. PROBLEM A, B & C were stations of a minor triangle ABC; C not being suitable for an instrument. A satellite station ‘S’ was therefore set up outside the triangle ABC in order to determine the angle at C. The distance of ‘S’ from C was 17.00 m. The length of AC = 16479 m and of BC = 21726 m. The angle SAC was found to be 63o 48' 00'' and the angle ASB 71o 54' 32''. Calculate angle ACB?

  18. Solution: Given Data: S' = 17.00 m Angle ASC = θ1 = 63o 48' 00'' Angle ASB = 71o 54' 32'' θ2 = CSB = ASB – θ = 71o 54' 32'' – 63o 48' 00'' θ2 = 08o 06' 32'' AC = l1 = 16479 m BC = l2 = 21726 m To Determine: Angle ACB =?

  19. Calculations: Using the relations: sinδ1 = (S' / l1) x sin θ1 = [17 x sin (63o 48' 00'')] / 16479 = sin-1[9.256 x 10-4] radians δ1 = 0o 3' 10.92'' degrees Similarly: Sinδ2 = (S' / l2) x sin θ2 = [17 x sin (08o 06' 32'')] / 21726 = sin-1[1.1037 x 10-4] radians δ2 = 0o 0' 22.77'' degrees

  20. Now: Since, angle ASB = θ - (δ1 + δ2) We have to find θ i.e., angle ACB. Angle ACB = Angle ASB + (δ1 + δ2) = 71o 54' 32'' + (0o 3' 10.92'' + 0o 0' 22.77'') = 71o 58' 5.69'' (Answer)

  21. Problem In a triangle ABC, was a church spire & could not be occupied. A satellite station was selected at 12.10 m from ‘C’ and inside the triangle ABC. From ‘S’ angle CSA = 135o 40' 30'' and ASB = 71o 29' 30'' were measured. And the lengths AC and BC were known to be approximately 2511 m and 1894 m respectively. Compute the angle ACB?

  22. SOLUTION Given Data: S' = 12.10 m Angle ASC = θ1 = 135o 40' 30'' θ = 71o 29' 30'' Angle BSC = θ2 = 360o (θ1 + θ) = 152o 50' 00'' AC = l1 = 2511 m BC = l2 = 1894 m To Determine: Angle ACB =?

  23. Calculations: Using the relations: sinδ1 = (S' / l1) x sin θ1 = [12.10 x sin (135o 40' 30'')] / 2511 = sin-1[3.367 x 10-3] radians δ1 = 0o 11' 34.5'' degrees

  24. Similarly: Sinδ2 = (S' / l2) x sin θ2 = [12.10 x sin (152o 50' 00'')] / 1894 = sin-1[2.916 x 10-3] radians δ2 = 0o 10' 1.66'' degrees

  25. Now: Since, Angle ACB = θ - (δ1 + δ2) We have to find ACB, so Angle ACB = 71o 29' 30'' - (0o 11' 34.5'' + 0o 10' 1.66'') = 71o 07' 53.84'' (Answer)

  26. Thanks

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