Trigonometric leveling

This is based on the surveying branch.. which shows 3 cases here.. for civil engineering students .. and as well as also who want to know about what is Trigonometric leveling..

1. Trigonometric Leveling Prepared By , Amit Bhoghara (CL 1309) Akash khunt (CL 1342) Siddhant Patel (CL 1343) Arpan Malaviya (CL 1344) Guided By, Hardik Sir

2. What is Trigonometric leveling? Definition: “ Trigonometric levelling is the process of determining the differences of elevations of stations from observed vertical angles and known distances. ” The vertical angles are measured by means of theodolite. The horizontal distances by instrument Relative heights are calculated using trigonometric functions. Note : If the distance between instrument station and object is small. correction for earth's curvature and refraction is not required.

3. Method of Observation Direct Method: Where is not possible to set the instrument over the station whose elevation is to be determined. Combined correction is required.

4. 2) Reciprocal method: In this method the instrument is set on each of the two station, alternatively and observations are taken. L BAC =  & L ABC  = 

5. Distance between A & B is Small AB' = AC = D L ACB = 90 Similarly, BA' = BC' = D L AC'B = 90 BC = D tan  AC' = D tan 

6. Distance between A & B is Large Cc & Cr required CB' = C'A' = 0.0673 D 2 True Difference A-B H=BB ' =BC + CB ' = D tan  + 0.0673 D 2 Depression angle B to A AC '=D tan  [ BC '= D ] True Difference A-B H=AA ' =BC + CB ' = D tan  - 0.0673 D 2 Adding Blue colour equation 2 H = D tan  + D tan  R.L of station B = R.L. of station A + H = R.L. of station A + D/2 [ tan  + tan  ]

7. METHODS OF DETERMINING THE ELEVATION OF A POINT BY THEODOLITE: Case 1. Base of the object accessible Case 2. Base of the object inaccessible, Instrument stations in the vertical plane as the elevated object. Case 3. Base of the object inaccessible, Instrument stations not in the same vertical plane as the elevated object.

8. Case 1. Base of the object accessible B A = Instrument station B = Point to be observed h = Elevation of B from the instrument axis D = Horizontal distance between A and the base of object h1 = Height of instrument (H. I.) Bs = Reading of staff kept on B.M. = Angle of elevation = L BAC h = D tan  R.L. of B = R.L. of B.M. + Bs + h = R.L. of B.M. + Bs + D. tan  If distance is large than add Cc & Cr R.L. of B = R.L. of B.M. + Bs + D. tan  + 0.0673 D 2

9. Case 2. Base of the object inaccessible, Instrument stations in the vertical plane as the elevated object. There may be two cases. Instrument axes at the same level Instrument axes at different levels. 1) Height of instrument axis never to the object is lower: 2) Height of instrument axis to the object is higher:

10. Case 2. Base of the object inaccessible, Instrument stations in the vertical plane as the elevated object. Instrument axes at the same level  PAP, h= D tan 1  PBP, h= (b+D) tan 2 D tan 1 = (b+D) tan 2 D tan 1 = b tan 2 + D tan 2 D(tan 1 - tan 2) = b tan 2 R.L of P = R.L of B.M + Bs + h

11. Instrument axes at different levels. 1) Height of instrument axis never to the object is lower:  PAP, h1 = D tan 1  PBP, h2 = (b+D) tan 2 hd is difference between two height hd = h1 – h2 hd = D tan 1 - (b+D) tan 2 = D tan 1 - b tan 2 -D tan 2 hd = D(tan 1 - tan 2) - b tan 2 hd + b tan 2 = D(tan 1 - tan 2) h1 = D tan 1

12. Instrument axes at different levels. 2) Height of instrument axis to the object is higher:  PAP, h1 = D tan 1  PBP, h2 = (b+D) tan 2 hd is difference between two height hd = h2 – h1 hd = (b+D) tan 2 - D tan 1 = b tan 2 + D tan 2 - D tan 1 hd = b tan 2 + D (tan 2 - tan 1 ) hd - b tan 2 = D(tan 2 - tan 1) - hd + b tan 2 = D(tan 1 - tan 2) h1 = D tan 1

13. In above two case the equations of D and h1 are, D h1

14. Case 3. Base of the object inaccessible, Instrument stations not in the same vertical plane as the elevated object. Set up instrument on A Measure 1 to P L BAC =  Set up instrument on B Measure 2 to P L ABC =  L ACB = 180 – (  +  ) Sin Rule: BC= b · sin  sin{180˚ - (  +  )} AC= b· sin  sin{180˚ - (  +  h1 = AC tan 1 h2 = BC tan 2

15. Thank You..!!..