TACHEOMETRY

1. TACHEOMETRY

2. OVERVIEW ● Rapid determination of distance and direction from a single instrument setup without the need for a tape has always been a top priority in surveying operations. ●Tacheometry was the traditional procedure by which distances and differences in elevation were determined indirectly using subtended intervals and angles observed with a theodolite on a graduated rod or scale. ● The most common tacheometric method practiced in the United States is stadia surveying, in which an engineer’s theodolite and graduated level rod are used.

3. STADIA METHOD Equipment for stadia measurements consists of: • A telescope with two horizontal cross hairs called stadia hairs. • A graduated rod called a stadia rod. • The process of taking stadia measurements consists of observing, through the tele­scope, the apparent locations of the two stadia hairs on the rod, which is held in a vertical position. • The interval thus determined, called the stadia interval or stadia reading, is a direct function of the distance from instrument to rod. The ratio of distance to stadia interval is 100 for most instruments.

4. STADIA HAIRS AND STADIA RODS a) Sub cross hair b) Stadia hair which go completely across the cross hair ring A conventional rod such as Philadelphia rod is adequate for stadia work where the sights does not exceed 60 m or 200 ft.

5. PRINCIPLES of STADIA Stadia hairs indicated by points a & b. distance between stadia hair is i . Apparent locations of stadia hairs on the rod are A & B, stadia interval is s.

6. CON’T • Rays from a passing through the optical center of the lens 0 and the focal point of the lens Fare brought to a focus at A.Similarly, the reverse is true, so that rays from A that pass throughFand 0 are brought to a focus at a. • Because ab = a/ b/, by similar triangles The horizontal distance from the principle focus to the rod is

7. CON’T In which is a coefficient called stadia intervalfactor Which is usually equal to 100 value For horizontal sight the distance from principle focus to rod is obtained by multiplying stadia interval factor by stadia interval D = Ks + ( f + c )= Ks + C C : distance from centre of instrument to principle focus.

8. CON’T

9. CON’T • In stadia surveying, most sights are inclined, and usually it is desired to find both the horizontal and the vertical distances from instrument to rod. The problem therefore resolves itself into finding the horizontal and vertical projections of an inclined line of sight. For convenience in field operations, the rod is always held vertical. • Figure (b) illustrates an inclined line of a sight. AB BEING the stadia interval on the vertical rod and A/ B/ being the corresponding projection normal to the line of sight. So the length of the inclined line of sight from centre of instrument is: (1)

10. CON’T For all practical purposes, the angles at A/ and B/may be assumed to be 90°. Let AB = s then A/ B/= s sin z = s cosα . Making this substitution in Equation (1), and letting K = f/i, the inclined distance is Di=Ks sinz+C(2a) Di=Ks cos α +C(2b) The horizontal component of this inclined distance is H= Ks sin2z+Csinz(3a) H = Ks cos2α+ Ccos α (3b)

11. CON’T • which is the general equation for determining the horizontal distance from center of instru­ment to rod, when the line of sight is inclined. In Equations (2) and (3), z is the zenith angle and α is the vertical angle measured at the instrument. The vertical component of the inclined distance is V= Ks sinz cosz + C cosz(4a) V= Ks cosα sinα+ C sinα (4b) The equivalents of sin z cos z and cos α sin α are conveniently expressed in terms of double the angles z and α, or V = 0.5 Ks sin2z + Ccosz (5a) V = 0.5KS sin2α+ Csinα(5b)

12. CON’T • which are the general equations for determining the difference in elevation between the center of the instrument and the point where the line of sight cuts the rod. To determine the difference in ground elevations, the height of instrument and the rod reading of the line of sight must be considered. Equations (4) and (5) are known as the stadia formulas for inclined sights.

13. PERMISSIBLE APPROXIMATIONS • vertical angles (α) are less than 3°, Equation (3) for horizontal distances may properly be reduced to the form H = Ks + C (6) This approximation should not be made for vertical angle greater than 20 Owing to unequal refraction and to accidental inclination of the rod, observed stadia intervals in general are slightly too large. To offset the systematic errors from these sources, frequently on surveys of ordinary precision, the constant C is neglected.

14. CON’T • in any ordinary case Eq. (3) may be expressed with sufficient precision, in the form H = Kssin2 z = Ks cos2 α(approximately) (7) Also, Eq. (5) often may be expressed with sufficient precision for ordinary work in the form V = 0.5Ks sin 2z = 0.5Ks sin 2α (approximately)(8)

15. CON’T Equations (7) and (8) are simple in form and most generally are employed. When K is 100, the common practice is to multiply mentally the stadia interval by 100 at the time of observation and record this value in the field notebook. This distance Ks is often called the stadia distance. Therefore, if the stadia interval were 7.375 m, the stadia distance recorded would be 737.5 m.

16. DIFFERENCE IN ELEVATION the instrument is at station A with a height of instrument, or h.i., above A equal to AB and the rod is held in a vertical position at station D. It is desired to determine the difference in elevation between points A and D given stadia observations and the zenith or vertical angles z or α

17. CON’T First, consider the problem when the difference in elevation between points A and D on the ground is desired. This situation occurs in stadia traversing and topographic surveying from a known point. The h.i. = AB, the stadia interval s, the zenith angle z or vertical angle α with the middle cross hair set on E, and the rod reading DE are observed and recorded. The difference in elevation between the telescope and E, which is equal to V, can be calculated by using Eq. (4a,b) or (8). The difference in elevation is Δel.= AB + V –DE or Δel.= h.i. +V- rod reading at D. Therefore, the elevation at D= elev. At A + (AB +V-DE)= h.i. at B+V-DE.

18. CON’T where the difference in elevation is desired between points B and F utilizing an intermediate instrument set up at G.

19. CON’T Assume that a backsight is taken on a rod held at B. The stadia interval, rod reading BC, and αB are recorded. Using these data, VBcan be calculated. Next, a foresight is taken on the rod held at F. The stadia interval, rod reading FE, and vertical angle -αFare observed and recorded so that VFcan be computed. A general expression for difference in elevation between two points B (the backsight) and F (the foresight) is (9)

20. CON’T In which sign on VB is opposite the sign of αB and the sigh of VF corresponds to the sign on αE. Therefore Δel.BF= BC – VB – EF In the revrese direction, if the backsight is on F and the foresight on B then Δel.= EF + VF +VB – BC

21. ERRORS IN STADIA Many of the errors of stadia are those common to all similar operations of measuring horizontal angles and differences in elevation. Sources of error in horizontal and vertical distances computed from observed stadia intervals are as follows: 1- Stadia Interval factor is not that assumed.This condition produces a systematic error in distances, the error being proportional to that in the stadia interval factor. The case is parallel to that of the tape which is too long or too short. 2- Rod is not of standard length. If the spaces on the rod are uniformly too long or too short, a systematic error proportional to the stadia interval is produced in each distance. Errors from this source may be kept within narrow limits if the rod is standardized and corrections for erroneous length are applied to observed stadia intervals.

22. CON’T Except for stadia surveys of more than ordinary precision, errors from this source usually are of no consequence. 3- Incorrect stadia Interval,The stadia interval varies randomly owing to the inability of the instrument operator to observe the stadia interval exactly. In a series of connected observations (such as a traverse) the error may be expected to vary as the square root of the number of sights. 4- Rod is not plumb. This condition produces a small error in the vertical angle. It also produces an appreciable error in the observed stadia interval and hence in computed distances, the error being greater for large vertical angles than for small angles. It can be eliminated by using a rod level.

23. CON’T 5- Unequal refraction.Unequal refraction of light rays in layers of air close to the earth’s surface affects the sight on the lower stadia hair more than the sight on the upper stadia hair and thus introduces systematic positive errors in stadia measurements. Whenever atmospheric conditions are unfavorable, the sights should not be taken near the bottom of the rod. 6- Errors in zenith or vertical angles. Errors in vertical circle readings are relatively unimportant in their effects on horizontal distances. For example, analysis of Eq. (7) shows that an uncertainty of 01’ in a zenith or vertical angle of 85° or 5°, respectively, yields discrepancies of 0.005 m in a 100-m sight

24. CON’T Similarly, an error of 01’ in a zenith or vertical angle of 75° or 15°, respectively, produces discrepancies of 0.02 m in a 100-m sight. With respect to differences in elevation, analysis of Eq. (8) reveals that an uncertainty of 01’ in zenith or vertical angles of 85° or 5°, respectively, results in discrepancies in elevation differences of 0.03 m for a l00-m sight. Uncertainty in the stadia interval, especially at higher angles of inclination, will have a more pronounced effect on elevation differences than errors in zenith or vertical angles.

25. TUTORIAL • EXAMPLE 1. The following data were obtained by stadia observation: zenith angle = 81°50/, stadia interval = 2.50 ft. The stadia interval factor is known to be 100 and distance from centre of instrument to principle focus = 0.75 ft. Calculate horizontal and vertical component of the inclined distance .

26. TUTORIAL • Solution By exact Eqs. 3a & 3b H = Ks sin2z+Csinz = (100)(2.50)(sin281°50/) + (0.75)(sin 81°50/) = (100)(2.50)(0.9798) + (0.75)(0.9899) = 245.0 + 0.7 = 245.7 = 246 ft. V= 0.5Ks sin 2z + C cos z = 0.5(100)(2.50)(0.28l2) + (0.75)(0.1421) = 35.2 + 0.11= 35.3 ft.

27. TUTORIAL If approximate Equations 7 and 8 are used and H=(100)(2.50)(sin2 81°50/) = (251)(0.9798) = 245 ft V = 0.5(100)(2.50) sin(2 x 8l°50/)= 35.2 ft.

28. TUTORIAL • EXAMPLE 2 The stadia interval was 1.372 m at a vertical angle of 20°32/. C = 0.30 m and the stadia interval factor is 100. Calculate H and V.

29. TUTORIAL Solution By Eq. 3b H= (100)(1.372)(cos2 20°32/) + (0.30)(cos 20°32/) = 120.6 m By Eq. 5.b V= 0.5(100)(1.372)(sin2(20o32/) + (0.30)(sin 20°32/) = 45.17 m

30. TUTORIAL • If approximate Equations 7 & 8 are used for this problem, H = 100(1.372) cos2α = 120.32 m V = 0.5(100)(1.372) sin (2x20°32/) = 45.06 m Use of the approximate equations for a solution with the large vertical angle resulted in an error of 1 cm in the difference in elevation. Had the telescope of the instrument been internally focusing, as in most modern transits and theodolites, C would be equal to 0 and approximate Equations 7 & 8 could be used with no danger of error.

31. TUTORIAL EX.3 A theodolite is used in locating points from a station the elevation of which is 233.39 m. The instrument constants are K = 100.0 and C = 0. The height of instrument above the station is 1.37m. Compute the horizontal distances and the elevations.

32. ASSIGNMENT a) A stadia interval of 2.515 m is observed with a theodolite for which the stadia interval factor is 100.0and C is 0.305 m. The zenith angle is 87°20/ 30// with the middle cross hair set on 1.37 m. If the instrument has a height of instrument (h.i.) of 1.52 m above the point over which it is set and the point has an elevation of 100.05 m, calculate the horizontal distance and elevation of the point sighted by the (a) exact stadia equations; (b) approximate stadia equations.