# TACHEOMETRY

## Transcription

TACHEOMETRY

TACHEOMETRY Unit 1 Tacheometry • Defined as a procedure of obtaining horizontal distances and differences in elevation based on the optical geometry of the instrument employed of the instrument employed • compared to taping and differential leveling, the g, distances and elevations obtained by tacheometric methods are usually of a lower methods are usually of a lower degree of accuracy Uses of Tacheometric Methods Uses of Tacheometric Methods • Check the more accurate taped distances to uncover p errors or mistakes • Determine differences of elevation between points • Carry lines of levels where low order accuracy is required • Measure the lengths of traverse lines M th l th f t li • Compile planimetric and topographic maps • Complete field survey for photographic map Complete field survey for photographic map • Locate details for hydrographic survey THE STADIA METHOD THE STADIA METHOD • this method employs p y the sighting telescope of an instrument in reading a small angle reading a small angle along a vertical plane and in determining the g length which the angle subtends on a graduated rod held graduated rod held vertical on the distant point • the word stadia denoted 600 Greek units, or 184 m 93 cm (606 ft 9 in) by present-day international standards THE STADIA METHOD the term is now applied to the cross hairs and rod used in making measurements, as well as to the method itself the equipment for stadia measurements consists of a telescope with two horizontal hairs called stadia hairs and a graduated rod called a stadia rod p y byy the distances can be measured veryy rapidly stadia method THE STADIA METHOD PRINCIPLE OF THE STADIA Since ab is equal to a’b’, by similar triangles q y g f/i = d/s And d = (f/i)s also D = d + (f + c) D = (f/i)s + C D = Ks + C STADIA CONSTANTS the stadia constant, the distance from the center of principal p focus the instrument to the p the quantity is composed of the the focal length (f), which remains constant, and the distance (c) from the center of the instrument to the center of the objective lens in the older instruments, instruments C varies from about 0.18 0 18 to 0.43 m in present present‐day day surveying instruments, C may be considered 0.30 for external‐focusing telescopes, 0 for internal‐focusing telescopes STADIA INTERVAL FACTOR the ratio f/I is called the stadia interval factor for any given instrument, the value remains constant and depends only on the spacing between the stadia hairs the most common value of K is 100 Sample Problem Sample Problem • Stadia Interval Factor Stadia Interval Factor A theodolite is set up at one end of a level base line 150 0m long The line is marked by stakes at line 150.0m long. The line is marked by stakes at every 30.0m and a stadia rod is held at each stake. The stadia intercept at each location is observed as follows: 0.302, 0.600, 0.899, 1.207, and 1.506 meters, respectively. Compute the stadia interval f factor (K) for each distance and also determine (K) f h di d l d i the average value of K. Solution k=D/s k1 = 30/ 0.302 = 99.3 k2 = 60/ 0.600 = 100.0 60/ 0 600 00 0 k3 = 90/ 0.899 = 100.1 k4 = 120/1.207 = 99.4 k5 150/1 506 = 99.6 99 6 5 = 150/1.506 kave = 99.7 Sample Problem Sample Problem • Horizontal Stadia Sights Horizontal Stadia Sights An automatic level with an internal focusing telescope was set up somewhere at mid‐length telescope was set up somewhere at mid length of of a long‐span steel bridge. The rod readings tabulated below were observed on a stadia rod held successively at the vicinity of the concrete abutments in the southern and northern approaches of the bridge. If the stadia interval h f h b id If h di i l factor of the instrument is 98.5, determine the length of the bridge length of the bridge. Sample Problem Sample Problem Rod Position Rod at Southern Approach Rod at Northern Approach Hair Readings Upper (a) 2.98 m Middle (c) 1.68 m Lower (b) 0.38 m 3 54 m 3.54 2 02 m 2.02 0 49 m 0.49 Solution ss= 2.98 = 2 98 ‐ 0.38 =2.60 0 38 =2 60 Ds= 98.5 * 2.60 = 256.1 m sn= 3.54 ‐ 3 0 9 30 0.49 = 3.05 Dn= 98.5 * 3.05 = 300.4 m D = 256.1 + 300.4 = 556.5 m INCLINED STADIA SIGHTS INCLINED STADIA SIGHTS INCLINED STADIA SIGHTS INCLINED STADIA SIGHTS ID = ks cos α + C (eq 2) ID = ks cos α + C (eq 2) 2 α + C cos α 2 HD = ks cos k C ( 3) (eq 3) VD = ks cos α sin α + C sin α (eq 4) DE = HI + VD ‐ RR Sample Problem Sample Problem The following data were obtained by stadia observations: vertical angle = +9º25’, upper stadia hair and lower stadia hair readings are 2.352m and 0.995m, respectively. The stadia interval factor is known to be 99.0 and C is 0.381m. The height of the instrument above the instrument station (point A) is 1.496m and rod reading is taken at 1.589m. Determine the g following: a) horizontal, vertical, and inclined distances by exact stadia formulas b) elevation of the point sighted (point B) is the elevation of b) elevation of the point sighted (point B) is the elevation of point A is 776.545m. c) difference in elevation between the two points.