TACHEOMETRYTACHEOMETRY

Unit 1

Tacheometry• Defined as a procedure of obtaining horizontal distances and differences in elevation based on the optical geometry of the instrument employedof the instrument employed

• compared to taping and differential leveling, the g,distances and elevations obtained by tacheometric methods are usually of a lowermethods are usually of a lower degree of accuracy

Uses of Tacheometric MethodsUses of Tacheometric Methods

• Check the more accurate taped distances to uncover perrors or mistakes

• Determine differences of elevation between points• Carry lines of levels where low order accuracy is requiredM th l th f t li• Measure the lengths of traverse lines

• Compile planimetric and topographic maps• Complete field survey for photographic map• Complete field survey for photographic map• Locate details for hydrographic survey

THE STADIA METHODTHE STADIA METHOD

• this method employs p ythe sighting telescope of an instrument in reading a small anglereading a small angle along a vertical plane and in determining the glength which the angle subtends on a graduated rod held

• the word stadia denoted 600 Greek units, or 184 m 93 cm (606 ft 9 in) bygraduated rod held 

vertical on the distant point

m 93 cm (606 ft 9 in) by present-day international standards

THE STADIA METHODthe term is now applied to the cross hairs and

rod used in making measurements, as well as torod used in making measurements, as well as tothe method itselfthe equipment for stadia measurementsthe equipment for stadia measurements

consists of a telescope with two horizontal hairscalled stadia hairs and a graduated rod called acalled stadia hairs and a graduated rod called astadia roddistances can be measured very rapidly by they p y y

stadia method

THE STADIA METHODTHE STADIA METHOD

PRINCIPLE OF THE STADIA

Since ab is equal to a’b’, by similar trianglesq y gf/i = d/s

And d = (f/i)s

also D = d + (f + c)D = (f/i)s + CD = Ks + C

STADIA CONSTANTSthe stadia constant, the distance from the center of

the instrument to the principal focusp pthe quantity is composed of the the focal length (f),

which remains constant, and the distance (c) from thecenter of the instrument to the center of the objectivelensin the older instruments C varies from about 0 18 toin the older instruments, C varies from about 0.18 to

0.43 min present‐day surveying instruments, C may bein present day surveying instruments, C may be

considered 0.30 for external‐focusing telescopes, 0 forinternal‐focusing telescopes

STADIA INTERVAL FACTORthe ratio f/I is called the stadia interval factorfor any given instrument, the value remainsfor any given instrument, the value remains

constant and depends only on the spacingbetween the stadia hairsbetween the stadia hairsthe most common value of K is 100

Sample ProblemSample Problem

• Stadia Interval FactorStadia Interval FactorA theodolite is set up at one end of a level base line 150 0m long The line is marked by stakes atline 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 (K) f h di d l d ifactor (K) for each distance and also determine the average value of K.

SolutionSolution

k=D/sk=D/sk1 = 30/ 0.302 = 99.3k 60/ 0 600 00 0k2 = 60/ 0.600 = 100.0k3 = 90/ 0.899 = 100.1k4 = 120/1.207 = 99.4k5 = 150/1 506 = 99 6k5    150/1.506   99.6kave = 99.7  

Sample ProblemSample Problem

• Horizontal Stadia SightsHorizontal Stadia SightsAn automatic level with an internal focusing telescope was set up somewhere at mid‐length oftelescope was set up somewhere at mid length 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 

h f h b id If h di i lapproaches of the bridge. If the stadia interval factor of the instrument is 98.5, determine the length of the bridgelength of the bridge.

Sample ProblemSample Problem

Rod Position

Hair Readings

Upper (a) Middle (c) Lower (b)Upper (a) Middle (c) Lower (b)Rod at Southern

2.98 m 1.68 m 0.38 mSouthern ApproachRod at 3 54 m 2 02 m 0 49 mRod at Northern Approach

3.54 m 2.02 m 0.49 m

SolutionSolution

s = 2 98 ‐ 0 38 =2 60ss= 2.98  0.38 =2.60Ds= 98.5 * 2.60 = 256.1 m

3 0 9 3 0sn= 3.54  ‐ 0.49 = 3.05Dn= 98.5 * 3.05 = 300.4 mD = 256.1 + 300.4 = 556.5 m

INCLINED STADIA SIGHTSINCLINED STADIA SIGHTS

INCLINED STADIA SIGHTSINCLINED STADIA SIGHTS

ID = ks cos α + C (eq 2)ID = ks cos α + C (eq 2)

k 2 C ( 3)HD = ks cos2 α + C cos α (eq 3)

VD = ks cos α sin α + C sin α (eq 4)

DE = HI + VD ‐ RR

Sample ProblemSample ProblemThe 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 following:g

a) horizontal, vertical, and inclined distances by exact stadia formulas

b) elevation of the point sighted (point B) is the elevation ofb) elevation of the point sighted (point B) is the elevation of point A is 776.545m.

c) difference in elevation between the two points.