Module 2 Tacheometry€¦ · Anallactic lens: A concave lens specially provided in a telescope between the object lens and eye piece to eliminate the additive constant (f + d) from

Module – 2 Tacheometry

G. Ravindra Kumar, Associate Prof, CED, Govt Engg College, Chamarajanagar Page 1 of 23

Tacheometry:

Basic principle, types of tacheometry, distance equation for horizontal and

inclined line of sight in fixed hair method, problems.

Tacheometry:

Tacheometric is also called

tachemetry or telemetry. It is the branch of

surveying in which both horizontal and

vertical distances between stations are

determined from instrumental observations.

Tacheometric is also called tacheometry. The

method is rapid and convenient. Although

the accuracy of tacheometry does not

compare favourable with that of chaining but

it is best adapted in obstacles such as steep

ground, deep ravines, stretches of water which make chaining difficult.

The instrument employed for tacheometric purpose, is

generally know as a tacheometer which is similar to theodolite

having diaphragm fitted with two additional horizontal wires, called

stadia hairs.

Situation where tacheometry can be used:

When obstacles like river, broken ground, stretches of water,

tacheometer gives speed and accuracy to work.

In rough country where measurement of horizontal and vertical distances are

difficult inaccurate and slow.

In locating contours and filling details in a topographic survey, this method is

fast and best.

Purpose:

The primary project of tacheometry is the preparation of contoured plans. It

is considered to be rapid and acurate in rough country and has thus been widely

used by engineers in location surveys for railways, canals, reservoirs, etc. Whenever

surveys of higher accuracy are carried out, tacheometer provide a good check on

distances measured with a tape or a chain.

Difference between Theodolite and Tacheometer

Theodolite Tacheometer

It is used for measurement of horizontal and vertical angle.

It is used for measurement of horizontal and vertical distances.

In Theodolite survey distances are measurement by chain or tape.

In Tacheometer survey direct measurement of distances are possible.

Suitable for plane and hilly area with less obstacles.

Suitable in case obstacles like never broken ground.

More statices are required in Theodolite survey

Less statices are required in Tacheometer survey

Diaphragms

Stadia hair

Theodolite



Systems of tacheometric measurement:

The various systems of tacheometric measurements may be classified as

under:

1. The stadia hair systems

2. The tangential systems

The Stadia hair systems: the stadia hair system may further be divided into two:

1. Fixed hair method 2. Moveable hair method.

Fixed hair method: In this, stadia hairs are kept at fixed interval and the intercept

on the levelling staff (or stadia rod) varies, depending upon the horizontal distance

between the instrument station and the staff. The intercept used in computation is

deduced by subtracting the lower stadia reading from the upper stadia reading.

When the staff intercept is more than the length of the staff, only half intercept is

read, which is equal to the difference between central stadia hair reading and the

lower/upper stadia hair reading.

This method can be suitable employed even when horizontal sight are not

possible. For inclined sights, reading may be taken by holding the staff either vertical

or normal to the line of sight. This is most common method of tacheometry and the

name stadia hair method generally refers to this method.

Movable hair method: In this method, the intercept on the levelling staff is

kept constant and the distance between the stadia hairs are variable. Targets on the

staff at a known distance apart are fixed and the stadia hairs are adjusted such that

the upper hairs bisects the upper target and the lower hair bisects the lower target,

In this case a provision is made for the measurement of the variable interval

between the stadia hairs. For inclined sights, readings may be taken by holding the

staff either vertical or normal to the line of sight as in the case of fixed hair method.

2. The tangential method: In this method, the stadia hairs are not used.

Readings on a staff are taken against the horizontal cross hair. To measure the

staff intercept, two pointing of the telescope are therefore, necessary. Full metre

value readings are taken to avoid the decimal part and also for simplification of

computations. This method is not adopted as two vertical angles are required to

be measured for one single observation.

Dist

anc

e

for

mul

a

for

hori

zon

tal sight:

A b

i

u

C F

f B

v

a

c

d

B’

A’

O S

D



Assuming ‘O’ is the optical centre of the object lens of the external focusing

telescope: a, b, c represent the three horizontal hairs. A, B, C represent respective

points on the staff which appear cut by three hairs, ‘ab’ is the length of the image on

the staff intercept AB.

Let, f = Focal length of the object lens

i = Stadia hair interval ab

S = staff intercept AB

D = Horizontal distance from the axis of the theodolite to staff

d = The distance between the optical centre of the object glass and the axis

of the theodolite.

With the basic knowledge of optics, it is clear that the rays from A and B which pass

through the exterior principal focus of the objective F, travel parallel to the principal

axis after refraction at A’ and B’.

Proof:

A’B’ = ab = stadia hair distance = i

From similar triangles ABF and A’B’F

B'A'

ABOF CF

B'A'

OF

AB

CF

AB = s, OF = f, A’B’ =i.

si

f CF

But D = CF + f + d = d f si

f

formula equation Distance icTacheometrd f si

f D

formula equation Distance icTacheometrC sK D

K and C are Tacheometric constants, K is multiplying constant and C = Additive

constant.

Determination of tacheometric constants (K and C) :

A 25 m

S4

0.924

S3 S2 S1

50 m 75 m 100 m

B

25 m 25 m 25 m 25 m



1. Measure a line AB 100 meters long on a fairly level ground and fix pegs at

25m intervals.

2. Set up the instrument at A and centre it over the ground point accurately.

3. Obtain the staff intercepts s1, s2, s3 and s4 by taking stadia reading on a staff

held vertically at each peg, keeping telescope horizontal by setting the vertical

circle verniers to read zero.

4. Substitute the different values of D and s in the tacheometric distance

formula, i.e. D= k s + C, to get four quadratic equation.

5. Solve the quadratic equations in pairs to get the values of tacheometric

constants.

6. Mean values are the required values of the constants.

d f S i

f C S K D

f = focal length of the object lens = distance between the object lens and the

plane of cross hairs,

d = distance between the object lens and the vertical axis of the theodolite.

i = Stadia interval.

Problem: (A.M.I.E, 1972 summer)

Determine the constants of a tacheometer from the following taken with it:

Distance of staff from the tacheometer vertical axis

Reading against stadia wires

Lower wire Upper wire

30m 1.086m 1.383m

60m 0.924m 1.521m

Solution:

Staff intercept (S1) = THR – BHR = 1.383 – 1.086 = 0.297m

Staff intercept (S2) = THR – BHR = 1.521 – 0.924 = 0.597m

D1 = KS1 + C ---- (1), D2 = KS2 + C ----(2)

S1

1.383

1.086

1.521

0.924

S2

30 m

60 m

THR = Top hair reading

MHR = Middle hair reading

BHR = Bottom hair reading



Subtracting eq (1) from eq (2)

0.3 K 30

C 0.297 K 30

C 0.597 K 60

100 0.3

30 K

C = 0.3

Problem: ®

Two distances of 20 m and 100 m were accurately measured out and the

intercepts on the staff between the outer stadia webs were 0.196 m at the former

distance and 0.996 at the latter. Calculate the tacheometric constants:

Solution:

Staff intercept (S1) = 0.196 m

Staff intercept (S2) = 0.996 m

D1 = KS1 + C ---- (1), D2 = KS2 + C ----(2)

Subtracting Eq (1) from Eq (2)

0.8 K 80

C 0.196 K 20

C 0.996 K 100

100 0.8

80 K

0.4 0.996 100 100 C

C 0.996 K 100

June 2012 / 10CV44 – 08 marks: ®

Sighted horizontally, tacheometric reads 1.645 and 2.840 corresponding to

stadia wires, are vertical staff 120 m away. The focal length of the object glass is

200 mm and the distance from the object glass to the trunnion axis is 150 mm,

Calculate the stadia interval.

Solution:

Given: D = 120m, S = 2.840 – 1.645 = 1.195 m, f = 200 mm = 0.2 m,

d = 150 mm = 0.15m. Stadia interval = i.

S1 =0.196 S2 = 0.996

20 m

100 m



d f S i

f C S K D

0.15 0.2 1.195 i

0.2 120

0.35 i

0.239 120

mm 1.99 m 10 x 1.99 0.35 - 120

0.239i interval Stadia 3-

Problem: ®

The stadia reading with horizontal sight on a vertical staff held 50m away

from a tacheometer were 1.284 and 1.780. the focal length of object glass way

25cm. The distance between the object glass trunnion axis of the tacheometer was

15cm.Calculate the stadia interval (A.M.I.E, 1981 Winter).

Solution:

Given: D = 50 m, S = 1.780 – 1.284 = 0.496 m, f = 25 cm = 0.25 m,

d = 15 cm = 0.15 m. Stadia interval = i.

d f S i

f C S K D

0.15 0.25 0.496 i

0.25 50

0.40 i

0.124 50

mm 2.5 m 10 x 2.5 0.40 - 50

0.124i interval Stadia 3-

Anallactic lens:

A concave lens specially provided in a telescope between the object lens and eye

piece to eliminate the additive constant (f + d) from the tacheometric distance

equations is known as an anallactic lens. It is fitted in external focussing telescopes

only.

Advantages of an anallactic lens:

The main advantage of an anallactic lens are:

i) By the introduction of anallactic lens the calculation of distances and heights, is

very much simplified. If the multiplying constant is 100 and additive constant is

0, the horizontal distance is obtained by simply multiplying the staff intercept

by 100.

ii) The anallactic lens is sealed against moisture or dust.

iii) The loss of sight can be compensated for by the use of slightly larger object

glass.



Disadvantages of an anallactic lens:

i) It increases absorption of light, decreasing the brightness of image.

ii) It also adds to the initial cost of manufacturing the telescope.

iii) It cannot be cleaned easily.

iv) In case of adjustable, it is a potential source of error unless proper field check

is made peridodically.

Distance and Elevation formulae for staff vertical: Inclined sight

Angle of elevation:

θ cosC θKScos (D) distance Horizontal 2

θ sinC 2

2θ sinKS (V) distance Vertical

S = Staff intercept = Top hair reading – Bottom hair reading

i) If Height of instrument axis is known

Elevation at B = Height of Instrument axis at A + V – r

ii) If a reading is taken on BM,

Height of Instrument axis = RL of BM + staff reading on BM (S1)

Elevation at B = Height of Instrument axis at A + V – r

iii) If the Elevation of instrument station at A is known

Elevation at B = RL at station A + h + V – r

Note:

If anallactic lens is fitted, C = 0

θKScos (D) distance Horizontal 2

THR = Top hair reading

MHR = Middle hair reading

BHR = Bottom hair reading

MHR

S1

BM

THR

Inst station

(A) D

MHR

BHR

V

r

S

B

Inst axis



2


Elevation of the staff station for angle of depression:

θ cosC θKScos (D) distance Horizontal 2

θ sinC 2


S = Staff intercept = Top hair reading – Bottom hair reading

i) If Height of instrument axis is known

Elevation at B = Height of Instrument axis at A – V – r

ii) If a reading is taken on BM,


Elevation at B = Height of Instrument axis at A – V – r

iii) If the Elevation of instrument station at A is known

Elevation at B = RL at station A + h – V – r

Note:

If anallactic lens is fitted, C = 0

θKScos (D) distance Horizontal 2

2


Dec 2014/ Jan 2015 /10CV44 – 10 marks: ®

BM Inst station

(A)

r BHR

MHR

B

D

S1

THR

V

S MHR



A tacheometer was setup at a station ‘A’ and the readings on a

vertically held staff at B were 2.255, 2.605 and 2.955. the line of sight being at

a inclination of +80 24’. Another observations on the vertically held staff at B.M gave

the readings 1.640, 1.920, and 2.200, the inclination of the line of sight being +10 6’.

Calculate the horizontal distance between A and B, and the elevation of B if the RL

of BM is 418.685 metres. The constants of the instruments were 100 and 0.3.

Solution:

i) Observation to BM:

K = 100, S = 2.200 – 1.640 = 0.560, C = 0.3, =10 6’ ,

central hair reading r1 = 1.920 m

θ sinC 2

2θ sinKS )(V distance Vertical 1

m 1.081 61 sin0.3 2

612 sin 0.560 100 )(V distance Vertical 0

0

1

Elevation of collimation at the instrument = R L of BM + r1 – V1

Elevation of collimation at the instrument = 418.685 + 1.920 – 1.081 = 419.524m.

ii) Observation to B:

K = 100, S = 2.955 – 2.255 = 0.700, C = 0.3, =80 24’


m 68.80 428 cos0.3 428cos 0.700 100 D θ cosC θKScos (D) B and A b/w distance Horizontal

002

2

θ sinC 2


80 24’

2.955

Inst station

(A) D

V1

2.605

2.255

10 6’

V2

r2

1.640

1.920

2.200

0.56

0.70

BM =

418.685

B

r1



m 10.160 42 8 sin0.3 2

42 82 sin 0.700 100 )(V distance Vertical 0

0

2

R L of B = Elevation of collimation at the instrument + V2 – r2

R L of B = 419.524 + 10.160 – 2.605 = 427.079 m.

June/July 2014/ 10CV44 – 12 marks ®

A tacheometer was set up at station A and the following readings were obtained on

a vertically held staff:

Station Staff station Vertical angle Cross–hair readings in m Remarks

A

B.M -20 18’ 3.225, 3.55, 3.875 RL of B.M.= 437.655 m

B +80 36’ 1.650, 2.515, 3.380

Calculate the horizontal distance from A to B and the RL of B if the constants of the

instrument were k = 100 and C = 0.4.

Solution:

i) Observation to BM:

K = 100, S = 3.875 – 3.225 = 0.650, C = 0.4, =20 18’ ,


θ sinC 2


80 36’

3.380

Inst station

(A) D

V1

B

2.515

1.650

20 18’

V2

r1

r2

3.225

3.875 3.550

0.65

1.73

BM = 437.655



m 2.625 18'2 sin0.4

2

18'22 sin 0.650 100 )(V distance Vertical 0

0

1

Elevation of collimation at the instrument station at A = R L of BM + r1 + V1

Elevation of collimation at the instrument station at A = 437.655 +3.550+2.625 =

443.830 m.

ii) Observation to B:

K = 100, S = 3.380 – 1.650 = 1.730, C = 0.4, =80 36’

Central hair reading r2 = 2.515 m

m 169.52 36' 8 cos0.4 36' 8cos 1.730 100 D θ cosC θKScos (D) B and A b/w distance Horizontal

002

2

θ sinC 2


m 25.64 36' 8 sin0.4

2

36' 82 sin 1.730 100 )(V distance Vertical 0

0

2

R

L of B = Elevation of collimation at the instrument at A + V2 – r2

R L of B = 443.830 + 25.64 – 2.515 = 466.955 m.

Problem ®

A tacheometer is set up at an intermediate point on a traverse course PQ and the

following observations are made on a vertically held staff:

Staff station Vertical angle Staff intercept Axial hair readings

P – 60 20’ 2.46 1.675

Q + 40 20’ 1.86 1.880

The instrument is fitted with an anallactic lens and the constant is 100. Find the

gradient of the line joining station P and Q.

Solution: Let the instrument station be ‘O’

40 20’

Inst station

(O) D2

V1

Q

1.88

60 20’

V2

r1

r2

1.675

2.46

1.86

P D1

1



i) Observation from instrument station O to P:

K = 100, S = 2.46, C = 0,( fitted with anallactic lens), =60 20’ ,


m 243 20' 6cos 2.46 100 D θ cosC θKScos )(D P and O b/w distance Horizontal

021

21

θ sinC 2


m 26.971

2

20'62 sin 2.46 100 )(V distance Vertical

0

1

ii) Observation from instrument station O to Q:

K = 100, S = 1.86, C = 0, =40 20’


m 14.014 20' 4cos 1.860 100 D θ cosC θKScos )(D Q and O b/w distance Horizontal

02

22

θ sinC 2


m 14.014

2

20' 42 sin 1.860 100 )(V distance Vertical

0

2

Vertical difference between PQ = V = r1 + V1 + V2 – r2

Vertical difference between PQ = V = 1.675 + 26.971 + 14.014 – 1.88

Vertical difference between PQ = V = 40.78 m

Horizontal distance between PQ = V = D1 + D2 = 243 +184.94 = 427.94 m

10.49

1

40.78

427.94

1

level in Difference

PQ b/w Distance

1PQ ofGradient

PQ b/w Distance

level in Difference PQ ofGradient

D1 =243m

1

40 20’

Inst station

(O) D2 = 184.94m

V1

Q

r2 =1.88

60 20’

V2

r1 =1.675

V

P

P

Q

D = 427.94m

V = 40.78m

10.49

1



June/July 2013/10CV44 – 10 marks: ®

The following are the observations taken from a tacheometer station, Find the

Gradient of line AB. Tacheometric constant 100 and 0.3.

Inst Station Staff station Bearing Vertical angle

Staff intercept

Axial hair reading

P A 400 35’ - 40 24’ 2.175 1.965

P B 1170 05’ - 50 12’ 1.985 1.865

Solution:

i) Observation from P to A:

K = 100, S = 1.965, C = 0.3, =40 24’, central hair reading r1 = 1.965 m

m 195.64 42 4 cos0.3 42 4cos 1.965 100 D θ cosC θKscos )(D A and P b/w distance Horizontal

0021

21

θ sinC 2


m 15.05 24' 4 sin0.3

2

24'42 sin 1.965 100 )(V distance Vertical 0

0

1

Elevation of collimation at the instrument at P = R L of staff station A + r1 + V1

Assuming, RL of staff station A = 100.000 m

A

Inst station

(P)

B D1

V1

40 24’

1.965 2.175

r1

50 12’

1.865

V2

r2

1.985

D2



Elevation of collimation at the instrument at P = 100 + 1.965 +15.05

=117.015 m.

ii) Observation from P to B:

K = 100, S = 1.985, C = 0.3, =50 12’,


m 197.17 12' 5 cos0.312' 5cos 1.985 100 D θ cosC θKScos )(D B and P b/w distance Horizontal

0022

22

θ sinC 2


m 17.94 12' sin50.3

2

12' 52 sin 1.985 100 )(V distance Vertical 0

0

2

R L of B = Elevation of collimation at the instrument at P – V2 – r2

R L of B = 117.015 – 17.94 – 1.865 = 97.21 m.

Difference in level between A and B=100.00 – 97.21=2.79 m

(B being at lower level)

Gradient b/w P and Q:

Bearing of PA = 400 35’

Bearing of PB = 1170 05’

Included angle APB = Bearing of PB - Bearing of PA

Included angle APB = 1170 05’ - 400 35’ = 760 30’ ()

Applying cosine rule,

θ cosPBPA2PB PA AB 222

m 243.19m 59,141.08 AB

m 59,141.08 30' 76 cos197.17 195.642197.17 195.64 AB

2

20222

87.16

1

2.79

243.19

1

level in Difference

ABb/w Distance

1

ABb/w Distance

level in Difference ABofGradient

Gradient of AB = 1 in 87.16 (Fall from A to B)

Problem: ®

Determine the gradient from a point P to another point Q from the following

observations made with a tacheometer fitted with an anallactic lens. The constant of

the instrument was 100 and the staff was held vertically.

Inst Station Staff station Bearing Vertical angle

Staff readings

R P 1300 + 100 32’ 1.255, 1.810, 2.365

Q 2200 + 50 06’ 1.300, 2.120, 2.940

Solution:

V1

1.255

1.810

2.365

1.11

P

r1 2.120 2.940

1.300

V2 r2

1.640

1170 35’

B

P

N 400 35’

A

760 30’

195.64

197.17



i) Observation from R to P:

K = 100, S = 2.385 – 1.255 = 1.110, C = 0 (anallactic lens)

=100 32’ , central hair reading r1 = 1.810m

m 107.29 23 01cos 1.110 100 D θ cosC θKScos )(D P and R b/w distance Horizontal

021

21

θ sinC 2


m 19.95

2

23102 sin 1.110 100 )(V distance Vertical

0

1

Elevation of collimation at the instrument at R = R L of staff station P + r1 – V1

Assuming, RL of staff station P = 100.000 m

Elevation of collimation at the instrument at R = 100 + 1.810 – 19.95 = 81.86 m.

ii) Observation from R to Q:

K = 100, S = 2.940 – 1.300 = 1.64, C = 0, =50 06’,


m 162.06 60 5cos 1.64 100 D θ cosC θKScos )(D Q and R b/w distance Horizontal

022

22

θ sinC 2


m 14.52

2

60 52 sin 1.64 100 )(V distance Vertical

0

2

R L of Q = Elevation of collimation at the instrument at R + V2 – r2

R L of Q = 81.86 + 14.52 – 2.120 = 94.26 m.

Difference in level between P and Q=100.00 - 94.26=5.74m (P being at higher level)


Bearing of RQ = 2200

Bearing of RP = 1300

Included angle PRQ = Bearing of RQ - Bearing of RP 2200’

R

900

N

1300’

162.06 P

107.29



Included angle PRQ = 2200 - 1300 = 900

Since it’s a right angled triangle,

m 194.36 162.06 107.29 RQ RP PQ 2222

33.86

1

5.74

194.36

level in Difference

PQ b/w Distance

1

PQ b/w Distance


1

Falling Gradient from P to Q = 1 in 33.86 (since P is at higher level)

Dec 2013/Jan 14/10CV44 – 08 marks: ®

The following observations were made using a tacheometer fitted with an anallactic

lens having the constant to be 100 and the staff held vertical.

Inst station

Height of Inst

Staff station

WCB Vertical angle

Hair Readings Remarks

O 1.550

A 300 30’ 40 30’ 1.155, 1.755, 2.355,

R.L of O =150.000 m B 750 30’ 100 10’ 1.250,2.000, 2.750

Calculate (i) horizontal distance AB (ii) RL of A and B (iii) Gradient from A to B

Solution:

i) Observation from O to A:

K = 100, S = 2.355 – 1.155 = 1.20, C = 0 (anallactic lens)

=40 30’ , central hair reading r1 = 1.755 m

30' 4cos 1.20 100 D θ cosC θKScos )(D A and O b/w distance Horizontal

021

21

m 119.26

θ sinC 2


V2

100 15’

1.250

2.000

2.750

1.25

0

B

r2

40 30’

1.755

Inst station (O)

150.000

2.355

1.155

V1 r1

1.200

A

D2 D1

1.550



m 9.386

2

30'42 sin 1.200 100 )(V distance Vertical

0

1

RL of station at A = RL of O + height of inst + V1 – r1

RL of station at A = 150.000 + 1.550 + 9.386 – 1.755 = 159.181 m

ii) Observation from O to B:

K = 100, S = 2.750 – 1.250 = 1.50, C = 0, =100 15’,


15' 10cos1.50 100 D θ cosC θKScos )(D B and O b/w distance Horizontal

022

22

m 145.25

θ sinC 2


m 26.27

2


0

2

RL of station at B = RL of O + height of inst + V2 – r2

RL of station at B =150.000 +1.550 + 26.27 –2.000 =175.82 m (B is at higher

level)

Difference in level between A and B =175.82 -159.181=16.639 m (B being at higher

level)

Horizontal distance and Gradient b/w A and B:

Bearing of OA = 300 30’

Bearing of OB = 750 30’

Included angle = Bearing of OB - Bearing of OA

Included angle AOB = 750 30’ - 300 30’ = 450


θ cosOBOA2OB OA AB 222

m 117.21

2

20222

m 13738.06 AB

m 13738.06 45 cos145.25159.1812145.25 159.181 AB

7.044

1

16.639

117.21

1

level in Difference

ABb/w Distance

1

ABb/w Distance

level in Difference B to A fromGradient

Raising Gradient from A to B = 1 in 7.044 (Since B is at higher level)

300 30’

O

A

450 N

145.25 B

159.181

750 30’



Dec 2012 /10CV44 – 10 marks: ®

Following observations were taken with a tachemeter fitted with an anallactic lens

having value of constant to be 100 and staff held vertical.

Inst Station Staff station Bearing Vertical angle Staff readings

O P N 370 W 40 12’ 0.910, 1.510, 2.110

Q N 230 E 50 42’ 1.855, 2.705, 3.555

Determine the Gradient between points P and Q.

Solution:

i) Observation from O to P:

K = 100, s = 2.110 – 0.910 = 1.200, C = 0 (anallactic lens)


m 119.36 21 4cos 1.200 100 D θ cosC θKscos )(D P and O b/w distance Horizontal

021

21

θ sinC 2

2θ sinKs )(V distance Vertical 1

m 8.76

2


0

1

V1

40 12’

0.910

1.510

2.110

1.20

P

r1

50 42’

2.705

Inst station

(O)

3.555

1.855

V2 r2

1.70

Q

D1 D2



Elevation of collimation at the instrument at O = R L of staff station

P + r1 – V1

Assuming, RL of staff station P = 100.000 m

Elevation of collimation at the instrument at O = 100 + 1.510 – 8.76 = 92.75 m.

ii) Observation from O to Q:

K = 100, s = 3.555 – 1.855 = 1.70, C = 0, =50 42’,


m 168.32 24 5cos 1.70 100 D θ cosC θKscos )(D Q and O b/w distance Horizontal

022

22

θ sinC 2


m 16.80

2


0

2

R L of Q = Elevation of collimation at the instrument at R + V2 – r2

R L of Q = 92.75 + 16.80 – 2.705 = 106.845 m.

Difference in level between P and Q=106.845–100=6.845m (Q being at higher level)


Bearing of P = N 370 W

Bearing of Q = N 230 E

Included angle POQ = Bearing of P + Bearing of Q

Included angle POQ () = 370 + 230 = 600


θ cosOQOP2OQ OP PQ 222

m 149.96m 22,487.7 PQ

m 22,487.7 60 cos168.32119.362168.32 119.36 PQ

2

20222

21.91

1

6.845

149.96

1

level in Difference

PQ b/w Distance

1

PQ b/w Distance


Raising Gradient from P to Q = 1 in 21.91 (since Q is at higher level).

Problem:

The elevation of a point P is to be determined by observation from two adjacent

stations of a tacheometric survey. The staff was held vertically upon the point and

the instrument is fitted with an anallactic lens, the constant of the instrument being

100. Compute the elevation of the point from the following data.

Inst station

Height of axis

Staff point

Vertical angle

Staff reading Elevation of station

A 1.42 P +20 24’ 1.230, 2.055, 2.880 77.750 m

B 1.40 P -30 36’ 0.785, 1.800, 2.815 97.135 m

Also calculate the distances of A and B from P.

Solution:

230

O

Q

168.32

P

119.36 370

N

E W

S

30 36’

2.815

1.800

V2

2.880

2.055 1.65 2.03

B = 97.135



i) Observation from A to P:

K = 100, s = 2.880 – 1.230 = 1.650, C = 0 (fitted with anallactic lens),


m 164.7 24' 2cos 1.650 100 D θ cosC θKscos )(D P and A b/w distance Horizontal

021

21

θ sinC 2


m 6.903

2


0

1

RL of P = R L of A + height of axis + V1 - r1 = 77.750 +1.42+6.903 – 2.055 =

84.018 m

ii) Observation from B to P:

K = 100, s = 2.815 – 0.785 = 2.03, C = 0, =30 36’, central hair reading r2 =1.8 m

m 202.100 36' 3cos 2.03 100 D θ cosC θKscos )(D P and B b/w distance Horizontal

022

22

θ sinC 2


m 12.73

2


0

2

RL of P = R L of B + height of axis – V2 – r2

RL of P = 97.135 +1.40 – 12.73 – 1.800 = 84.005 m

84.00584.018 P of Elevation Average m 84.0122

1



Distance and Elevation formulae for staff Normal: Inclined

sight

a) Line of sight at an angle of Elevation:

θ sinr θ cos CKS (D) distance Horizontal

θ sin CKS (V) distance Vertical

If the Elevation of the instrument station at A is known:

Elevation of staff station B = Elevation of instrument station at A + h + V – rcos

If a reading is taken on BM,


Elevation at B = Height of Instrument axis at A – V – rcos

Elevation of the staff station for angle of depression:

MHR Inst axis

S1

BM

Inst station

(A) D

V

L cos

r sin

h

r cos

S

B

L r

D

B

MHR Inst axis

S1

BM

Inst station

(A)

V

L cos

r sin

h

r cos

S

L



θ sinr θ cos CKS (D) distance Horizontal

θ sin CKS (V) distance Vertical Elevation of staff station B = Elevation of instrument station at A + h – V – r x cos

If the Elevation of the instrument station at A is known:

Elevation of staff station B = Elevation of instrument station at A + h – V – rcos

If a reading is taken on BM,


Elevation at B = Height of Instrument axis at A – V – rcos

Problem: ®

In a tacheometer survey made with an instrument whose constants are 100

and 0.5, the staff was inclined so as to be normal to the line of sight for each

reading. Two sets of readings were as given below. Calculate the gradient between

the staff stations P and Q and the reduced level of each if that of R is 41.800 m.

Instr Station Height of Inst axis

Staff station Bearing Vertical angle

Stadia Reading

R 1.600

P 850 + 40 30’ 1.000, 1.417, 1.833

Q 1350 - 40 00’ 1.000, 1.657, 2.313

Solution:

K = 100, C =0.5, r1 = 1.417, r2 = 1.657,

S1 = 1.833 – 1.00 = 0.833, S2 = 2.313 – 1.000 = 1.313,

1 = 40 30’, 2 = 4

0 0’, R.L = 41.800m

V1

1.417

V2 40 00’

1.000

1.657

2.313

Q r2 cos

40 30’

Inst station

(R)41.800

1.833

1.000

1.60

P

D2

D1

r1 cos



Sight from R to P:

m 83.65 30' 4 ins1.417 30'cos40.50.833100 D

θ sinr θ cos CKS )(D distance Horizontal

001

111

m 6.57 30' 4 sin0.50.833100V

θ sin CKS )(V distance Vertical

01

11

Elevation of P:

Elevation of staff station P = Elevation of instrument station at A + h + V1 – r1cos

Elevation of staff station P = 41.800 + 1.60 + 6.57 – 1.417 x cos 40 30’ = 48.56 m.

Sight from R to Q:

m 131.36 4 sin1.657 4 cos0.51.313100 D

θ sinr θ cos CKS )(D distance Horizontal

002

222

m 9.19 4 sin 0.51.313100V

θ sin CKS )(V distance Vertical

02

22

Elevation of Q:

Elevation of staff station Q = Elevation of instrument station at A + h – V1 – r1cos

Elevation of staff station Q = 41.800 + 1.60 – 9.19 – 1.657 x cos 40 = 32.56 m.

Difference in elevation b/w P and Q = 48.56 – 32.56 = 16 m.


Bearing of P = 850

Bearing of Q = 1350

Included angle PRQ = Bearing of RQ – Bearing of RP

Included angle POQ = 1350 – 850 = 500


θ cosRQRP2RQ RP PQ 222

m 100.63m 10126.53 PQ

m 10126.53 50 cos131.3683.652131.36 83.65 PQ

2

20222

6.29

1

16

100.63

1

level in Difference

PQ b/w Distance

1

PQ b/w Distance


Falling Gradient from P to Q = 1 in 6.29 (since P is at higher level).

850

R

83.65 N

E W

S

Q

131.36

P

1350

Documents

Module 2 Tacheometry€¦ · Anallactic lens: A concave lens specially provided in a telescope between the object lens and eye piece to eliminate the additive constant (f + d) from