SURVEYING I

UNIT – 5

TACHEOMETRIC SURVEYING

Tacheometric systems – Tangential, Stadia and Subtence methods – Stadia Systems –

Horizontal and Inclined Sights – Vertical and Normal Staffing – Fixed and Movable

Hairs – Stadia Constants – Anallactic Lens – Subtense Bar.

Two Marks Questions Answers

1. Define Tacheometric Surveying.

Techeometry is the branch of angular surveying in which horizontal and

vertical distances of points are obtained by optical methods as opposed to the

slower process of measurement by chain or tape. The method is very rapid and

continent.

2. Explain scope of Tacheometric Surveying.

Although accuracy of tachometric general compares unfavorably with

that of chain or tape it is best adopted in steep hills, deep ravines, across water

bodies, swamps, etc where chaining is impossible or difficult.

3. What is Tangential Tacheometry?

In tangential method the stadia hair is not used, the readings are taken

against the horizontal cross hair. To measure the staff intercept, two points are

noted on the staff along with in their corresponding vertical angles. This

necessitates the measurement of vertical angles twice for one observation.

4. What is Stadia Tacheometry?

Fixed hair method observations are made with the help a stadia

diaphragm having stadia wires at fixed or constant distance apart. The readings

on the staff corresponding to all the three wires are taken. Stadia wires will

therefore depend on the distance between the staff and the instrument station.

5. Explain Subtense Method.

Subtense method is similar to the fixed hair method except that the stadia

interval is variable. Suitable arrangement is made to very the distance between

the stadia hairs so as to set them against the two targets on the staff kept at the

point under observation.

6. What are the merits and demerits of movable hair method?

Advantage of movable hair method is its greater accuracy than stadia

method, since only targets are to be bisected.

A disadvantage is that it has longer time in field for observation and in

computation of distance.

7. Describe the conditions in which tacheometric surveying is advantages.

The primary object of tacheometry is the preparation of contour maps or

plans requiring both vertical and horizontal controls. Also on surveys of higher

accuracy, it is used during preliminary or reconnaissance survey or to check

distances measured using tapes.

8. What are the advantages of an anallactic lens used in a tacheometer?

Provision of anallactic lens forms the vertex at the vertical axis and its

position is always fixed irrespective of the staff position. The anallactic lens is

usually provided in the external focusing telescope and not in the internal

focusing telescope since the later is virtually anallactic due to the very small

additive constant.

9. Show the arrangement of lens in an anallactic telescope.

10. Differentiate stadia and tangential tacheometry.

In stadia method observations are made with the help a stadia diaphragm

having stadia wires at fixed or constant distance apart. The readings on the staff

corresponding to all the three wires are taken.

In tangential method the stadia hair is not used, the readings are taken

against the horizontal cross hair. To measure the staff intercept, two points are

noted on the staff along with in their corresponding vertical angles. This

necessitates the measurement of vertical angles twice for one observation and

has better accuracy compared to stadia tacheometry.

11. Differentiate stadia tacheometry and subtense bar method.

In stadia method observations are made with the help a stadia diaphragm

having stadia wires at fixed or constant distance apart. The readings on the staff

corresponding to all the three wires are taken.

Subtense method is similar to the fixed hair method except that the stadia

interval is variable. Since subtense bar is used staff reading are not required. It is

more accurate than stadia and tangential method but consumes more time for

observation and calculation.

12. Name the essential features of a tacheometer?

A tacheometer must incorporate the following features.

The multiplying constant should have a nominal value of 100 and the

error contained in the value should not exceed 1 in 1000.

The axial horizontal lines should be exactly mid way between the other

two lines.

The telescope should be truly anallactic.

13. What is a Subtense Bar?

subtense bar consist of two targets supported at a fixed distance (2m or

3m) apart on an invar rod with a spirit level mounted on a tripod stand.

14. What are the points to be considered while using a subtense bar?

The points to be considered while using a subtense bar are

The bar should be made of material whose length does not vary with

difference in temperature.

It should be centrally mounted and should be perfectly horizontal.

It should be perpendicular to the line joining the theodolite station.

15. Name the Stadia Tacheometric Constants.

Horizontal distance D = Ks + C

Multiplying constant K = f / i

Additive constant C = f + d

Where,

s is staff intercept

f is focal length of objective

i is interval between stadia hairs

d is distance of vertical axis from instrument station.

16. Name the types of telescopes used in stadia tacheometry.

The telescope used in stadia tacheometry may be

External focusing telescope

External focusing telescope with anallactic lens

Internal focusing telescope.

17. Find multiplying constant of a anallactic telescope if staff intercept for a

distance of 35m is 0.700m?

For stadia tacheometry,

Horizontal distance D = Ks + C

Where M is multiplying constant

s is Staff intercept

C is Additive constant

i.e. 35 = (M x 0.700) + C

Since Telescope is Anallactic, C = 0

Hence Multiplying Constant, M = 35 / 0.700

M = 50

18. What are the different systems of tacheometric survey?

Different systems of tacheometric measurements are

1. The stadia system.

Fixed hair method or stadia method.

Movable hair method or subtense method.

2. The tangential method.

3. Measurements by means of special instruments.

19. Define analytic lens.

The vertex of the measuring triangle falls at the exterior prinicipal focus

of the objective and not at the vertical axis of the instrument. In 1840, porro

devised the external focusing anallactic telescope, the special feature of which is

an additional lens, is called an anallactic lens.

16 Marks Questions And Answers

1. What is tacheometric survey? Explain the instruments used.

Tacheometric survey:

Techeometry is the branch of angular surveying in which horizontal and

vertical distances of points are obtained by optical methods as opposed to the

slower process of measurement by chain or tape. The method is very rapid and

continent. Although accuracy of tacheometric general compares unfavorably

with that of chain or tape it is best adopted in steep hills, deep ravines, across

water bidies, swamps, etc where chaining is impossible or difficult.

The primery object of tacheometry is the preparation of contour maps or

plans requiring both vertical and horizontal controls. Also on surveys of higher

accuracy, it is used during preliminary or reconnaissance survey or to check

distances measured using tapes.

Instruments used:

An ordinary theodolite fitted with a stadia diaphragm is generally used in

tacheometric survey. The stadia diaphragm consists of one stadia hair above and

another stadia hair an equal distance below the cross hair being mounted in the

same vertical plane as the vertical and horizontal cross hair. The telescope used

in stadia tachemetry may be

External focusing telescope

External focusing telescope with anallactic lens

Internal focusing telescope.

An tacheometer must incorporate the following features.

The multiplying constant should have a nominal value of 100 and the

error contained in the value should not exceed 1 in 1000.

The axial horizontal lines should be exactly mid way between the other

two lines.

The telescope should be truly anallactic.

2. Explain the methods employed in tacheometric surveying?

Different systems of tacheometric measurements are

1. The stadia system.

Fixed hair method or stadia method.

Movable hair method or subtense method.

2. The tangential method.

3. Measurements by means of special instruments.

Fixed hair method:

In this method, observations are made with the help a stadia diaphragm

having stadia wires at fixed or constant distance apart. The readings on the staff

corresponding to all the three wires are taken. The staff intercept i.e. is the

difference of readings corresponding to the top and the bottom stadia wires will

therefore depend on the distance between the staff and the instrument station.

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

read. For inclined sights readings may be taken by keeping the staff either

vertical or normal to the line of sight. This is the most common method in

tacheometry.

Subtense Method:

This method is similar to the fixed hair method except that the stadia

interval is variable. Suitable arrangement is made to very the distance between

the stadia hairs so as to set them against the two targets on the staff kept at the

point under observation. Thus in the case the staff intercept i.e., the distance

between the two targets is kept fixed while the stadia interval, i.e., the distance

between the stadia hairs is varied. As in the case of fixed hair method inclined

sights may also be taken.

The Tangential Method:

In this method the stadia hair is not used, the readings are taken against

the horizontal cross hair. To measure the staff intercept, two points are noted on

the staff along with in their corresponding vertical angles. This necessitates the

measurement of vertical angles twice for one observation.

The Anallactic Lens:

The word ‘anallactic’ means ‘unalterable’ or ‘invarible’. By the

provision of anallactic lens the vertex Is formed at the vertical axis and its

position is always fixed irrespective of the staff position. The anallactic lens is

usually provided in the external focusing telescope and not in the internal

focusing telescope since the later is virtually anallactic due to the very small

additive constant.

The Subtense Bar:

For measuring comparatively shorter lines in a traverse a subtense bar

may be used as a subtense base. The subtense bar is usually 2m or 3m long and

is mounted on a tripod. The bar is centrally supported on a leveling head for

accurate centering and leveling. A clamp and slow moving screw is also

provided to rotate the bar about its vertical axis. Either a pair of sights or a small

telescope is provided at the center of bar to align it perpendicular to the line of

sight from the theodolite.

The angle AOB at O is usually measured with a theodolite, preferably by

method of repetition. It should be noted that the difference in elevation between

the theodolite station and the subtence bar station does not affect the magnitude

of the angle AOB.

3. To determine the multiplying constant of a tacheometer, the following

observations were taken on a staff held vertically at distance, measured

from the instrument.

Observation Hori. Distance in “m” Vertical angle Staff intercept

1 50 + 3° 48′ 0.500m2 100 + 1° 06′ 1.000m3 150 + 0° 36′ 1.500m

The focal length of the object glass is 20cm and the distance from the object

glass to trunnion axis is 10cm. The staff is held vertically at all these

points. Find the multiplying constant.

Solution:

Additive constant C = ( f+d ) =0.20 + 0.10 = 0.30m

(1) First observation:

Horizontal distance D = Ks cos2θ + C cosθ

50 = K x 0.500 cos2 3°48’ + 0.30 cos3°48’

K = 99.84

(2) Secound Observation:

100 = K x 1.00 cos2 1°6’ + 0.30 cos1°6’

K = 99.74

(3) Third Observation:

150 = K x 1.50 cos2 0°36’ + 0.30 cos0°36’

K = 99.81

Average value of multiplying constant K = 13 (99.84 + 99.74 + 99.81)

K =99.80

4. To determine the elevation of station P in a tachometer survey, the

following observations were made with the staff held vertical. The

instrument was fitted with an anallactic lens and its multiplying constant

was 100.

Instrument

station

H.I

(m)

Staff

station

Vertical

angle

staff readings (m)

O 1.45 B.M. - 6°00′ 1.335, 1.895, 2.460

O 1.45 C.P. + 8°30′ 0.780, 1.265, 1.745

P 1.40 C.P. - 6°30′ 1.155, 1.615, 2.075

If R.L. of B.M. is 250m, calculate R.L. of P.

Solution:

Instrument at O and staff B.M.

S1 = 2.460 – 1.335 = 1.125m

H.I = 1.450m, K = 100, θ = 6°00′ (depression)

V1 = Ks1

2 Sin2θ1 + C Sinθ1

= 100 x 1.125

2 Sin12°00′ + 0.00 x Sin6°00′

V1 = 11.695m

R.L. of plane of collimation = R.L. of B.M. + h1 + V1

= 250.00 + 1.895 + 11.695

= 263.590m

Instrument at O and staff at C.P.

S2= 1.745 – 0.780 = 0.965m

H.I = 1.450m, θ = 8°30′ (elevation)

V2 = Ks2

2 Sin2θ2 + C Sinθ2

= 100x 0.965

2 Sin17°00′ + 0.00 x Sin8°30′

V2 = 14.100m.

R.L. of C.P = R.L. of B.M. plane of collimation + V2- h2

= 263.590 + 14.106 – 1.265

= 276.431m.

Instrument at P and staff at C.P.

S3= 2.075 – 1.155 = 0.920m

H.I = 1.400m, θ = 6°30′ (depression)

V3 = Ks3

2 Sin2θ3 + C Sinθ3

= 100 X0.920

2 Sin13°00′ + 0.00 x Sin6°30′

V3 = 10.348m.

R.L. of station P = R.L. of C.P. + h3 + V2- H.I

= 276.431 + 1.615 + 10.348 – 1.400

= 286.993m.

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

observations made with a tacheometer fitted with an analytic lens. The

constant of the instrument was 100 and staff was held vertical.

Instrument

station

Staff

station

Bearing Vertical

angle

staff readings (m)

RP 130° + 10°32′ 1.255, 1.810, 2.365

Q 220° + 5°6′ 1.300, 2.120, 2.940

Solution:

Instrument at R and staff at P

θ1 = 10°32′, S1 = 2.365 – 1.255 = 1.110m

Horizontal distance at RP = D= KS1 cos2θ1 + C cosθ1

= 100 x 1.110 x cos210°32′ + 0

=107.290m.

V1 = Ks1

2 Sin2θ1 + C Sinθ1

= 100 x 1.110

2 sin21°4′ + 0

=19.950m.

(Alternatively, V = D tanθ1 = 107.290 x tan10°32′ = 19.950m)

Let the R.L. of R be X

Then R.L. of P =R.L. of R + H.I + V – h

= X + 19.950 – 1.810

= X +18.140

Instrument at R and staff at Q

θ2 = 5°6′, S2 = 2.940 – 1.300 = 1.640m

Horizontal distance at RQ = D= KS2 cos2θ2 + C cosθ2

= 100 x 1.640 x cos25°6′ + 0

= 162.70m.

V2 = Ks2

2 Sin2θ2 + C Sinθ2

= 100x 1.640

2 sin10°12′ + 0

= 14.520m.

R.L. of P =R.L. of line of collimation + V – h

= X + 14.520 – 2.120

= X +12.400

Difference of level between P and Q

= (X + 12.400) – (X + 18.140)

= 5.740m fall

Distance PQ

Angle QRP = bearing of RQ – bearing of RP

= 220° - 130°

= 90°

PQ2 = 107.2902 + 162.7002

= 194.890m.

Gradient = 5.740 / 194.890

= 1 in 33.950m fall.

6. A tacheometric 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 +8°36′ 2.350 2.105

Q +6°06′ 2.055 1.895

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

Compute the length of PQ and reduced level of Q, that of p being 321.50m.

Solution:

(a) Observation from the instrument to P:

θ = 8°36′, S = 2.350m

Horizontal distance at OP = D= KS cos2θ + C cosθ

= 100 x 2.350 x cos28° 36′ + 0

= 229.750m

V = Ks2 Sin2θ + C Sinθ

= 100 x 2.350

2 sin17°12’+ 0

= 34.745m

Difference in elevation between P and the instrument axis

= 34.745 – 2.105

= 32.640m (P being higher)

(b) Observation from the instrument to Q:

θ = 6°6′, S = 2.055m

Horizontal distance at OQ = D= KS cos2θ + C cosθ

= 100 x 2.055 x cos26°6′ + 0

= 203.180m

V = Ks2 Sin2θ + C Sinθ

= 100 x 2.055

2 sin12°12’+ 0

= 21.713m

Difference in elevation between Q and the instrument axis

= 21.713 – 1.895

= 19.818m (Q being higher)

Since the tacheometer is set up at an intermediate point on the line PQ, the

distance PQ = 229.750 + 203.180 = 432.930m.

Difference in elevation of P and Q

= 32.640 – 19.818 = 12.822 (P being higher)

R.L. of Q= R.L. of P – 12.822

= 321.500 – 12.822

= 308.678m.

7. A theodolite has a tacheometric multiplying constant of 100 and an additive

constant of zero. The centre reading on a vertical staff held at point B was

2.29m when sighted from A. if the vertical angle was +25° and the

horizontal distance AB 190.326m, calculates the other staff readings and

show that the two intercept intervals are not equal. Using these values,

calculate the level of B if A is 37.950m angle of depression and the height of

the instrument 1.35m.

Solution:

D = KS cos2θ + C cosθ

S = D / K cos2θ

S = 190.326 / 100 cos225°

S = 2.317m.

Inclined distance MC = L = D secθ = 190.326 x sec25° = 210.002m.

Now 2S0 = L / 100

S0 = L / 200 =210.002 / 200 = 1.050m.

By sine rule, S1 = (S0 Cosβ/2) / (Cos (θ + β/2)

Where β/2 = 0°17’11.35”

S1 = (1.050 x Cos 0°17’11.35”) / Cos25°17’11.35”

S1 = 1.161m

Similarly, by sine rule,

S2 = (S0 Cosβ/2) / (Cos (θ - β/2)

S2 = (1.050 x Cos 0°17’11.35”) / Cos24°42’48.65”

S2 = 1.156m

Hence staff reading are:

Upper : 2.292 + 1.161 = 3.453

Lower : 2.292 – 1.156 = 1.136

Check: S =2.317m

Now V = D tanθ = 190.326 tan25° =88.750m

R.L.of B = 37.950 + 88.750 + 1.350 - 2.292

= 125.758m.

8. Explain how a subtense bar is used with a theodolite to determine the

horizontal distance between two points. The horizontal angle subtended at a

theodolite by a subtense bar with vanes 3m apart is 15’40”. Compute the

distance between the instrument and the bar. Deduce the error of

horizontal distance if the bar 2° from the normal to the line joining the

instrument and bar station.

Solution:

β = 15’40” = 940”

D = (206265 S / β) = (206265 x 3) / 940

D = 658.293m.

The above distance was calculated on the assumption that the bar was

normal to the line of joining the instrument and bar station. If, however, the bar

is not normal, correct horizontal distance is given by

D' = D Cosβ = 658.293 x cos2° = 657.892m.

Error = D' – D = 658.293 – 657.892 = 0.401m.

Ratio of error = e / D' = 0.401 / 657.892 = 1 in 1640.628.

9. A tacheometer was setup at station A and the following readings were

obtained on a vertically held staff.

Station Staff station Vertical angle Hair readingRemarks

AB.M. -2°18′ 3.225, 3.550, 3.875 R.L.of B.M.

is 437.655m B +8° 36′ 1.650, 2.515, 3.380

Calculate the horizontal distance from A to B and the R.L. of B, if

the constants of the instrument were 100 and 0.4.

Solution:

(a) Observation to B.M.:

V = Ks2 Sin2θ + C Sinθ

Here K = 100; C = 0.40m; S = 3.875 – 3.225 = 0.650m; θ = 2°18′

= 100 x 0.650

2 sin4°36’ + 0.40 x Sin2°18′

= 2.606 + 0.016

= 2.622m.

Elevation of collimation at the instrument = 437.655 + 3.550 + 2.622

= 443.827m.

(b) Observation to B.M.:

θ = 8°36′, S = 3.380 - 1.650 = 1.730m

Horizontal distance at AB = D= KS cos2θ + C cosθ

= 100 x 1.730 x cos28°36′+ 0.40 x cos8°36′

= 169.132 + 0.396

= 169.528m

V = Ks2 Sin2θ + C Sinθ

= 100 X1.730

2 sin17°12’ + 0.40 x Sin8°36′

= 9.611 + 0.060

= 9.671m.

R.L.of B = 443.827 + 9.671 - 2.515

= 450.983m.

10. How to you calculate the horizontal and vertical distances between an

instrument station and a staff station when the line of collimation is inclined

to the horizontal and staff is held vertically.

To determine distance and elevation when staff is held vertical:

As the staff is held vertical, the staff intercept AB is not normal to the

line of sight OC, draw a line A'B' passing through C and perpendicular to OC,

cutting OA at A' and OB at B'.

From right angle triangle OFC

OFC = 90° - θ

BCB' = θ (As CB' is perpendicular to OC)

A'CA = B'CB = θ

Let the stadia – hairs subtend an angle α, then

COA' = α2

COA = 90°- α2

CA'A = 180° - (90°- α2 )

= 90° + α2

The value of α/2 (its value being 17’11” for K = 100) is very small. Hence,

triangles AA'C and BB'C may be assumed to be right angled triangles. Thus,

A'B' = A'C + B'C

= AC cosθ + BC cosθ

= (AC + BC) cosθ

= S cosθ

Inclined distance OC,

L = KA'B' + C

L = KS cosθ + C

But, D = L cosθ

= (KS cosθ + C) cosθ

D = KS cos2θ + C cosθ

V = FC = L sinθ

= (KS cosθ + C) sinθ

= KS cosθ sinθ + C sinθ

= Ks2 sin2θ + C sinθ

Elevation of staff station for angle of elevation = H.I + V – h

Elevation of staff station for angle of depression = H.I - V – h

11. Explain the procedure of estimating the horizontal and vertical distances

where the line of collimation is inclined to the horizontal and the staff is

held normal to the line of collimation.

Solution:

To determine distance and elevation when staff is held normal;

As the staff at E is held normal to the line of sight AC. Therefore the

staff intercept AB is normal to the line of sight OC.

Line of sight at an angle of elevation:

Let, AB = S= Staff intercept,

CE = h = central hair reading

θ = angle of elevation, and

OC = L = inclined sight.

Drop perpendicular CF' to horizontal OF. Then,

L = KS + C

OF' = (KS + C) cosθ

D = OF' + F'F

= (KS + C) cosθ + h sinθ

Elevation of the staff station = OC sinθ

= L sinθ

= (KS + C) sinθ

Elevation of the staff station = H.I + V - h cosθ

Line of sight at an angle of depression:

L = KS + C

OF' = (KS + C) cosθ

D = OF' - FF' =OF' - EE'

= (KS + C) cosθ - h sinθ

Elevation of the staff station = OC sinθ

= L sinθ

= (KS + C) sinθ

Elevation of the staff station = H.I - V - h cosθ

The distance elevation formulae deduced in the foregone are applicable

only for external focusing telescope fitted with an anallactic lens. In the

instruments equipped with an internal focusing telescope, the systems of lenses

of object glass and an internal lens do not have constant focal length, since the

distance between these two is varied to achieve focusing and since the anallactic

lens has to be placed at a constant distance from the object glass. Therefore, an

internal lens in the internal focusing telescope cannot be regarded as a substitute

for an anallactic lens. But by suitably designing the optical system, the anallactic

point may be brought approximately to the centre of the instrument, thereby

reducing the additive constant to so small figure that it may be disregarded. The

variation in the focal length is very small if the sights are more than 30m and

practically the same formulae may be used as in the case of an external focusing

telescope. For smaller sights, an additive constant of 0.2m, if used gives results

with an error of about 1%, except for sights less than 5m. it is impossible to

proportion the focusing lens system to serve as an anallactic lens but some

companies have produced internal focusing telescopes which they claim to be

perfectly anallactic.These have three lenses between the objective and

diaphragm and therefore necessitates larger apertures to compensate for the loss

of illumination.

12. Explain how a subtense bar is used with a theodolite to determine the

horizontal distance between two points.

In this method, the base AB is kept in a horizontal plane and the angle

AOB is measured with the help of the horizontal circle of the theodolite.

Thus, in fig, let AB be the horizontal base of a length S and let O be the

position of the instrument meant for measuring the horizontal angle AOB. If the

line AB is perpendicular to the line OC, where C is midway between A and B,

we have from ΔOAC,

D = 0.5 S cot β2

D = S / (2tanβ2 )

This equation is the standard expression for the horizontal distance

between O and C. if β is small, we get

tan β2 =

β2 , where β is in radians

=β

2x 206265 , if β is in seconds (since 1 radian = 206265

seconds)

Substituting in equ, we get

D = s x 206265

β where β is in seconds.

The accuracy of the expression depends upon the size of angle. For

similar angles (say upto 1°), expression may be taken as exact enough for all

practical purposes. The angle at O is generally measured with the help of a

theodolite, while a subtense bar is used to provide the base AB.

13. Explain different errors that may arise in stadia surveying.

The various sources of errors which arise in tacheometry may be divided into

three heads:

i. Instrumental errors.

ii. Errors due to manipulation and sighting.

iii. Errors due to natural causes.

Instrumental errors: They consist of:

a) Errors due to imperfect adjustment of the tacheometer

The effects of in adjustments of various parts on the accuracy have already been

discussed in the chapter on theodolite. However, with reference to tacheometric

observations, the accuracy in the determination of distances and elevations are

dependent upon: (a) the adjustment of altitude, (b) the elimination and

determination of intex error, and (c) accuracy of reading to the vertical circle.

Since all these three have serious effects on the elevation, proper care should be

taken to adjust the altitude bubble and to see that the altitude bubble is in centre

of its run when observations are taken.

b) Errors due to erroneous divisions on the stadia rod

Since the accuracy in the determination of staff intercept depends on the

graduations, the latter should be bold, uniform and free of errors. The stadia rod

should be standardized and corrections for erroneous length should be applied if

necessary.

c) Errors due to incorrect value of multiplying and additive constants

To eliminate the errors due to this, the constants should be determined from time

to time, under the same conditions that occur in the field.

Errors due to manipulation and sighting:

They consist of errors due to:

Inaccurate centering and bisection.

Inaccurate leveling of the instrument.

Inaccurate reading to the horizontal and vertical circles.

Focusing

Inaccurate estimation of the staff intercepts.

Incorrect position of the staff.

Errors due to natural causes:

They comprise errors due to:

Wind

Unequal refraction.

Unequal expansion.

Bad visibility.

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

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

former distance and 0.996m at latter. Calculate the tacheometric constants.

Solution:

Let the constants be K and C

For the first observation 20 = K x 0.196 + C ---------(1)

For the second observation 100 = K x 0.996 +C ----------(2)

Subtracting (1) and (2), we get K(0.996 – 0.196) = 100 – 20

From which K = 100

Substituting in (1), we get C = 20 – 0.196 x 100

C = 0.40m.

15. Find up to what vertical angle, sloping distance may be taken as horizontal

distance in stadia work, so that the error may not exceed 1 in 400. Assume

that the instrument is fitted with a anallactic lens and that the staff is held

vertically.

Solution:

Let the angle be θ

True horizontal distance D = KS cos2θ

Sloping distance L = KS

Sloping distanceHorizontaldistance =

LD =

KSKS cos2θ = sec2θ ---------(1)

If the error is 1 in 400, we have

LD =

400+1400 =

401400 ----------(2)

In the limiting case, equating (1) and (2), we get

sec2θ = 401400

Maximum vertical angle = 2°52”16. Following observations were taken from two traverse stations by means of a

techeometer fitted with an anallactic lens. The constant of the instruments

is 100.

Instrument

station

Staff

station

Height of

instrument

Bearing Vertical

angle

Staff reading

A C 1.38 226°30’ +10°12’ 0.765,1.595,2.425

B D 1.42 84°45’ -12°30’ 0.820,1.840,2.860

Co - ordinates of station A 212.3N 186.8W

Co – ordinates of station B 102.8N 96.4W

Compute the length and gradient of the line CD, if B is 6.5m higher than A.

Solution:

a) Observation from A to C:

θ = 10°12′, S = 2.425 – 0.765 = 1.660m

Horizontal distance at AC = D= KS cos2θ + C cosθ

= 100 x 1.660 x cos210°12′ + 0

= 160.80m.

V = Ks2 Sin2θ + C Sinθ

=100 X1.660

2 sin20°24′ + 0

= 28.931m.

Let the elevation of A = 100.00m

R.L.of C = 100.00 + 1.380 + 28.931 – 1.595 =128.716m.

b) Observation from B to D:

θ = 12°30′, S = 2.860 – 0.820 = 2.040m

Horizontal distance at BD = D= KS cos2θ + C cosθ

= 100 x 2.040 x cos212°30′ + 0

= 194.40m.

V = Ks2 Sin2θ + C Sinθ

= 100 X2.040

2 sin25°00′ + 0

= 43.107m.

R.L.of B =100.00 + 6.50 = 106.50m

R.L.of D = 106.50 + 1.420 – 43.07 – 1.84 = 62.973m.

c) Length and gradient of CD:

Length of AC = 160.80m; R.B of AC = S46°30’W

Hence AC is in the third quadrant.

Latitude of AC = -160.80cos 46°30’ = -110.70

Departure of AC = -160.80sin 46°30’ = -116.60

Length of BD = 194.40m; R.B of BD = N84°45’E

Hence BD is in the first quadrant.

Latitude of BD = 194.40cos 84°45’ = +17.80

Departure of BD = 194.40sin 84°45’ = +193.60

Now, Total latitude of A = +212.30 Total departure of A = -186.80

Add latitude of AC = - 110.70 Add departure of AC = -116.60 ------------- ------------

Total latitude of C = +101.60 Total departure of C = -303.40

Similarly, Total latitude of B = +102.80 Total departure of B = - 96.40

Add latitude of BD = + 17.80 Add departure of BD = +193.60 ------------- ------------

Total latitude of D = +120.60 Total departure of D = + 97.20

Thus, the total co – ordinates of the points C and D are known.

Latitude of the line CD = Total latitude of D – Total latitude of C

= 120.60 -101.60

= + 19.00

Departure of the line CD = Total Departure of D – Total Departure of C

= 97.20 – (-303.40) = + 400.60

The line CD is, therefore, in the fourth quadrant.

Length CD2 = 192 + 400.602

= 401.10m.

Gradient of CD = 128.716−62.973

401.100

= 1 in 6.1 (falling)