CE2259 Manual

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DEPARTMENT OF CIVIL ENGINEERING

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Department of Civil Engineering-CCET

SURVEYING II

List of Experiments

1) Determination of constant of Tacheometer.

2) Determination of elevation of point by Tacheometric surveying.

3) Determination of elevation of point and horizontal distance between

them by Tacheometric survey.

4) Determination of gradient of given length if road by Tacheometric

survey.

5) Setting out of simple circular curve by offset from chord produced

method.

6) Setting out of simple circular curve by Rankine method of tangential

angle.

7) Setting out of simple transition curve by tangential angle method.

8) Study of stereoscope.

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Experiment No- 1 Aim: Determination of the Multiplying and additive constant of given Tacheometer

Apparatus : A tacheometer with tripod, tape, leveling staff, wooden pegs, ranging rods

etc.

Figure: Formulae:

When the line of sight is horizontal, then

D = KS + c

Where,

D = Horizontal distance between instrument station and staff station.

K = Multiplying constant of a tacheometer

S = Staff intersect i.e. difference between top and bottom stadia hair

reading.

When line of sight is inclined and staff vertical then:

D = KS cos2 + c cos

Where,


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S = Staff intersect i.e. difference between top and bottom stadia hair reading.

= The inclination of the line of collimation to the horizontal.

c = The additive constant of the tacheometer

Theory: PRINCIPLE OF STADIA METHOD

The stadia method is based on the principle that the ratio of perpendicular to the

base is Constant in similar isosceles triangles.In fig let two rays OA and OB be

equally inclined to the central ray OC. Let A2B2, A1B1 and AB be staff intercepts.

Evidently OC2

OC1 OC cons tan tk 1 Cot A2 B2 A1 B1 AB 2 2

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This constant k entirely depends upon the magnitude of the angle . If is made

equal to 3422, the constant k=1/2Cot 1711=100. In this case the distance between

the staff and the point o will be 100 times the intercept. In actual practice, observation

may be made with inclined line of sight. in the later case, the staff may be kept either

vertically or normal to the line of sight. We shall first derive the distance elevation

formulae for the horizontal sights.

Horizontal Sight:- Considering fig in which o is the optical centre of the objective of an

external focusing telescope.

Let A,C and B= The point cut by the three lines of sight corresponding to the three

wires.

b, c and (Top, axial and bottom )hairs of the diaphragm.

a b= i= interval between the stadia hairs (stadia interval)

AB=s=Staff intercept.

f=focal length of the objective.

f1=Horizontal distance of the staff from the optical centre of the objective.

f2=Horizontal distance of the cross-wires from O.

d= Distance of the vertical axis of the instrument from O.

D= Horizontal distance of the staff from the vertical axis of the instrument.

M= Centre of the instrument, corresponding to the vertical axis.

Since the rays Bob and AOa pass through the optical centre they are straight so

that s

AOB and aob are similar. Survey-II

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Department of Civil Engineering-CCET f1

s f 2 i

--------------------------------(1)

Again, since f1 and f2 are conjugate focal distances, we have , from lens formula 1

1

f f 2

1 --------------------------(2)

f1 Multiplying throughout by ff1, we get

f1 f1 f f f 2

Substituting the values of f1 f 2

s

in the above, we get i

f s

1 i f f --------------------------(3)

The horizontal distance between the axis and the staff is D f1 d

D f i s ( f d ) -------------------(4)

D ks C Equation (4) is known as the distance equation. In order to get the horizontal distance,

therefore, the staff intercept s is to be found by subtracting the staff reading

corresponding to the top and bottom stadia hairs.

The constant k f is known as the multiplying constant or stadia interval factor and the i

constant (f+d)=C is known as the additive stadia if the instrument.

Determination of Constant k and C

The values of the multiplying constant k and the additive constant C can be

computed by the following methods:

1st Method :- In this method ,the additive constant C=(f +d) is measured from the

instruments while the multiplying constant k is computed from field observations Survey-II

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1) Focus the instruments to a distant object and measure along the telescope the

distance between the objective and crosshair.

1

1

1

f f1 f 2

Since f1 is very large in this case, f is approximately equal to f2 i.e. equal to the

distance of the diaphragm from the objective.

2) The distance d between the instrument axis and the objective is variable in case

of external focusing telescope, being greater for short sights and smaller for long

sights. It should, therefore, be measured for average sight. Thus, the additive

constant (f +d) is known.

3) To calculate the multiplying constant k, measure a known distance and take

the S1 on the staff kept at that point , the line of sight being horizontal. using

equation

D1 KS1 C

K D1 C

S1

For the average value, staff intercepts, s2, s3 etc., can be measured corresponding

to distance d2, d3 etc., and mean value can be calculated.

2nd Method:-

In this method, both the constants are determined by field observations as under:

1) Measure a line, about 200 m long on fairly level ground and drive pegs at some

intervals, say 50 meters. Survey-II

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1


2) Keep the staff on the pegs and observe the corresponding staff intercepts with

horizontal sight.

3) Knowing the values of d and s for different points, a number of simultaneous

equations can be formed by substituting the values of d and s in equation (1.1).

The simultaneous solution of successive pairs of equations will give the values of k

and c, and the average of these can be found.

If s1 is the staff intercept corresponding to distance D1 and s2 corresponding to D2,

we have,

D1 KS1 C and-----------------------------------------------(1)

D2 KS 2 C ----------------------------------------------------(2)

Subtracting (1) from (2) we get,

K D2 D1 -----------------------------------------------------(3) S 2 S1

Substituting the values of k in (1) we get,

C D1 D2 D1

S S 2 S1

C D1 S 2 D1 S1 D2 S1 D1 S1

S2 S1

C D1 S 2 D2 S1 -------------------------------------------------(4)

S 2 S1

Thus, equations 3 and 4 give the values of K and C. Survey-II

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1) Tacheometry: It is a branch of angular Surveying in which horizontal and

vertical distance of point are obtained by instrumental observation. )

Tacheometer: It is a transit theodolite having a stadia telescope i.e. telescope

fitted with stadia diaphragm.

A leveling staff can be used for sighting purpose up to 100m distance.

Procedure:

1) Select an instrument station A on a fairly leveled ground and fix a peg.

2) Do the temporary adjustment over A.

3) With vertical circle to the left of the observer and reading 0000000 bisect staff

held at 10m, 20m, and 30m from A along straight line.

4) Note down the staff reading against top and bottom stadia hair on staff held

at 10m, 20, 30m from A.

5) In case of inclined line of sight the same procedure as stated above is

followed step by step with a vertical angle of 0500000 in the vertical circle of

the theodolite. In this case, the vertical circle is held to the left of the observer

and with the reading 0500000 in the circle the staff is bisected at 10m, 20m,

and 30m from A along straight but inclination line of collimation.

Observation Table:

Instrument station

Staff station

Distance Vertical angle

Stadia hair Reading Remark

Top Center Bottom

A

D1 D2 D3

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Department of Civil Engineering-CCET Calculation:

D = Ks + c

For three staff stations,

D1 = Ks1+c ------- (1)

D2 = Ks2+c ------- (2)

D3 = Ks3+c ------- (3)

As ; s1, s2, s3 can be known solving (1) &(2), (2) & (3) , (1) & (3) to get 3 values

of m & c ,then average of three values is required answer.

D = Ks cos2 + c cos

For, three station the equations are;

D1 = Ks1 cos2 1 + C cos 1 ------- (1)

D2 = Ks2 cos2 2 + C cos 2------- (2)

D3 = Ks3 cos2 3 + C cos 3------- (3)

As ; s1, s2, s3 can be known solving (1) &(2), (2) & (3) , (1) & (3) to get 3 values

of K & C ,then average of three values is required answer.

Result: a) For horizontal line of collimation;

1) The additive constant c for a given tacheometer is found out to be ---------

2) The multiplying constant m for a given tacheometer is found to be ---------

b) For inclination line of collimation;

1) The additive constant c for a given tacheometer is found out to be ----------

2) The multiplying constant k for a given tacheometer is found to be ---------

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Experiment No-2 Aim: Determination of Elevation of points by Tacheomentric surveying.

(the R.L.s of First Floor and Second floor of Civil Engg.).

Apparatus: A tacheometer with tripod, tape, leveling staff, wooden pegs, ranging rods

etc.

Figure:

S1

h1

V1

Horizontal line of sight

O

Formulae:


V KS Sin 2 2

c sin

D KSCos 2 CSin

Where, K= Multiplying constant =100

C= Additive constant

S= Staff intercept.

V =Vertical distance measured from horizontal line of straight to

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Central stadia hair reading on staff.

H = Central stadia hair reading on staff.

= vertical angle

Theory:- The Tacheometer is an instrument which is generally used to determine

the horizontal as well as vertical distance . it can also be used to determine the

elevation of various points which cannot be determine by ordinary leveling.

When one of the sight is horizontal and staff held vertical then the RLs of staff

station can be determined as we determine in ordinary leveling .But if the staff

station is below or above the line of collimation then the elevation or

depression of such point can be determined by calculating vertical

distances from instrument axis to the central hair reading and taking the angle

of elevation or depression made by line of sight to the instrument made by line

of sight to the instrument axis.

Distance and Elevation formula for staff vertical.

Let P= Instrument station

Q= Staff Station

M= Position of instrument axis.

O= Optical centre of the objective.

A,C,B= Point corresponding to the readings of the three hairs.

S=AB= Staff intercept Survey-II

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I =Stadia interval

= Inclination of the line of sight from the horizontal.

L= Length MC measured along the line of sight.

D=MQ= Horizontal distance between the instrument and the staff.

V= Vertical intercept, at Q between the line of sight and the horizontal line

H= Height of the instrument

R= Central hair reading

= Angle between the extreme rays corresponding to stadia hairs.

Draw a line ACB normal to the line of sight OC.

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From ACA, AC= ACcos

AB=AB cos =scos

Since the line AB is perpendicular to the line of sight OC,

Hence, we have

MC=L=KAB+C=KScos L= Length MC measured along the line of sight.


V= Vertical intercept, at Q between the line of sight and the horizontal line





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Similarly V= Lsin

(KS cos.+c )sin

= KScos.sin+csin

V KS sin 2

C sin ------------------------------------------------------(2)

(a) Elevation of the staff station for angle of elevation

If the line of sight has an angle of elevation , as shown in fig.

Elevation of staff station= Ele of instrument +h+ v-r

(b) Elevation of the staff station for the angle of depression

Elevation of Q= Elevation of P +h-v-r.

Procedure:

1) Set up the instrument in such a way that all the point should be visible from the

instrument station.

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2) Carryout the temporary adjustment and set vernier zero reading making line of

sight horizontal.

3) Take the first staff reading on Benchmark and determine height of instrument.

4) Then sight the telescope towards the staff station whose R.Ls are to be

calculated. Measure the angle on vernier if line of sight is inclined upward or

downward and also note the three crosshair readings.

5) Determine the R.Ls of various points by calculating the vertical distance.

Observation Table:

Instrument station

Staff station

Vertical angle

Stadia hair Reading Remark Top Center Bottom

A BM 0000000 R.L.= 100.00m

G.Floor First Floor Second

Floor

Third Floor

Calculation: D = KS cos2 + C cos

1) For ground floor:- V1 = (K1S1sin2 )/2 + C sin

R.L of ground floor = RL of BM + h + V1-h1 Result: The RLs of Various points are found as follows.

S.No POINTS R.Ls 1 G.Floor 2 First Floor 3 Second Floor 4 Third Floor

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Experiment No-3 Aim: Determination of elevation of points and horizontal distance between them by

Tacheomentric survey.

Apparatus : A tacheometer with tripod, tape, leveling staff, wooden pegs, ranging rods

etc.

Figure:

S1 S2

h1 h2

V1

Fourth floor

V2

Horizontal line of sight

O

Formula:

B.M.

OP D1 KSCos 2 cCos

OQ D2 KSCos 2 cCos

V1 KS Sin2 2

V 2 KS Sin2 2

cSin cSin

When line of sight is inclined and staff is held vertically, then.

Where,

V = Vertical distance measured from horizontal line of sight to central hair reading

on staff.

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Distance between P&Q (D) = D12 D2 2 2D1D2Cos

D= Distance between one corner to other corner of college building.

OP = Horizontal distance between instrument station and IV floor of one end of

college building.

OQ = Horizontal distance between instrument station and IV floor of other end of

college building.

= Horizontal angle between OP and OQ.

Distance and Elevation formula for staff vertical.

Let P= Instrument station

Q= Staff Station

M= Position of instrument axis.

O= Optical centre of the objective.

A,C,B= Point corresponding to the readings of the three hairs.

S=AB= Staff intercept

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I =Stadia interval

= Inclination of the line of sight from the horizontal.

L= Length MC measured along the line of sight.


V= Vertical intercept,at Q between the line of sight and the horizontal line





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Department of Civil Engineering-CCET R= Central hair reading



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Department of Civil Engineering-CCET Procedure

1) The instrument was setup in such way that the two point whose horizontal

distance in to be determined were visible and the temporary adjustment were

done.

2) The height of the instrument was determined by holding the staff vertically on

any selected BM and the R.L. of B.M. was taken as 100.00

3) The staff were held at first point &Second point & reading of three hairs were

taken with line of sight horizontal.

4) Set the 000 on vernier A and 180 on vernier B interest the staff at station

one , the upper plate clamping screw and lower plate clamping should be

tight during intersecting the first staff station .After taking the reading ,loosen

the upper plate clamping screw & turn the telescope clockwise intersect the

staff at 2nd station .Tighter the upper plate clamping screw take the staff

reading of cross hair &the reading of vernier A & vernier B. Again, loosen the

lower plate clamping screw &turn the telescope to intersect the staff at first

station. Then repeat the procedure same as above at least three times.

Observation Table:

Instrument

station

Staff

station

Vertical

angle

Horizontal

angle


Top Center Bottom

O

P 0000000 0000000

R.L.= 100.000m Q

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Calculation:

V1= Ks1 sin 2 1

2

V2= Ks2 sin 2 2

2

+ C sin 1 + C sin 2

Where,

S1= Staff intercept between top and bottom stadia hair at IVrt floor of first

end of building. = ---------m.

S2 = Staff intercept between top and bottom stadia hair at IVrt floor of other

end of building. = ---------m.

V1 = Vertical distance between measured from horizontal line of sight to

central stadia hair reading on staff at IVrt floor of first end of building.

V2 = Vertical distance between measured from horizontal line of sight to

central stadia hair reading on staff at IVrt floor of other end of building.

1 = Vertical angle on staff at IVrt floor of first end of building.

2 = Vertical angle on staff at IVrt floor of other end of building.

K & C = are multiplying & additive constant of the tacheometer

respectively.

OP = Horizontal distance between one corner to other corner of college

building.

OQ = Horizontal distance between instrument station and IVrt floor of other

end of college building.

= Horizontal angle between os and oe. Survey-II

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PQ = Horizontal distance between instrument station and IVrt floor of first

end of college building.

(Note :- Use +ve sign when angle elevation is measured.

Use -ve sign when angle depression is measured.)

R.L. of IVth floor of first end of college building is

= R.L. of B.M. + back sight on B.M. + V1-h1.

R.L. of IVrt floor of other end of college building is

= R.L. of B.M. + back sight on B.M. + V2-h2.

Results:-

1)The distance between two station points is found to be--------------.

2) R.Ls of staff station P=-------------.

3) R.Ls of staff station Q=-------------.

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Experiment No- 4 Aim: Determination of gradient of given length of road by Tacheomentric survey

( Horizontal distance between two points)

Apparatus: A tacheometer with tripod, tape, leveling staff, wooden pegs, ranging rods

etc.

Figure:

S 1

h 1

V 1

H o r i z o n t a l l i n e o f s i g h t

S 2 O

h 2

V 2

B . M . Formulae: When the line of sight is horizontal, then

D = Ks + C

Where,


K = f/i=100 Multiplying constant of a tacheometer

C= (f+d)=0 additive constant of a tacheometer



D = KS cos2 + C cos

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Where,




= The inclination of the line of collimation to the horizontal.

C = The additive constant of the tacheometer

Distance between P&Q (D)= D12 D2 2 2D1D2Cos

Gradient = (RLs of P-RLs of Q)/Length

Theory:

Trigomentrical levelling is the branch of surveying in which the relative elevations

of the points are determined from the observed vertical angles and known

horizontal distance. Whereas in ordinary leveling the difference in elevation is

ascertained by running a line between the given points.

Let P be the point whose reduce level is to determine and M be the reading on

bench mark.The instrument is set up at any convenient point B in the vicinity of

the object. The telescope is kept horizontal and staff reading M on the bench

mark is taken. The point A sighted and the angle of elevation is observed. The

reduced level of a point P can be ascertained as follows

Let D be the horizontal distance between the instrument and the object, H be the

height of point P . Be the angle of elevation, then we have

H=Dtan Survey-II

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RLs of P=RLs of B.M.+h+V-HI

RLs of Q= RLs of M+HI+V+h

Procedure:

1) Setup the instrument station A and level it carefully with respect to plate bubble

tube. First the centering be done.

2) Select the staff station at a convenient place and held it properly.

3) Release the vertical circle clamping screw and bisect the staff by making the

horizontal line of sight.

4) At the same time the horizontal vernier should read to (00 000) and clamp it in

position.

5) Take three staff readings on the staff station and find the stadia intercept.

6) Release the upper plate clamping screw and bisect the another staff again

and take all three staff readings and then determine the staff intercept

7) Determine the horizontal angle between these two staff station and note the

value .

8) Now determine the horizontal distance and vertical distance between staff

station and instrument station respectively.

9) Determine the horizontal distance between two staff station by applying cosine

rule.

10)Determine the of two staff station.

11) Finally determine the gradient of given length by given formula

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Department of Civil Engineering-CCET Observation Table:

Instrument

station

Staff

station

Vertical

angle


Top Center Bottom

A

P R.L.=

100.00m

Q

Result: The gradient of given length of two staff station is found to be by

tachometric survey.

Survey-II

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Experiment-5 Aim:- Setting out of simple circular curve by offsets from chord produced method

Problem Two straight intersect at chainage (30+10), the deflection angle being

44. Calculate the necessary data for laying out a curve by the method of offsets from

the chord produced. The degree of the curve being 7. The peg interval being equal to

20meters. Procedure for setting out of curve

1) Locate the tangent points T1 and T2 on the straights AB and CB.

2) Cut T1D1 equal to the length of the first sub chord (C1) already calculated along

the tangentT1B. Survey-II

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3) With T1 as centre and T1D1 radius, swing the chain or tape such that the arc D1D=

calculated offset O1, thus fixing the first point D on the curve.

4) Keep the chain along T1D and pull it straight in the forward direction of T1D until

the length DE1 becomes equal to second C2 (i.e the length of normal chord).

5) With D as centre and DE1 as radius, swing the chain such that the arc

E1E=calculated offset O2, thus fixing the second point E on the curve.

6) Continue the process repeating the point (d) and (e) until that end the curve is

reached. The last point so fixed must coincide with the previously located points

T2 (the last curve tangent point ) if not,find out the closing error. If it is small (say

with in 2m) it should be distributed to all the points by moving them sideways by

an amount proportional to the square of their distances from the point T1,

otherwise the whole curve should be set out again.

Solution:- Given degree of curve,D=7

Deflection angle,=42

Radius of curve R 1718.87

245.55m 7

Tangent length = R Tan /2 =245.55x0.4040=99.20m

Length of Curve = l xRx 1800

=188.57m

Chainage at the point of intersection = (30+10) chains = 30x20+10=610m

Chainage at 1st tangent point=610.00-99.20=510.80m (25+10.80) chains

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2 2

2

Department of Civil Engineering-CCET Chainage at end of curve or second tangent point = 510.80+188.57 = 699.37m

(34+19.37) chains

Note:-20m chain used.

Length of 1st Sub- chord =(26 +00)-(25+10.80) =9.20m

Number of full chord =34-26=8

Length of last sub-chord = (34+19.37)-(34+00)=19.37m

Check: Length of Curve = 1ST sub chord +Full chord + last sub chord

= 9.20+8x20+19.37 =188.57m

Now from equation length of first offset, O C1 =

9.20 0.77m 1 2R 2 X 245.55

Length of second offset O C2 (C1 C2) = 2 2R

20(9.20 20.00) 2 X 245.55

1.19m

Offsets from O3 to O8 are given by equation

O toO

C =

20 2 1.63m 3 8 R 245.55

Last offset O Cn (Cn 1 Cn )

n 2xR O10 19.07(20.00 19.37)

2 X 245.55 1.55m

Results:- By offsets from chord produced method the simple circular curve was plotted

on the ground

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Experiment No 06 Aim : Setting out of simple circular curve by Rankine method of tangential angle.

Problem

Two tangent intersect at chainage (180+10) the defection angle being 20.Calcuate all

the data necessary for setting out a 3 simple circular curve by method of deflection

angle (tangential angle) .The Peg interval may be taken as 30m. Procedure for setting out of curve

1) Locate the tangent points T1 and T2 on the straights AB and CB.

2) Set up the theodolite at the beginning of the curve T1.

3) With the vernier A of the horizontal circle set to zero, direct the telescope of the

ranging rod fixed at the point of intersection B and bisect it. Survey-II

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4) Unclamp the vernier plate and set the vernier A to the first tangential angle 1,

the telescope being thus directed along T1D.

5) Measure along the line T1D, the length equal to first sub-chord (C1) thus fixing first

point D on the curve.

6) Unclamp the vernier plate now and set the vernier A to the second total

tangential angle 2 , the line of sight is now directed along T1E.

7) With the zero end of chain or tape at D1 and with a arrow held at distances of

D1E=C2 (second chord or say normal chord),swing the chain about D1 until the

line of sight bisects the arrow, thus fixing the second point Eon the curve.

8) Repeat the process until the last point T2 is reached.

Field Notes

The record, of various total tangential angles and angles to which the theodolite

readings are to be set, is given in tabular from as under.

TABLE OF TANGENTIAL ANGLES

Point Chainag

e in

meters

Chord

length in

meters

Tangential

angle ()

Total

Tangential

angle

Actual

theodolite

reading

Remarks

0 , 0 , 0 ,

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0

Department of Civil Engineering-CCET Solution:-

Deflection angle,=20

Degree of the curve =3

Radius of curve R 1719

1719

573m D

Tangent length T R tan 2

3

573 tan100 101.02m

The length of the curve, l R

180

3 .14 x 573 x 20 180

=200.02m

Chainage of the point of intersection=180x30+10=5410m

Chainage of the first tangent point T1=5410-101.02=5308.98m (176+26.98)

Chainage of the end tangent point T2=5308.98+200.02=5509.00m(183+19)

Length of 1st sub-chord =(177+00)-(176+28.98)=1.02m

Number of full chord =183-177=6

Length of last sub-chord =(183+19)-(183+00)=19m

Check length of 1st sub-chord +length of 6full chord +length of last chord = length of

curve (1.02+6x30+19)=200.02m

From equation

The tangential angle, n 1718.9 Cn R

forthe1stsubchord 1718.9 1.02 min utes 1 573

(033.60)

2 3 3 7

for full chord =1718.9x 30 573

89.99 minutes (12959.46)

Survey-II

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lastchord 1718.9x 19 57 minutes (0570) 8 573

The tangential angles for various chord are as follows.

1=1=0033.60

2= 1+1=(0033.60)+(12959.40)= 1333

3= 2+2=(1333)+(12959.40)= 332.40

4= 3+3=(332.40)+(12959.40)= 4331.80

5= 4+4=(4331.80)+(12959.40)= 631.20

6= 5+5=(631.20)+(12959.40)= 73300.60

7= 6+6=(73300.60)+(12959.40)= 9300

8= 7+7=(9300)+(12959.40)= 100000

Check :- 8=1/2 =1/2(20)=10


Point Chain age in meters

Chord length in meters

Tangential angle ()

Total Tangential angle

Actual theodolite reading

Remarks

0 , 0 , 0 ,

T1 176- 26.98

----- --- -- -- -- --- --- --- --- ----

1 177+00 1.02 0 3 3.6 0 3 3.6 0 3 00 2 178+00 30 1 29 59.4 1 33 3 1 33 00 3 179+00 30 1 29 59.4 3 3 2.4 3 3 00 4 180+00 30 1 29 59.4 4 33 1.8 4 33 00 5 181+00 30 1 29 59.4 6 3 1.2 6 3 00 6 182+00 30 1 29 19.4 7 33 0.6 7 13 00 7 183+00 30 1 29 59.4 9 00 9 3 00 T2 183+19 19 0 57 00 10 00 00 10 00 00

Results:- By tangential angle method the simple circular curve was plotted on the ground

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Experiment No 07 Aim : Setting out of simple transition curve by tangential angle methods Problem

Two straight interest at chainage (102+10). The deflection angle of two straight is

42.it is proposed to interest a right handed circular curve of 500m radius and

transition curve at each end, the road bend is to be designed for maximum speed

of 90km/hour and for maximum rate of change of acceleration of 30cm/sec3. make

all the necessary calculation for setting out the combined curve by tangential

deflection method. the peg interval for transition curve be taken as 15m and for

circular curve as 30m. Survey-II

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3

3

Department of Civil Engineering-CCET Procedure for setting out a transition curve and circular curve by tangent offsets.

1) Locate the tangent point T1 .

2) Calculate the offsets for the transition curve by formula perpendicular offset

' y' x

Where x is measured along the tangent T1B and Tangential offset

6RL

' y' x

Where l is measured along the curve.

6RL

3) Starting from the point T1, with the help of tape and chain calculated

perpendicular offsets or tangential offsets such as y1,y2,y3----------at a length of

x1,x2,x3-----------or l1,l2,l3---------are laid at the end of the chainages x1,x2------- or

l1,l2------- likewise the junction point F is located. The number of chord in which the

length of transition curve is divided depends on the length of the transition curve.

Check :- at x=L/2, the transition curve bisects the shift and at x=L the perpandiclar

offset=4s.

4) The process is repeated to layout the other transition curve from T2.

5) The circular curve between the point F and H laid in the same manners as

explained in laying simple circular curve.

Note:- The peg interval for the transition curve may be from 10 to 15 meters, while

that for the circular it may be 20meters or 30meters.

Solution :-=42. R=500m,v=90km/hour= 90x1000

25m / sec 60x60

, r=0.30m/sec2

1) length of transition curve L V

253

104.17m rR 0.3 X 500

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2 2


2) shift of curve, S L

(104.17) 0.904m 24R 24 X 500

L

3) Total tangent length T1 B (R S ) tan 2 2

(500 0.904) tan 210 104.17 244.38m 2

L 4) Spiral angle, 1 2R

104.17 0.10417radian 2x500

0.1047 x180 deg rees 5.97 0 50 58'12 ''

5) Length of circular curve = R( 2 1)

262.31m 1800

6) Chainages:

1) Chainage of intersection point B=102+10=102X30+10=3070m

2) Chainage of tangent point T1=3070-Total tangent length

= 3070-244.38=2825.62m (94+5.62)

3) Chainage of the junction point F of the transition curve and the circular

curve=2825.62+104.17=2929.79m (97+19.79)

4) Chainage of the junction point H of the circular curve with that of second

transition curve = 2929.79+length of circular curve

=2929.79+262.31 =3192.10m (106+12.10)

5)Chainage of last tangent point T2=3192.10+104.17=3296.27m (109+26.27)

Check:- Chainage of T2 = Chainage at T1+2XL+ length of transition curve

=2825.62+2x104.17+262.31=3296.27m

7) Tangential deflection angles for the transition curve are calculated as below.

Chainage of T1=94+5.62

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2

2

2

2

2

Department of Civil Engineering-CCET Chainage of1st point on the transition curve=(94+15)

Length of the 1st point on transition curve from tangent point T1 =(94+15)-

(94+5.62)=9.38m

Or L1=9.38m (as the peg interval is 15m for the transition curve)

Length for the 2nd point (L2)=9.38+15=24.38m

Length for the 3rd point (L3)= 24.38+15=39.38m

Length for the 4th point (L4)=39.38+15=54.38m

Length for the 5th point (L5)= 54.38+15=69.38m

Length for the 6th point (L6)=69.38+15=84.38m

Length for the 7thpoint (L7)=84.38+15=99.38m

Length for the 8th point (L8)= 99.38+15=104.17m

573l 2 Now tangential deflection angle min utes

RL

For 1st point 1 573x9.382 ,

0.968 00 0 ' 58" 500x104.17

For 2ndpoint 2 573x24.38

,

6.592 00 6 ' 35" 500x104.17

For 3rd point 3 573x39.38

,

17.06 00 17 ' 4" 500x104.17

For 4th point 4 573x54.38

,

32.53 00 32 ' 32" 500x104.17


,

52.94 00 52 ' 56" 500x104.17


,

78.30 10 18 ' 18" 500x104.17

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2

2



,

108.60 10 48 ' 36" 500x104.17


,

119.40 10 59 ' 24" 500 x104.17

Check:- 8 1 5

0 58 '

12" 10 59 ' 24" 3 1 3

8) Tangential deflection for the circular curve:

Length of the sub-chord =(98+00) (97+19.79)=10.21m

Number of full chord =106-98=8chains

Length of last last sub-chord (106+12.10)-(106+00)=12.10m

Check = length of curve =8x30+10.21+12.10=262.31m

Tangential deflection angle for the circular curve:

1718.9Cn min utes n R

1st tangential deflection angle n 1718.9xC1 R

1718.9 X 10.21

35.10 min utes 500

= 0-35-6

Tangential deflection angle for full chord 2 to 9 =

1718.9x30 500

min utes 103.134

Tangential deflection angle for last sub-chord = n

=0-41-36

Total tangential deflection angles

= 1-43-8

1718.9x12.10

41.60 min 500

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0

Department of Civil Engineering-CCET 1=1=00-35-6

2= 1+2=(00-35-6)+(1-43-8)= 2-18-14

3= 2+3=(2-18-14)+(1-43-8)= 4-1-22

4= 3+4=(4-1-22)+(1-43-8)= 5-44-30

5= 4+5=(5-44-30)+(1-43-8)= 7-27-38

6= 5+6=(7-27-38)+(1-43-8)= 9-10-46

7= 6+7=(9-10-46)+(1-43-8)= 10-53-54

8= 7+8=(10-53-54)+(1-43-8)= 12-37-2

9= 8+9=(12-37-2)+(1-43-8)= 14-20-10

10= 9+10=(14-20-10)+(0-41-36)= 15-1-46

Check= 2 1 42 (2

0 58 ' 12" )2 300

3' 36" 150 1' 48"

2 2 2

9) Tangential deflection angles for the second transition curve:

The second transition curve is set out from the point of tangency T2 therefore

tangential deflection angles are calculated in the same manner as in the first

case but from point T2.

Now chainage at T2=(109+26.27)

Chainage at the first point on the 2nd transition curve=109+15

Length for the first point L1=(109+26.27)-(109+15)=11.27m

As peg interval for transition curve =15m

Length for the 2nd point from T2L2=11.27+15=26.27m

Length for the 3rd point from T3L3=26.27+15=41.27m

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2

2

2

2

2

2

2

2


Length for the 4th point from T4L4=41.27+15=56.27m


Length for the 6thpoint from T2L6=71.27+15=86.27m


Length for the 8th point from T2L8=104.17mi.e of junction point H

573l 2 Now tangential deflection angle min utes

RL

For 1st point '1 573x11.27

,

1.40 00 1' 24" 500x104.17

For 2ndpoint ' 2 573x26.27

,

7.59 00 7 ' 35" 500 x104.17

For 3rd point ' 3 573x41.27

,

18.73 00 18 ' 43" 500 x104.17

For 4th point ' 4 573x56.27

,

34.83 00 34 ' 50" 500x104.17


,

55.87 00 55 ' 52" 500x104.17


,

81.87 10 21' 52" 500x104.17

For 7th point , 7 573x101.27

,

112.83 10 52 ' 50" 500x104.17

For 8th point ' 8 573x104.17

,

119.40 10 59 ' 24" 500x104.17

Check:- 8 1 5

0 58 '

12" 10 59 ' 24" 3 1 3

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Table:1Tangential Deflection Angle for the (1st Transition Curve)

Point Chainage

in meters Chord length in meters

Tangential Deflection angle


Remarks

0 , 0 , T1 94+5.62 ----- --- -- -- --- --- ---- 1 94+15 9.38 0 0 58 0 1 0 2 95+15 24.38 0 6 35 0 6 40 3 95+15 39.38 0 17 4 0 17 0 4 96+00 54.38 0 12 32 0 32 20 5 96+15 69.38 0 12 56 0 53 00 6 97+00 84.33 1 18 18 1 18 20 7 97+15 99.38 1 48 36 1 48 40 T2 104.17 104.17 1 59 24 1 59 20

Table:2 Tangential Deflection Angle for the (2ndt Transition Curve)

Point Chainage

in meters Chord length in meters

Tangential Deflection angle


Remarks

0 , 0 , T2 109+26.27 - - - - - - - 1 109+15 11.27 0 1 24 1 1 20 2 109+00 26.27 0 7 35 0 7 40 3 108+15 41.27 0 1 43 0 18 10 4 108+00 56.27 0 34 50 0 34 49 5 107+15 71.27 0 55 52 0 56 00 6 107+00 86.27 1 21 52 1 22 00 7 106+15 101.27 1 52 50 1 53 00 H 106+12.0 104.17 1 59 22 1 59 20


Poin t

Chainage in meters

Chord length in meters

Tangential angle ()

Total Tangential angle


Remarks

0 , 0 , 0 , F 97+19.76 8 98+00 10.21 0 35 6 0 35 6 0 35 6 9 99+00 30.00 1 43 8 2 18 14 2 18 20 10 100+00 30.00 1 43 8 4 1 22 4 1 20 11 101+00 30.00 1 43 8 5 44 30 5 44 40 12 102+00 30.00 1 43 8 7 27 38 7 27 40 13 103+00 30.00 1 43 8 9 10 46 9 10 49 14 104+00 30.00 1 43 8 10 53 54 10 54 0 15 105+00 30.00 1 43 8 12 37 2 12 37 0 16 106+00 30.00 1 43 8 14 20 10 14 20 0 H 106+12.10 12.10 0 11 3 15 1 46 15 4 40

Result:- By transition curve and circular curve by tangent offsets was plotted on the ground Survey-II

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STEREOSCOPE

Stereoscope in an instrument used of viewing stereopairs. Stereoscopes are

designed for two purposes

1. To assist in presenting to the eyes the images of a pair of photographs so that

the relationship between convergence and accommodation is the same as

would be in natural vision.

2. To magnify the perception of depth.

There are two basic types of stereoscopic for stereoscopic viewing of

photographs:

A) Mirror stereoscope B) Lens stereoscope.

a b N e g a t i v e 1

b ' a

N e g a t i v e

E Y E E Y E

C a m e r a L e n s C a m e r a L e n s

P H O T O 2 A P H O T O 2

B ( A ) S t e r e o p a i r o f n a i l

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R E T I N A

E Y E E

R E T I N A

E ' E Y E

M m M '

A

b a

P r i n t 1 a ' b '

P r i n t 2

B

A) Mirror stereoscope

The mirror stereoscope shown diagrammatically consist of a pair of small

eye-piece mirrors mirrors m and a pair of larger wing mirrors, M and M each of

which is oriented at 45% with the place of the photographs. Show a nail mounted

on a block of timber , and is being photographed by two camera position .The

camera lens is placed fist in the position of left eye and then in the position of

right eye, and separate photographs are taken in each position of left eye and

then in the position of right eye and separate photographs are taken in each

position. It will be noted that the head of the nail is to the left in the left film and to

the right in the right film. AB are the images of the nail AB in the two films.

Contact prints from these negatives are placed in the mirror stereoscope

as show in where only images of the nail are drawn .

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The dour mirrors transfer the light to the eyes exactly(exactly for accommodation)

as if it had come from nail as shown by dotted line. The convergence and retinal

disparity are sufficient for the observer to see the nail in three dimensions.

The total distance b M m E or M m E from the eye to the pane of the

photographs varice 30 cm to 45cm, in order that the unaided eye may

comfortably view the photographs. The angle is determined by the separation

of photographs that give the most eye comfort, and is compatible with the

distance bMme. If this distance is to be reduced, a pair of magnifying lenses is

placed at E and E each magnifier has a focal length slightly smaller than the

distance bMme. Some types of mirror stereoscope have a set of removable

binoculars which are place at the eye positions E and E.

Show a wild ST 4 mirror stereoscopes with a parallax bar manufactured by

Ms Wild Heerbrugg Ltd. It is used for spatial observation of stereo photographs

upon a maximum model size of approximately 18cmcm. The distance between

the central point of mirrors is 25cm for all interpupillary distances. The whole

model area can be seen through the two lenses provided for correction of the

bundle of rays and for accommodating. A removable set of eyepieces with 3 X

magnification can be swung in over these lenses for closer examination of parts

of the model and study of details. A pair of eyepieces particularly useful when

selecting tie points in aerial triangulation. The two inclined binocular eyepiece

tubes are adjustable for interpupillarly distance of 56 to 74mm and have eye-

piece adjustments for focusing the separate images .

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The greatest single advantage of the mirror stereoscope is the fact that

photographs may be completely separated of the mirror stereoscope is the fact

that the photographs may be completely separated for viewing , and the entire

overlap area may be seen stereoscopically without having to slip the

photographs.

B) Lens stereoscope

A lens stereoscope consists of a single magnifying lens for each eye, and

no mirrors. The two magnifying lenses are mounted with a separation equal to the

average interpupillary distances of the human eyes, but provision is made for

changing this separation to suit the individual user.

The distance between the nodal point of the lens and the plane of the

photographs depends upon the focal length of the lens. The two photographs

can be brought so closet the eyes that proper convergence can be maintained

without causing the photographs to interfere with each other as show in since the

photographs are very close to the eyes, the images occupy larger angular

dimensions and therefore appear enlarged show a lens stereoscope.

The lens stereoscope is apt to cause eye strain as accommodation is not in

sympathy with convergence and the axes of the eyes are forced out their normal

conciliation and can be slipped in ones pocket this type being called a pocket

stereoscope. Because of larger size, mirror stereoscope is not so portable as is the

pocket stereoscope. Survey-II

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Documents

CE2259 Manual