laboratory manual for -...

Department of Civil Engineering

Survey-II

Laboratory Manual

For SURVEYING-II

FIFTH SEMESTER B.E. CIVIL

DEPARTMENT OF CIVIL ENGINEERING

G.H. Raisoni College of Engineering Hingna Road, Digdoh Hills, Nagpur – 16


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SURVEYING –II A) List of Practical

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 methode. 8) Study of stereoscope.

B) SURVEY CAMP (On any of the following topics)

1) Road Project. 2) Irrigation Project. 3) Water Supply Project


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Experiment No- 1

Aim: Determination of the Multiplying and additive constant of given Tachometer Apparatus : A tachometer 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. M = Multiplying constant of a tachometer 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 θ + c cosθ Where,

D = Horizontal distance between instrument station and staff station. K = Multiplying constant of a tachometer S = Staff intersect i.e. difference between top and bottom stadia hair reading. θ = The inclination of the line of collimation to the horizontal. c = The additive constant of the tachometer

Theory: PRINCIPLE OF STADIA METHOD The stadia method is base 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 andAB be staff intercepts.


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Evidently

221tan

11

1

22

2 βCottkconsABOC

BAOC

BAOC

====

β

This constant k entirely depends upon the magnitude of the angle β. If β is made equal to 34’22”, the constant k=1/2Cot 17’11”=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 kine 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 α (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.


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is

ff=

2

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

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

12

111fff

+= --------------------------(2)

Multiplying throughout by ff1,we get

fffff +=

2

11

Substituting the values ofis

ff=

2

1 in the above, we get

ffisf +=1 --------------------------(3)

The horizontal distance between the axis and the staff is dfD += 1

)( dfsifD ++= -------------------(4)

CksD += 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 ifk = is known as the multiplying constant or stadia interval factor and the

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

1) Focus the instruments to a distant object and measure along the telescope the distance between the objective and crosshair.

21

111fff

+=

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.


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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

CKSD += 11

1

1

SCDK −

=

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. 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,

CKSD += 11 and-----------------------------------------------(1) CKSD += 22 ----------------------------------------------------(2)

Subtracting (1) from (2) we get,

12

12

SSDDK

−−

= -----------------------------------------------------(3)

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

112

121 S

SSDDDC

−−

−=

12

11121121

SSSDSDSDSDC

−+−−

=

12

1221

SSSDSDC

−−

= -------------------------------------------------(4)

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

1) Tachometry: It is a branch of angular Surveying in which horizontal and vertical distance of point are obtained by instrumental observation. ) Tachometer: 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.


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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 00000’00” bisect staff

held at 10m, 20m,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 05000’00” 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 05000’00” in the circle the staff is bisected at 10m, 20m, and 30m from A along straight but inclination line of collimation.

Observation Table:

Stadia hair Reading Instrument station

Staff station

Distance

Vertical angle Top Cent

er Bottom Remar

k

D1 D2 A D3

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 θ + c cosθ For, three station the equations are; D1 = Ks1 cos2 θ1 + C cosθ1 ------- (1) D2 = Ks2 cos2 θ2 + C cosθ2------- (2) D3 = Ks3 cos2 θ3 + C cosθ3------- (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 techeometer 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 tachometer with tripod, tape, leveling staff, wooden pegs, ranging rods etc. Figure:

Horizontal line of sight

V1

S1

h1

O

θ

Formulae: When line of sight is inclined and staff vertical then:

θθ sin2

2 cSinKSV +=

θθ CSinKSCosD += 2 Where, K= Multiplying constant =100 C= Additive constant S= Staff intercept.

V =Vertical distance measured from horizontal line of straight to Central stadia hair reading on staff.

H = Central stadia hair reading on staff. θ = vertical angle

Theory:- The Tachomentry is an instruments 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 aixs.


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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 I =Stadia interval

Ø= 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 β = Angle between the extreme rays corresponding to stadia hairs. Draw a line A’CB’ normal to the line of sight OC. <AA’C=900+β/2, being the exterior angle of the∆COA’. Smillarly,from ∆COB’, <OB’C=900-β/2, Since β/2,is very small (its value being equal to 17011’for K=100), <AA’C and <BB’C may be approximately taken equal to 900 <AA’C = <BB’C =900

From ∆ACA’, A’C= ACcos Ø A’B’=AB cos Ø=scos Ø Since the line A’B’ is perpendicular to the line of sight OC, Hence ,we have MC=L=KA’B’+C=KScos L= Length MC measured along the line of sight. D=MQ’= Horizontal distance between the instrument and the staff.

Θ

θ


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V= Vertical intercept,at Q between the line of sight and the horizontal line H= Height of the instrument R= Central hair reading β = Angle between the extreme rays corresponding to stadia hairs. Draw a line A’CB’ normal to the line of sight OC. <AA’C=900+β/2, being the exterior angle of the∆COA’. Smillarly,from ∆COB’, <OB’C=900-β/2, Since β/2,is very small (its value being equal to 17011’for K=100), <AA’C and <BB’C may be approximately taken equal to 900 <AA’C = <BB’C =900

From ∆ACA’, A’C= ACcos Ø A’B’=AB cos Ø=scos Ø Now, the horizontal distance D= LcosØ =(kscosØ+c)cosØ D=KScos2Ø+CcosØ---------------------------------(1) Similarly V= LsinØ (KS cosØ.+c )sinØ = KScosØ.sinØ+csinØ

φφ sin2

sin CKSV += ------------------------------------------------------(2)

(a) Elevation of the staff station for angle of elevation If the line of sight has an angle of elevation Ø, 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.

θ


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Procedure:

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

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:

Stadia hair Reading Instrume

nt station

Staff station Vertical angle Top Center Bottom Remark

A BM 00000’00” R.L.= 100.00m G.Floor First Floor

Second Floor

Third Floor Calculation: D = KS cos2θ + C cosθ 1) For ground floor:- V1 = (K1S1sin2θ)/2 + C sinθ 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 tachometer with tripod, tape, leveling staff, wooden pegs, ranging rods etc.

Figure: θ θ

V1

S1

h1Fourth floor

O

B.M.

V2

h2S2

Horizontal line of sight

Formula:

θθ cCosKSCosDOP +== 21 θθ cCosKSCosDOQ +== 22

θθ cSinSinKSV +=221

θθ cSinSinKSV +=222

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. Distance between P&Q (D)= αCosDDDD 21221 22 + 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 IVrt floor of other end of college building. ∝ = Horizontal angle between OP and OQ.


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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 I =Stadia interval

Ø= 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 β = Angle between the extreme rays corresponding to stadia hairs. Draw a line A’CB’ normal to the line of sight OC. <AA’C=900+β/2, being the exterior angle of the∆COA’. Smillarly,from ∆COB’, <OB’C=900-β/2, Since β/2,is very small (its value being equal to 17011’for K=100), <AA’C and <BB’C may be approximately taken equal to 900 <AA’C = <BB’C =900

From ∆ACA’, A’C= ACcos Ø A’B’=AB cos Ø=scos Ø Since the line A’B’ is perpendicular to the line of sight OC, Hence ,we have MC=L=KA’B’+C=KScos 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

Θ

θ


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H= Height of the instrument R= Central hair reading β = Angle between the extreme rays corresponding to stadia hairs. Draw a line A’CB’ normal to the line of sight OC. <AA’C=900+β/2, being the exterior angle of the∆COA’. Smillarly,from ∆COB’, <OB’C=900-β/2, Since β/2,is very small (its value being equal to 17011’for K=100), <AA’C and <BB’C may be approximately taken equal to 900 <AA’C = <BB’C =900

From ∆ACA’, A’C= ACcos Ø A’B’=AB cos Ø=scos Ø Now, the horizontal distance D= LcosØ =(kscosØ+c)cosØ D=KScos2Ø+CcosØ---------------------------------(1) Similarly V= LsinØ (KS cosØ.+c )sinØ = KScosØ.sinØ+csinØ

φφ sin2

sin CKSV += ------------------------------------------------------(2)

(a) Elevation of the staff station for angle of elevation If the line of sight has an angle of elevation Ø, as shown in fig.

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

1) The instrument was setup in such way that the two pts 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 0º0’0” on vernier A and 180º 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 thev 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 Staff Vertical Horizontal Stadia hair Reading Remark


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station station angle angle Top Center Bottom P 00000’00”

O Q 00000’00” R.L.= 100.000m

Calculation:

V1= Ks122sin 1θ + C sinθ1

V2= Ks22

22sin θ + C sinθ2

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. θ1 = Vertical angle on staff at IVrt floor of first end of building. θ2 = Vertical angle on staff at IVrt floor of other end of building. K & C = are multiplying & additive constant of the tachometer 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. 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 as--------------. 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 tachometer with tripod, tape, leveling staff, wooden pegs, ranging rods etc. Figure:

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

V 1

S 1

h 1

O

B . M .

V 2

θ

θ

h 2

S 2

Formulae: When the line of sight is horizontal, then D = Ks + C Where, D = Horizontal distance between instrument station and staff station. K = f/i=100 Multiplying constant of a tachometer C=(f+d)=0 additive constant of a tachometer 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 θ + C cosθ Where, D = Horizontal distance between instrument station and staff station. K = Multiplying constant of a tachometer S = Staff intersect i.e. difference between top and bottom stadia hair reading. θ = The inclination of the line of collimation to the horizontal. C = The additive constant of the tachometer Distance between P&Q (D)= αCosDDDD 21221 22 + 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


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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 . ø Be the angle of elevation, then we have H=Dtanø 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 0’00”) 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 ø. 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

Observation Table:

Stadia hair Reading Instrument station

Staff station

Vertical angle Top Cent

er Bottom Remark

P R.L.= 100.000m A Q Result: The gradient of given length of two staff station is found to be ___________ by tachometric survey.


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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 44º. Calculate the necessary data for laying out a curve by the method of offsects from the chord produced. The degree of the curve being 7º. 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 snb chord (C1) already calculated along

the tangentT1B. 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


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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=7º Deflection angle,Ø=42º

Radius of curve mR 55.2457

87.1718==

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

Length of Curve = 0180ΦΠ

=xRxl =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 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,RCO2

21

1 = = mX

77.055.2452

20.9 2

=

Length of second offset RCCC

O2

( )2122

+= = m

X19.1

55.2452)00.2020.9(20=

+

Offsets from O3 to O8 are given by equation

RCtoOO

2

83 = = m63.155.245

202

=

Last offset xR

CCCO nnnn 2

)( 1 += −

mX

O 55.155.2452

)37.1900.20(07.1910 =

+=

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 20º.Calcuate all the data necessary for setting out a 3º 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. 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

δ3

δ2

λ3

λ2

δ1=λ1


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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

Tangential angle (δ)

Total Tangential angle

Actual theodolite reading

Remarks Point Chainage in meters

Chord length in meters

0 , ” 0 , ” 0 , ” Solution:- Deflection angle,Ø=20º Degree of the curve =3º

Radius of curve mD

R 5733

17191719===

Tangent length mRT 02.10110tan5732

tan 0 ===φ

The length of the curve,180

2057314.3180

0xxRl ==φπ =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,RC

n n9.1718=δ

utesstsubchordforthe min573

02.19.171811 =δ (0º3’3.60”)

7332 δδδδ −−−−−== for full chord = 99.89573309.1718 =x minutes (1º29’59.46”)

57573199.17188 === xlastchordδ minutes (0º57’0”)

The tangential angles for various chord are as follows. ∆1=δ1=00º3’3.60” ∆2= ∆1+δ1=(00º3’3.60”)+(1º29’59.40”)= 1º33’3” ∆3= ∆2+δ2=(1º33’3”)+(1º29’59.40”)= 3º3’2.40” ∆4= ∆3+δ3=(3º3’2.40”)+(1º29’59.40”)= 4º33’1.80” ∆5= ∆4+δ4=(4º33’1.80”)+(1º29’59.40”)= 6º3’1.20”


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∆6= ∆5+δ5=(6º3’1.20”)+(1º29’59.40”)= 7º33’00.60” ∆7= ∆6+δ6=(7º33’00.60”)+(1º29’59.40”)= 9º3’00” ∆8= ∆7+δ7=(9º3’00”)+(1º29’59.40”)= 10º00’00” Check :- ∆8=1/2 ∆=1/2(20º)=10º TABLE OF TANGENTIAL ANGLES





Chord length in meters

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 42º.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 as30m.

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

RLxy

6''

3

= Where x is measured along the tangent T1B and Tangential offset

RLxy

6''

3

= Where l is measured along the curve.

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 alength 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.


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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 :-Ø=42º. R=500m,v=90km/hour= sec/256060

100090 mxx

= , r=0.30m/sec2

1) length of transition curve mXrR

VL 17.1045003.0

2533

===

2) shift of curve, mXR

LS 904.050024

)17.104(24

22

===

3) Total tangent length 22

tan)(1LSRBT ++=

φ

m38.2442

17.10421tan)904.0500( 0 =++

4) Spiral angle, radianxR

L 10417.05002

17.10421 ===Φ

''00 12'58597.5deg1801047.0==

Πreesx

5) Length of circular curve = mR 31.262180

)12(0 =−Π φφ

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 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


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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

Now tangential deflection angle utesRLl min573 2

=α

For 1st point "',

02

5800968.017.104500

38.95731 −−===xxα

For 2ndpoint "',

02

3560592.617.104500

38.245732 −−===xxα

For 3rd point "',

02

417006.1717.104500

38.395733 −−===xxα

For 4th point "',

02

3232053.3217.104500

38.545734 −−===xxα

For 5th point "',

02

5652094.5217.104500

38.695735 −−===xxα

For 6th point "',

02

1818130.7817.104500

38.845736 −−===xxα

For 7th point "',

02

3648160.10817.104500

38.995737 −−===xxα

For 8th point "',

02

2459140.11917.104500

17.1045738 −−===xxα

Check:- "'0"'0

1 245913

12585318 −−=

−−== φα

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:

utesRCn

n min9.1718

=δ

1st tangential deflection angle utesXRxC

n min10.35500

21.109.17189.1718 1 ===δ

= 0º-35’-6”

Tangential deflection angle for full chord 92 δδ to = 134.103min500

309.1718=utesx

= 1º-43’-8”


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Tangential deflection angle for last sub-chord = min60.41500

10.129.1718==

xnδ

=0º-41’-36” Total tangential deflection angles

∆1=δ1=00º-35-6” ∆2= ∆1+δ2=(00º-35-6”)+(1º-43’-8”)= 2º-18’-14” ∆3= ∆2+δ3=(2º-18’-14”)+(1º-43’-8”)= 4º-1’-22” ∆4= ∆3+δ4=(4º-1’-22”)+(1º-43’-8”)= 5º-44’-30” ∆5= ∆4+δ5=(5º-44’-30”)+(1º-43’-8”)= 7º-27’-38” ∆6= ∆5+δ6=(7º-27’-38”)+(1º-43’-8”)= 9º-10’-46” ∆7= ∆6+δ7=(9º-10’-46”)+(1º-43’-8”)= 10º-53’-54” ∆8= ∆7+δ8=(10º-53’-54”)+(1º-43’-8”)= 12º-37’-2” ∆9= ∆8+δ9=(12º-37’-2”)+(1º-43’-8”)= 14º-20’-10” ∆10= ∆9+δ10=(14º-20’-10”)+(0º-41’-36”)= 15º-1’-46”

Check= "'0"'0"'00

1 481152

363302

2)12582(4222

−−=−−

=−−−

=− φφ

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 Length for the 4th point from T4L4’=41.27+15=56.27m Length for the 5th point from T2L5’=56.27+15=71.27m Length for the 6thpoint from T2L6’=71.27+15=86.27m Length for the 7th point from T2L7’=86.27+15=101.27m Length for the 8th point from T2L8’=104.17mi.e of junction point H

Now tangential deflection angle utesRLl min573 2

=α

For 1st point "',

02

' 241040.117.104500

27.115731 −−===xxα

For 2ndpoint "',

02

' 357059.717.104500

27.265732 −−===xxα

For 3rd point "',

02

' 4318073.1817.104500

27.415733 −−===xxα

For 4th point "',

02

' 5034083.3417.104500

27.565734 −−===xxα


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For 5th point "',

02

' 5255087.5517.104500

27.715735 −−===xxα

For 6th point "',

02

' 5221187.8117.104500

27.865736 −−===xxα

For 7th point "',

02

, 5052183.11217.104500

27.1015737 −−===xxα

For 8th point "',

02

' 2459140.11917.104500

17.1045738 −−===xxα

Check:- "'0"'0

1 245913

12585318 −−=

−−== φα

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

Tangential Deflection angle



Chord length in meters 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)

Tangential Deflection angle



Chord length in meters 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


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TABLE OF TANGENTIAL ANGLES




Remarks Point

Chainage in meters

Chord length in meters 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


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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.

N e g a tiv eN e g a tiv e 1

B

A

ab 'ba

C a m e ra L e n s C a m e ra L e n s

(A ) S te re o p a ir o f n a il

E Y E

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

E Y E


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B

A

E '

P r in t 2P r in t 1

M '

b 'a 'ab

mM

E

R E T IN A

E Y EE Y E

R E T IN A

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 . 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 ϕis 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 are 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’.


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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 18cm×cm. 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 . 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 one’s pocket this type being called a pocket stereoscope. Because of larger size, mirror stereoscope is not so portable as is the pocket stereoscope.


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B) SURVEY CAMP (On any of the following topics) 1) Road Project. 2) Irrigation Project. 3) Water Supply Project Field book Page

(Simple, Reverse, Compound, Transition)

By offsets from chord produced method BY Rankine method of tangential angle. Transition curve and circular curve by tangent offsets.

Cross Hair Reading

S. N.

Station point

Vertical Angle

Top

Center Bottom

S V Height of Instrument

Reduced Level


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1)Road Project. Longitudinal Section Gradient Cutting and Filling Cross-Section Mass Curve Design Curve (Simple,

2) Irrigation Project.

Preliminary Survey Topographical Survey Counter survey Water Table Survey Soil Classification Catchments and command area calculation

3) Water Supply Project

Location Survey Gradient computation Source of Water supply System Intake Tank(Structure) Pumping System Collection in Reservoir Pumping System For Conveyance to water treatment plant Treatment Plant to E.S.R. Distribution System

laboratory manual for -...

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