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10CV44 Surveying- II

Abdul Rehaman

Dept. Of Civil Engineering Page 2

1. The 3 foot screens are brought to the certain of their run

2. By adjusting the legs the head plate is made horizontal by eye judgment

3. The plate level is brought parallel to any of the two foot screens and the corresponding foot

screens are turned inwards or outwards simultaneously till the bubble is in the center of the

run

4. Plate level is turned perpendicular to its earlier position and the 3rd

foot screen is turned

inwards or outwards till the bubble comes to the centre.

5. The steps 3 & 4 are repeated till the bubble is at the centre for any direction

Elimination of Parallax: It is an adjustment in which the image of the bisected object is made

to fall on the plane of cross hair.

It is done through the following steps:

1. Looking through the eyepiece lye piece is turned clockwise or anticlockwise till the cross

hairs are seen dark.

2. Telescope is turned to a far off object and looking through the eye piece the focusing screen

is turned till the clean image of the object is seen.

c) Measurement of Horizontal angle by repetition method (June 2012, July 2013)

Horizontal angles are measured in 2 method

A) Method of repetition:

It is a method in which the angle between 2 points on objects in measured repeatedlyfor n no. of times, in different formats the actual angle in each format will be.

The method is adopted

i) When there are few objects between which angle is required

ii) Very accurate value of the angle is required.

Tabulation

Object

Bisected

Face:-

Scale A Scale B Mean No. of

repetition

P 0 0 0 0 0 0 0 0

1Q 42 25 20 26 40 42 26 00

P 42 25 20 26 40 42 26 002Q 84 48 40 48 20 86 48 30

P 84 48 40 48 20 86 48 303Q 126 22 20 21 0 126 21 40

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PR

Q

Procedure:

1. Instrument is fixed and centered at R (the reference point)

2. Releasing UCS and LCS the horizontal plate reading is made 00 0 0the clamp screws are

tightened.

3. Releasing UCS the telescope is turned to Q and UCS is clamped.

4. With Upper tangential screw bisection of point is made.

5. The reading of scale A and scale B are interred in the corresponding column of the tabular column.

This completes 1st

repetition.

6. Releasing LCS the telescope in turned back to of PLCS is clamped after

Bisection with this the same reading which was at Q now will be at P.

7. The above procedure listed in step 3 and 4 is repeated to the required No. of repetitions

The accurate angle PRQ will be equal to the final reading/No. of repetitions

d) (Dec2014, July2013)

Face: It is a condition that tells informs the side or position of the vertical circle to the observer

Face left: If the vertical circle is to the left of the observer it is called face left observation

Face right: Its a condition when the vertical circle is to the light of the observer

e)

i) Transiting : Its a process of rotating the telescope about horizontal axis along the

vertical plane through 1800.

ii) Swinging the telescope : It is a direction of rotation of the instrument about vertical axis in the

horizontal plane. When the instrument is rotated in the clockwise direction it is called right swing. When

the instrument is rotated in the anticlockwise direction it is called left swing. (Dec2014)

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f. The various temporary adjustments of a theodolite: (Jan2014)

i) Setting Up and centring ii) Levelling up and iii) Elimination of parallax

g) The procedure of measuring horizontal angle by repetition method and errors eliminated by

this method

Method of repetition:

It is a method in which the angle between 2 points on objects in measured repeatedly for no. of times, in

different formats the actual angle in each format will be. The method is adopted

1. When there are few objects between which angle is required

2. Very accurate value of the angle is required.

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Tabulation

Object

Bisected

Face:-

Scale A Scale B Mean No. ofrepetition

P 0 0 0 0 0 0 0 0

1Q 42 25 20 26 40 42 26 00

P 42 25 20 26 40 42 26 002Q 84 48 40 48 20 86 48 30

P 84 48 40 48 20 86 48 303Q 126 22 20 21 0 126 21 40

P R

Q

Procedure:

1. Instrument is fixed and centered at R (the reference point)

2. Releasing UCS and LCS the horizontal plate reading is made 0 0 0 0the clamp screws are

tightened.

3. Releasing UCS the telescope is turned to Q and UCS is clamped.

4. With Upper tangential screw bisection of point is made.

5. The reading of scale A and scale B are interred in the corresponding column of the tabular column.

This completes 1st

repetition.

6. Releasing LCS the telescope in turned back to of PLCS is clamped after bisection

With this the same reading which was at Q now will be at P.

7. The above procedure listed in step 3 and 4 is repeated to the required No. of repetitions

The accurate angle PRQ will be equal to the final reading/No. of repetitions (Dec2014)

h) Measuring direct angles, deflection angles, prolonging a straight line, running of a straight lines

between two points, locating point of intersection of two straight lines, laying of horizontal angle,

laying of angle by repetition. (June2014)

i) Prolonging a line is an important supplementary work carried out by theodolite.

a) When theodolite is in adjustment: (June 2014)

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This procedure helps in continuing a line during base line measurement in laying out

pipe line or roadway, setting out curves etc. the work is accurate and is faster.

Procedure:

1. Let AB be the line to be prolonged

2. Instrument is placed at pt. B with all temporary adjustments

3. Releasing the UCS and LCS telescope is turned about the vertical axis till the peg or arrow

at pt. A is bisected then the plate ensues i.e. (UCS & LCS) are clamped.

4. Telescope is plunged there by line of sight shifts after B in line with AB.

5. Telescope is now rotated to bisect the tip of the ranging rod at a convenient distance. This

gives point C.

6. Now again the instrument is shifted to point C and the procedure in step 3 & 4 us repeated to

get further pints D, E, F etc. and hence the line is prolonged.b) When the theodolite is not in adjustment (Poor adjustment)

This method is adopted when the instrument is not in adjustment.

Procedure:

1. Let AB be the line which is to be continued on prolonged.

2. Instrument is centered about point B with all temporary adjustments.

3. The arrow kept at A is bisected and the horizontal plate is clamped

4. Telescope is transited and a pt. C on other side of B is bisected. Releasing the horizontal

plate, pt. Ais again bisected by swinging the telescope and the horizontal plate is clamped

5. Transiting the telescope to bisect the earlier pt. C, when we dont get C we bisect another

point Cin the same line (Now instrument is said to be not in adjustment.

6. Distance CC is measured pt. C is obtained by measuring distance CC or CC = CC/2

Instrument is now shifted to point Cwhich is in line with AB and the above procedure

From step 2 to 5 is repeated

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Unit 2:a) For a dumpy level to be in proper adjustments, what are the conditions that are to be satisfied? (June

2012)

Desired relation

i.

Collimation axis perpendicular to axis of bubble tube

ii. Axis of bubble tube perpendicular to vertical axis

iii. Vertical axis perpendicular to horizontal cross hair

b) List of permanent adjustments of a transit, spire test (June 2012, Dec 2012, June

2011, July 2013)

Adjusting the horizontal axis (Spire Test):

Aim: Horizontal axis perpendicular to V.A

Necessity: If the horizontal axis is not prospered to V.A when the telescope is transited we do

not get the vertical plane but it will be scant thus during prolonging a line or while

transferring foundation points we do not get the designed straightness.

Testing:

1. Instrument is established nears to high raised object like transmission town or multistoried

building with all temporary adjustments.

2. Top of the object is bisected and horizontal plate is clamped. Telescope is rotated down to get a

point on the ground mares to the station pt. (B)

3. Telescope is made to transit & the instrument swing back to bisect B again

4. Clamping the horizontal plate telescope is lifted up to bisect earlier point A if bisected

instrument is said to be another pt. C is noted and in adjustment if not corrections has to be

applied.

Rectification or Adjustments:

1) Looking through the telescope in the last stage trunnion screw of the vertical frame

Is released and the telescope is physically adjusted till the midpoint of AC is bisected.

2) The tasting procedure is repeated again with the corrections applied at the end of tartly till point

A is bisected. When telescope is transited in the different face.

c) List the conditions to be satisfied by a Vernier theodolite to be in permanent adjustments?

(July 2013)

Following are the desired relationship b/w the various axes for the proper

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Functioning of theodolite.

1. Plate level axis must be perpendicular to vertical axis: If this relation does not exist vertical axis

will not be truly vertical and hence we get error in centering of the instrument which gives a

very wrong observed value.

2. Horizontal axis must be perpendicular to vertical axis: This is essential to get accurate horizontal

angle measurements in any face.

3. Horizontal axis shall be perpendicular to axis of collimation (AOC): This is important while

prolonging a line or to eliminate the index error in the vertical circle.

4. Altitude bubble shall be parallel to axis of collimation (AOC): This relationship will reduce the

index error in the vertical forms as well as in getting concurrent values in different face

readings.

C

d. AC=40.76m (Dec 2012)70

BC=46.08m

60 50

A B

H1=23.23m

H2= 25.04m

RL of C = 20+2.5+23.53=46.03m

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e. The fundamental lines of a transit theodolite and state the desired relationships between

them. (Dec 2014)

1. Horizontal/ trunnion axis: It is an axis parsing though the centre of vertical circle and A. frame.

Telescope is supported and rotated about this axis in the vertical place.

2. Vertical axis: Its an axis passing through the centre of the level plates. Instrument is rotated about

this axis in the horizontal place.

3. Axis of collimation: It is an axis passing through the centre of cross hair of the eyepiece of

the and the objective. This should run along the centre of the telescope tube.

4. Axis of plate level: It is an axis passing tangentially to the bubble of the spirit tube of the horizontal

plate, when the instrument is leveled.

f. True difference in elevation between A & B = H= 2.250-2.025=0.225

Apparent difference in elevation between A & B=H=1.875-1.670=0.205

Calculated reading on B at the same level of the staff = 1.875-0.225=1.650

Collimation error for 100m=1.670-1.650=0.020m

Correction on A = 20/100*0.020=0.004m Correction on

B=120/100*0.020=0.024m

Correct reading on staff A=1.875-0.004=1.871m

Correct reading on staff B=1.670-0.024=1.646m

True difference = 1.871- 1.646=0.225m (July 2013, Jan 2014, June 2014)

g. (Dec 2014)

Adjusting the horizontal axis (Spire Test):

Aim: Horizontal axis perpendicular to V.A

Necessity: If the horizontal axis is not prospered to V.A when the telescope is transited we do not get

the vertical plane but it will be scant thus during prolonging a line or while transferring foundation points

we do not get the designed straightness.

Testing

I) Instrument is established nears to high raised object like transmission town or multistoried

building with all temporary adjustments.

II) Top of the object is bisected and horizontal plate is clamped. Telescope is rotated down

to get a point on the ground mares to the station pt. (B)

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III) Telescope is made to transit & the instrument swing back to bisect B again

IV) Clamping the horizontal plate telescope is lifted up to bisect earlier point A if bisected

instrument is said to be another pt. C is noted and in adjustment if not corrections has to be

applied.

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Unit 3:

Q 3. a) Derivation of expression (D & h) (June 2014)

When the base of the object is Inaccessible (single plane method)

i. When the perpendicular inst. Is at a lower level than 2

nd

inst. StData collected: Instrument at A

O1Vertical angle to top of object/tower

S1Staff reading on BM Inst.

At B

O2Vertical angle to top of object/tower

S2staff reading on BM

bHorizontal distance between st A and st B

Calculation:

RL of tower or height of tower = HIA + h1

h1 = D tan 1 ----------- (1)

h2 = (b + D) tan 2 ---------- (2) (2)(1)

D = [S + b tan 2] / [tan 1 - tan 2]

This is a condition when the object whose

height is required will be in a thick forest onmid of a pond and we cannot reach the base of

the object. In such case, we take the help of

the data collected from 2 inst. Stations and

calculate the height of the object.

Procedure

1. Let Q be the top of the lower whose height of is required

2. Instrument is set at a convenient point with all temporary adjustments.

3. Bisecting the tip, the vertical angle to the object and staff reading on the BM is observed as Q1

and S1 respectively.

4. When the top of the object is bisected horizontal plates are clamped

5. Telescope is transited and a ground point B for the second instrument station is bisected.

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6. Instrument is shifted and centered over B.

7. With all temporary adjustments vertical angle Q2 to the top of the object and staff reading with

telescope horizontal on BM as S2 are observed.

8. These datas are utilized to find the height of the object as shown in the

Calculations.

Data Collected:

Inst. At A: Q1 vertical angle to top of the lower h1-ht of top of tower above LOS.

Inst. At B:Q2 vertical angle to top of lower h2 ht of top of lower above LOS

b- Distance between 1st

and 2nd

inst.

Calculation:

h1 = D tan 1

h2 = (D + b) tan 2

But h1- h2 S1- S2 = S

(D + b) tan 2 D tan 1 = h1- h2 =S D (tan 2- tan 1) + b tan 2 = S

D = (Sb tan 2) / (tan 2 tan 1) Substituting D is h1 we get

H1 = (Sb tan 2) tan 1 / (tan 2 tan 1)

R.L of top of tower = S1 + BM + h1

b) List of advantages of total station (June 2012, Dec 2013) Advantages of total station.

1. Reduce error

2. Time saving

3. Accurate outcome

4. Precise data

c. The expression for the horizontal distance, vertical height and the elevation of an inaccessible

object by single plane method. (July 2013, Dec 2014)

When the base of the object is Inaccessible (single plane method)

i. When the perpendicular inst. Is at a lower level than 2nd

inst. St

Data collected: Instrument at A

O1 Vertical angle to top of object/tower

S1 Staff reading on BM Inst. At B

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D = [S + b tan 2] / [tan 1 - tan 2]

This is a condition when the object whose height is required will be in a thick forest on mid of a pond

and we cannot reach the base of the object. In such case, we take the help of the data collected from 2

inst. Stations and calculate the height of the object.

Procedure

1. Let Q be the top of the lower whose height of is required

2. Instrument is set at a convenient point with all temporary adjustments.

3. Bisecting the tip, the vertical angle to the object and staff reading on the BM is observed as Q 1 and

S1 respectively.

4. When the top of the object is bisected horizontal plates are clamped

5. Telescope is transited and a ground point B for the second instrument station is bisected.

6. Instrument is shifted and centered over B.

7. With all temporary adjustments vertical angle Q2 to the top of the object and staff reading with

telescope horizontal on BM as S2 are observed.

8. These datas are utilized to find the height of the object as shown in the calculations.

Data Collected:

Inst. At A:Q1 vertical angle to top of the lower h1-ht of top of tower above LOS.

Inst. At B: Q2 vertical angle to top of lower h2 ht of top of lower above LOS

b- Distance between 1st and 2nd inst.

Calculation:

h1 = D tan 1

h2 = (D + b) tan 2

But h1- h2 S1- S2 = S

(D + b) tan 2 D tan 1 = h1- h2 =S D

(tan 2- tan 1) + b tan 2 = S

D = (Sb tan 2) / (tan 2 tan 1)

Substituting D is h1 we get

H1 = (Sb tan 2) tan 1 / (tan 2 tan 1) R.L

of top of tower = S1 + BM + h1

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d) 1 = 280

42' 2 = 180

6' b = 100 m (June 2012)

S1 = 1.67

S2 = 2.550

S = 2.5501.670 = 0.88 m

D = (b + S cot 2) tan 2 / (tan 1 tan 2) D = 152.13 m

h1 = tan 1 = 152.32 tan 280 42'

h1 = 83.29 m

h2 = (b + )tan 2 => h2 = 82.40 m

R.L of the top of hill

= R.L of B.M + S1 + h1

= 345.580 + 83.29 + 1.675 = 425.539 mor 345.580 + 82.40 + 2.555 = 425.539 m

e. Height of vane about instrument axis is equal to Dtan=2000tan9030=334.68m

CC=0.06728D2

=0.06728*4=0.27 (July2013)

Height of vane above instrument axis = 334.68+0.27=334.95m

RL of vane= RL of BM+BS+Height of vane=50.217+0.880+334.95=386.047m

RL of Q=386.047-4.00=382.047m.

f) Distance D=bsin2/sin(1+ 2)=60*sin68018/sin(60

030+ 68

018)=71.532m

Height h1=dtan1 = 71.532tan100

12=12.87m (July2012, Dec 2014)

RL of Q= 264.910m

Height h2= bsin1 / sin(1+ 2)tan2=60sin60030*tan10

018/sin(60

030+68

018)

= 12.78m

RL of Q = 264.910m

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Unit 4:

a) Determining the constants of a tacheometer (June 2012, June2013)

Tachometric constants K and C are determined in the field with the following procedure.

1. Instrument is set at a convenient point on a level ground withal temporary adjustments.

2. Horizontal plate of the instrument is clamped and along the plane ground points are marked at regular

intervals from the instrument ground point.

3. Staff is held at all the marked points say A,B,C . And the corresponding staff intercept (Upper hair

heading lower hair reading) say SA, SB, SC,.are determined.

4. Knowing the distance to each staff point form the instrument the constant K & C are determined as

shown in the calculations

We have, D1= K SA+C

D2= K SB+C

D3= K Sc+C

Solving above equations

Equation2equation 1 gives

D2- D1 = (KSB+C) - (KSA+C)

= KSB - KSA D2- D1 = K(SB SA)

K1 = D2 D1 / SB SA

K2 =D2 D1 / SC - SA

K3 = D3 D2 / SC - SB

True value of K = (K1 + K2 + K3) / 3

In the same way values are substituted in corresponding equations to get value of C1, C2, C3True value

of C is obtained by taking the average

C = (C1 + C2 + C3) / 3

b)pitch p = 1/100 = 0.01 cm, K = f/p = 2250 (Jan 2014)

m = 3.425 + 3.930 = 7.355 c = f + d = 0.425 m

= KS/m + C S = 3 m

= 2250 x 3/7.355 + 0.425

= 918.17 m

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c) K = 100, C = 0, = KS cos2 (Jan 2014)

CA = 100 x cos2 40 20' V = (KS sin2) /2CA = 61.64 m --> V1 = 4.671 m

CB = 61.99 m --> V2 = 0.192 mAB = ( CA2 + CB2 2CA.CB cos 35020')

AB = 37.25 m

Assuming RL of instrument axis as 100.0

R.L of A = RL of instrument axis + V1 R1 = 103.061

R.L of B = 98.782 m

Difference in elevation between A & B = 103.06198.782

= 4.279 m

Gradient AB = 1 in 8.77

d) Definition + Systems of tacheometry (Dec 2012)

Also called as indirect leveling to find the height of object / RL of object only with one instrument

station RL is found & is calculated. Compared to previous measuring methods this is very fast.

Tachometric surveying is a type of surveying in which we determine the height or elevation of the objects

similar to leveling work.

In this method, the horizontal distance to the object base is not measured but calculated from the

observed data. Hence the method is fast easy and convenient. Thus the method is suitable to find out the

elevation sin hilly areas, river valley, rough terrain where the distance to the object from the instrument

cannot be measured.

As all the calculations depends only on observed data. The method may not be accurate. The staff

man should be able to reach the point whose elevation is required. In tachometric surveying and

instrument called tachometer is adopted this is nothing but normal venire theodolite fitted with stadia

haired.

Stadia hairs are the additional cross hairs placed one above & one below the regular

horizontal cross hair.e. A B angle APB=7630 (June2012)

PA=216.51m

V1= 16.66m PB=197.16m

P V2=17.94m

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AB=256.55m, Gradient=1/216.8m

f) Also called as indirect leveling to find the height of object / RL of object only with one instrument

station RL is found & is calculated. Compared to previous measuring methods this is very fast.

Tachometric surveying is a type of surveying in which we determine the height or elevation of the

objects similar to leveling work.

In this method, the horizontal distance to the object base is not measured but calculated from the

observed data. Hence the method is fast easy and convenient. Thus the method is suitable to find out the

elevation sin hilly areas, river valley, and rough terrain

Where the distance to the object from the instrument cannot be measured.

As all the calculations depends only on observed data. The method may not be accurate. The staff

man should be able to reach the point whose elevation is required. In tachometric surveying and

instrument called tachometer is adopted this is nothing but normal venire theodolite fitted with stadiahaired.

Stadia hairs are the additional cross hairs placed one above & one below the regular

horizontal cross hair.

Following are the different arrangements of stadia hairs. (June2013)

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RL of instrument axis = 150.00+1.550=151.550m

RL of A = 151.550.9.386-1.755= 159.181m

RL of B = 151.550+26.265-2.0=175.81m

Distance between A B =104.050m

Difference of level between A and B = -16.635

Gradient from A to B= 1 in 6.25

i) Tacheometric equation for distance and elevation (when the staff is held normal to the line of sight)

(Dec2014)

a. When the LOS is normal upwards

When the staff is held normal the staff intercept Swill be normal to the inclined distance along the

inclined line of sight.

Therefore we can write

L = KS+C

But D = the horizontal distance between the inst. & object

D = P1D

1

D = P1C

1+ C

1D

1

From the figures

In le PCC1 = cos x P1C1 / L

P1C

1= Lcos & C1D1 = r sin

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In le CC1D1

sin = C1D1/ r

Substitutes the value of L (1) from we get

D = (KScos ) +rsin

or

D = KS cos + Ccos +rsin For

telescope with analectic lens And, we

know that.

V = Lsin

(KS+C) sin

V = KSsin +C sin

For telescope with anallactic lens

Elevation of Pt D = HI +Vr cos

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Part B

Unit 5:

Q 5. a) Definition (June 2013)

1. Mid ordinateMp (O0) this is the line joining centre of curve to mid-point to the long chord.

O0 = R (1 - cos /2)

2. Vertex distance/external distance: PV= (V)

This is the distance between centres of the curve to intersection angle

VP = R (sec /21)

3. Point of tangency: T2 (P.T):

This is the point where the curve ends and changes to straight stretch

b) By successive bisection of chord: (June2012)

In this method the major chord is bisected successively to get the midpoints of small curves. These

midpoints when jointed will give the curvature of simple circular curve.

Procedure:

1. after getting T1, T2 they are connected to get the major chord.

2. The midpoint is identified and mid-ordinate is erected to get P.

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3. T1 P is taken as an independent curve for which T1 P is the major chord and the midpoint M1

is identified.

4. At M1 the midordinate O1 further curve is constructed

5. With this T1 P1 and P1 P will now be 2 independent curves for which we set out midordinates

to get curve points P2 P3 .

6. This process can be continued to get number of mid-ordinate points joining which, we get smooth

curve.

Calculations

Midordinate MP = R (1 - cos /2)= R - (R

2(L/2)

2)

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Ordinate M1P1 = R(1cos /4) = R - (R2 (L/4)2)

c) O= R - (R2 (L)2) (June2012)

O1 = 400 - (4002

02

) = 0

O2 = 0.125m

O3 = 0.50m O4

= 1.12m O5 =

2.00m O6 =

3.13m

d) Various methods of setting out simple curves (June2012) Simple circular

curves can be set in two ways.

a. Linear method: Here, only chain and tape is used as a major device for setting out curves.

The method is adopted for small curves and in roadway construction.

b. Angular / Instrumental method: Here theodolite is used as a major device along with tape

and chain the method is adopted in setting out large curves, in railway line construction

and where accuracy is preferred.

e) O0 = R- (R2-(L/2)2); L = 100m, O0 = 5m, x = 10m (June 2012)

R = 252.5 m

Ox = (R2

-x2)(R-O0)

x 10 20 30 40 50

Ox 4.8 4.21 3.21 1.81 0

f) Tangent length = Rtan(/2) = 97.48m (June 2013)curve length = (/180) R x = 188.50 m

Chainage of T1 = 119097.48 = 1092.52

02

Chainage of T2 = 1281.02m

n = 1718.89 Cn/R n = n-1 + n

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Peg No Chainage (m) Chord Length(m) Tangential Angle Deflection Angle Actual ThreadReading

T1 1092.52

1 1110 17.48 10

40' 09'' 10

40' 09'' 10

40' 0''

2 1140 30 20

51' 53'' 40

32' 02'' 40

32' 0''

3 1170 30 20

51' 53'' 70

23' 55'' 70

24' 0''

4 1200 30 20

51' 53'' 100

15' 48'' 100

15' 40''

5 1230 30 20

51' 53'' 130

07' 41'' 130

07' 40''

6 1260 30 20

51' 53'' 150

59' 34'' 150

59' 40''

7 1281.02 21.02 20

0' 26'' 180

0' 0'' 180

0' 0''

Check = 7 = 36/2 = 180

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g. Offsets from chord produced (Deflection distance method) (July 2013, Jan 2014, Dec 2014)

This method is an important one which is used to set out large curves and cures on high ways or

road constructions.

Procedure:

1. The total curve length is divided into no. of equal parts such that joining these parts we get sub

chords and the sub chord length will be equal to the corresponding curve length ie., T1A = T1A

2. The length of 1st

chord C1 is measured along tangent to get A1 At A1 an arc of length o1 is

drawn to cut an arc of radius C1 drawn from T1 this will give point A on the curve

3. The chord T1A is produced to B1 so that A B1 = C2 At B1 and ordinate of length O2 is cut

through arc to cut an arc of radius C2 from pt. A this gives pt. B on the curve.

4. Step 3 is repeated till we reach T2by getting points on the curve

Data and Calculation:

T1v = Rare tangent/ Backward tangent

T1A = T1A = C1 = First sub chord

A B1= AB = C2 = Normal chord

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1

O1 = A1a = First ordinate = T1A* 8

O2 = B1b = Second ordinate

= On-1

VT1A = =AA1 T1= B1A B2

T1OA = 2

Let the first sub chord T1A Make an angle with the back tangent (race tangent)

Angle V T1A== T1A A1

Now the arc length A1A = 01 = T1A x

Here

T1A = T1O x 2

T1OA=2A T1 A1

=C1 / 2R

O1 = C2

/ 2R

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Point Chainage Chord

length

Deflection

angle

Total

deflection

angle

T1 1183 - - -

P1 1190 7 0 488 0 488

P2 1210 20 2 1731 3 539

P3 1230 20 2 1731 5 2310

P4 1250 20 2 1731 7 4041

P5 1270 20 2 1731 9 5812

P6 1290 20 2 1731 12 1543

P7 1310 20 2 1731 14 3314

T2 1313.89 3.89 0 2645 14 5959

' =C2 (C1 + C2) / 2R

On = Cn (Cn + Cn-1) / 2R

h) Degree of curve is expressed in terms of either (June2014)

a. Radius : (Adopted is Britain)

b. Degree of curvature : (Adopted in India, USA, Germany) To

express the degree of curvature there are 2 definitions

b. Arc definition: According to this degree of curve is the central angle subtended at the centre by an

arc of 30m length.

D0 = 1718.87 / R (degree)

c. Chord definition: According to this degree of curve is the angle subtended at the centre by a chord of

30M or 10M length. R = 1719 / D (Dc2014)

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Unit 6:

a) (Dec2014)

R1 = 350m R2 = 500m

Length of common tangent

R1tan 55/2 + R2 tan 25/2 = PQ

PQ = common tangent = 293.04 m

length of First tangent = 125.75 + 350 tan 55/2

T1I = 307.95m

Length of second tangent = 243.74+500 tan 25/2

T2I = 354.59m

I Arc Length = (/180) 350 x 550

= 335.93

II Arc length = 218.14m

b) Sin (/2) = V/L = 25/220 (June 2012)

= 130 03'

V = 2R(1- cos ) R = 484m L = 110m

Setting of I arc offsets from long chord

X m 0 10 20 30 40 50 55

Ox 3.14 3.04 2.73 2.21 1.48 0.55 0

O0 = R- (R2-(L/2)

2) O0 = 3.14m

Ox = (R2

-x2)(R-O0)

c) Compound curves: A kind of circular curves which has more than one are of a circle having

different radio with centers of the arc on one side of common tangent.

Length of the common tangent = R=(R1tan

1/2+ R

2tan

2/2)

Length of the arc = R1 1/180 (Dec2012)

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d) Reverse curve: Its a kind of circular curve which has more than one arc of a circle with different o

equal radii having the centers of are on either side of the common tangent.

(July2013)

They are used when either the two straights are parallel or their angle of intersection is too small.

They are used in hilly terrains, in railway sidings, on highways when low speed is required. When r1

and r2 are the radius of two circular arms total angle of deflection is the central angle is 1 and 2.

1 and 2 are the angle between the straights.

R=d/(tan1/2 + tan2/2)

e) 1= 1800

- 1500

= 300

(Dec2014)

2 = 1800- 140

0= 40

0

=300+400=700

Length of the tangent for first arc = 200tan150

=53.58m

Length of the tangent for second arc = 300tan200

=109.19m

Common tangent length = 162.77m

Chainage of T1= 950-164.92=785.08m

Curve length = 104.72m

Chainage of T2= 785.08+104.72= 889.80m

Long curve length = 209.44m

Chainage of T3=1099.24m

Deflection angle of full chord = 1718.9*20/200=205153 Deflection angle of sub-chord=

1718.9*4.12/200 = 004034 Deflection angle of full chord for 30m= 1718.9*30/300=205153

Deflection angle of final sub-chord = 1718.9/300*29.44=204841

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Unit 7:

Q7. a) Transition curve (June 2012)

Characteristics of Transition Curve

1. In order to fit in the transition curve at the ends a circular imaginary curve of slightly greater

radius has to be shifted towards the center. The distance through which the curve is shifted is known

as shift (S) of the curve equal to L2/24R, where L is the length of each transition curve and R is the

desired circular curve

2. Length of the combined curve is equal to

= (R + S) tan /2 + L/2

3. Spiral angle 1 = L /2R radians

4. The central angle for the curve = - 21

5. Length of circular curve is equal to R(- 21) / 1800

6. Length of combined curve = R(- 21) / 1800

+ 2L

7. Chainage of beginning of combined curve = chainage of intersection pointtotal length for combin

curve

8. Chainage of junction point of transition curve and circular curve = chainage of tangent point + lengt

of transition curve

9. Chainage of other junction point of the circular curve and the other transition curve is equal tochainage of E + length of circular curve.

10.Chainage of the end point of the combined curve = chainage of T + length of combined curve

11.The deflection angle for any point on the transition curve distant l from the beginning of

combined curve = L/6R radians

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Length of Transition Curve

1. Length may be assumed on the basis of experience and judgment.

2. Length may be such that super elevation is applied at a uniform rate 1 in 300 to 1 in 1200.

3. The length of the transition curve may be such that super elevation is applied at an arbitrary time

rate of a cm/sec. The super elevation attained

= (L/v) x a = h.

4. The radial acceleration on the circular curve is L = v3/ CR, where v is the speed in m/sec , C is

the rate of change of radial acceleration in m/sec2

, R is the radius of curve in meters.

b) Length of vertical curve (Dec2012, Jan 2014)

Vertical curves are introduced at changes of gradient to avoid impact and to maintain good visibility.

They are set out in a vertical plane to round off the angle and to obtain gradual change of gradient.

They are also called as summit curves if they have convexity upwards and valley curves if they have

concavity upwards.

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c) V = 50kmph = 13.88m/s = 0.5m/s R = 300m (June 2012)

1. length of transition curve = V3

/(R) = 17.867m

2. Super elevation BV2/(9R) = (8 x 13.88

2)/(9.81 x 300) = 0.5243

3. Shift =l2/(24R) = 17.8692/(24 x 300) = 0.044m

d) V= 60 kmph= 16.66m/s (Dec 2012, Jan2014)

K=0.3m/s3

R=200m

Length of transition curve = v3/KR = 77m

e) Vertical curves (July 2013, Dec2014)

Vertical curves are introduced at changes of gradient to avoid impact and to

maintain good visibility. They are set out in a vertical plane to round off the angle and to obtain gradua

change of gradient. They are also called as summit curves if they have convexity upwards and valley

curves if they have concavity upwards.

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The most common type of a planimeter is the polar planimeter. The area is computed by utilizing

the relationship between the tracing arm point, moved over the outline of the figure, and the connected

recording wheel which records the displacement.

A planimeter essentially consists of two bars hinged together. At the extreme end on one of the

bars, a weight is suspended over a 'needle point' or 'anchor point' which is used to anchor the bar outside

the area to be measured. The other bar known as the tracing arm, has a 'tracing point' at its extreme end.

The tracing point is moved as desired, about the needle point. At the other end of the tracing arm there

is a roller which rolls on the surface of the plan as the pointer is moved. Thus, when in use the planimeter

has three contact points on the surface of the plan, the anchor point, the tracing point, and the roller

circumference. A fixed Vernier is attached to the roller drum. A disk is also connected to the roller, by a

work drive such that one revolution of the roller turns the disk trough one part.

In large planimeters the tracing arm is made adjustable. This arrangement has two distinct

advantages. Firstly, the arm can be adjusted to the unit area of the plan; and secondly, the planimeter may

be tested for any unit area, and if found incorrect , the error can be eliminated by adjusting the arm.

Consider a moving line AB, of length L as shown in the fig. the ends of which move in a given loci. The

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ends A and B, of the line AB, are displaced to A1 and B2 respectively. Let the normal displacement of

end A be dz and the rotation of B1 as d.

The area swept ABB2A1,

dA' = Ldz + LL d

or dA' = Ldz + L2

d

Now suppose a wheel F fixed upon AB, its plane being perpendicular to that line, so that in the

displacement of AB, the wheel rolls when the point F moves perpendicularly to AB. Let dbe the angle

through which the wheel turns upon its axis in passing from F to F'. If r is the radius of the wheel, rdis

the length of the arc applied to the paper. This length is equal to dz + the arc L'd

rd= dz + L'd

Eliminating dz from Eq.(1)

dA' = rLd+ (L2/2LL')d

dA' = rLd+ (L2/2LL')d

When the directing curve AA1 is exterior to the area X,

rLd= rL= Lz (where z = r)

d= 0, since AB returns to its original position without having made a circuit about O.

Therefore integrating Eq.(4)A'=A = Lz

e. Cross Sectional area A=(b+nh)h (Dec2014)

A0=10.2sqm,

A1=14.84sqm,

A2=28.43sqm,

A3=34.38sqm,

A4=23.63sqm,

A5=16.23sqm,

A6=9.58sqm.

Trapezoidal rule =5096.4 cub.m

Prismoidal rule = 5142.9 cub.m

Total area= 6975m2

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f) Cross Sectional area A=(b+nh)h (June 2014)

A0=10.2sqm,

A1=14.84sqm,

A2=28.43sqm,

A3=34.38sqm,

A4=23.63sqm,

A5=16.23sqm,

A6=9.58sqm.

Trapezoidal rule A=d/2(O0+On+2(O1+O3+.On-1)

V=5096.4 m3

Simpsons rule A=d/3(O0+On+4(O1+O3+.)+2(O2+O4+)

V=5142.9 m3

Documents

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