# Surveying

July 29, 2017 | Author: Courtney Ward | Category: Surveying, Geometric Measurement, Geometry, Geography, Geomatics

Surveying...

#### Description

CE401/17

1

SURVEYING UNIT-3A OMITTED MEASUREMENTS 3.1. CONTENTS: 1. AIMS 2. OBJECTIVES 3. INTRODUCTION 4. OMMITTED MEASUREMENTS 5. ADDITIONAL PROBLEMS 6. S.A.Q. 7. SUMMARY 8. ANSWERS TO S.A.Q. 3.2. AIMS: To compute the length or bearing of one or two sides of a closed traverse field work. 3.3. OBJECTIVES: To determine a) length or bearing of one line b) length or bearing of two sides adjacent to each other c) length or bearing of sides not adjacent to each other d) length and bearing of any line in someway related to the given traverse. The student is expected to solve mainly numerical problems of different cases. 3.4. INTRODUCTION: In traversing work, a) distances between the stations, i.e. line AB, BC, CD, DE, EA are measured using a chain or tape, and b) bearings of lines AB, BC, CD, DE, EA or included angles theodolite.

A,B,C,D,E in the traverse shown in figure are noted using a compass or

CE401/17

2

Fig.3A.1 Traverse From the values of lengths ( l ) and Bearing ( ) Latitudes and Departures of survey lines are calculated. Latitude is a coordinate length of a line measured parallel to an assumed meridian. L = l cos

. Northward latitudes are taken as positive and Southern latitudes are termed

negative. Departure is a coordinate length of a line measured at right angles to the assumed meridian direction. D = l sin . Departures measured eastward are positive and westwards are taken as negative. A closed traverse satisfies the following conditions. a)

Sum of the interior angles of a closed polygon is equal to (2n – 4) right angles; where n = number of sides.

b)

i) Algebraic sum of the latitudes equals to zero or sum of the (+) Northings = sum of (-ve) Southings i.e.. l cos

= 0.

ii) Algebraic sum of the departures equal to zero or sum of (+) Northings = Sum of (-ve) Westings i.e. . l sin

= 0.

These rules are useful to determine and adjust the closing errors of a closed traverse in a systematic way. 3.5. OMITTED MEASUREMENTS: While taking the lengths and bearings of main sides of a closed traverse in the field, a surveyor may omit to take either length or bearing of one or two sides. Such missing quantities may be computed as the conditions of a closed traverse are known. Few standard cases are studied with the help of worked out examples. 3.5.1. Case-I : When length or bearing of one line is required: Example 1. : A four sided traverse A B C D has the following lengths and bearings. Find the length and bearing of closing line DA.

Solution:

Side

Length (m)

Bearing

AB

248

500

BC

320

1600

CD

180

2300

CE401/17

3

(a) Draw a rough sketch of traverse ABCD (b) Calculate latitudes and departures of all survey lines

(c) Length of closing Line =

(

Latitude ) + ( departure ) 2

2

(d) Bearing of the closing line is obtained from the relationship tan

=

Departure Latitude

Fig 3A.2 Table – Traverse Computations for Example 1: (1) Line

(2)

(3)

(4)

Length Whole

Reduced

(m) (l) circle

bearing

Bearing

(5)

BC CD

248 320 180

500 1600 2300

(7)

Latitude

(8) Departure

( ) Northing +

AB

(6)

N500 E S 200 E S 500 w

Southing

Easting

248 cos 50

248 Sin 500

= 159.41

=189.98 320 cos 200

320 Sin 200

= 300.70

=109.45

180 cos 500

Westing

180Sin500

CE401/17

4

=115.70 159.41

416.40

=137.89 299.43

137.89

Algebraic sum of the latitudes = Difference between Sum of Northings and Sum of Southings = 159.41 – 416.40 = (-) 256.99. Algebraic sum of the Departure = Difference between sum of Eastings and Sum of Westings = 299.43 – 137.89 = (+) 161.44.

(256.99)2 + (161.44)2

Length of line D A = tan =

= 303.49m

161.44 = 0.628 256.99

i.e., reduced bearing of DA = N 320 81w (whole circle bearing of line) DA = 360 – 320 81 = 3270521 .

Note: (i): The missing line DA is in North west quadrant as northern Latitude and western departure of the line DA make the sum total of latitudes equals to zero and the sum total of departure equals to zero. (ii) As per the coordinate system , AD2 = AZ2 + ZD2 where AZ = departure and DZ = latitude of line DA.

3.5.2. Case II: When length or bearing of two sides adjacent to each other are missing. Example 2: Lengths and Bearings of a theodolite traverse are shown below. Calculate the missing lengths RS and ST. Line

PQ

QR

RS

ST

TP

Length(m)

980

675

--

--

700

Bearing

00

N 250121 w

S 75 0 61 w

S 560 241 E

N 35 0 361 E

CE401/17

5

Fig 3A.3. Solution: (a) Assuming PQRT to be a closed traverse as in example 1 , solve for the length and bearing of RT. (b) Use the Sine rule and obtain the sides of triangle RST. Line

Latitude

Departure S

PQ QR

E

+ 980 Cos 0

+ 980 sin0 0

1

- 675 Sin 250121

+ 675 Cos 25 12

RS

-RS cos 75061

ST

- ST cos 560241

TP

W

+700 cos 350361

- RS Sin 75061 + ST Sin 560241 + 700 sin 350361

Algebraic Sum of the latitudes = 2159.93 Algebraic Sum of the departures = 120.08. Length of R T =

2159.932 + 120.082 = 2162.2m

Reducing bearing of RT = tan- 1

120.08 2159.93

Reduced bearing of RT = S 20331w In triangle RST , SRT = S 75061w – S 20331w = 720331

CE401/17

6

RST = (900 - 75061) + (900 – 560241) = 480301 STR = 180 – (720331 + 480301) = 580571 and RT = 2162.2m. Using Sine Rule

SR ST RT = = Sin STR Sin SRT Sin RST Hence RS = 2162.2

ST = 2162.2

Sin 580571 = 2473.39 Sin 48030 sin 720331 = 2759.76m. sin 480301

Example3: Calculate the length of DE and bearing of CD from the following observations

Line

Length(m)

Reduced bearing

AB

1000

O0 O1

BC

550

N 220151w

CD

2500

---

DE

---

S 580101w

EA

700

N360301 E

Solutions:

(a) Calculate the length and bearing of CE as in the previous examples. (b) In the Triangle CDE , CD , CE and CDE ,

DEC are known. Using Sine rule , obtain the angle

DCE length DE are calculated.

(c) Calculate the bearing of CD with the help of Bearing of DE and included angle CDE.

CE401/17

7

Fig 3A.4 (a)

latitude = 1000 cos 0 + 550 cos 220 15 + 700 cos 360301 = 2071.37. departure = 1000 Sin 0 – 550 sin 220151 + 700 sin 360301 = 208.13. Length of CE = tan =

L2 + D 2 = 2081.8m.

D 208.13 = 0.1005 = L 2071.37

Bearing of CE = S50441w = 1850441. In a closed traverse ABCE as L must be zero , bearing of CE must have Southern (-Ve) latitude and as

D must be zero bearing of CE must also have negative i.e., western

departure). (b)

DEC = Bearing of ED + Bearing of EC = 58010 W + N50 441E = 630541 CD = 2500 and CE = 2081.8m.

Using Sine Rule , CDE = 680011 and

ECD = 180 – (630541 + 680011) = 48051

CE401/17

8

DE =

CD SinDEC × SinECD

2500 = 3741.38. Sin 63 541 Sin 48051 0

(c) Bearing of CE = Bearing of DE – included angle CDE = 1210501 – 680011 = 530491

Example 4: Calculate the bearings of RS and ST from the data given below Line

Length(m)

Bearing

PQ

980

O

QR

675

3340481

RS

2500

---

ST

2750

---

TP

700

350361

Solution: (a) As in example 2 , calculate the length and bearing of RT. (b) As three sides of triangle RST are known , calculate the area of the triangle. =

s (s a )(s b )(s c ) where a , b , c are sides and s =

(c) From the relationship area of triangle =

1 (a + b + c). 2

1 1 1 ab sin C = bc sin A = ca Sin B 2 2 2

obtain the angles A , B , C of the triangle. (d) Calculate the bearings of lines RS and ST as the required bearings and included angles are known. (a) length of RT = 2162.2m bearing of RT 1820331 (b) Area of triangle RST =

(c) Sin S =

(3706.1)(1206.1)(956.1)(1543.9)

2× = 0.7472 RS . ST

Sin R =

2 = 0.9504 SR . RT

Sin T =

2 = 0.8639 ST . TR

= 2568626.4 Sqm

CE401/17

9

S = 480.35 ,

R = 710.88 ,

T = 590. 76

(d) Bearing of RS = Bearing of RT + included angle SRT = 182033 + 710531 = 2540261 Bearing of ST = Bearing of SR + included angle RST = 740261 + 480211 = 1220471 . 3.5.3. Case III: When length or bearing of sides not adjacent to each other are required. Example 5. Find the omitted bearings of lines BC and DE in the given traverse data Line

AB

BC

CD

DE

EF

FA

Length(m)

850

350

750

450

1070

1500

Reduced bearing

N410301w

---

S63000w

---

S260301E

N530301E

Solution:

Fig 3A.5 (a) Shift a side whose bearing is not known , to adjacent position of another side whose bearing is also not known. i.e., draw a line from C., Parallel to DE and equal to DE. Join the new point D1 with B. In the traverse ABD1 EFA , length and bearing of D1B is only unknown. (b) In the C D1 B obtain 2

CD1B using the formula.

CD1 + D1B 2 CB 2 obtain also angle CBD1 Cos CD B = 2CD1 × D1B

CE401/17

10

(c) Calculate the bearing of line D1C , using the values of included angle C D1B and bearing of D1B. Bearing of DE is same as that of CD1 . (d) Calculate the bearing of B C , using the values of bearing of BD1 and included angle CBD1 . (a)

Line

latitude

Departure

AB

+850cos 410301

BD1

l cos

l sin

D1E

- 750 cos 630

- 750 sin 410301

EF

- 1070 cos 260301

+ 1070 Sin 260301

FA

+ 1500 cos 530301

+ 1500 sin 530301

-850 Sin 410301

latitude = 230.77

departure = 451.73

BD1 = 507.26m , bearing of BD1 = S 620571w b. Cos C D1 B =

450 2 + 507.262 350 2 = 0.7389 2 × 450 × 507.26

CD1B = 420221 cos C B D1 =

350 2 + 507.262 450 2 = 0.4993 2 × 350 × 507.26

CBD1 = 60031 c. Bearing of D1C = Bearing of D1B -

CD1B = N 620571E - 420221 = N 200351E

Bearing of DE = Bearing of CD1 = S 200351W = 2000351 d. Bearing of BC = Bearing of BD1 +

CBD1 = S620571W + 60031 = 3030 .

3.6. Additional Problems: Example 6: Find the bearing of side AB and length of side CD from the date given below.

Side

length(m)

bearing

AB

550

roughly east

CE401/17

11

BC

320

1700

CD

???

2500

DA

300

200

Solution:

Fig 3A.6. 1

1e

(a) Draw AC 11 and equal to BC Draw CC1 111e and equal to BA. (b) Calculate length and bearing of C1D from the triangle AC1D. (c) Solve the triangle DCC1 using sine rule obtain

DCC1 and length C D.

Bearing of CC1 is equal to bearing of BA. (e) Northern latitude of C1D = 300 cos 200 – 320 cos 100 = 33.23 Western departure of C1D = 300 sin 200 + 320 sin 100 = 158.15 C1 D =

33.232 + 158.152 = 161.03m.

Bearing of C1D = N 78061W (f)

CDC1 = Bearing of DC1 – Bearing of DC (g) = (180 – 78061) – (2500 – 1800) = 310541 DCC1 = 80541

CD 161.03 550 = = 1 1 SinDCC SinDCC Sin310 54

CE401/17

12

Length of CD = 680.11m Bearing of AB = Bearing of C1C = Bearing of C1D +

DC1C

= 2810541 + 1390121 = 610061 Example 7: Calculate the distance between a point P on AB 20m from A and a point Q on C D 50m from C , from the traverse details given below. Line

Length(m)

Bearing

AB

60

N 600 W

BC

100

N 300 E

CD

75

S 300 E

Fig 3A.7 Solution: PBCQ is a closed traverse. Obtain the length and bearing of PQ as in Example 1. latitude = 40 cos 600 + 100 cos 300 – 50 cos 300 = 63.5 departure = 40 sin 600 + 100 sin 300 – 50 sin 300 = 61.8 Distance PQ =

(63.5)2 + (61.8)2

3.7. Self Assessment Questions:

= 88.6m.

CE401/17

13

Example 8: A traverse is run from A to E where there are certain obstacles in the way and the following observations are noted. Line

Length(m)

W.C.B

AB

420

3600

BC

500

3450501

CD

610

3000201

DE

400

300301

It is necessary to mark a point which is exactly midway between A and E. Calculate the length and bearing of this point from Station .C. 3.8. Summary: A closed traverse (polygon) satisfies two conditions viz ,

latitude = 0 ,

departure = 0.

Missing lengths or bearings are obtained in different cases using the conditions of a closed figure. Two relationships of closed traverse that are quite useful are (I) missing length =

(

latitude ) + ( departure ) and (ii) tan 2

2

departure . latitude

A rough sketch helps in analyzing the problem. Application of solution of triangles is required in a number of cases. In addition to using formula

(Lat )2 + (Dep )2

to find the length

‘1’ of the survey line or the side of a traverse , Since length = Lat . x Sec. R.B. Dep. X Cosec R.B., The length may also be calculated from either of the above two relationships , using Lat , or Dep. Whichever is grater. Thus in the solution of example 1 on page 3 , using Dep which is grater than Lat., length = 256.99 sec. 320081 which gives 303.48. However , it may be seen that each of these three formulas gives almost the same value. When L and D happen to be large as in example 2 and only four figure log tables are used for calculations , the above two formulas serve well.

***

CE401/17

14

SURVEYING UNIT - 3B PERMANENT ADJUSTMENTS OF A THEODOLITE 3.1 Contents: Aims Objectives Introduction Plate Level Adjustment Collimation Adjustment Spire Test Two Peg Adjustment Vertical Axis Adjustment SAQ Summary 3.2. Aims: To know the fundamental relationships between various lines of a transit theodolite and to adjust the instrument if standard relationship are not maintained , so that errors in observation are eliminated. 3.3. Objectives: Precise observations are to be noted with a theodolite. Some instrumental errors are eliminated by making permanent adjustments of the instrument.

Student is expected to be familiar with parts of a theordolite , various axes or lines and the relationships among them. Whether the instrument is in order or nor , can be verified. If any adjustment is required to be made , it may be done with care.

CE401/17

15

Fig 3B.1 Transit Parts 1. Horizontal or trunnion axis about which telescope (vertical circle) rotates. 2. Axis of level tube 3. Vertical axis about which upper and lower plates rotate 4. Line of sight or line of collimation 5. Plate level (bubble) 6. Altitude (bubble) level 7. Diaphragm screws 8. Clip screw 9. Levelling screws 10. Vertical motion screw. In a transit theodolite , the following conditions be satisfied. 1. Axis of the Plate level must be perpendicular to vertical axis. 2. Line of collimation must be at right angles to Horizontal axis. 3. Horizontal axis must be perpendicular to vertical axis. 4. Line of collimation must be parallel to altitude level bubble line. 5. When plate bubble and altitude bubble are centered , fixed vernier of the vertical circle must read zero.

CE401/17

16

3.5. Axis of Plate level perpendicular to vertical axis. Procedure: 1. Set up the theodolite and make temporary adjustments. 2. Keep the plate bubble parallel to pair of levelling screws. Rotate the instrument about vertical axis through 1800 . If the plate bubble remaining central , that means plate level is perpendicular to vertical axis and vertical axis is truly vertical. 3. If not , bring the bubble to its centre of run , adjusting half the deviation by means of capstain headed screws at the end of tube and the other half by means of respective levelling screws. 4. Repeat the test until bubble traverse for other position. Alternative method: 1. Keep the telescope horizontal and clamp the vertical circle. Revolve the instrument such that the bubble is parallel to line joining two foot screws. Adjust the same two levelling screws for bringing the altitude bubble to the centre.

CE401/17

17

2. Keep the telescope over the third foot screw and bring the bubble to the centre by adjusting the third screw only. See that the bubble traverse for all positions. Otherwise repeat the procedure. 3. Sight an object with altitude and plate bubble in adjustment in one position. Transit the theodilic , with changed face bisect the same object. Bubble should remain central. 4. if not , note the deviation. Correct half the divisions by using clip screw of vertical circle and the other half by same set of levelling screws. 5. Keep the telescope over third foot screw and adjust the bubble. Repeat the procedure to make perfect adjustment for other positions. 3.6. Collimation line and optical axis of telescope should coincide. The objective of this adjustment is to place the intersection of crosshairs in the optical axis of telescope. Horizontal crosshair must be in the plane of motion of optical centre of object glass. Line of collimation must be perpendicular to horizontal axis. Horizontal hair adjustment is required for measuring vertical angles. Vertical hair adjustment is for measuring horizontal angles between points of different elevation and also required when the instrument is transitted.

Fig 3B.2 Procedure: 1. From a theodolite at station P , keeping the telescope horizontal , take readings on Staff held at A as Ad , and at B as Ba.

CE401/17

18

Fig 3B.3.

CE401/17

19

Procedure: 1. Set theodolite at P clamp both horizontal motions. Sight object at A 100m , away. 2. Turn the telescope and locate point R , on one side of line A A1 , about 100m from P. Transit the telescope , unclamp the vernier plate , swing through 1800 and again bisect the object at A , with the telescope reversed. Clamp the upper motion. 3. Transit the telescope. If R is again bisected by the cross hairs , the instrument is in good adjustment. R may be out of sight. 4. In that case mark a point S , on the other side of R , as shown in figure. Mark an arbitrary point at

1 of angular 4

distance in RPS. Adjust the horizontal diaphragm screws and move the diaphragm , until the vertical hair coincides with the new mark. 3.7. Spire Test: The required condition is that horizontal axis must be perpendicular to vertical axis. The line of sight revolves in a plane perpendicular to horizontal axis (when the instrument is levelled properly) i.e. in a vertical plane. This is required for all movements of telescope in altitude. Procedure: 1. Set up the instrument and make temporary adjustments. Sight a flag pole on the top of a building about 10m. away. Depress the telescope after clamping both the horizontal motions. Mark the point coinciding with the cross hairs on the wall near the base in the line of sight as A. 2. Unclamp the horizontal motion. Transit the telescope. Swing through 1800 and sight the same flag poletop. Clamp the horizontal motion. Lower the telescope and mark a point coinciding with the cross hairs on the wall as B. 3. Mark a point C , half way in between A and B on the wall. Sight C from the telescope and clamp the horizontal motion. Raise the telescope upwards. Line of sight may not strike the top of flagpole , if the theodolite requires adjustment. 4. Adjust the horizontal axis by means of screws called trunnion screws near the top of the standard A frame so that the line of sight passes through the top of flag pole. Repeat the test , few times to make perfect adjustment. 3.8. Two Peg Adjustment: Line of collimation should be horizontal when the telescope bubble traverses. Axis of telescope level must be parallel to the line of collimation. This adjustment should be satisfied for all observation regarding vertical angles.

CE401/17

20

CE401/17

21

CE401/17

22

(i) instrument stations in the same vertical plane as object and instrument axis at same level at both the stations. (ii) Instrument axes at different levels but the two instrument stations in the same vertical plane as the object. (iii) Same as above but instrument axes at very different levels. ELEVATION OF OBJECT: Base of object inaccessible , object and the instrument stations not in the same vertical plane. WORKED OUT EXAMPLES , ASSIGNMENT PROBLEMS. SUMMARY: 4.2. Aim: To determine the relative elevation of stations (or points) with the help of (i) observed vertical angles (may be angle of elevation or depression) and (ii) measured or computed horizontal distances. 4.3. Objectives: In unit 1 of this course the student has studied about , the use of theodolite and measurement of vertical and horizontal angles with the same. After going through this chapter the student should be able to solve the field problems wherein he has to determine the difference in elevation between different stations and also the horizontal distances between different stations wherever required in the problem. The observations required are the measurement of vertical angles and also horizontal angles in certain cases. 4.4. Introduction: Trigonometrical levelling is that branch of surveying in which the use of trigonometrical functions is made for determining the elevation of an unknown point. The main observation is either an elevation angle or a depression angle (with the help of a theodolite) to a known point on a staff or a pole. Knowing the horizontal distance between the point sighted (staff) and the vertical angle , the vertical component of the difference in elevation between the point sighted on the staff (or pole) and the instrument axis , can be determined. This is added to or subtracted from the R.L. of the instrument axis to get the R.L. of the point sighted. From this value the height of the point sighted above the ground level is subtracted to get the R.L. of the corresponding ground point. 4.5. Elevation of Object , Base of Object Accessible:

Fig 4.1. P = Instrument Station

CE401/17

23

Q = is the position of the object PQ = D = known distance as Q is accessible Q1 = Point sighted on the object Q2 = The point on the vertical through Q where the horizontal through the instrument axis ‘O’ meets. Q1O Q2 is the angle of elevation observed. ‘S’ is the staff reading on a B.M. when the line of collimation through ‘O’ is horizontal. R.L. of Q1 is required.

h = tan D h = D tan

4.1

R.L. of Q1 = R.L. of B.M + S + h = x(say)

4.2

R.L. of Q = X - Q1Q 4.6. Base of Object Inaccessible: This problem can be solved in a number of ways depending on the particular situation (general topography of the ground) and position of the instrument stations with reference to the object. In all the different cases the instrument has to be set up at two places. 4.6.1.: Case 1: Instrument axis at stations P and Q at the same level (a rare possibility) and also the instrument stations and the elevated object in the same vertical plane.

Fig 4.2. The instrument stations P and Q and the object station R are in the same vertical plane. R.L1s. of the two instrument axis (A and B) same. Distance between the two instrument stations P and Q can be measured. Let it be ‘d’. and

are the vertical angles measured at P and Q.

D and R.L. of R to be determined.

CE401/17

24

h = D tan h = (D + d) tan D tan D (tan D=

= (D + d) tan - tan

) = d tan

d tan (tan tan

4.3

)

R.L. of R1 = R.L. of B.M + S + h (R.L of Instrument axis) + h

4.4

R.L of R (Ground point) = R.L. of R1 - R1R. 4.6.2.1. Base of Object Inaccessible Instrument Stations in the same Vertical Plane as the Object Instrument Axes at Different Levels (Figs 4.3.1. and 4.3.2):

Fig 4.3.1 P , Q Instrument Stations ‘d’ distance between the instrument stations. D distance between the instrument station ‘P’ and the object. R staff or object Station. A and B level of instrument axes. S1 and S2 readings on staff held on B.M. B1B1 = (S2 - S1) cos and

= S cos

vertical angles measured from P and Q at levels A and B.

h1 = D tan h2 = (D + d) tan h1 - h2 = S1 - S2 = S(difference in levels of instrument axis)

CE401/17

25

D(tan

- (D + d) tan

= D(tan

- tan ) – d tan

- tan ) = S + d tan

S + d tan (tan tan

D=

= D tan

4.5

)

h1 = D tan =

(S + d tan ) tan (tan tan )

4.6

h2 = (D + d) tan =

S + d tan tan tan

=

S + d tan tan tan

+ d tan tan

4.7

Alternatively By extending the line R1B to meet the horizontal through A at B1 . h1 = D tan = (D + d + S cos ) tan D tan

= (D + d + S cos ) tan

D(tan

- tan ) = (d + S cos ) tan .

D=

(d + S cos ) tan (tan tan )

h1 = D tan

h1 = h2 =

4.8

and h2 = (D + d) tan .

(d + S cos ) tan tan

tan

(d + S cos ) tan (tan tan )

tan + d tan

4.9

4.10

If the instrument axis at P is at a higher level than at Q a similar formula for D , h1 and h2 can be arrived at and the student is advised to do it as an assignment (fig 4.3.2).

CE401/17

26

Fig 4.3.2. 4.6.2.2.: Base of the object Inaccessible:- Instrument stations and object in the same vertical plane. Difference in levels of instrument axes considerable.

Fig 4.3.3.

CE401/17

27

Fig 4.3.4. If the difference in levels of the instrument axes is considerable this difference S1 - S2 = S cannot be obtained by observing the same staff kept on a B.M from the two different instrument stations as S1 - S2 = S is greater than the size of the staff used for the purpose. In such a case the following procedure is adopted. The horizontal distance D between the instrument station P and the object is obtained in terms of ‘S’ (difference in levels of instrument axes) as in 4.6.2.1. above , and the value of S calculated as shown below is substituted there (fig 4.3.4). From the instrument station at P after sighting the object at R , the telescope is transited and station Q fixed. After measurement of angle angle

x(to point R on object at P(station) the instrument is shifted to Q , from Q

to the point R1 on the object measured. A staff is kept at P and angle ‘ ’ to a known point. S on the staff

(staff reading ‘r’) measured. Before shifting the instrument from ‘P’ height of the instrument at P is measured.

h = tan d

h = d tan

= h - r = (d than - r) = Difference in level between the instrument axis at Q and the ground point at P. To this difference the height of the instrument at P(h1) (when the instrument was set up there at P to measure ) was added to get ‘S’. h1 = D tan h2 = (D + d) tan (h2 - h1) = S = (D + d) tan D tan D (tan

- D tan

+ S = (D + d) tan - tan ) = d tan B - S.

CE401/17

28

D=

d tan S (tan tan

D=

d tan (tan

h1 =D tan

)

(d tan tan

)

d tan (tan

=

r)

4.11

(d tan tan

)

r)

tan

4.12

R.L of R1 = R.L of A + h1 R.L of R = R.L. of A + h1 - R1R R.L. of A = S1 + S R.L of R = S1 + S + h1 - R1R

4.13

(w.r.t B.m) 4.6.3. Base of Object Inaccessible. Instrument Stations Not in the same Vertical Plane as the Elevated Object.

Fig 4.3.5. P , Q are the instrument station (Fig 4.3.5) R , the position of the object R1 the point sighted on the object from the instrument stations P and Q. P Q = d measured P R = D to be determined A and B levels of instrument axes at P and Q. 1

and

3

= 180 - (

2

horizontal angles measured at P and Q in the ground triangle PQR. 1

+

2 ).

From Sine Rule ,

PR PQ QR = = Sin 2 SinPRQ Sin 1

CE401/17

D Sin

29

= 2

PR = D =

Sin(180

d

(

1

+

2

))

d sin 2 Sin( 1 + 2 )

4.14

d sin 1 Sin( 1 + 2 )

QR . H1 = D tan =

d Sin sin ( 1 +

2 2

)

2

)

tan

4.15

h2 = Q R tan =

d Sin sin ( 1 +

1

4.16

R.L. of R1 = B.M + S + h1

4.17

R.L of R = B.M + S + h1 - R1R

4.18

The same can be checked from the observations at Q (i.e., B). 4.7. Worked our Examples: 4.7.1.: To determine the elevation of the top of a flag staff the following observations were made Instrumentation Station

Angle of elevation

Remarks

P

1.265

100 . 481

R.L. of B.M 248.360

Q

1.085

0

1

7 . 12

Stations P and Q and the top of the flag staff are in the same vertical plane. If the distance between P and Q is 50m find the elevation of the top of the flag staff. Solution: S = S1 - S2 = 1.265 - 1.085 0.180

hd = cot 100481 hd = 0.18 cot 100481 S

CE401/17

30

Fig 4.4 h2 = (D + d) tan h2 = (D + hd) tan (D + d) tan D(tan D=

= (D + hd) tan

- tan ) = d tan - hd tan

d tan tan

hd tan tan

(S cot )

(

)

=

50 × tan 7 0121 0.18 cot 100 481 tan 100.481 tan 100.481 tan 7 0121

=

6.316 0.18 6.136 = 95.875. = 0.19 0.126 0.064

(

)

h2 = (D + d) tan = (95.875 + 50) tan 70 121 = 145.875 x 0.126 = 18.428m. R.L. of top of flag = R.L of B.M + S2 + h2 = 248.360 + 1.085 + 18.428 R.L. of top of flag staff = 267.875m

4.7.2. A theodolite was set up at a distance of 200m from a tower. The angle of elevation to the top of the tower was 80181 while the angle of depression to the foot of the tower was 20241. The staff reading on the B.M. with R.L. 248.360 with the telescope horizontal was 1.285. Find the height of the tower at the R.L. of the top of the tower.

CE401/17

31

Fig 4.5 h1 = tan 80181 200 h1 = 200 tan 80 181 = 200 x 0.14588 = 29.18 h2 = 200 tan 20 241 = 200 x 0.0419 = 8.38 h1 + h2 = height of the tower = 200(tan 80 .181 + tan 20.241) = 200(0.14588 _ 0.0419) = 37.56. R.L of the top of the tower = 248.360 + 1.285 + 29.18 = 278.825 m.

4.8. For Practice Exercise: 4.8.1. To determine the height of a factory chimney the following observation were taken from two instrument stations P and Q which are in the same vertical as the chimney.

Instrument At P Q

Reading on a staff kept on a B.M 1.150 1.510

Angle of Elevation 180 . 361 100 . 121

Remarks. Distance between P and Q = 50m.

Find the height of the chimney. 4.8.2. : The top (T) of a church spire was sighted from two stations P and Q at very different levels , the stations P and Q being in line with the top of the spire. The angle of elevation from P to the top of the spire was 360121 as that from Q to the top of the spire was 160 81. The angle of elevation from Q to a vane 1.5m above the foot of the staff held at P was 80241. The heights of instruments at P and Q were 1.755 and 1.555m respectively. The horizontal

CE401/17

32

distance between p and Q was 100m and the R.L. of Q was 250.00. Find the R.L. of the top of the spire and the horizontal distance from P to the foot of the spire. 4.8.3. : Find the R.L of the top of church spire R from the following observations taken from two stations P and Q 50 meters apart (the top of the spire and the two stations being in different planes). Q P R = 600

P Q R = 500

Angle of elevation form P to the top of the spire = 300 Angle of elevation from Q to the top of the spire = 280 Staff reading from P on a B.M of R.L. of 250.00 = 3.50 Staff reading from Q to the same B.M.

= 1.00.

SUMMARY: The common procedure in practice for determining the heights of objects (such as towers , poles , or chimney) and also the reduced levels of various object stations knowing the R.L of a B.M was explained. Two problems on the above principles were worked out and 3 problems were given for the practice of the student as an assignment. Solutions for the problems in assignments will be supplied at the time of contact programme. Note: Units 4 and 5 are independent of the first 3 units. So an assignment exclusively independent from unit 3 was set for the unit 4. Actually it can be clubbed with unit 5). *** SURVEYING UNIT - 5 TRIGONOMETRICAL LEVELLING 5.1. Contents: Aim Objectives Introduction Geodetical observations Combined correction for curvature and Refraction Axis signal correction Determination of Difference in elevation Single observation (i) for angle of elevation (ii) for angle of depression. Reciprocal observations Worked out examples Assignment Problems. Summary. 5.2. Aim:

CE401/17

33

To know the different corrections such as Refraction , curvature , and Axis signal to be applied , calculation of the same when the distances involved are geodetic and large and then to determine the difference in elevation between stations observed. 5.3. Objectives: The student should be able to apply judiciously the principles studied in this chapter such as applying curvature , refraction and Axis signal corrections as per the guidelines indicated in the lesson and then he must be able to determine the difference in elevations, between the various stations observed and must be able to determine the elevation of a particular station given the elevation of another station. 5.3. Introduction: In Geodetic observations of trigonometrical levelling where the distances involved are large ordinary principles of plane surveying are not applicable. For smaller distances the corrections can be applied in linear measure. For larger distances the corrections for curvature and refraction are applied in angular measure directly to the observed angles. For the benefit of the student linear corrections are reviewed briefly here.

Fig 5.1. ABC is the level line through the instrument axis at A. AC is the horizontal line through A. AO plumb line (vertical line through A) CBO plumb line through B. As the line of sight through A is horizontal it bisects the vertical through B at C. d = distance between the two stations in km. correction fur curvature is BC = Cc. The effect of curvature is that the objects appear to be lower than what they really are because of higher reading recorded. AO = Radius of the earth = R = 6370 km assumed. AO2 + AC2 = OC2 = (OB + BC)2 = OB2 + BC2 + 2OB.BC = R2 + 2 x Rcc (BC is very small when compared with OB). AC = AB = d R2 + d2 = R2 + 2R.Cc curvature correction Cc = d2 / 2R km or

d2 x 1000 meters = 0.0785d2 meters. 2R

CE401/17

34

The effect of Refraction is that the ray of light when travels through atmosphere bends and the reading recorded would be lower than what it should be and so the objects appear to be higher than what they really are i.e., the effect of refraction is opposite to that of curvature and this value is found to be

correction for refraction =

1 of that of curvature. 7

1 d2 = 0.0112d2 × 7 2R

combined correction (-ve) = (0.0785 - 0.0112) d2 = 0.0673 d2 where d is in kilometre.

Relation between the observed angles and corrections (curvature and Refraction): It is required to determine the difference in elevation between two points P and Q separated by horizontal distance ‘d’ P1P0 and Q Q10 are the plum lines (vertical lines) drawn through the two ground points P and Q and ‘O’ their intersection point i.e., the centre of earth. General topography of the ground is shown with the usual convention , chord P1Q = Arc PQ1 = Chord PQ1 = Arc PQ1 = d Q1 is the projection P on the plumb line through Q P1 is the projection of Q on the plumb line through P Arc PQ1 is the level line through P Arc QP1 is the level through Q (To review the memory of the student A level surface is concentric with the surface of the earth which is assumed to be a spheroid. A level line is a line on a level surface. Every point on the line is equidistant from the centre of the earth. A horizontal line is tangential to a level line. A horizontal line through a point is perpendicular to the plumb line or vertical line or Radius of earth drawn through that point). PO1 = Horizontal drawn through P Q2 QO1 = Horizontal drawn through Q. PQ = direct lines of sight from P to Q. But due to terrestrial refraction the line of sight bend and it takes a curved path as shown in fig 5.2. For Q to be visible from P the telescope at P should be directed along PP1 tangential to the curved line of sight. Similarly for P to be visible from Q the telescope at Q should be directed along QQ1tangential to the curved line of sight drawn through Q. PP1 and QQ1 are the apparent lines of sight through P and Q (because of the effect of refraction) as against the direct line from P to Q. The vertical angles measured at any point are always with reference to the horizontal lines drawn through the corresponding ground points.

CE401/17

35

Fig 5.2. P1PO1 =

1

= Measured or observed angle of elevation from P to Q after applying correction for the

difference in height of signal at Q and the height of the instrument at P. Q2QQ1 =

1

= Observed angle of depression from Q to P after applying correction for the difference of the

height of the signal at P and the height of the instrument at Q. P1PQ = Q1QP = r = angle of refraction are angular correction for refraction. Refraction correction is assumed to be same at both stations. = angle subtended at the center of the earth by the distance PQ1 (d) over which the observations were made. O1PQ1 the angle between the horizontal through P and the chord PQ1 . =

2

= correction for curvature at P.

= Q2 QP1 =

2

correction for curvature at Q.

CE401/17

36

Coefficient of refraction m is the ratio of the angle of refraction and the angle subtended at the centre of earth by the distance over which observations are taken

r

m=

or

r=m .

In the absence of refraction the measured angle at P should have been QPO1 . Because of refraction the measured angle = P1 PO1 =

1

.

True angle after applying refraction correction = QPO1 = P1PO1 - P1PQ =

1

-r

In the absence of curvature the angle should have been measured with reference to PQ1 . PQ1 would have coincided with PO1 the horizontal through P. Correction for curvature = O1 PQ1 True angle after applying Refraction and curvature corrections = QPO1 + O1PQ1 =

-r+

1

5.1

2

Similarly true angle of depression at Q after applying curvature and refraction corrections =

-

1

+r

2

P1 Q

5.2

parallel to PQ1

P1QP = QPO1 =

-r+

1

2r =

+

2 1

-

= 1

1

-

2

+r

.... (5.3. (applicable for large M and small d).

r=m 2m =

+

1

=

1

+

- 2m

1

=

1

+

(1 - 2m).

1

-

1

5.4

Value of m varies roughs between 0.06 and 0.08. An average value of 0.07 may be taken in the absence of accurate data. At a given place greatest value of m occurs in the early morning. It diminishes until 9 or 10 A.M. after which it remains fairly constant upto about 4 ‘O’ clock after which it commences to increase. So value of (1- 2m) is always a positive value. That means observed angle of depression (between two points) always exceeds the angle of elevation by the amount

(1 - 2m).

CE401/17

37

The student may arrive at the above relation by equation the interior angles to the external angle for the QPO1 . (This may be verified as an assignment) Rewriting 5.3 (for large d and small H) 2r =

-(

1

-

1)

1

r=

1

2

2

for large ‘d’ and small ‘H’ both are angles of depression changing the sign of

2 1

2

2m = +

1

)

2

m =

1

(

1

r=

1.

1

-( =

1

+ 2

+

1

1)

(1 - 2m).

5.4a

5.5. Combined correction for Curvature and Refraction: Correction for curvature is additive for angle of elevation and subtractive for angle of depression and correction for refraction is subtractive for angle of elevation and additive for angle of depression as seen from the angles reproduced below (5.1 and 5.2). True angle of elevation =

1

-r+

2

vide

5.1

vide

5.2

True angle of depression =

1

+r-

2

The combined correction for angle of elevation =

2

- r and the combined correction for angle of depression = -

2

+ r. The magnitude of curvature correction is much more than refraction correction and so the combined correction is positive or additive to angles of elevation and subtractive (or negative) for angles for depression. The combined angular correction to be applied is of very small magnitude and is usually arrived at in seconds , as shown below:

CE401/17

38

Fig 5.3 d is the horizontal distance between the stations. R = Radius of earth

is the central angle subtended by the two stations from which observations are taken.

R =d =

180

degrees

d 180 d radian = × degrees R R As the corrections are very small they are expressed in seconds =

=

d 180 d radians = × × 60 × 60 seconds R R

d ( ) seconds R

= value of

=

2

1 sin 111

d md ;m = 11 R sin 1 R sin 111

=

d 2 R sin 111

Combined angular correction for curvature and refraction

CE401/17

=

2

=

2

=

2

39

-r

m =

2m 2

(1 – 2m)

d (1 – 2m) seconds 2R sin 111

=

5.5

which should be added for angles of elevation and should be subtracted from angles of depression. 5.6. Axis Signal Correction: (Also called eye and object correction): While explaining the details of fig 5.2. where in the correction for curvature and refraction are explained in detail it is mentioned that

1

or

1

are the observed or measured angles of elevation and depression after applying

axis signal correction. 1

or

1

= observed angle ± Axis signal correction.

Fig 5.4.

CE401/17

40

Normally in all observations the height of the signal is not the same as the height of the instrument axis above the station and so a correction known as Axis signal correction (or eye and object correction) has to be applied to the observed angles. h1 and h2 heights of instruments at P and Q respectively. S1 and S2 heights of signals at P and Q respectively BQ = difference in height of signal at Q and height of instrument at P = (S2 - h1) d = Horizontal distance between P and Q Arc PP1 is the level line through P. QQ1 is the level line through Q and the details with reference to this line are not required in this explanation and so not shown in the Fig. Q station observed. PA is the horizontal line through P BPA =

= observed angle of elevation uncorrected for axis signal.

= observed angle of depression corrected for axis signal (not shown in the figure 5.4). 1

= angle of elevation corrected for axis signal.

1

= angle of depression corrected for axis signal.

BPQ = 2

1

= axis signal correction (angular) at P.

axis signal correction at Q (not shown in fig)

Construction: Produce PQ upto C and draw perpendicular BC x to PB at B In

BPO

BPO = + 90

BPA +

APO

0 0

POB =

PBO = 180 – (90 + ) 90 -

-

= 900 - ( + )

As per construction

PBC= 900

QBC = 90 – [90 – ( + )] = 9/ 0 - 9/ 0 + ( + ) =

As

1

+ .

is very small PCB can be approximately taken as 900

BC = cos QBC = cos QBC = cos ( + ) BQ BC = BQ cos ( + )

CE401/17

41

= (S2 – h1) cos ( + ) The value of

i.e., the correction for axis signal has to be determined.

1

BC = tan PB

1

BC determined above. Now PB has to be determined for that , take the PP1B = 90 +

BPP1 =

+

P1 B

2 2

PBP1 = 90 – ( + ) Applying sin Rule

PB PP1 = sin PP1B sin PBP1 Before applying sine rule carefully observe what is required and how to make use of the available data. PB is required PP1 = d known ; all the three angles known PB = PP1

sin PP1B sin PBP1

sin 90 + =d

sin[90

(

2 +

)]

cos

=d From

2 cos ( +

PBC Tan

tan Usually

)

1

=

1

=

(S h )cos ( + BC = 2 1 PB d cos / cos ( + 2

(S 2

h1 ) cos 2 ( + d cos / 2

is small compared to

)

) )

(exact value)

5.6

and may be ignored.

In such a case

tan

1

=

(S 2

h1 )cos 2 d

5.7

CE401/17

42

The correction is evidently subtractive If observations are taken from Q to P (angle of depression) it can be shown that

tan

(S1

=

2

h2 ) cos 2 d

tan

1

=

tan

2

=

After calculating

1

2

1

or

5.8 are very small it can be calculated with sufficient accuracy

=

S 2 h1 seconds d sin 111

5.9

=

S1 h2 seconds d sin 111

5.10

and

2

angular corrections for axis signal for angles of elevation and depression.

1

= observed angle corrected to axis signal =

-

1

= observed angle of depression corrected to axis signal = +

1 2

DETERMINATION OF DIFFERENCE IN ELEVATION: After determining the correct angles (angle of elevation or depression) the next stage is to determine the difference in elevation between the instrument station and the observed station. This can be done either by (1) observation from a single station or (2) by reciprocal observation i.e., taking observations from both the stations. 5.7.1. Single observation: The observations are made and from one station say (P) to Q. The corrections to be applied are (1) Curvature (2) Refraction (3) Axis signal. Since sign of correction depends on sign of the observed angle whether angle of elevation or depression , elevation and depression angles are considered separately for determining the difference in elevation. 5.7.1.1. Angle of elevation: = observed angle of elevation to Q 1

=

-

1

-= observed angle corrected for axis signal.

CE401/17

43

Fig 5.5.

S 2 h1 seconds d sin 111

=

Arc PQ1 = chord PQ1 = PA = d P1PQ = r = m QQ1 = H = difference in elevation in

PQQ1

CE401/17

44

PQ1Q = 90 +

QPQ1 =

1

2

–r+

2

PQQ1 = 180 – (90 +

= 90 -

-

2

= 90 – (

1

+r-

1

2

)–(

1

–r +

2

)

2

–r+ )

Applying sine rule = 90 - (

1

-m + )

QQ1 PQ1 = sin QPQ1 sin PQQ1 QQ1 = H = PQ1

sin =d

=d

Substituting

m +

1

2 m +

sin [90

(

sin

1

m +

1

m +

cos( =

1

2

)]

)

d R sin 111

( sin & ' H=d cos Note

sin QPQ1 sin PQQ1

(1 – 2m)

1

d % "# 2 R sin 1 \$ d 1 + (1 m ) R sin 1"

+ (1 2m )

d d and (1 – m) in seconds 2R sin 1" R sin 1"

For approximate expression when

is small.

H = tan (corrected angle of elevation) d = d tan (corrected angle of elevation) = d tan (

1

m +

2

)

5.11

CE401/17

= d tan

45

( & '

1

+ (1 2m )

d % 2 R sin 1" #\$

This formula is to be used and when

5.12

is small.

5.7.1.2. For angle of Depression: = observed angle of depression to P 1

= observed angle of depression corrected to axis signal

Fig 5.6 Arc QP1 = Level line through Q QB = horizontal through Q P1P = H = difference in elevation between Q and P Arc QP1 = chord QP1 = d = horizontal distance In

QPP1 P1QP =

1

-

QP1P = 90 -

2 2

+r=

1

-

2

+m

CE401/17

46

P1PQ = 180 – (90 -

= 90 + = 90 90 – (

-

2 1

+

-

1

1

+

2

2

)–(

1

-

2

+m )

-m

-m +m )

PP1 P1Q = sin P1QP sin QPP1

sin sin P1QP H = PP1 = P1Q1 =d sin QPP1 sin 90

( d sin & ' H= ( cos & '

1

1

1

(

2 1

d % "# 2 R sin 1 \$ Exact d % (1 m ) R sin 1" #\$

(1

+m +m

)

2m )

5.13

For small values of P1P = H = P1Q tan P1QP = d tan (

= d tan

( & '

1

+m -

1

(1

2

)

2m )

% d 2 R sin 1" #\$

5.14

5.7.2. Reciprocal Observations: By reciprocal observations for determining the difference in elevation the refraction correction can be eliminated. The principle involved here is that observations should be taken simultaneously from both the stations i.e., the vertical angles (elevation or depression) to the other station from each station has to be taken simultaneously and the refraction effect at each station is assumed to be same and it will be nullified as will be seen from the calculations below. If it is not possible to take simultaneous observations. Observations should be taken at the same time at both the stations on different dates , at the time when refraction variation is minimum. Refraction is less variable between 10 A.M. and 3 P.M. The results obtained by this method are more accurate than from signal station observations.

CE401/17

47

Fig 5.7 After having studied the different figures in this chapter (unit 5) till now , the notation used need not be explained. P1Q is parallel to PQ1 1

–r+

2

=

1

+r-

2

(5.1 and 5.2)

Corrected angle of elevation = Corrected angle of depression. If

A=

B

CE401/17

48

A+ B 2

Then each angle =

similarly 1

–r+

=

2

1

r+

1

+r-

+

2

= 1

+ 2

=

1

+ 2

1

In

+r

2

1

+m =

+ 2

1

1

–r+

+

1

1

Thus each corrected angle =

1

1

2

=

Also

2

(written like this for convenience)

+ 2

1

2

QPQ1 QPQ1 =

2

=

1

-m +

2

Substituting these values in the above equation PQQ1 = 90 – (

1

+ r)

from fig 5.7)

QQ1 PQ1 = sin QPQ1 sin PQQ1 H = QQ1 = PQ1

sin =d

sin QPQ1 sin PQQ1

m +

1

2 1 + m )]

sin[90

(

sin

m +

=d

`H = d

1

cos(

1

sin

1

cos(

+m

2

)

m + 1

+m

2

)

CE401/17

49

But d1 - m +

1

2 + 2

1

+m =

1

= 1

+

+ 2

1

2

Substituting these values in the above equation 1

d sin H= 1

cos If however

2

+ 2

+ 2

1

+ 2

1

1

5.15

+

2

is small 1

H = d tan If both

and

1

1

5.16

are angles of depression the expression for H can be obtained by changing the sign of

1

in the above equation 1

d sin

2

H= 1

cos

1

1

2

5.17

+

2

The general expression for H is 1

sin H=

( cos & '

use plus sign for (-) minus sign for

±

1

2 1

± 2

1

% + # 2\$

= angle of elevation. = angle of depression.

If the value of H obtained from the above expression is positive (+) Q is higher than P and if H is negative (-) Q will be lower than P. 5.8. Worked out examples: 5.8.1. Given the following data with reference to two survey stations P and Q , find the difference in levels between the points P and Q are the reduced level of Q. d = Horizontal distance between P and Q = 5600m. -

= Angle of depression from P to Q 10 281 OO11

S2 = Height of signal at Q 3.90m.

CE401/17

50

h1 = height of instrument at P 1.55m m = coefficient of refraction 0.07 R sin 1 “= 30.876m R.L. of P = 1265.00. Solution: Axis signal correction (angular) for the difference in height of the instrument at Q and the height of the instrument at P. tan

=

S 2 h1 (equation 5.9) d sin 1"

3.90 1.55 × 206265 5600

=

(Value of

1

1 = 206265) = 86.55 seconds 01”26”. 55 for depression angle correction is additive. sin 1"

10 281 00 1

=

+

=

01 26.55 1 29 26.55 0

= observed angle corrected to axis signal

1

= central angle =

d R sin 111

5600 = 181.37 = 0311.1137 30,876

=

2

= curvature correction = 90.”68 = 011 30.” 68

r = refraction angle = m = 0.07 × 181.37 = 12.69 seconds. H = (for angle of depression)

d sin cos(

1

1

+m 2 +m

)

= 10 291 26.55 1

+m -

2

=

1

+ 12.69 29 39.24

-

01 30.68

0

(vide equation

5.13)

CE401/17

51

10 29 39.24 - 03 01.37 = 0 1+m 2 1 26 37.87

H = 5600 ×

( (

) )

0.02566 sin 10 281 08.56 = 5600 × = 143.63m 0 0.9997 cos 1 26 37.87

Difference in levels between P and Q R.L. of Q = 1265 - 143.63 = 1121.37m. Worked out example 2: 5.8.2. The following reciprocal observations were made from two points P and Q d Horizontal distance between P and Q = 4860m. angle of elevation of Q at P 10 051 2111 angle of depression of P at Q 10 001 5011 h1 height of instrument at P 1.35m S1 Height of signal at P 6.10m h2 Height of instrument at Q 1.38m S2 height of signal at Q 6.21m. Find the difference in level between P and Q and the coefficient of refraction. Take 11

R sin 1 = 30.88m. Solution: A Axis signal correction at P 1

=

(Value of

=

S 2 h1 d sin 111

(refer equation 5.9)

6.21 1.35 = 206265 = 206.2611 4860

1 = 206265 | = 031 2611 . 26 (negative) 11 sin 1

Axis signal correction at Q (equation 5.10)

S1 h2 6.10 1.38 (206265) = d sin 111 4860

= 200.32 secs = 031201132 (+ve) Note: Student is advised to repeat the calculations for axis signal correction using equation 5.7 and 5.8 Central angle

=

d 4860 = 11 R sin 1 30.88

= 157.1138 = 02.13711.38

CE401/17

52

= 011.181169

2

R sin111 = 30.88 given. This can be calculated and verified. R = radius of earth = 6370km. 1

=

1

= 100512111 - 03 26.26

1

=

1

=

1001154.74 +

2

= 10 0015011 + 03 20.32

1

1

=

+ 2

10 041101132 1

= 1001154.7411 + 10 04110.3211

(2

0

)

00 05.06 / 2

10 03 2.53 = 1

+ 2

1

+

2

1

2

= 10 03

2.53

+ 01

18.69

10 04 21.22 =

1

+ 2

1

+

2

H = difference in elevation in between P and Q 1

d sin ( cos & '

1

+ 2

+ 2

1

% + # 2 \$

1

(5.15) or (5.18)

( ) ( ) sin (1 .0507 ) = 4860 × cos(1 .0725) = 4860

sin 10 031 2115 cos 10 04 211122 0

0

= 4860 ×

0.01834 = 89.15m. 0.9998

H = 89.15m

For determining coefficient of refraction m use equation 5.3.

CE401/17

53

2r =

+

1

-

1

(r = m 2m = r=

or

2

+

+

-

1

1)

1

2

1

2

1

1

2 1

= 10 04 10.32

1

2

- 10 02 54.74 (0 02 15.58)/2 = 0110711.79 1

2

1

2

= 0 01 18.69 - 0 01 07.79 0. 00 10.90 = 10.90

r = 10.911 = 157.78

r

=m=

m = 0.069

10.9 157.38 Ans.

SUMMARY: When the distances involved are large the principles of plane surveying are not valid and the effect of curvature of earth and the refraction effect on light rays have to be taken into account particularly for determining the difference in elevation of various ground stations. The above principles were explained in detail in the above pages and the application of the same was also explained with the help of two numerical examples. *** SURVEYING UNIT - 6 TACHEOMETRY 6.1. Contents: 6.2. Aim 6.3. Objective 6.4. Introduction

CE401/17

54

6.5. Systems of Tacheometry survey 6.6. Stadia system 6.7. Angular Tacheometry (inclined sights) 6.8. Tacheometer 6.9. Anallatic lens 6.10. Theory of Anallatic lens. 6.11. Worked out Examples 6.12. Assignment problems 6.13. Summary. 6.2. AIM: To study the method of finding the horizontal and vertical distances by Instrumental observations , eliminating the direct linear measurement by chain and tape. 6.3. OBJECTIVES: After completing this lesson the student should be able to calculate the horizontal and vertical distances using the observations taken by instruments where principles of optics is being adopted. Different formulae connected with these observations have been explained in detail in this chapter. 6.4. Introduction: Tacheometry (or Telemetry) is a branch of angular surveying in which both the horizontal and vertical positions of points are determined from the instrumental observations , the labour of chaining being entirely eliminated. The principle of optics is adopted here. The usual instrument employed is called a Tacheometer and is nothing more than a transit theodolite adopted for distance measuring by the provision of special fittings. The field work can be executed with considerable rapidity more so in rough and undulating ground where ordinary levelling is tedious and chaining is both slow and inaccurate. The underlying principle common to different systems of tacheometry is that the horizontal distance between an instrument station A and a point B as well as the elevation of B relative to the instrument can be deduced from 1. The angle at A subtended by a known short distance at B and 2. The vertical angle from A to B. The various tacheometric methods available employ the principle in different ways and differ from each other in methods of observation and reduction and are classified as follows: 6.5. Different Systems of Tacheometric measurement: (i) The stadia system in which the necessary observations to the point are secured with one pointing of the telescope. (a) Fixed hair method (b) Movable hair method. (ii) The tangential method (Two pointings of the telescope are required here). (iii) Measurement by means of speical instrument.

CE401/17

55

Fig 6.1. d is the distance from object glass to the vertical axis of the instrument. D = Horizontal distance from the vertical axis of the instrument to the staff.

CE401/17

56

The rays Aoa and Bob passing through ‘O’ are straight lines so that triangles AOB and aob are similar , whence

f1 S = f2 i Q f1 and f2 are conjugate focal distances,

1 1 1 = + f f1 f 2 multiplying throughout by f f1 f1 = f + f . Substituting

f1 f2

S f for 1 i f2

f1 =

fS + f i

Adding d to each side D = f1 + d = D=

fS + f +d i

fS + f +d i

6.1

The intercepts ‘S’ is observed as the difference of the stadia hair readings. The quantities

f and f + d for i

the particular instrument must be known to calculate D.

f i

and (f + d) for a particular instrument are constant.

D = KS + C K=

f i

, called multiplying constant

6.1a

C = (f + d) , called additive constant. The above two constants

f and (f + d) are called tacheometric constants and the equation 6.1 or 6.1a as i

the distance equation in tacheometry. 6.6.3. Determination of Tacheometric constants: (1) Measuring accurately two distance D1 and D2 on the ground (distance between the instrument and the staff) and the corresponding staff intercepts S1 and S2 D1 = KS1 + C (a) D2 = KS2 + C (b) Solving the two above simultaneous equations the tacheometric constants K and C can be determined.

CE401/17

57

(2) By direct measurement on the telescope: (a) Set up the tacheometer and sight to a far off object and focus properly. (b) Measure the distance between the centre of the object glass and the diaphragm (f2) (c)

1 1 1 = + f f1 f2

f1 is large and

(fig 6.1)

1 is every small and can be neglected. f2

1 1 + f f2 or

f = f2 (measured).

(d) ‘d’ distance between the instrument axis and the objective is not constant strictly speaking , but the variation is very small for different lengths of sights. So it can be measured directly on the instrument after focussing to a far off object. (e) Knowing f and d the constant c i.e., additive constant (f + d) is known. (f) i (stadia interval) being very small cannot be measured accurately. The value of then from equation

f has to be computed first and i

f = K knowing f and K , ‘i’ can be determined. i

(g) Focus the telescope to a staff kept at a known distance. (h) Measure the distance D1 between the instrument and the staff and the corresponding staff intercept ‘S’. D1 = KS1 + C find K knowing D1 , S1 , and C K=

f , f already known find i. i

6.7. Angular Tacheometry (Inclined Sights): In this case the staff may be held either vertically or normal to the line of sight , the former method being generally preferred. 6.7.1. Inclined Sights (staff vertical): This is adopted when the ground is undulating and when it is not possible to make observations with line of collination horizontal. Observation to be taken: Angle of elevation or depression and three readings of staff corresponding to three horizontal hairs. Let

be the angle of elevation (or depression) of the line of sight from the horizontal. The inclined distance

‘L’ from the trunion axis G to the point C on the staff could be obtained directly from the formula 6.1 (or 6.1a) if the observed intercept is normal to GC. S is the (AB) observed intercept ie difference between top and bottom hair readings. A1 B1 drawn perpendicular to GC at C.

CE401/17

58

A1 B1 = S1 = S cos L= =

f 1 S + (f + d) i f S cos i

+ (f + d)

6.2a AB = S: A1B1 = S1 = S cos : C is the mid point of AB or A1B1 6.2b AB = S: A1B1 = S1 : CE = central hair reading = h Fig 6.2 (a) & (b). Since the required horizontal distance D = L cos D={ D=

f S cos i

f S cos2 i

+ (f + d)} cos

+ (f + d) cos

6.2

CE401/17

59

= KS cos2

+ C cos

6.2a

V = L sin V = [KS cos = KS cos V = KS V=

+ C] Sin

Sin

+ (f + d) Sin

sin 2 + C sin 2

6.3a

f sin 2 S + ( f + d ) sin 2 i

6.3

If E is the foot of the staff and CE = h is the centre hair reading in the staff corresponding to vertical angle at G. Reduced Level of E = R.L. of height of Instrument Axis. (foot of staff) + v - h. In case of depressed sight , R.L. of E = R.L. of Height of instrument axis - V - h.

Fig 6.2c In General R.L. of bottom of staff (staff point) = R.L. of Instrument axis ± V - h Use + Sign for angle of elevation - Sign for angle of depression. 6.7.2. Inclined sights: Staff Normal to line of Collination: In fig 6.3a General arrangement of lens instrument axis and the staff are shown. For clarity for calculation refer fig 6.3b and 6.3c.

CE401/17

60

In this case when the staff is held normal to line of sight. AB = S is perpendicular to GC line of sight. L = GC =

f S + (f + d) i

D = Horizontal distance = Distance from the instrument axis to the foot of the staff , 1

E = GF + F1 F. for angle of elevation and D = GF1 - F1F for angle of depression. GF1 = L cos =

f S cos i

+ (f + d) cos

Fig 6.3a

CE401/17

61

Fig 6.3 (b & c) Let r = central reading (reading corresponding to central hair). CC1 = Horizontal projection meeting the vertical through E , the foot of the staff at ‘C’. As per construction CC1 = FF1 (fig 6.3b) or fig 6.3c. CC1 = FF1 = r sing D={

f S cos i

Use + r Sin - r sin For vertical distances V = L sin

+ (f + d) cos } ± r sin

for angle of elevation for angle of depression

6.4

CE401/17 f S + (f + d)] Sin i

=[ =

62

f S sin i

+ (f + d) Sin

If h is the projected reading on the vertical through the foot of the staff. h = r cos

where r is the central hair reading

R.L. of the staff point E = R.L. of Instrument axis + V - r cos . R.L. of staff point. E = R.L. of Instrument axis ± {

f S sin i

+ (f +d) sin } – r cos

6.5

Use + V for angle of elevation and - V for depression angle E = R.L. of Instrumentation axis + V - h

6.5a.

6.8. TACHEOMETER: Although an ordinary transit fitted with stadia hairs or points can be employed for tacheometry , accuracy and speed are promoted if the instrument is specially adopted for the work. The requirement of a good tacheometer (used for tacheometry) are: (1). The telescope should be powerful having a magnification of at least 20 diameters (preferable 20 to 30 diameters). (2) For sufficiently bright image the effective diameter of objective should be at least 35mm (preferable 35 to 45mm). (3) The multiplying constant (

f ) should have a value of 100. i

(4) The telescope should be truly Anallatic (explained below in 6.9). 6.9. The Anallatic Lens: To eliminate the additive constant (f + d) or (c) from the distance equation. D= or

f S + (f + d) i

KS + C

For Tacheometric telescopes an additional lens known as an anallatic lens is sometimes provided in the telescope between the object glass and the eye piece , which makes the additive constant vanish thus simplifying the computations. Principle: D=

f S + (f + d) shows that the staff intercept is proportional to D - (f + d) the distance between the staff i

station and the exterior principle focus of the objective. The latter point there- fore forms the apex of a constant visual angle between the sides of which the quantity ‘S’ is intercepted. If this apex were situated on the vertical axis

CE401/17

63

of the instrument (point ‘G’ in the earlier figures of this chapter) the term (f + d) would vanish and D would be proportional to ‘S’. This is accomplished by the introduction in the telescope of an additional convex lens called an ‘ANALLATIC LENS’ placed between the eye piece and object glass and at a fixed distance from the object glass. The anallatic lens is generally provided in external focussing tacheometer telescopes. Its use simplifies the reduction of observations but it is open to the objection that it increases the absorption of light in the telescope with consequent reduction in brilliancy of the image. If the additive constant is zero and the multiplying constant = 100 (by the introduction of anallatic lens). Then D = KS = 100 S 100 times the staff intercept. Self Assessment Questions: (1) What is an Anallatic Lens ? (2) What are the advantages of providing an Anallatic Lens ? (3) Mention the disadvantages if any of using an anallatic lens ? 6.10. THEORY OF ANALLATIC LENS: Let

O = Optical centre of the objective N = Optical centre of the Anallatic lens G = Position of vertical axis of the instrument F1 = Exterior Principal focus of Anallatic lens. A , B points on the staff corresponding to the stadia wires i = ba is the position of the actual image i1 = b1a1 is the image which would be produced if no anallatic lens were interposed. f1 and f2 the conjugate focal lengths of the objective

D = Distance of the staff from the vertical axis of the instrument.

CE401/17

64

Fig 6.4 d = distance of the vertical axis (G) from the objective f1 = focal length of Anallatic lens f = focal length of the objective m = distance of diapharagm from objective n = distance of the anallatic lens from the objective. The rays emanating from A and B corresponding to stadia wires along AG and BG are refracted by the object glass and meet at a point F1. The distance between the Anallatic lens and the object glass is so fixed that the point F1 happens to be the exterior principal focus of the anallatic lens. Hence the rays passing through F1 will emerge in a direction parallel to the axis of the telescope after being refracted by the anallatic lens. Then ba is the inverted image of AB of the staff the points ‘b’ and ‘a’ correspond to the stadia wires. If the anallactic lens was not interposed the rays would have formed a virtual image b1a1 at a distance f2 from the object glass. From the conjugate relationship for the objective

1 1 1 = + f f1 f2

6.6a

Since the intercept and the image (S and 1) are proportional to their distance from ‘O’.

S i1

=

f1 f2

6.6b

For the anallatic lens b1a1 and ba are conjugate and their distances (f2 - n) and (m - n) from N are connected by

1 = f1

1 f2

n

+

1 m

n

6.6c

CE401/17

65

The negative sign being required for (f2 - n) since b1a1 and ba are on the same side of N. The length of b1a1 and ba are proportional to their distance from N so that

i1 f 2 = i m

n n

6.6d

An expression for D can now be obtained by eliminating f2 , m and i1 , from the above equations. Multiplying 6.6b and 6.6a , we get

S f f = 1. 2 i f2 m But

f1 f = 1 f2 f =

f1

also

f2 =

And

f2 m

n n

1

(from 6.6a

f f ff1 f1

f

n f2 n = +1 n f1 =

substituting for

(from (6.6c)

n+ f1

f2

f1

f1 f and 2 m f2

n in 6.6c n

n+ f1

S f f f = 1 × 2 i f

f1

( ff 1 % n + f1# & f1 f & f1 f # = # f & f1 & # #\$ '& on further simplification

f1 =

ff 1 S i f + f1

(

f1

f n f + f

n

1

) n

The distance between the instrument axis and the staff = f1 + d D = (f + d) = This is of the form D = KS + C

ff 1 f + f

1

n

×

S i

(

f1

f n f + f

1

) +d

n

6.6f

CE401/17

where K =

66

ff 1 1

f + f

n

(

×

f1

f n

C=d-

f + f

1 i

1

) n

Now the condition that D should be proportional to S requires that the 2nd and 3rd terms in 6.6f should vanish. i.e.,

d=

(

f1

f n f + f

1

)

n

which is secured by placing the anallatic lends N so that

n= f1 +

fd f +d

6.6g

under these conditions the apex of the tacheometric angle is situated at G , the centre of the trunnion axis. The value of f1 and i must be so arranged that the multiplier

ff 1

(f + f

1

1

n

)× i

is a suitable round number say 100.

If all these conditions are fulfilled then the Distance equation with an anallatic lens reduces to D = KS = 100 S

6.7

The formula for inclined sights on either the fixed or the movable hair (to be studied in next unit) system are similarly modified by omitting the term involving the second constant i.e., D = L cos

= KS cos2

V = L sin

=

KS sin 2 etc..... 2

The Anallatic lens is usually provided with means for adjusting its position in the telescope so that if the constant is found to differ from 100 , the distance n may be adjusted until the desired value is obtained.

6.11. Worked out Examples: 6.11.1. A tacheometer has an object glass with 15cm focal length and stadia interval 3mm. If the object glass distance from the trunnion axis is 25cm , arrive at the tachemetric formula for the distance in terms of the staff intercept. Solution: Data given f = 150 mm i = 3mm d = 25mm Multiplying constant

f 150 = 50 = 3 i

Additive constant (f + d) = 15 + 25 = 40 cm = 0.4 metres.

CE401/17

D=

67

f s + (f + d) i

D = (50 S + 0.4) metres.

Ans.

6.11.2.: To determine the tacheometric constants a levelling staff was kept at distance of 60 and 80 meters from a theodolite station and the corresponding staff intercepts recorded wre 0.6m and 0.80 meters respectively. Find the constants D = KS + C 80 = 0.8K + C

(1)

60 = -.6K + C

(2)

(1) - (2) = 20 = 0.2K. K = 100 Substituting the value of K in (1) 80 = -.80 x 100 + C Ans. K = 100 , C = 0. 6.11.3. (JNTU 1984 Sept): The following tacheometric data pertain to an intermediate point on AB Station

A

B

Staff intercept

2.47

Axial hair reading Bearing Vertical angle

2.27

3.34 0

1

3.13 0

218 37

1

38 37 0

1

- 50 30

- 600 201

The instrument fitted with anallatic lens has a constant as 100. Compute the length AB and R.L. of B. R.L of A is 223.80 m and the observations are based on a vertically held staff. Solution: Note to student: In all survey problems the student is advised to draw as many sketches as are required to get a clear idea about the problem and then transfer the given data on to the sketches. Here if the instrument station is ‘O’ and the staff stations are A and B , if the whole circle bearings are indicated in the diagram it will be clear from the diagram that the staff stations A and B and the instrument station ‘O’ are in the same vertical plane in a straight line (though sloping on either side from the instrument station) as is evident from the minus vertical (depression) angles.

CE401/17

68

Fig 6.5

As the staff at A and B is held vertical D = KS cos2

K = 100 C=0

+ C cos

as the instrument is fitted with an anallatic lens

D = 100 S cos2 For distance OA1 , S = 2.47 ,

1

= 500.5

OA1 = 100 x 2.47 cos2(500.5) = 247 x (0.636)2 = 99.91 OB1 = 100 x 2.27 cos2 (600.333) = 227 x (0.495)2 = 55.62. Distance AB = A1O + OB1 Horizontal distance between A and B =

S 1 = 2.27 2

= 600.333

CE401/17

69

A1O + OB1 = 99.91 + 55.62 = 155.53 V1 = KS1 =

sin 2 2

1

=100 x 2.47 x sin (2 x 500 .5) / 2 =

247. sin 790 = 121.28 2

similarly V2 =

(

100 × 2.27 sin 2 × 60 0 201 2

(

79 )

)

590 201

)

=

227 sin 180 0 2

=

227 × sin 59 0 201 = 97.61 2

(

247 sin (180 2

)

R.L. of A given as 223.8 R.L. of instrument axis = 223.8 + 3.34 + 121.28 = 348.42 (Axial hair reading V1 at A) R.L. of B = R.L. of instrument axis - V2 - central hair reading at B = 348.42 - 97.61 - 3.13 = 247.68. Answers Horizontal distance AB = 155.53 meters R.L. of B = 247.68m.

6.11.4. (JNTU Aug / Sept 83): In a tacheometric survey , conducted with an instrument whose constants are 100 and 0.5m the staff was inclined so as to be normal to the line of sight for each reading. Two sets of readings were as given below. Calculate the gradient between the staff stations P and Q and the reduced level of each if that of R = 41.800 m. Instrument

Ht of Instrument

Staff Station

Bearing

Station R

Vertical

angle 1.600

P

0

85

+ 40301

1.000 1.417 1.833

Q

0

135

0

-4 .00

1.000 1.657 2.313

Solution:

CE401/17

70

Fig 6.6 From the given data and the above figures it is evident that the included angle (Horizontal between RP and 0

RQ = 50 . From R to P it is angle of elevation and from R to Q it is angle of depression. The observation sketches for the two lines are shown separately below. L1 = KS1 + C

K = 100 C = 0.5

S1 = 1.833 - 1.000 = 0.833.

(if necessary the student is advised to refer the corresponding sketch given near the derivation).

Fig 6.7 L1 = 100 x 0.833 + 0.5 = 83.3 + 0.5 = 83.9m. D1 = R1 C1 + C1 P1 = L1 cos

1

+ r1 Sin 0

1

1

= 83.8 Cos (4 . 30 ) + 1.417 x sin (40 . 301) = 83.8 x 0.9969 + 1.417 x 0.0785 = 83.54 + 0.1112. D1 = 83.651

CE401/17

71

V1 = L1 sin

1

= 83.8 sin (40 . 301) = 83.8 x 0.0785 = 6.5783. h1 = component of central hair reading along the vertical = r1 cos

1

= 1.417 x cos (40 301)

= 1.417 x 0.9969 = 1.4126 R.L. of R = 41.800 R.L. of P = R.L. of R + Ht. of instrument + V1 - h1 = 41.8 + 1.6 + 6.5783 - 1.4126. R.L. of P = 48.5657. In the next stage calculate the distance RQ and then the R.L. of Q.

Fig 6.8 L2 = inclined distance along line of collination = KS2 + C

K = 100 C = 0.5

S2 = 2.313 - 1.000 = 1.313

L2 = 100 x 1.313 + 0.5 = 131.3 + 0.5 = 131.80 D2 = R1 C1 - Q1 C1 = L2 cos

2

- r2 sin

2

0

= 131.80 cos 4 - 1.657 sin 40 = 131.8 x 0.9976 - 1.657 x 0.0698 = 131.484 - 0.1157 = 131.368. D2 = 131.368 V2 = L2 sin

2

= 131.368 sin 40 = 131.368 x 0.0698 = 9.169m. h2 = projection of central hair reading r2 on the vertical axis = r2 cos

2

= 1.657 cos 40 = 1.657 x 0.9976 = 1.653.

R.L. of Q = R.L. of R + Height of Instrument- V2 - h2

CE401/17

72

= 41.8 + 1.60 - 9.169 - 1.653. R.L. of Q = 32.578 Difference in R.L. of P and Q = 48.565 - 32.578 = 15.985 say 16m Horizontal distance between P and Q is given by

Fig 6.9 PQ =

PR 2 + RQ 2

2 PR.RQ cos R

=

(83.651)2 + (131.368)2

=

6997 + 17258 14127

( )

2 × 83.51 × 131.368 cos 50 0

= 10128 = 100.63 Gradient from P to Q =

16 1 = 100.63 6.28

Answers:R.L. of P = 48.57 R.L. of Q = 32.59 Gradient from P to Q 1 in 6.28. 6.12. Problems for Practice: 6.12.1. A tacheometer has a diaphragm with three cross hairs spaced at distances apart of 0.65m. The focal length of

the object glass is 23.4cm. A staff is held normal to the line of sight on a Bench Mark whose R.L. 106.365m. The reading on the staff are 2.465 , 2.010 , 0.155m with the telescope inclined at an angle of 50201 downwards.

Find the distance of the instrument station from the Bench Mark at its reduced level if the height of the instrument above the ground is 1.35m (UPSC 1974). 6.12.2. Tacheometric readings were taken from a survey station ‘S’ to a staff held vertically at two pegs A and B

and the following readings were recorded.

CE401/17

73

Point

Horizontal Circle

A

0

Vertical Circle

1

1

62 00

1

+ 4 10 30

0

152 00

B

1

0

1

1

- 5 05 00

0.881 1.881 2.880

The multiplying constant of the instrument was 100 and the additive constant zero. Calculate the horizontal distance from A to B and the height of peg A above the axis level of the instrument. 6.12.3.: A line was levelled tacheometrically with a tacheometer fitted with an anallatic lens , the value of the constant being 100. The following observations were made with the staff held vertical on each station. Instrument

Ht. of axis

Staff Station

Vertical angle

1.60

B.M

-20 . 181

1.650 , 2.500 , 3.350

Station P P Q

1.60 1.50

0

1

Q

+8 .36

R

0

1

+10 .42

1.720 , 2.670 , 3.620 1.055 , 2.055 , 3.055

R.L. of B.M = 250.250. Determine the Gradient of the line QR (AMIC Winter 1983).

SUMMARY:In this chapter measurement of distances using optical principles was explained. This method is useful particularly in sloping and undulating terrain where the measurement by chain or tape is difficult. Different methods adopted under different situations were explained above. For quick computation of distances (Horizontal and Vertical) a lens called anallatic lens is introduced in the telescope of theodolites to make the multiplying constant 100 and addition constant zero , in the distance equation. The principles of the same was explained in detail along with numerical examples worked out at the end. *** SURVEYING UNIT - 7 7.1. CONTENTS: Introduction , Principles of Subtense Method Distance equation , Subtense Bar - Problems , Tacheometric tables. 7.2. AIM: The Principle of Subtence Method of Tacheometry Types of subtense method and subtense bar are explained in this chapter. The use of tacheometric tables is also explained. 7.3. OBJECTIVES:

CE401/17

74

After going through this chapter the student should know the difference between the fixed hair stadia method and the movable hair stadia method and the basic difference between the instruments used in the fixed and movable hair methods. The student should also know the use of tacheometric tables for quick reduction of data when a large number of observations is involved. 7.4. INTRODUCTION: Stadia method is one of the methods of tacheometric surveying. Stadia metod is subdivided into (i) fixed hair method and (ii) movable hair method. In the fixed hair method as the name indicates the distance between the stadia hairs (one above and the other below the central hair) remains constant and the staff intercept varies. In the movable hair method also knwo as SUBTENSE METHOD the stadia interval i.e., the distance between the two stadia hairs is not constant. The intercept ‘S’ (distance between given targets) is constant and the distance between the stadia hairs for a particular intercept varies depending on the distance of the staff or the targets on the staff from the instrument. Depending on how the targets are placed one over the other (vertical) or in a horizontal plane (side by side) (horizontal) the methods are known as VERTICAL SUBTENSE METHOD OR HORIZONTAL SUBTENSE METHOD. The terms ‘subtense method’ is now more or less exclusively applied to horizontal base subtense method. Both the methods are explained in the chapter besides working out some numerical examples. The use of tacheometric tables is also explained.

7.5.1. PRINCIPLE OF SUBTENSE METHOD: For the subtense method of tacheometry the theodolite must be fitted with an arrangment to alter the distance between the stadia hairs for the given intercept. A theodolite with such arrangement is known as subtense theodolite. For moving the upper and lower wires above or below the central fixed hair , micrometer screw arrangement (fig 7.1) is made. Two micrometer graduated drums are provided one for moving the upper wire and the other for moving the lower wire. The whole number of revolutions made by the micrometer drum are recorded on the vertical dial while the fractional portion of the revolutions for the upper and lower wires are read on the upper and lower drums respectively. The distance of the respective wires from the central fixed wire is first read in terms of number of revolutions made on the drum and then converting the number of revolutions into linear distance by multiplying the number of revolutions with the pitch of the screw. Pitch of the screw is the linear distance traversed by the wire for one complete revolution of the drum.

CE401/17

75

Fig 7.1. Micrometer screw arrangement for knowing stadia interval. If n1 and n2 are the readings of the number of revolutions read on the upper and lower micrometer drums respectively the total number of revolutions of the drum n = n1 + n2 and if ‘p’ is the pitch of the screw , Distance between the stadia wires = np = i 7.5.2. DISTANCE EQUATION: In the fixed hair method of tacheometry distance equation D=

f S + (f + d) i

Here f/i and (f + d) are constants , K and C respectively D = KS + C In the movable hair method as ‘i’ is variable f/i is not constant i = np D=

f S + (f + d) np

p = pitch of the screw is constant

f = constant say K1 p D=

K1 S +C n

7.5.2.1.

When S = 0 distance between stadia hair = 0. If there is an index error , ‘e’ in the initial reading it should be added or subtracted appropriately. In that case equation 7.5.2.1. becomes

CE401/17

76

K1 S +C D= n e

7.5.2.2

For inclined sights and staff held vertical

K1 S Cos2 + C cos D= n V=

K1 S n

7.5.2.3

Sin2 + C Sin 2

7.5.2.4

When there is no index error. 1

or

D=

KS Cos 2 + C cos n e

V=

K1 S n e

Sin2 + C sin , 2

When there is index error of ‘e’ and S = staff intercept (the base) or the distance between the targets. 7.5.3. SUBTENSE BAR: It is a horizontal wooden bar about 2 to 3 meters long fitted with two circular vanes at a constant distance of 2m apart supported on a tripod stand. In the center of the horizontal bar there is a sight rule placed at right angles to direction fo the bar for aligning the bar at right angles to the line of sight. The sight alidate can be turned on its pivot so that it is parallel with the horizontal bar when not in use. The fixed circular targets are coloured in red and white diagonally on one side and black and white on the other side. When it is intended to fix the two targets at 3m distance the red and white targets are kept towards the observer and the targets fixed in slots which are enactly 3 meter center to center. For 2m center to center between the slots the back side of the targets is kept towards the observer. The colour enables the observer (from a distance) to observe the distance apart between the targets.

Fig 7.2(a) Subtense Bar

CE401/17

77

Fig 7.2 (b) 7.5.4. PROBLEMS: The stadia intercept read by means of a fixed hair instrument on a vertically held staff is 1.05 meters , the angle of elevation being 50 361. The instrument constants are 100 and 0.3. What would be the total number of turns registered on a movable hair instrument at the same station for a 1.75m intercept on a staff held on the same point the vertical angle in this , case being 50 241 and the constants 1000 and 0.5. Solution: (a)

D=

f S cos2 i

= KS1 cos2

1

1

+ (f + d) cos

+ C cos

1

with fixed hair instrument

1

Given K = 100 C = 0.3 ,

1

= 50 . 361 = 50. 6

S1 = 1.05 D = 100 × 1.05 cos2 50.6 + 0.3 cos 50.6 = 104 + 0.298 = 104.298. (b) For the same position of the staff and intercept with movable hair instrument D=

Given

2

K1 S2Cos 2 n

2

= 50 . 241 = 50 . 4

S2 = 1.75 C = 0.5 K1 = 1000

+ C1 cos

2

CE401/17

78

104.298 =

1000 × 1.75 2 cos 5.4 + 0.5 cos 50 . 4 n

1734.5 + 0.4978. n 1734.5 = 104.298 - 0.4978 = 103 n

=

n = number of turns registered on a movable hair instrument. =

1734.5 = 16.7 103.8

= 16 full turns + 0.7 times a turn

7.5. PROBLEM FOR EXERCISE: (Part of problem UPSC 1976): A theodolite (with a multiplying constant of 100 and additional constant zero) was set up at station A and observation were taken to a horizontally held subtense bar at station D. The horizontal angle subtended by the end targets of the subtense bar at the theodolite at A was 00 5112011 . Distance between the end targets of the subtense bar was 4m. Vertical angle at A to the central target on the subtense bar = 160 241 1811 . Find the horizontal distance between the instrument station (A) and the subtense bar position (D). and also the difference in elevation between A and D. 7.6. TACHEOMETRIC TABLES: For survey of large extent where the number of points observed in a tacheometric survey are large reduction of stadia notes can be done quickly with the help of tacheometric tables. In the tacheometric tables the corrections to be applied to the horizontal distance and the vertical component as a multiple of staff intercept are given for different values of . A simple form of such a table for an angle of elevation up to 30 is shown below and a numerical example explaining its use is also given. To explain the use of tacheometric table given below: Let S = Staff intercept = 1.5m in a particular observation = 20 201 Additive constant (f + d) = c = 0.4. TACHEOMETRIC SURVEYING: 00

Minutes

0 2 4 6 8 10 12 14 16

Hor. Corr. 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

10 Diff. Elev 0.00 0.06 0.12 0.17 0.23 0.29 0.35 0.41 0.47

Hor. Corr. 0.03 0.03 0.03 0.04 0.04 0.04 0.04 0.05 0.05

20 Diff. Elev 1.74 1.80 1.86 1.92 1.98 2.04 2.09 2.15 2.21

Hor. Corr. 0.12 0.13 0.13 0.13 0.14 0.14 0.15 0.15 0.16

30 Diff. Elev 3.49 3.55 3.60 3.66 3.72 3.78 3.84 3.89 3.95

Hor. Corr. 0.27 0.28 0.29 0.29 0.30 0.31 0.31 0.32 0.32

Diff. Elev 5.23 5.28 5.34 5.40 5.46 5.52 5.57 5.63 5.69

CE401/17 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 C = 0.2m C = 0.3m C = 0.4m

79

0.00 0.00 0.00 0.00 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.01 0.02 0.02 0.02 0.02 0.02 0.02 0.03 0.03 0.03 0.00 0.00 0.00

0.52 0.58 0.64 0.70 0.76 0.81 0.87 0.93 0.99 1.05 1.11 1.16 1.22 1.28 1.34 1.40 1.45 1.51 1.57 1.63 1.69 1.74 0.00 0.00 0.00

0.05 0.05 0.06 0.06 0.06 0.07 0.07 0.07 0.07 0.08 0.08 0.08 0.09 0.09 0.10 0.10 0.10 0.11 0.11 0.11 0.12 0.12 0.00 0.00 0.00

2.27 2.33 2.38 2.44 2.50 2.56 2.62 2.67 2.73 2.79 2.85 2.91 2.97 3.02 3.08 3.14 3.20 3.26 3.31 3.37 3.43 3.49 0.01 0.01 0.01

0.16 0.17 0.17 0.18 0.18 0.19 0.19 0.20 0.20 0.21 0.21 0.22 0.22 0.23 0.23 0.23 0.24 0.25 0.26 0.26 0.27 0.27 0.00 0.00 0.00

4.01 4.07 4.13 4.18 4.24 4.30 4.36 4.42 4.47 4.53 4.59 4.65 4.71 4.76 4.82 4.88 4.94 4.99 5.05 5.11 5.17 5.23 0.01 0.01 0.02

0.33 0.34 0.34 0.35 0.36 0.37 0.37 0.38 0.38 0.39 0.40 0.40 0.41 0.42 0.43 0.44 0.44 0.45 0.46 0.47 0.48 0.49 0.00 0.00 0.00

5.75 5.80 5.86 5.92 5.98 6.04 6.09 6.15 6.21 6.27 6.32 6.38 6.44 6.50 6.56 6.61 6.67 6.73 6.79 6.84 6.90 5.96 0.01 0.02 0.02

To find the horizontal distance between the staff and the instrument and the difference in elevation between the instrument axis and axial reading on the staff (vertical component). From the table corresponding to 20 201 the following information is noted: Horizontal correction

Difference in elevation

0.17

4.07

0.00

0.02

Distance reading = 100 x 1.5 = 150m Horizontal Distance = 150 - (0.17 x 1.5) + (0.4 - 0.00) 150.00 - 0.255 + 0.4 = 150.145m. Horizontal distance = 150.145m. V = 4.07 x 1.5 + 0.02 = 6.105 + 0.02 = 6. 125m. ***

SURVEYING UNIT-8

CE401/17

80

8.1. CONTENTS: Introduction, Tangential method (Different cases) , Problems , Special Instruments , (i) Beam Stadia Arc , (ii) Direct Reading or Auto-Reduction (Self-reducing) tacheometers. (a) Jeffcott Direct Reading Tacheometer (b) Szepessy Direct Reading Tacheometer and (c) Auto reduction (self reducing) Tacheometer. 8.2. AIM: The tangential method of tacheometry and the use of special instrument for quick reduction of Tacheometric data are explained with the help of examples. 8.3. OBJECTIVES: As this is the concluding chapter on Tacheometry the student is expected to know thoroughly the difference between different methods used in this type of survey and also to know in detail the different special instruments (mentioned in this chapter) used for quick reduction of tacheometric data. 8.4. INTRODUCTION: When the telescope of a theodolite is not fitted with a stadia diaphragm the TANGENTIAL METHOD of tacheometry is used. In this method the horizontal and vertical distance from the instrument to the staff station are computed from the observed vertical angles to two vanes fixed at a constant distance ‘S’ apart upon the staff. The vanes are bisected every time with the axial hair and necessarily two vertical angles are to be observed corresponding to each vane and recorded. There may be three cases of the vertical angles: (i) Both angles - angles of elevation (ii) Both angles - angles of depression (iii) One angle of elevation and one angle of depression. 8.5. TANGENTIAL METHOD: 8.5.1. Both the observed angles being angles of elevation: P1 represents the axis of the instrument (fig 8.1) P position of the instrument Q Staff station R and T the two vanes angle of elevation to T from P1 angle of elevation to R from P1.

CE401/17

81

Fig 8.1. D = Horizontal distance between the instrument on the staff h = Height of lower vane above the foot of the staff at Q From fig 8.1, V + S = D tan

(A)

V

(B)

= D tan

(A) - (B) = S = D(tan

D=

(tan

S tan

- tan )

)

8.5.1.1

V = D tan

V=

S tan tan tan

8.5.1.2

Equation 8.5.1.1. gives the horizontal distance between the instrument station and the staff. Equation 8.5.1.2 gives the difference in elevation between the instrument axis and the lower vane. Elevation of staff station Q = Elevation of Inst. Axis + v - h. 8.5.2. Both the observed angles being angles of depression:

CE401/17

82

Fig 8.2. V = D tan

(A)

V - S = D tan A - B = S = D(tan

(B) - tan )

D = S / (tan V = D tan

- tan ) =

8.5.2.1

S tan (tan tan

8.5.2.2

)

Elevation of staff station Q = R.L. of Inst. axis p1 - V - h. 8.5.3. One of the obserbed angles is an angle of elevation and the other an angle of depression:

Fig 8.3 S - V = D tan

(A)

V = D tan

(B)

A + B = S = D(tan D=

(tan

S + tan

+ tan )

)

8.5.3.1

CE401/17

83

V = D tan =

S tan (tan + tan

8.5.3.2

)

Elevation of staff station Q = Elevation of Instrument axis p1 - V - h. The amount of calculation work is practically the same in both the tangential and stadia methods. For each observation in tangential method vertical angle is to be observed. This method is considered inferior to the stadia method. The stadia method (fixed hair method) with the staff held vertical is the common used method of Tacheometry. 8.6. PROBLEMS: 8.6.1. An ordinary transit theodolite without stadia hairs was set up at station p and the vertical angles observed to upper and lower vanes (T and R) of a staff held at Q are respectively 60 ,301 and 20 241. The distance between the vanes of the stadia rod was 2m. R.L. (Elevation) of station was 10.25m. Height of instrument at p was 1.650m. Height of lower vane above ground is 1.05m. Calculate the horizontal distance PQ and the R.L. of staff station Q. Solution:

Fig 8.4 0

V + 2m = D tan 6 30 V

1

0

(1)

1

= D tan 2 24 0

(2) 1

0

1

(1) - (2) = 2 = D(tan 6 30 - tan 2 24 ) = D(0.1139 - 0.0419) = 0.07202 D. D = 2 / 0.07202 = 27.768 m. 0

Ans.

1

V = D tan 2 , 24 = 27.768 x 0.0419 = 1.164m. R.L. of Instrument axis = R.L. of p + ht. of Instrument

CE401/17

84

= 10.25 + 1.65 = 11.90 m. R.L. of Q = R.L. of instrument axis + V - h = 11.9 + 1.164 - 1.05 = 12.014 m. 8.6.2. Exercise: An intercept of 3m was marked on vertical pole (Q). The angle of elevation to the top of the intercept is 0

1

3 .30 and to the bottom of the intercept is 20.301, from an instrument station p. Height of Instrument axis at p is 1.52m and the R.L. of the instrument station is 20.50. If the bottom of the 3m intercept is at a height of 1.10m from the bottom of the pole (Q). Calculate the reduce level of the bottom of the pole. (Ans. 19.67 m). 8.7. SPECIAL INSTRUMENTS: 8.7.1. THE BEAMAN STADIA ARC: It is a mechanical device fitted to the vertical circle of a theodolite (or to the telescopic alidata of a plane table) to reduce rapidly an inclined stadia distance (L) to the corresponding horizontal distance (D) and the vertical component (V) (i.e., the difference in elevation) without measuring vertical angles and without much calculation or even without the use of tables etc. Principle Used: The reductions would be simplified if only the values of

used were those for which either cos2

or

1 sin 2 is a convenient figure. The former (cos2 ) varies too slowly for the small angles usually required , but a 2 list of values of

for which

1 sin 2 = 0.01 , 0.02 etc. can be prepared as follows: 2

‘ ‘ to the nearest second

1 sin 2 2 0.01

8

341

2311

0.02

1

08

46

0.03

1

43

12

0.04

2

17

39

0.05

2

52

11

0.06

3

26

46

‘ ‘ to the nearest second

1 sin 2 2 0.07

40

011

2611

0.08

4

36

12

CE401/17

85

0.09

5

11

06

0.10

5

46

07

etc.

etc.

If these particular angles were used the vertical component ‘V’ for a multiplying constant of 100 and no additive constant would be S , 2S , 3S etc. This is the principle used in preparing the graduations of the Beaman Stadia Arc. The Beaman Stadia Arc is fixed to the vertical circle and consist of a scale engraved with the above angles on either side of zero upto about 260 331 5411 for which The scale is figured in terms of 100

sin 2 2

1 sin 2 = 0.40. 2

and is read against a fixed index mark. The arc carries two

scales H and V having their central points marked 0 and 50. To avoid possible confusion between elevation and depression the zero is commonly marked 50 so that 50 must be subtracted from every reading. No fitting is required to enable fractional parts of the scale to be read since for every sight a graduation of the scale is to be brought opposite the index by means of the vertical circle tangent screw (fig 8.5). To facilitate the calculation of horizontal distance the stadia arc also carries a scale of percentage reduction to be applied to the distance reading

f ×s. i

The example below will explain the use of Beaman stadia arc. Its scales are shown in fig 8.5.

CE401/17

86

Fig 8.5. Example: Observed Data: Central hair reading: 1.86 Reading on V scale : 56 Reading on H scale : 2 Staff intercept

: 1.5m

Elevation of instrument axis: 80.00 Assuming constants of the instrument as 100 and 0 , determine (i) the horizontal distance between instrument and the staff and (ii) the elevation of the staff. Solution: Stadia arc reading = 56 - 50 = 6. V = 6 x 1.5 = 9.0 Elevation of the instrument being 80.00,

CE401/17

87

elevation of the staff = 80 + 9.0 - 1.86 = 87.04m Horizontal correction = 2 x 1.5 = 3.0m. Horizontal distance = 100 x 1.5 - 3.0 = 147.0m. 8.7.2. JEFFCOTT DIRECT READING TECHEOMETER: Invented by the late Dr. Jeffcott , this direct reading tacheometer involves an automatic variation in the stadia interval so continued that multiplying the observed stadia intercept read on the staff by a fixed constant gives the reduced horizontal distance and also the vertical component in the same way. In this instrument the horizontal and stadia hairs of the diaphragm of the ordinary tacheometer are replaced by three platinum or platinum iridium pointers of which the two outer ones were movable. Both the horizontal and vertical components of the line of sight could be obtained directly without having to measure the vertical angle.

Fig 8.6. The Jeffcott Direct Reading Tacheometer When the telescope is elevated or depressed the two outer pointers were actuated by a system of levels and cams in such a manner that the staff interval read between the fixed central pointer and the movable right hand pointer when multiplied by 100 (200 in some instruments) gives the reduced horizontal distance and the staff intercept read between the center pointer and the left had pointer when multiplied by 10 gives the vertical component. 8.7.3. Another type of direct reading tachemeter also was invented by a Hungarian going by the name the ‘Szepessy direct reading tacheometer’ based on the principle of tangential system using the percentage angles. A scale of percentage of vertical angles is engraved on a glass arc which is fixed to the vertical circle cover. By means of prisms , this scale is reflected in the view of the eye-piece and when the staff is sighted , the image of the staff is seen along side that of the scale. The scale is divided to 0.005 and numbered at every 0.01 so that the graduation 10 corresponds to the angle whose tangent is 0.10 or 10%.

CE401/17

88

The read the staff (i) a numbered division say 15 is brought opposite the horizontal cross hair by means of the vertical circle tangent screw and the staff reading at this division (axial reading) is noted. (ii) The staff intercept between the short 0.005 divisions immediately , above and below the numbered division is read. This intercept multiplied by 100 gives the horizontal distance D , while the vertical component V is obtained by multiplying the intercept by the number marking the division brought opposite the horizontal cross hair. Suppose the staff intercept is 1.96 and the number is 15. Then the horizontal distance (D) is 196 and the vertical component is 1.96 x 15 = 29.4. Direct reading tacheometers have the following defects viz. (i) Pointers are inconvenient to read with (ii) Half intercepts cannot be measured and (iii) Effect of parallax is unavoidable.

8.7.4. Auto - Reduction (Self reducing) , Tacheometers: In this type of tacheometer the movable pointers are replaced by special curves engraved on rotating glass circles. In the kern DKR self reducing tacheometer the engraved circle is mounted at right angles to the axis of the telescope with the face on which the curves are engraved almost in contact with the ordinary diaphragm carrying a vertical line. The circle rotates as the telescope is elevated or depressed and the appropriate parts of the four curves are seen in the eye piece against the center of the staff when the fixed vertical line is directed to it. For the Hammer Fennel type of Auto reduction tacheometer , in the field of view are seen four curves marked by the letters N , E , D and d. (Fig 8.7). The ‘N’ curve is the zero curve, ‘E’ curve is for reading distance ‘D’ curve is to be used for angles upto + 0

14 while ‘d’ curve is for angles upto + 470 . The curve D and d are marked +ve for angles of elevation and for angles of depression. The multiplying constant for the distance curve is 100 , that for the height curve (D) , is 10 while for the height curve (d) , it is 20. To take a reading the zero curve is (N) made to bisect the specially marked zero point of the staff by bringing the perpendicular edge of the prism into line with the staff. The staff readings are taken with the distance curve and the height curve. The distance curve reading multiplied by 100 gives the horizontal distance D while the height curve reading multiplied by the corresponding multiplying constant gives the vertical component V. Auto reduction tacheometers permit both the distance and the difference of altitude to be read by a single reading of a vertically held staff , thus reducing the tacheometric operation as simple as that of ordinary levelling.

CE401/17

89

Fig 8.7 Special Auto Reduction Device (Hammer Fennel) 8.7.5. Electronic Tacheometers: A recent development in tacheometers was made by Zeiss bringing into use two electronic tacheometers. (i) SM. 11 electronic tacheometer and (ii) Reg Elta , 14 electronic recording tacheometer. Both these instruments look like large theodolities. The former is one with digital read-out of distances and circle read - output by means of a scale reading microscope and the latter is an electronic recording tacheometer offering not only digital readout of angles and distances , but also direct recording of the data , on founched paper. Any further elaboration of these instruments is beyond the scope of the present study. *** CURVE SURVEYING 9.1 Contents: Introduction, Types of curves, Simple Curves, Designation of a curve, Elements of a curve. Setting out simple curves; Linear methods-Offsets from long chord, Offsets from tangents, Offsets from Chords produced. Angular method: Rankine’s method of deflection angles, two theodolite method, Tacheometric method, problems, exercises. 9.2. Aim: Different types of curves and the condition under which each type is adopted with emphasis on simple curves is explained in this Unit. Different methods of calculating the date required for setting out simple curves in the field (both linear and angular) are explained with numerical examples at the end of the unit. 9.3. Objectives:: As it is difficult to get a long continuous stretch of straight approach connecting two places on the surface of the earth, curves are a must in communication lines to over come the obstacles on the way. The student is expected to know the different methods of setting out the curves connecting the straight portions. This chapter suggests some of the popular methods in different situations.

CE401/17

90

9.4. Introduction: The ranging of curves is required in the location of various kinds of public works. For example in road, railway or pipe line construction two straights will normally be connected by a curve whenever there is a change in direction. As a general rule a curve which is a circular arc is required to connect two straight lengths and they must be tangential to the curve in order that there shall be no abrupt break at the junctions. Different types of curves are adopted in different situations and each is dealt with separately here in different units. 9.4.1. Simple Curve or a Circular Curve: It is a single arc connecting two straights. This curve is dealt with in

detail in Unit 9.

Fig.9.1.(a)

Fig.9.1.b

9.4.2. Compound Curve: Consists of two arcs of different radii bending in the same direction and lying on the same side of their common tangent, their centers being on the same side of the curve. (Unit 10). 9.4.3. Transition curve: A curve of variable radius inserted between a straight and a circular curve in order that the centrifugal force may build up in a gradual and uniform manner and this way the lateral shock is minimised (Unit 11)

FIG.9.1.c 9.4.4. Reverse Curve:

CE401/17

91

Consists of two arcs of same or different radii curving in opposite directions with the centers on opposite sides of the curve and on opposite sides of the common tangent (Unit 12)

FIG.9.1.d 9.5. Simple Curves: 9.5.1. Designation of a Curve: A curve may be designated either in terms of its radius or with reference to the angle subtended at the center by an arc or chord of a particular standard length. Designation by radius is adopted in U.K. and designation by central angle is adopted in U.S and India. The standard arc or chord length in 30m in India (100 feet in U.S). The angle which the standard chord (or arc) subtends at the centre is called the Degree of theCurve. Let R = Radius in meters D = Degree of the curve Let AB = arc of 30m length Du = D ×

180

= D = Central angle R = RD = Arc length Du = D

R .×

R=

×

180 D = 30m (arc length) 180

180 × 30 1718.9 = meters. D D

CE401/17

92

Alternatives: If the degree of the curve is defined in terms of the chord length and the chord AB in Fig.9.2.

the Fig.9.2. is 30m let E be the mid point of this chord. Draw E Perpendicular to AB

sin

D AE 30 / 2 D = = = 2 OA R 2

When D is small, Sin )

D = 2

D 2

15 15 where D in radians = D D/2 × 2 180 15 × 360 1718.9 R= = D D

R=

9.5.2. Elements of a Curve: Let AB and BC be the two straight lines intersecting at B and let the curve touch them at the two tangent points T1 and T2, fig(9.3). If the chainage is carried out in the directions T1 and T2, T1 is called the Point of curve and T2 is called the Point of tangent. From T1 if the curve turns to left it is called a left hand curve. The radii of curve at T1 and T2 will be at right angles to AB and BC respectively. If these perpendiculars are drawn their intersection at ‘O’ locates the centre of the circle of which the curve T1 E T2 is a part . The line joining T1 and T2 is known as the LONG CHORD.

CE401/17

93

Fig.9.3 From the properties of the circle

B T1 T2 =

B T2 T1

Tangent distance B T1 = B T2 If a line be drawn connecting O and B it will intersect the curve at E and the long chord T1 T2 at D (its mid point). The point E (Mid point of the curve) is called the CROWN POINT or APEX or SUMMIT of the curve. The intercept D E on the line OB between E and the Point of intersection of O B and the long chord T1 T2 (the Point D on T1 T2 ) is known as the VERSED SINE OF THE CURVE

F B C ( = ) ) is known as the

deflection angle of the curve and is equal to the sum of the two interior and opposite angles of the triangle B T1 T2 i.e.

F B C = 2 B T1 T2 OR

B T1 T2 = B T2 T1 =

T1 O T2 =

) 2

) = FBC

T1 B T2 1 is known as the intersection angle I. Deflection angle From

F B C = 180 - I = )

B T1 O

tangent length B T1 ) B T1 = = tan = OT1 2 R R Tangent length = R tan

B T1 = B T2 = R tan

) 2

) 2

where R is the radius of the curve. If R and

the positions of T1 and T2

)

are known tangent distances can be calculated and hence

CE401/17

94

Apex distance B E = OB - OE

)

O B = O T1 Sec

2

= R Sec

) 2

OE=R

B E = R (Sec

) 2

- 1)

D E = Versed sine of the curve =O E - O D = R - R Cos

= R (1 - Cos = R Ver Sine

) 2

) 2

)

) 2

Length of the curve = R )

)

is the deflection angle R is the radius of the curve.

9.5.3. Setting out simple curves: 9.5.3.1. Linear Methods: a) Offsets from long chord b) Offsets from tangents c) Offsets from chords produced 9.5.3.2. Angular or Instrumental Methods: a) Rankine’s method of deflection angles b) Two theodolite method c) Tacheometric method 9.5.3.1. Linear Methods : While plotting the curve on the ground it is rather a series of chords than arcs because the difference in chord and arc length is negligible particularly for chords not greater than

1 th of radius of the curve. 20

Length of Unit chord = 30m for flat curves 20m for sharp curves 10m for very sharp curves For plotting any curve whatever may be the method adopted the following preliminary calculations should be done to locate the tangent points T1 and T2. i)

The direction of the two tangents is known as they are the two directions to be connected by a curve. Producing the tangent line, the intersection point B can be got.

ii)

Setting the theodolite at B, the angle I can be measured.

CE401/17

95

iii)

180 - I = ) = deflection angle (Fig.9.3).

iv)

Tangent length B T1 = B T2 = R tan

v)

From B measuring backwards and forwards the tangent lengths along the tangents T1 and T2 can be fixed.

) 2

and so tangen t length can be calculated .

Chainage of 1st tangent point T1 = Known chainage of B - Tangent length B T1

)

vi)

Length of curve = R

vii)

Chainage of T2 = Chainage of T1 + length of curve (R

viii)

Chainage of T1 = p Chains + r links

))

Chainage of 1st point on the curve = P + 1 chain length of 1st chord = 100 - r links (assuming 100 links for each chain). All subsequent points are at unit chord (Unit chord may be 30m or 20m) ix)

Length of last chord is again less than unit chord i.e. chainage of T2 - previous full chains upto the previous of penultimate point on the curve.

x)

First and last chords are called sub-chords.

9.5.3.1.a. (Linear Method) Offsets from long chord. Fig.9.4.

Small curves for street kerbs etc may be set out by means of offsets from the long chord. A formula for the ordinate at any distance ‘x’ from the center D of the long chord can be deduced as follows. Let L be the length of the long chord T1 and T2 its extremeties. DE the versed sine (O0) and O the center of the curve. Ox is the ordinate from long chord at distance x from D. From

O T1 D 2

O T 1 = O D2 + T1 D2

CE401/17

96

= (OE - DE)2 + T1 D2 R2 = ( R - O0)2 + (

R - O0 =

L 2 ) 2

L 2 2 R - ( ) 2

O0 = D E = R -

L 2 2 R - ( ) 2

from which O0, the versedsine can be calculated when R and L are known Draw H H1 Parallel to T1 T2 and HG Parallel to OE From

OHH1 OH2 = OH12 + HH12 R2 = (OD + Ox)2 + x2 OD + Ox = Ox =

=

R 2 - x2

2 2 R - x - OD

R 2 - x2 - (R - Oo)

If the radius of the curve is very large in comparision with the length of the chord T1 T2 the ordinate at x (H G) is very nearly equal to the radial length H G1 so that T1G × GT2 = HG (2R - HG) HG =

T1 G × G T2 2R

HG2 being the square of a small quantity can be neglected.

Ox1 = x1

(L - x1 ) 2R

x1 measured from T1 ( and not from D) To set out the curve as per the method indicated above, T1 T2 must be located as indicated in 9.5.3. Then divide the long chord T1 T2 into number of parts and calculate the offsets at those points as per any of the above two methods. Erect perpendiculars along the long chord. The tops of all offsets will be along the required curve. 9.5.3.1.b. Offsets from Tangents: In this method the offsets are set out from the two tangents B T1 and B T2 (Fig.9.4. and 9.5.) From Fig.9.5.a (radial offsets) Let the Offset a a1 ( = Ox ) at a distance Tl a ( =x ) from the tangent point T1 be set out radially as in Fig.9.5.a.

CE401/17

97

FIG.9.5. a

T12

= a a1 ( a a1 + 2R) = 2R × a a1

(neglecting squares of small quantities)

a T1 2 aa1 = 2R

x2 = (approximately) 2R

Alternatively AT2 = O a2 - OT12 x2 = (R + Ox)2 - R2 (R + Ox)2 = R2 + x2 Ox =

2

R + x

2

- R (exact)

For flat curves the center O is often inaccessible and the ordinates are then set out without any great error at right angles to the tangents BT1 and BT2 and at equal distances along these lines. This entails that the points so fixed on the curve are at unequal distances apart which is occassionally an inconvenience and the curve though closes approximately to a circle is strictly speaking parabolic. Join O a1 (Fig.9.5.b) anda draw a1 a11 parallel to BT1 and r to O T1 (O a1)2 = O a112 + a1 a112 R2 = (R - Ox)2 + x2 R - Ox =

R 2 - x2

Ox = R -

R 2 - x2

To locate E the summit of the curve the value of x1 required = e E = O E Sin

) 2

= R Sin

) 2

9.5.3.1.c. Offsets from Chords produced: This method is usually employed when an angle measuring instrument is not available and when it is necessary to resort to chain and tape. The method is as follows. It is generally desirable

CE401/17

98

in railway and other works that the chainage shall be carried continuously throughout the whole line so that if the chainage at T1 is n chains + m links the first chord length will be the remaining portion of the chain length i.e.(100m) links. The end of the m th link which falls at T1 is held there while the front end of the chain is swung round from a in line with T1 and B through the calculated distance aa1 (as explained below), thus forming the point a1 i.e. (n + 1) chains from the commencement of the work. (Fig.9.6). The chain is next pulled in the direction of T1 a1 produced until a1 b = 1 chain and the rear end of the chain being held at a1. It is swung round this point as centre moving the fore end from b to b1 through the 2nd calculated offset distance b1b1 . a1b1 is then prolonged to C . b1 C being the length of the chord required and the point C1 found by swinging the chain round b1 as center through a distance equal to the 3rd calculated offset CC1.

FIG.9.6.

Derivation of formula for offsets from the chord Let the angle a T1 a1 between the tangent a T1 and the chord T1 a1 be center ‘O’ by T1 a1 is equal to 2 T1 a1 = R × 2 =

T1 a1 2R

Similarly arc aa1 is nearly equal to the chord aa1. a a1 = T1 a1 =T1 a1 x

=

T1 a1 2R

T1 a1 2 C1 2 = = O1 2R 2R

where O1 is the 1st offset and C1 is the length of the first chord. At a1 draw a tangent to the curve cutting T1 a1 in a2 and b b 1 in b2. a2 T1 = a2 a1 since both are tangents to the circle from a2.

a 2 T1 a 1 =

a 2 a 1 T1

radians. Then the angle subtended at the

CE401/17

Also

99

a 2 a1 T1 =

b a1 b2 (vertically opposite)

The triangle a T1 a1 and b a1 b2 are approximately similar as both are nearly isosceles

b b2 a a1 = a1 b T1 a1 i.e. b b2 =

=

a1 b a b x a a1 = 1 x O1 T1 a T1 a C2 C2 x 1 2R C1

=

C2 C1 2R

Also b2 b1/2 being the offset from the tangent a1 b may be written as

C 22 . 2R

The second offset bb1 is therefore equal to b b2 + b2 b1 i.e. O2 =

C 2 C1 C2 C ( C1 + C 2 ) + 2 = 2 2R 2R 2R (C 2 + C3 ) Similarly O3 = C3 from which 2R when C2 and C3 are equal as would usually be the case. O3 = C3 x

C3 2 2 C3 = 2R R

The remaining offsets are all equal to this when equal chords are employed, except the last one (which should finish at the tangent point T1). As the last chord length is probably an uneven number of links is On =

Cn ( Cn - 1 + Cn ) 2R

9.5.3.2. Antular (Instrumental) Method: a) Rankine’s Method of Tangential Angles:

This method involves the use of one theodolite and a chain or tape. This method is most frequently adopted for setting out railway or other important curves.

The theodolite is set up at the tangent point T1 with the vernier adjusted to 3600 and the telescope is

directed to B and clamped. If the chainage at T1 is n chains + m links, the length of the first chord will usually be (100 - m) links. The first tangential angle BT1 a1 is calculated from the formula derived below and the vernier of the instrument fixed to read this angle the telescope being then directed towards a1. The end of the m th link is retained at T1 and the remaining (100 - m) links swung round this point as a centre, while the person looking through the telescope indicates to the leader at a1 the direction in which he must move in order to obtain exact coincidence with the cross hairs.

CE401/17

100

FIG.9.7. a1 having been located and the instrument still remaining at T1, the vernier is altered to read the 2nd tangential angle B T1 b1 and reclamped. The telescope is consequently directed along the line T1 b1 and the second chord a1 b1 is set out by swinging the chain round a1 as a center until the image of b1 coincides with the cross hairs. The remaining chords are set out in a similar manner. e.g. the position of C1 is fixed by swinging the chain round b1 as a center and getting coincidence with the cross hairs of the diaphragm when the telescope is directed along the line T1 C1 and the vernier reads the third tangential angle B. T1 C1 . Derivation of formulae for Tangential Angles: The first tangential angle (

1

) in the angle B T1 a1 between the tangent T1 B and the chord T1 a1 and is

equal to half the angle subtended by T a1 at the centre of the circle so that

T1 O a1 = 2

1.

When the length of the chord is small compared with the radius of the curve, the chord T1 a1 is approximately equal to arc T1 a1 and its length

l1=2

R

2 1 where 2 360

circle.

360 l1 90 l × = × 1 degrees 2 2 R R 60 x 90 l = × 1 minutes R l1 i.e. 1 = 1718.9 minutes. R 1

=

Similarly the tangential angle

2

of a1 b1 from the tangent at a1 i.e.,

R is the circumference of the complete

CE401/17

101

b a 1 b1

1718.9 ×

is

l2 minutes R

b a 1 b1 between the tangent b a1 and the chord a1 b1 is equal to the angle in the opposite segment. i.e.,

b a 1 b1 =

a1 T b1

B T1 b1 the total tangential angle of b1 i.e. C2 is equal to tangential angle of C1 i.e.

3

is equal to

The total tangential angle should evidently be equal to

n

1

+

2

+

3

1

+

2

. Similarly

where C3 = 1718.9

for the last chord (i.e.

1

+

2

+

B T1 C1 , the total

l3 minutes R 3

+ ---------------------- +

n

)

B T1 T2

9.5.3.2.b. Two Theodolite Methods:

FIG.9.8. To set out a curve over a rough ground, dispensing with chain and tape two theodolites can be used one each simultaneously at T1 and T2 . The method is not economical as two experienced surveyors are required instead of one as in the method mentioned before in 9.5.3.2.a. If the chainage at the tangent point T1 is n chains + m links, the first point a1 on the curve will be (100 - m) links from T1 and the tangential angle B T1 a1 i.e.

1

is calculated as

1

=

1718.9

( 100 - m ) minutes . (R R

is expressed in links). The angle B T1 a1 in equal to the angle in the opposite segment i.e. total tangential angle

2

i.e.

B T1 b1 is equal to

T 1 T2 a1 and similarly the second

T1 T2 b1 .

If a theodolite is set up at T1 with its vernier reading 3600 and sighted to B while another instrument with its vernier reading 3600 is fixed at T2 and sighted to T1, then by fixing the same reading

on each instrument, the

intersection of two lines of collimation gives a point on the curve.

The same total tangential angles calculated in the earlier method (9.5.3.2.a) can be used in the two theodolite method. The calculated total angles should be rounded off to be a multiple of the least count of the instrument before setting the angle on the instrument vernier.

CE401/17

102

9.5.3.2.c. Tacheometric curver ranging:

FIG.9.9. This method can be used for rough and undulating ground avoiding chain and tape but this is not so accurate as the other instrumental methods. T1 is the position of the Tacheometer

1,

2,

T1 D = D1 = 2 R Sin

1

T1 E = D2 = 2 R Sin

2

T1 F = D3 = 2 R Sin

3

3,

etc. are the total deflection angles for the points D, E and F respectively. i.e. D1, D2, and D3 can

calculated by knowing the radius of the curve and the total deflection angle upto that point D1 =

or

f S1 + (f + d) horz.line of signals. i f S1 cos i

+ (f + d) Cos

for inclined signal knowing the instrument constants

f and f + d and D1, S1 can be calculated i similarly S2 etc. ------Sn . Sighting to B, vernier is set to 3600. Then the first total deflection angle w.r.t. T1 B is set i.e. the instrument (Telescope) is directed in the direction of T1 D. A staff is moved along T1 D till the calculated intercept S1 on the staff is read. Similarly S2, S3, ------Sn. i.e. staff moved along T1 E till an intercept of S2 is got and point E fixed and staff moved along T1 F till an intercept of S3 is got and point F fixed and the procedure repeated till Sn. 9.6.Worked out Examples: Example 9.6.1.: Two straights AB and BC of a communication line intersect at a chainage (200 + 15), the angle of deflection being 1200. calculate the chainages of the point of commencement and the point of tangency if the radius of the right handed circular curve is 200 meters. SOLUTION:

F B C = Deflection Angle

CE401/17

103

=

T1 O T2 central angle = 1200 =

T1 B = Tangent length = R tan

) 2

R = 200 meters T1 B = 200 tan 60 = 346.41 meters as R is substituted in meters. b)

length of the curve = R) Substitute ) in radius = 200 x 200 x

c)

180

= 419.05 meters as R is in meters.

chainage of the point of intersection is given as 200 + 15 200 inchains. 15 may be meters or links. Put the final answer in the same units on the given chainage.

Assume also value of the chain 20m or 30m if not given and if it is required. Assuming 30m chain, Tangent length T1 B = 346.31m = 11chains + 16.31m Length of the curve = 419.05 meters = 13 chains + 29.05m Chainage of intersection point B = 200ch + 15m - (Tangent length) Ans: Chainage of point of commencement =

- (11 + 16.31)

188 + 28.69m

= + (13 + 29.05m)

CE401/17

104

Ans: Chainage of point of Tangency

=

202 + 27.74m

Example 9.6.2.: Two straights AB and BC meet at a point B. The bearings of BA and BC are 2000 and 680 . These two are to be connected by a simple curve of radius 25 chains. The chainage of the pointed of intersection is (320 + 50). Tabulate the values required to set out the curve by deflection angles. The least count of the theodolite to be used is 20” . Take the standard length of chord as 20m (Note 20m chain is being used and it is divided into 100 links i.e. each link is 0.2 meters). SOLUTION:

Fig. R = 20 chains = 20 x 20 = 400 m B T1 = R tan

) 2

= 400 x tan 240 = 178.09m = 8 chains + 18.09m = 8 chains + 90.45 links

Length of the curve = R ) = 400 x

180

x 48 = 335.23m = 16 ch + 15.23 m = 16 ch + 76.15 links

Chainage of intersection points = 320 + 50

- (8 + 90.45) 311 + 59.55

Subtract Tangent length

=

=

16 + 76.15

Chainage of T2

=

328 + 35.7

CE401/17

105

i.e. pt. of tangency

Chainage of point of commencement = 311 + 59. 55 Chains links To set out the curve by deflection angles length of 1st chord = 100 - 59.55 = 40.45 links = 40.45 x 0.2 = 8.09 meters All other chords are of 20m length. 1

l1 8.09 = 1718.9 x min utes R 25x20

= 1718.9 x

= 27.81 minutes = 27’ 48. ‘’6

l2 = S3 - - - - - Sn-1 R 20 = 1718.9 x = 68.756 minutes 25 x 20

2=

1718.9 x

= 10 08” 45. ‘’36 Total deflection angles 1 2

=

1 +

=

1

= 0o 27' 48. ' '6

= 00 27' 48.' '6 + 1 08 45.36

2

= 10 36 "34 * 3

=

1

+

2 0

+

"

"

10 36 ' 40 ' '

3

= 1 36 ' 34 ' ' + 1 08 ' 45.36 0

To set on theodolite 00 27 ' 40 ' '

2 0 45 ' 20' '

= 2 0 45 ' 19.36 ' '

and so on. The student is advised to calculate

n-1

and

n.

To summerise the above problem to set the curve in the field theodolite reading should be set to ‘

(1st tangential angle) 2= 3

0

1

= 20

36 ‘ 40 ‘’ 45’ 20 ‘’

(2nd total tangential angle) (3rd total tangential angle)

9.7. Exercise: 9.7.1.: Explain the following with reference to a simple curve i) Designation of a curve ii) Versed Sine

1

= 00 27 40 ‘

CE401/17

106

b) Calculate the data required for setting out a simple curve of radius 100 meters. Deflection angle is equal to 600. i) by offsets from long chord ii) Two theodolite method

least count of the instrument = 20” chainage of the intersection point is equal to 350 chains + 20 meters. 30 meter chain being used. iii) by offsets from chords produced. 9.7.2.: Explain the procedure for setting out a simple curve by ordinates from long chord method.

b) Calculate the ordinates at 5m distances for a circular curve having a long chord of 60m and a versed sine of 3m. 9.8.SUMMARY: 9.8.1. Offsets from long chord O0 = versed sine = R -

R

2

L 2

2

Ox = Ordinate at x from the mid point of long chord 2

R +x

=

2

- (R - O 0 ).

9.8.2 Offsets from tangents

2

x i) Ox = aa1 = 2R =

2

R +x

3(approximate)

2

- R (exact)

ii) Perpendicular offsets Ox = R -

R

2

x

2

9.8.3. Offsets from chords produced O1 =

C12 2R

O2 = C2 O3 = O4 =

( C1 + C 2 ) 2R On - 1 =

C32 R

( Cn -1 + C n ) 2 R l = 1718.9 1 minutes (Tangential angle) R

On = Cn 9.8.4.

1

****

SURVEYING

CE401/17

107

UNIT - 10 CURVE SURVEYING 10.1 Contents:

INTRODUCTION, Obstacles in locating simple curves: Intersection point inaccessible Buildings or other objects obstructing the line of sight, COMPOUND CURVES, Relationship between the different parts of a compound curve, setting out compound curve, problems. 10.2 Aims: Obstacles that we come across in general in curve setting and how to over come them is explained in this unit. Also about the compound curve the relationship between its different elements is explained and two numerical examples have been worked out. 10.3.Introduction: While connecting two places separated by some distance, the communication line has to take a bend to overcome an obstacle. The bend or the transition from one straight to another should be smooth, hence the introduction of the curves in communication lines. Even while negotiating a curve if further obstacles come, the curve has to take a further bend, thus changing the radius of the curve, necessiating the introduction of a compound curve. The student should be able to know the difference between different curves and under what circumstances each type is adopted. 10.4: From a detailed study of the unit 9 it is clear that for setting out a curve the targent points T1 and T2 and the intersection point B must be located first irrespective of the method adopted for plotting the curve. In certain cases the intersection point B may be inaccessible to measure back or forward the tangent lengths BT1 and BT2 to locate respectively T1 and T2. Some times some obstruction may come in the path of the curve and in certain cases T2 is farther off from T1 and it may not be possible to set up the entire curve from T1 only. This unit deals with the explanation of overcoming the different types of obstacles (hurdles) indicated above, besides compound curves and the relations between its different elements. Two problems have been worked out. 10.5.1. Obstacles in Curve Location Intersection point Inaccessible: To find the positions of the two tangent points T1 and T2 for a curve of radius R when the directions of the two tangents AB and BC have been decided upon but the point of intersection B is inaccessible.

CE401/17

108

Fig.10.1 On the tangents BA and BC respectively fix any two points E and F and measure the angles AEF and EFC. By subtracting each of these values from 1800 the angle BEF and BFE in the tringle BEF are found, and the deflection angle ) is equal to BEF + BFE. The length EF is next measured and the distances BE and BF calculated

BF =

EF x sin BEF sin EBF

BE =

EF x sin BFE sin EBF

The Radius of the curve R and the deflection angle ) being known the tangent distances BT1 and BT2 are calculated from BT1 = BT2 = R tan

) 2

To fix the point T1 on the ground a length equal to BT1 - BE is chained off (measured) from E and similarly T2 is found by chaining a distance (BT2 - BF) from F after which the curve may be set out by any of the known methods.

CE401/17

109

10.5.2. Buildings or other objects obstructing the line of sight: If it is not possible on account of buildings or other objects intervening in the line of sight, to set out all the points on a curve by means of tangential angles from one position of the instrument T1 the instrument may be moved to another point on the curve as described in the procedure below and the work continued. Fig.10.2. Suppose c1 is the last point which can be seen from T1 and that the tangential angle is

c. while the

instrument is still stationed at T1 and directed towards c1 set out another point in the same line T1c1 produced (fig.10.2). Move the instrument to c1, fix the vernier at 3600 and direct the cross hairs on to C3. If nc1 c4 be the tangent at c1, nc1 = nT1 and the angle nc1 T1 = c, consequently the vertically opposite angle c3 c1 c4 must be equal to c, so that if the vernier of the theodolite is again set at c the telescope will point along the tangent c1 c4, The angle c4 c1 d1 between this tangent and the next chord c1d1 is equal to the angle c1 T1 d1 in the opposite segment (i.e. 4). Hence by adding c to each , c3 c1 d1 will be equal to d. Similarly the angle e1 c1 d1 = and c3 c1 f1 =

e1 T1 d1 and consequently c 3 c1 e1 =

e

f etc.

This way the instrument can be moved from the tangent point to any other oint on the curve and the staking out proceeded with from that point using the same calculated tangential angles as before. 10.5.3: Then T1 T2 is large, part of the curve can be set from T1by tangential angles and the remaining from T2 using the same method.

CE401/17

110

10.6.1 COMPOUND CURVE:

Fig.10.3 AB and BC are the initial straights. Because of some obstruction both the straights are to be connected by a compound curve T1 DT2. The two branches of the curve join at D. With their centres at O1 and O2 and Radial R1 and R2 respectively, R2 denoting the greater whether it comes first or second in the direction of chainage. The two end straights as in simple curves, on being produced meet at B, the intersection point. ) is the deflection angle and it is the intersection angle. The tangent lengths T1B and T2B are necessarily unequal and will be distinguished as T11 and T21 of which T21 is greater and is adjacent to the arc of greater radius. The common tangent at D meets these end tangents at E and F and makes with them the angles )1 and )2 which in turn are equal to the respective central angles subtended by the arcs. For the two simple curves ET1 = ED = R1 tan FT2 = FD = R2 tan

)1 2

)2 2

= t1

= t2

a) The centres O1 and O2 are in the same straight line with D since radii O1D and O2D are both perpendicular to the common tangent. b) ) = )1 + )2 since ) is the exterior angle to the triangle BEF. 3)

BE =

EF x sin )2 sin)

T1 = t1 + (t1 + t2)

sin ) 2 sin )

... ... ... ... (A)

CE401/17

111

Similarly

T2’ = t2+ (t1 + t2)

sin )1 ... ... ... ... (B) sin )

length of the compound curve = R1)1 + R2)2 10.6.2 Relationship between the parts of a compound curve: The seven quantities involved in a compound curve of two branches), )1, )2, R1, R2, T1 and T2 are inter related in a manner deducible from the expressions (A) and (B) above

T1' = t1 + (t1 + t 2 )

= R1 tan

sin ) 2 (fig.10.3) sin)

)1

) ) sin )2 + (R1 tan 1 + R 2 tan 2 ) 2 2 2 sin)

1 - Cos)1 or )2 ) or ) 2 put tan 1 = 2 sin)1 or ) 2 sin ) 2 1 - cos )1 1 - cos )1 1- cos)2 T1' = R1 + (R1 + R2 ) sin )1 sin )1 sin )2 sin ) 1- cos )1 T1 sin ) = R1 ( Sin) + Sin) 2 ) + R 2 (1- cos) 2 ) sin )1 Substituting sin ( ) - )1) for sin )2 this reduces to T11 sin ) = R1 (sin ) sin )1 + cos ) cos )1 - cos )) + R2 ( 1 - cos )2) = R1 [cos ()-)1) - cos )] + R2 ( 1 - cos )2) which may be expressed as R1 [ (1 - cos )) - ( 1 - cos )2) ] + R2 ( 1 - cos )2) (The candidate has to do the simplification in between steps and check the above simplification ) or T11 sin ) = R1 versin ) +. (R2 - R1) versin . )2 . . .(C) similarly T2’ sin ) = R2 versin ) - (R2 - R1) versin )1 . . . (D) Expressions (C) and (D) in conjunction with the relationship ) = )1 + )2 provide three simultaneous equations, the solution of which will determine three of the seven parts of the curve if the remaining four form the data

Setting out of compound curve: The two branches of the compound curve (which are individual simple curves) can be set out as per the methods explained in unit 9 for setting out simple curves. Example 10.7.1: Given R1 = 20 chains

R2 = 40 chains.

T1 = 17.5 chains ) = 630 30’ measured chainage at the commencement of smaller curve (1st arc) = 250.5 chains,

CE401/17

112

Determine T2, chainage of point of compound curvature (point D in fig 10.3) and chainage of point of tangency of the 2nd curve. SOLUTION: T1’ sin ) = R1 versin ) + (R2 - R1) versin )2 = R1 (1 - cos ) ) + (R2 - R1) (1 - cos )2) 17.5 sin 630.5 = 20 (1 - cos 630.5) + (40 - 20) (1 - cos )2) on simplifying )2 = 390.57 ( 390 34’ 12’’) )1 = ) - )2 = 63.5 - 39.57 = 230.93 T2’ sin ) = R2 versin ) - (R2 - R1) versin )1 T12 sin 630.5 = 40 (1- cos 63.5) - 20 (1 - cos 23.93) on simplifying T21 = 22.83 chains

Arc T1D = Arc T2D =

)1 R1

22 23.93 x 20 x = 8.35 chains 180 7 180 22 39.57 x 40 )2 R 2 = x = 27.64 chains 180 7 180 =

chainage of T1 = 250.5 chains. chainage of D = 250.5 + Arc T1D = 250.5 + 8.35 = 258.85 chains Add Arc length DT2

+ 27.64

286.49 chains

Chainage of T2 Answer: Second tangent length T2’

= 22.83 chains

chainage of point of compound curvature (D)

= 258.85 chains = 286.49 chains

chainage of T2

Example 10.7.2 : Two straights AI and BI meet at I on the far side of a river. On the near side of the river a point E was selected on the straight AI and a point F on the straight BI and the distance from E to F measured and found to be 85m. The angle AEF and found to be 1650 36’ and the angle BEF 1680 44’. If the radius of a circular curve joining the straight is 500m, calculate the distance along the straight from E to F to the tangent points. Solution IEF = 1800 - 1650.36’ = 140.24’00’’

IFE = 1800 - 1680 44’ = 110 16’00’’ Deflection angle ) = 140 . 24’ + 110 . 16’’ = 250 40’ EF =

85m given

CE401/17

113

) = 140 24’ + 110 16’ = 250 40’ Fig.

From

IEF

EF 85 0 x sin 11 16' = x 0.195 = 38.27m 0 sin 25 40' 0.433 85 x 0.2487 85 0 x sin 14 24' = = 48.82 m FI = 0 0.433 sin 25 40' 250 401 ) = 500 tan AI = BI = R tan 2 2 EI =

AI = BI = 113.90 m AE = AI = EI = 113.90 - 38.27 = 75.63 m FB = BI - IF = 113.90 - 48.82 = 65.08 m 10.8.Exercise: 10.8.1:

Two straights AB and BC meet at an inaccessible point B. They are to be fixed by a circular curve of

radius 450m in length. Two points P and Q were selected respectively on AB and BC. The following observations were made.

APQ = 16 00

CQP = 1450.

Distance PQ = 125m and chainage of P = 1500.00. Calculate the chainages of intersection point B, point of curvature and point of tangency. 10.8.2. A railways siding is to be curved through a right angle. In order to avoid buildings the curve is to be compound, the radius of the two branches being 8 chains and 12 chains. The distance from the intersection point of the end straight to the tangent point at which the arc of 8 chains radius leaves the straight is to be 10.08 chains.

CE401/17

114

Obtain the 2nd tangent length or the distance from the intersection point to the other end of the curve and the length of the whole curve. 10.9 SUMMARY: For a compound curve Length of the back tangent T1 = [ R1 versin ) + (R2 - R1) versin )2] / sin ) Length of the forward tangent T2 = [R2 versin ) - (R2 - R1) versin )] / sin ) *****

SURVEYING UNIT : 11 TRANSITION CURVES 11.1 Contents: Introduction, requirements of a transition curve, super elevation (cant) calculation of length of transition curve different methods, equation for ideal transition curve, shift of the curve, determination of total length of the curve - problems. 11.2 Aim: To study about curves of varying radius known as transition curves. 11.3 Objectives: The student is expected to know from this chapter the necessity and requirements of transition curves and the salient features connected with the setting out of the same. 11.4.1 Introduction: A transition curve is a curve of varying radius introduced between a straight (tangent) and the circular curve to have smooth transition of direction. It is also called an easement curve. The object of introducing transition curve is i) To accomplish gradually the transition from the tangent to the circular curve and circular curve to the tangent. ii) To obtain a gradual increase of curvature from zero at the tangent point to the specified quantity at the junction of the transition curve with the circular curve thereby increasing the superelevation from zero at the junction point to a specified value at the junction of the circular curve and the transition curve. 11.4.2 Requirements of a Transition Curve:

CE401/17

115

i) It should meet the original straight and the circular curve tangentially. ii) The radius of the transition curve at the direction of the circular curve should be same as that of the circular curve . iii) Rate increase of curvature along the transition curve should be the same as that of increase of superelevation. iv) Length of transition curve should be such that full superelevation is attained at the junction with the circular curve. 11.5.1(a) Superelevation: Raising of the outer edge of the road or rail above the inner one is called superelevation or cant. The effect of centrifugal force is to push the vehicle off the road (rail or track). In order to counteract this action the plane of the rails or road surface is made perpendicular to the result (R’) of the centrifugal force (P) and the weight of the vehicle (W).

Fig.11.1 W = Wt. of the Vehicle , P = Centrifugal force, v = Speed in meter/sec. g = acceleration due to gravity. R = Radius of the curve in meters. h = Superelevation in meters. b = Width of the road in m G = Distance between rails.

Wv 2 P = gR

CE401/17

But

116

P v2 = W gR P dc h = tan = = W ac b 2 P h v = = g.R W b h =

bv 2 gR

in case of highways

(.11.1. a )

=

Gv 2 gR

for railways.

( 11.1. b )

p i.e. ratio of the centrifugal force and the weight of the vehicle is called centrifugal ratio. The W 1 1 maximum value of centrifugal ratio is taken as for roads and for railways and this ratio is useful in 4 8 The ratio

arriving at the length of the transition curve. 11.5.1(b) Equilibrium Cant and Cant Deficiency: In case of railways if the cant is provided as per the equation

GV 2 where h is superelevation, the load carried by both the wheels will be the same, the spring will be h = gR equally compressed and the passengers will not tend to lean in either direction. Such cant is known as the “equilibrium cant”. If the cant provided is less than this more weight will be carried by the outer wheels, the outer springs will be more highly compressed than the inner and the passengers will tend to lean outward. The track under these conditions will have cant deficiency. 11.5.2 Determination of length of transition curve. (Of the various methods some are mentioned below) 11.5.2 (a) Arbitrary Method: Based on uniform variation of superelevation over a length as

1 in n l = nh n = rate of canting L = Length of transition curve.

CE401/17

117

h = Superelevation. 11.5.2 (b) Based on rate of change of Radial Acceleration: The rate of change of radial acceleration should be such that the passengers should not experience any sense of discomfort when the vehicle is moving over the curve. L = Length of transition curve in meters = rate of change of radial acceleration (0.3 m/sec3 will pass unnoticed) v = Max.Speed in meters/second V = Max. Speed in Kmph. t = time taken to travel over transition curve =

acceleration attained in this time =

L m/sec2 v

x

Radial acceleration for circular curve =

L seconds v

,2 R

m / sec2

L v2 = v R L = length of transition curve =

v3 L = R

.....

v

3

R ( 11.3a )

3

=

v 0.3R

.....

v =

V x 1000 60 x 60

L =

1 V x 1000 3 ( ) 0.3 R 60 x 60

-

V3 14 R

.....

( 11.3b )

( 11.3c )

(Note 11.3 b can be used when v is in m/sec are 11.3c when v is in m/sec and 11.3c when V is in Kmph).

CE401/17

118

11.5.2 (c) Based on Centrifugal Ratio:

P v2 1 = = W gR 4 gR 9.81 R = = 2.452 R 4 4 v3 L = (as per 11.3 b) 0.3R ( 2.452 )3/2 R 3/2 = = 12.80 R ..... 0.3 R for highways v =

( 11.4 a )

Similar expression can be derived for railways by taking

P v2 1 = = W gR 8 L = 4.526 R

........

( 11.4b )

11.5.3 Equation for Ideal Transition Curve: (Clothoid spiral). Fig.11.2

T = beginning of the transition curve E = Junction of transition curve with circular curve M = any point on the transition curve . = radius of transition curve at M. R = radius of circular curve.

CE401/17

119

)1 = inclination of the tangent to the transition curve at M to the initial tangent TB ) = the angle between the tangent TB and the tangent to the transition curve at the junction point E. ( This angle is known as the spiral angle ) L = Length of the transition curve. The fundamental requirement of the spiral curve is that its radius of curvature at any point shall vary inversely as the distance ‘ l ’ from the beginning of the curve.

1 l

. or for all curves

1

. d) dl

= ml = curvature =

d) =

1

. 1

.

x dl

d) = ml dl Integrating we get

)

2

=

ml 2

Constant of integration is zero since

)

= o when

l = o

At the junction point E

l= L

. = R )1 = 1

.

= mL

or m =

1 1 = at the junction point . L RL

CE401/17

120

1 l2 l2 ml 2 x = = RL 2 2RL 2 2 l ) = 2 RL l = 2 RL

) =

l = k )

( 11.5 )

where k = 2 RL L 1 L2 ) = spiralangle = x ( at the junction point l = L ) = 2R RL 2 L .............. (11.6 ) Spiral angle ) = 2R 11.5.4 Shift of the Curve: To introduce transition curves between the straights and the circular curve the main

2

curve is shifted inwards and this displacement ‘S’ is generally known as shift whose value is

L 2 4R

FIG.11.3 11.5.5. To determine the shift of the circular curve, length of the combined curve and the total tangent length T1 I is

the original back tangent. L = Length of the transition curve DE = X and T1 D = Y are the co-ordinates of E with reference to the point of of the curve i.e. T1. R = Radius of the circular curve.

commencement

CE401/17

121

) = Spiral angle EB = Portion of circular curve beyond E EN = tangent at E Draw E parallel to T1 I to meet OT1’ at M By definition

DNE = ) =

NEM

=

EOM

Arc EB = R ) = Rx

L 2R

=

L 2

( Q ) = spiral angle =

L ) 2R

But EC is approximately equal to EB

L 2

EC =

Hence the shift T’1 B bisects the transition curve at C Shift S = T11 B = T11 M - B M = DE - (OB - OM) = X - (R - R cos ) ) = X - R(1 - cos ) ) = X - R x 2 sin2

= X - 2R(

) 2

)

) 2

Since sin ) = ) for small values of ).

3

l For a cubic parabola the co-ordinates of any point are represented by X = 6RL Substituting

l = L and ) =

L 2R

L3 L 2 Shift S = - 2R ( ) 6RL 4R

CE401/17

122

2

=

2

L L - 2R 6R 16R 2 2

Shift S =

2

1 1 L ( ) = R 6 8

L 24R

...

( 11.7 )

Hence the shift of the main curve is directly proportional to the square of the length of the transition curve and inversely proportional to the radius of the circular curve. Referring to fig.11.3 Let T1 E and E T2 be the transition curves I = Point of intersection = Deflection angle S = Shift ) = Spiral angle Total tangent length T1 I = T1 T’1 + T’1 I T’1 I = ( R + S ) tan

T1 T’1 =

2

L 2

Total tangent length = ( R + S ) tan

+

2

L 2

(11.8)

Length of combined curve Central angle for the circular curve = length of circular curve =

R (

- 2)

- 2 )) 180

(11.9)

Total length of the combined curve = L +

= 2L +

R (

- 2 )) 180

- 2 ))

R ( 180

0

+ L

(11.10)

CE401/17

123

11.6 Problems: 11.6.1 : A transition curve is proposed for a circular curve of 400 m radius the gauge being 1.5m between rail centers and maximum superelevation restricted to 12 cm. The transition is to be designed for a velocity such that no lateral pressure in imposed on the rails and the rate of gain of radial acceleration is 30 cm/sec3. Calculate the required length of the transition curve and the design speed.

(JNTU

Sept.1984)

1.18 V 12

h =

or

2

R.h = 1.18

V =

Design speed

= 63.77 km ph.

3

L =

400 x 12 = 63.77 kmph 1.18

V = 14 R

63.77 x 63.77 x 63.77 = 46.3 meter 14 x 400 3

V (The student is advised to arrive at the formula L = 14R = 0.3 m/ sec2 before substituting the various values in the formula).

using 11.6.2:

Two straights AB and BC intersect at chainage 2500 m the deflection angle being 600 . It is proposed to

insert a circular curve at 250 m radius with two transition curves 75m long at each end. Calculate the shift of the main curve, the spiral angle of the transition curve and the chainage at the point of commencement of the curve. Data given L = 75 m Length of transition curve. R = 250 m Radius of circular curve = 600 Deflection angle. Chainage at pt. of commencement = 2500 m

2

L Shift S = = 24R Spiral angle ) =

L 2R

75 x 75 24 x 250

= 0.9375 m

75 = 0.15 radians 2 x 250

= =

0.15 x 180 o = 8 . 59 T' 80 35’ 24”

CE401/17

124

Tangent length of the combined curve = (R + S) tan

2

+

= (250 + 0.9375) tan

L 2 60 75 + 2 2

= 250.9375 x tan 300 + 37.5 = 250.9375 x 0.5773672 + 37.5 = 144.8 + 37.5 = 182.38 m Chainage pt. of intersection

= 2500.00

Total tangent length

= 182.38 -------------

Chainage pt. of commencement = 2317.62 m 11.7 Exercise: 11.7.1.: Two straights intersect at chainage (100 + 15), the intersection angle being 1000. The centrifugal ratio is 0.25 and the maximum allowable speed is 50 kmph. Rate of change of acceleration is 0.3m/sec3 Calculate the radius of the circular curve, length of transition curve and the chainage of the beginning and end of the combined curve and the Junction of the transition curve with the circular curve. 11.8.Summary: Centrifugal ratio

P W

=

v2 gR 2

h = superelevation =

bv for highways. gR 2

G V = for railwys. gR Length of transition curve (i) Aribitrary method L = n h (ii) Based on rate of change of radial acceleration

v3 L = , 0.3 R

v in m/sec.

CE401/17

125

=

V3 , 14 R ,

V in kmph

(iii) Based on centrifugal ratio

P V2 = W gR

=

12.8 /R

L =

v2 gR

=

L = Spiral angle )

1 for highways. 4

=

1 for railways. 8 4.526 /R

L 2R 2

Shift of circular curve =

L 24 R

Total tangent length = (R + S) tan

Total length of combined curve

= 2L +

R (

2

+

L 2

- 2) )

1800

*****

SURVEYING UNIT - 12 REVERSE CURVES AND VERTICAL CURVES 12.1. Contents: Reverse curves, vertical curves, Types of vertical curves, length of vertical curve, computations for

setting out vertical curves. Problems on reverse curves and vertical curves, Exercises. 12.2. Aim: To study about reverse and vertical curves, where and how they are used and some points on the setting

out of the same. 12.3. Objectives: The student is expected to know from this chapter the necessity and requirement of reverse and

vertical curves and some methods connected with calculation of salient points and setting out of the same. 12.4. Introduction:

CE401/17

126

12.4.1. Reverse curves: A reverse curve consists of two circular arcs curving in opposite direction with a common tangent at their junction. The radius of the two curves may or may not be equal . The points at which the two arcs meet is called the point of reverse curvature or point of contraflexure. Reverse curves are a must when two parallel straights are to be connected particularly in railway sidings or when two straights meet at a small angle. 12.4.2. Vertical Curves: When two different or contrary gradients meet they are connected by a curve in a vertical plane called a vertical curve. The main purpose of providing this curve in roads and railways is to obtain a gradual change in grade avoiding abrupt change at the apex. The vertical curve may be a circular arc or a parabolic curve. Parabolic curve is generally preferred. 12.5. Theory: 12.5.1. Reverse curves: Components of the reverse curve bend in opposite directions and so the curve direction also changes at the junction point. Reverse curves should be avoided on main high ways or Railways where speeds are necessarily high for the following reasons:

1. Because of sudden change of cant (Super elevation) from one side to the other if proper care is not taken at the junction point accidents are likely to occur. 2. At the junction point steering is very dangerous in the case of highways. It is always preferable whenever practicable to insert a short straight length between the two branches of a reverse curve. A few cases of reverse curves are explained below (i) Reverse curve connecting non parallel straights (ii) connecting parallel straights. 12.5.2. Reverse curve connecting Non-parallel straights:

Fig.12.1

CE401/17

127

Of the various elements of a reverse curve knowing certain quantities for example the central angle 2

and the length of the common tangent DEF to find the common radius R (equal Radius)

1

and

and the chainages

of T1, E, and T2 if that of B is given. DF = Common tangent of length ‘d’ O1E = O2E = R since T1D and DE are tangents to the first arc they are equal in length. Similarly EF and FT2 with reference to the second curve

BDE =

EFC =

=

1

DT1 = DE = R tan FT2 = FE = R tan

1/2 2 /2

DF = d = DE + EF = R tan

d = R (Tan

tan

From

+ tan

1

and

) =

2

1

2 2

-

+ tan

1

2

+ R tan

2

2

2 )

2

d

R =

Knowing R,

1

2

2

........

( Eq.12.1 )

2

2

, the lengths of the two arcs can be determined 1

BDF, BD =

DF x sin sin )

2

BT1 = BD + DT1 (distance of the intersection point from the beginning of the curve) =

d sin 2 + R tan 1 ........... sin ) 2

Chainage of T1 = Chainage of B - T1B Chainage of E = Chainage of T1 + length of first arc Chainage of T2 = Chainage of E + length of 2nd arc.

( Eq.12.2 )

CE401/17

128

The first branch of the curve can be set out from T1 and the 2nd branch from E by the method of Tangential angles. 12.5.3. Reverse curve connecting parallel straights: Given the radius of the two branches and the corresponding central angles of the two curves to determine the perpendicular distance between the straights (v) and the direct distance between T1 and T2 i.e., L and the projected distance between T1 and T2 i.e., distance between the perpendicular drawn at T1 and T2. AT1 and T2C are the two parallels to be joined at T1 and T2 R1 smaller radius and R2 larger radius. (or vice versa). 1

central angle corresponding to R1

2

central angle corresponding to R2

E = Point of reverse curvature.

Fig.12.2. Through E draw a line BE parallel to the two tangents. O1T1 &&el to O2T2

Q Perpendicular to two parallel tangents. 1

=

2

0 O1E = R1 0 O2E = R2 O1O2 = R1 + R2

CE401/17

129

To determine v v = T1B + T2D T1B = O1T1 - O1B = R1 - R1 cos

(

= R1 ( 1 - cos

1

=

2

=

)

)

= R1 ver sin T2D = O2T2 - O2D = R2 - R2 cos

2

= R2 ( 1 - Cos = R2 ver sin

2

)

2

v = T1B + D T2 = R1 versin

1

=

2

since

1

+ R2versin

( 12.3 )

2

v = ( R1 - R2 ) versin

( 12.3a )

To determine the direct distance between T1 and T2 i.e., L

1

T1E = 2 R1 sin

2

T2E = 2 R2 sin 2

2

L = 2 (R1 + R2) Sin

sin

2

=

2

v L

from which L =

L = 2(R1 + R 2 )

2v ( R1 + R 2 ) - - - - - - - -

h = BE + ED = R1 sin

v L

( 12.4 )

+ R2 sin

( R1 + R2 ) sin If R1 = R2 = R V = ( R + R2 ) versin

( 12.5 )

CE401/17

130

= 2R versin = 2R ( 1 - cos

). . . . . .

L = 2 ( R1 + R2 )

v 4Rv = L L

( 12.6 )

4Rv

or L =

( 12.7 )

h = 2R sin

( 12.8 )

12.6. Problems: 12.6.1. A reverse curve is to be set between two parallel railway tangents 30m. apart. If the direct distance between the tangent points T1 and T2 is 150m and if the two arcs of the curve are to have the same radius, calculate the radius. Solution:

As R1 = R2 = R (refer Fig.12.2.) L = direct distance between T1 and T2 = 150m = v

4Rv

= 30m

4R x 30

150 =

4R x 30 = 1502 R =

150 x 150 4 x 30

=

750 = 187.5m. 4

Note: As an extension of the problem if the student is required to calculate the data required for setting out of the two simple curves (forming the reverse curve) by any of the methods, (offsets from long chord or deflection angles) he must be able to do so, making use of the principles, and formula explained in earlier units. 12.6.2. Two Roads AB and CD meet at a point Q. They are to be connected by a reverse curve such that A and D are on opposite sides of the common tangents BEC

ABC = 120o

BCD = 120o

The common tangent BC = 25 chains of 20m length i.e., 500m. Given the chainage of B as 2000 meters determine the common radius R and the chainage of the Point D. Solution:

= 180 - 1200 = 600 = 180 - 1200 = 600\

d

R= tan

2

+ tan

= 2

500 60 60 tan + tan 2 2

CE401/17

131

Fig.12.3 =

500 2 tan 30

=

500 = 250 x 1.32 = 433 meters 1 2x 3

B A = R tan

2 60 = 433 tan 2 = 433 tan 30 = 433 x

Arc AE = R

= 433 x 30 x

Also Arc ED

= 226.6m

Chainage of B

= 2000m

Tangent length BA

= 250m

Chainage of A

= 1750m

=

Chainage of E Add Arc length ED Chainage of D

1 3 180

= 250 meters. = 226.60m.

226.6

= 1976.6 =

226.6

= 2203.2m

12.7. Vertical Curves: 12.7.1. General: All the curves discussed earlier are related to movement in horizontal plane. On highways or railways changes in vertical motion occurs due to difference in elevation between points on ground. Movement must be smooth for a vehicle whenever there is an elevation change in vertical motion. Hence vertical curves are

CE401/17

132

intended to join two intersecting grades for the safety and convenience of movement whenever there is an elevation or depression along the direction of curve. Parabolic curves are prefered as the gradient on this curve changes from point to point uniformly throughout. This ensures smooth movement of vehicles along the track. 12.7.2. Terminology: a) Grade or slope of the curve (g): The increase or decrease in elevation along the direction of motion is known as grade or gradient of a road. It is a positive grade (up grade) when elevation increase and negative grade (down grade) when elevation decreases from point to point on the straight or curve.

Gradient is expressed as a percentage (2% or 3%) or as 1 vertical to ‘n’ horizontal (1 in 200). b) Rate of change of grade (r): is also important for smooth travel along the curve. For a parabolic (vertical) curve, the general equation is y = ax2 + bx + c. Slope of the curve is

dy = 2ax + b and the rate of change of gradient is ( d2y/dx2) = 2a. In this case ‘r’ is constant dx

(usually about 0.03 to 0.06 per 20m interval ), whereas ‘g’ varies from point to point.

Fig.12.4 c) Tangent correction: It is the vertical offset distance from the point P on the tangent to the point Q on the curve. PQ,MN are tangent corrections as shown in the figure. Elevation of point P is known from the values of g and r. Point Q on the curve is obtained using the calculated value of tangent correction PQ. d) Chord gradient: It is the elevation difference CN between two adjacent station points on the curve Q and N as

CE401/17

133

shown in the figure

Elevation of station points on the curve are calculated using the values of corresponding

gradients and the elevations of previous station points starting from the first station nearer to tangent point. 12.7.3. Types of vertical curves:

Fig.12.5. 12.7.4. Length of vertical curve: It can be obtained from the known values of g1 and g2 and the rate of change of grade ‘r’ (as length is equal to

g in a given case ) i.e., r g1 - g 2 chains (12.9 ) r g - g2 = 1 x 20 or 30m depending upon the interval taken r for example if g1 = + 1%

L =

g 2 = - 0.8% and

r

= 0.18% for 20m chain

length of the curve =

(+1) - (- 0.8) = 10 chains or 200 meters. 0.18

CE401/17

134

In all the types length of curve is always considered as the horizontal projection, distances along the curve are measured horizontally and offsets from the tangents to the vertical curve are measured vertically. 12.7.5. Setting out vertical curve by tangent correction method: Equation of the parabola can be written as Y = ax2 + bx

dy = 2ax + b dx At

dy = dx

x = 0,

g1 = b hence y = ax 2 + g1x ....( i )

+ g1

tangent correction h = PQ = PR - QR = g1x - y

......( ii )

from ( i ) and (ii ) h = -ax2 or h

2

(2n) . .....( iii )

where 2n = no. of chords each of length

l as shown in the fig.12.4.

The constant of proportionality k in the equation h = k (2n)2 Let e1 and e2 are corresponding increase (or decrease ) in elevation per chord length ‘ l ’. For the total curve change in elevation from point A to C consisting of two parts ne1 and ne2 algebraically added to obtain the height between A C points. Hence k (2n)2 = n(e1 - e2) k =

( e1 - e2 ) 4n

( 12.10 )

From this tangent correction values (h) for different points are calculated to trace the vertical curve as follows: a) Chainage of starting point A = Chainage of apex point B -

1 length of curve (i.e.nl) 2

Chainage of last point on the vertical curve = Chainage of apex point + nl Elevation of starting point A = Elevation of apex point - ne1 Elevation of last point on the curve = Elevation of apex point + ne2 ( Note: e1 and e2 are known as g1 and g2 are given ) Elevation of E =

ne1 + ne 2 2

CE401/17

135

Elevation of F = Elevation of E + Kn2 or Elevation of B - Kn2 b) At h = Kn2, tangent corrections are obtained for the values of n = 1, n = 2 etc., c) Elevations of corresponding stations on the tangent and on the curve are obtained at regular intervals of 20m, 40m. etc., (1 Chain, 2nd Chain, 3rd Chain etc., in the horizontal direction on AE line) using the above computed tangent correction values. Chord gradient method: As e1 and e2 correspond to rise or fall of tangents per chord length ' l ' , Difference in elevation between two points on the curve are obtained (chord gradient) as e1 -k, e1 -3K, e1 -5k etc. The value of k is

e1 - e2 and hence elevation of all points on the curve are computed starting from 1st 4n

station point. Evaluation of a point = Elevation of previous point ± chord gradient. 12.7.6. Sight distance: While driving a vehicle on a vertical curve the driver may not be in a position to see objects in front of him over a summit portion. It is quite possible that a driver has to (i) feel about an obstruction of the traffic and (ii) react mentally and physically and apply the brakes. Depending upon the speed of the vehicle it stops after moving some distance along a highway.

Sight distance is the length of road way visible to the driver. For a speed of 50 to 60 km/hr. normal stopping distance may be around 60 to 80m. Care has to be taken when this sight distance is entirely on the vertical curve. Fig.12.6

In the fig.12.6. ABC is the line of sight of the driver tangential to the curve. h1 and h2 are the heights of driver and the object position on the highway. p and q are distances on the curve on up and down grades totalling to a length of sight distance S. h1

p2

and h2

q2

CE401/17

136

the constant of proportionality is

g1 - g 2 L 1 ( ) ( 2 ) 100 2 L

s

where L is the length of the vertical curve.

hence s =

10 2 L g1 - g2

[ h1

+

( h1 + h2

h2 )

]

2

L=

s ( g1 - g2 ) 200 ( h 1 + h 2 )2

( 12.11)

For a reasonably minimum value of sight distance and the usual height of the observer on a road the length of the vertical curve may be calculated from the above formula. 12.8.PROBLEMS: 12.8.1: A vertical parabolic curve 120m long is introduced to connect two grades -1.5% and + 1.5% R.L. of point of intersection 300.00m and its chainage is 550m. Calculate the elevation of points on curve at 20m interval (i) by tangent correction method of (ii) by chord gradient method for few stations. Solution: a) Tangent correction method. No. of stations on either side of apex = 3 e1 = change of elevation for 20m in 1st half of curve = (g1/100) x 20 = -(1.5/100) x 20 = -0.30m. e2 = +

1.5 x 20 = 100

+ 0.3m

Elevation of starting point of curve = 300 + ( 3 x 0.3 )m = 300.9m. Tangent correction value h =

e1 - e 2 2 N 4n -

h =

1.5 - 1.5 2 N = 4 x 3

-3 2 2 N = 0.25 N 12

i.e., h = -0.25N2 1st station point: tangent elevation = 300.9 -0.3 =300.6m

CE401/17

137

tangent correction = 0.25 curve elevation = 300.6 + 0.25 = 300.85 IInd station point: tangent elevation = 300.6 -(2xO.3) = 300.00 tangent correction = 0.25(22) = 1m curve elevation = 300 +1 = 301m IIIrd station point: tangent elevation = 306.6 - ( 3 x 0.3 ) = 299.70 tangent correction = 0.25 (32) = 2.25m curve elevation = 299.7 - 2.25 = 301.95m Chord gradient method: Any chord gradient value is e1 -( 2N - 1 ) k at 1st point chord gradient = -0.3 - ( 2 - 1 ) ( -0.25 ) = -0.05 at 3rd point gradient = -0.30 - ( 6-1 ) ( -0.25 ) = +0.95 Elevation of 1st point on curve = 300.9 - 0.05 = 300.85 3rd point on curve = 301 + 0.95 = 301.95 (All other points may be calculated by the student as an assignment). 12.8.2.: Calculate the length of the vertical curve if the height of the driver’s eye in 1.96m above the ground.. Height of an obstacle is 0.16m on the highway ahead of the vehicle, value of sight distance to be observed is 200m Grades on the vertical curve are g1 is +1.2% and g2 is -1.8%. Solution: Length of the curve =

200 2 (1.2 + 1.8) = 333.3m 200( 1.96 + 0.16 ) 2

12.9.Exercise: (Problems for practice) 12.9.1. A reverse curve AB is to be set out between two parallel railway tangents 12m apart. If the two arcs of the curve are to have the same radius and the distance between the tangent points A and B is 96m. Calculate the radius. The curve is to be set out from AB at 8m intervals along that line. Calculate the offsets. 12.9.2.: Design a vertical curve connecting two gradients 2% to 1.5% at a summit (R.L. 70.50 chainage 850.00m). The curve is to be such that two points 300m apart and 1.25 above the curve are intervisible.

CE401/17

138

SUMMARY Reverse Curves:

d

(i) Connecting non-parallel straights R =

tan

1

+ tan 2 2 2

BT1 = distance of the intersection point from the beginning of the curve = [ d sin

2/

sin ) ] + R tan

1/2

(ii) Curve connecting parallel straights v = (R1+R2) Versin

( for R1 = R2 = R)

= 2R versin

L = 2 (R1 + R2) sin =

L =

2

2v (R1 + R2 )

4Rv (for R1 = R 2 = R)

h = (R1 + R2) sin h = 2R sin

(for R1 = R2 = R)

Vertical Curves: From the given values of g1, g2, r i) the elevation of tangent at various intervals, ii) tangent corrective values, iii) chord gradient values, are computed using the following equations, k =

e1 e 2 4n

h = k n2 (for tangent correction values) Chord gradient values are e1 - k, e1 - 3k, e1 - 5k etc., From these values, elevation of various points on the parabolic vertical curve is determined. Length of vertical curve, from the considerations of minimum value of sight distance is obtained using the formula: L=

s 2 (g1

g2 )

200 ( h1 + h 2 ) 2

where h1 and h2 are the heights of observer and obstacle respectively. *****

SURVEYING UNIT - 13 TRIANGULATION SYSTEM

CE401/17

139

Contents: 1. Aims and Objectives 2. General 3. Triangulation figure and frame 4. Strength of a figure 5. Field work 6. Intervisibility of stations 7. Signals and Towers 8. Summary. 13.1.Aims and Objectives: In geodetic surveying,

Triangulation is followed for all computations.

General

principles connected with triangulation systems, signals, stations and procedures for conducting field job are described in brief. This chapter gives a basic knowledge on the essentials of triangulation networks. 13.2.General: Triangulation is the process of measuring angles of a triangle, in addition to one side

( a base line )

and other sides are computed. The vertices of individual triangle are known as triangulation stations. Stations form a system of control points in the field which is the basis of network of triangles. Triangulation is more accurate than theodolite traversing and so the former is preferred. Ground features are located from these survey stations and survey lines. 13.2.1.The purpose of a triangulation survey is (i) to establish ground control points for plane and Geodetic surveys of large size field areas . (ii) to locate engineering works accurately (iii) to fix up control points for photogrammetric surveys . Triangulation systems are classified on the basis of precision requirements and to suit for a purpose. (i) Primary triangulation: This is the most accurate and highest grade of triangulation system. It is used for determining the size and shape of earth’s surface. Subsidiary triangulation stations are connected with the precise control points. All precautions must be observed in linear and angular measurements and their computation work. (ii) Secondary triangulation: In this system, two primary series are connected. Hence control points are nearer. Accuracy is not as in primary triangulation system. (iii) Tertiary Triangulation: It is the system to provide control points between the stations located by primary and secondary systems. These form the immediate control points for topographical surveys, by the Survey departments. Details of three systems are shown in the table: _______________________________________________________________________ S.No.

Specification

Primary

Secondary

Tertiary

________________________________________________________________________ 1. Length of base

8 to 12 Km

2. Length of side

16 to 150 Km

2 to 5 Km 10

to 25 Km

100 to 500 Km 2 to 10 Km

3. Average triangular error permitted

1”

3”

12”

CE401/17 4.

140

Maximum station closure

5.

3”

Actual error in base

8”

15”

1 in 50,000

1 in 25,000

1 in 10,000

1 in 50,000

1 in 20,000

1 in

6. Probable error of distances

5,000

________________________________________________________________________ 13.3.Triangulation Figure and Frame: An arrangement of series of triangles is known as layout.

(a) Simple Triangles

(c) Centered triangles and polygons

(d) grid iron system

(e) Centered system.

CE401/17

141

Figs.13.1 Systems of triangulation Fig.13.1 (a) shows a simple layout of triangles useful in a narrow strip of land. Though the system is simple and provides for rapid work, observational checks may not be adequate as work is proceeded on one route. Fig. 13.1 (b) shows a system of quadrilateral with diagonals observed.

This provides good checks on the

observation system as lengths are computed from various combinations of sides and angles. Fig.13.1 (c) is a combination of centered triangles and polygons. For covering large areas, this system is adopted. Progress of work is likely to be slow as more stations are to be set up. Fig. 13.1 (d) is a grid iron system, used for a primary triangulation work, consisting of a series of chain triangles running through the country. Distance between two stations may be around 200 km. Area in between and enclosed by the triangle may be filled up with secondary and tertiary triangulation systems. This has been adopted in India, France, etc. Fig. 13.1 (e) is a centered system consisting of a network of triangles only, extending in all directions with an initial base line set up at the centres of a country. This has been found useful for moderate areas and is adopted in U.K. 13.4.Strength of a figure:This is a measure of accuracy in the system. In a triangle (or a system of triangles) one side and two angles are observed. Other sides of a triangle are computed. Well shaped and conditioned triangles offer more precision. Alternate routes are planned and the best route giving will shaped figures and networks of system is adopted. Accuracy also depends upon the number of observations, number of trignometric and geometric conditions, magnitudes of angles, etc. Accumulation of errors must be avoided, with suitable intermediate checks. Well conditioned triangle is one, in which an error in angular measurements, has a minimum effect upon computed sides. Sides are equally affected if they are equal. In the

ABC,

a = c

sin A sin C

Let an error A introduce a, in side or

Fig.13.2

CE401/17

a1 = c i.e.

142

cos A A sin C

a1 = A cot A ________________________________ (1) a Let an error C introduce a 2 inside, similarly a2 = - C cot C _________________________________ (2) a Assuming, A = B = , and C = 180o - (A + B) = 180 - 2A in an isosceles triangle, a = a

cot 2 A + cot 2 2A __________________________ (3)

This means

a is minimum if a

4 cos2 A + 2 cos2 A - 1 = O or A = 56o14’ i.e. the best shape of a triangle is an isoscales triangle with base angles 56o 14 each. For all practical purposes, equilateral triangles are well conditioned triangles. Angles less than 30o and more than 120o should be avoided in the chain of triangles. 13.5.Field work: Field work of triangulation consists of (a) reconnaissance (b) Erection of signals (c) Measurement of base line (d) measurement of horizontal and vertical angles. (a) reconnaissance means preliminary inspection of the field to decide on the best method and ways of carrying out field work. The entire terrain is properly examined. Suitable site is chosen for running a base line. Position of triangulation stations are located. Accurate topographical maps or aerial photographs may be quite useful in the preliminary survey. If they are not available, plane table or theodolite traversing is carried out in the field. Suitable system of triangulation is organised only after computing the reconnaissance survey. 13.5.1.Location of Stations: Triangulation stations must be intervisible. This should be ascertained first. Elevated places are chosen, and signals erected at high places to facilitate easy observations from station to station. Easy access and approach must be available for a station. All the stations in one system should form well conditioned triangles. Detailed surveys are carried from these stations for locating the salient features. There should be always a possibility of future expansion of a base and triangulation system for further work. Clear line of sight to signals is preferred.

CE401/17

143

Stations are permanently marked on the ground on a stable foundation.

Marks must be clear and

indestructable, preferably on a rock or pipes embedded in concrete. The station is located with reference to two or three permanent identification marks on the field. 13.6.Intervisibility of Stations: (D) Distance of visible horizon from a station of known elevation. (h is the height of station above datum ) above mean seal level is given by h = 0.06735 D2 where h is in meters and D is in Kilometres.

(a)

(b)

Fig 13.3 Intervisibility of Stations In case (a) shown in fig.13.3, where there is no intervening ground obstruction, D1 =

h1 and D2 = 0.06735

h2 and D = D 1 + D2 . 0.06735

Required elevation of one station is calculated from the other known values. In case (b), shown in fig.13.3, when there is an intervening ground obstructs, the elevation of obstruction must be calculated using the Mc Cows, formula as ,

h =

1 1 d 2 2 ( h 2 + h1 ) + ( h 2 - h1 ) - ( S + d ) x 0.06735 2 2 s

when the elevation of intermediate object is known, whether the line of sight joining the two stations goes above the intermediate point or not, is ascertained. Example: Stations A and B are 50 km apart and their elevations are 220 m and 250 m. respectively. The intervening ground is at a mean elevation of 200 m. Find the minimum height of signal to be erected at B so that the line of sight may not graze the ground less than 3 m. Ans:

CE401/17

144

Minimum elevation of line of sight = 200 + 3 = 203 (Datum level) ht. of A (h1) = 220 - 203 = 17 m or D1 =

17 = 15.89 0.06735

D2 = 50 - D1 = 34.11 h2 = 0.06735 x (34.11)2 = 78.36m elevation of line of sight at B = 203 + h2 = 281.36 m. Minimum height of signal above ground at B = 281.36 - 250 = 31.36 m. 13.7.Signals and Towers : Signals are used to define the correct position of a triangulation station on the ground. They are erected truly vertical. Bisecting the centre of a signal is a very important operation, to avoid errors. Signals are classified as Luminous signals and opaque signals. Some of the opaque signals used for daytime observations have been shown in figs (a), (b), (c) and (d). Signals are painted in black and white in alternate so that they are conspicuous in observation. They are held vertically by tripod stands.

(a) Pole Signal

(b) target signal

(d) Beacons

(c) Stone Cairn

CE401/17

145

Figs.13.4 Signals Pole and target signals are used for an observation upto 5 km. Cairn is a signal of about 3 to 5 m height above stones arranged for a support. Beacons are signals in red and white colours. Three poles are provided for easy Centering over a station mark. They are useful when observation are taken from two stations at a time. In a variety of Luminous signals, heliotropes, reflect the Sun’s rays towards the station for observation. These are used in clear weather and Sky. Heliotrope consists of a circular plane mirror and a sight vane carrying cross wires. Mirror can be rotated in any direction. Mirror has a small hole at the centre so that the sight vane can be adjusted (looking through the hole) from the flashes given at the station. Heliotrope is not distrubed when the flash falls at the centre of cross. Mirror is rotated such that the shadow of the hole falls on the exact centre of cross of sight vane. Reflected rays are observed at the station. Heliotropes are useful when observation stations are at an elevated place. Night signals: For sights less than 80 km, oil lamps with reflectors are used, and for higher distances acetylene lamps are used. An ideal signal should be clearly visible from a distance against any background. It should be properly centered over the station mark. Line of sight should accurately bisect the signal centre. When an opaque signal is bisected by the observer, observer may see only partly illuminated portion of the signal and bisect that portion only. This is not the true centre line of the whole signal circle. This error is known phase of signal. Angles of vision are reduced to the centre of the signal by applying a phase correction, to the observed angles depending upon the relative positions of sun and signal. Elevated Towers are useful for fixing up the positions of instrument stations,.

This facilitates easy

visibility from station point to signals. They must be sufficiently rigid and stable as at higher elevations survey party and instruments are working. Various types of signals and towers are constructed before the commencement of triangulation work at suitable points. 13.8.Summary: In this chapter, systems and layouts of triangulation, and procedures for location stations have been discussed. Always, intervisibility of triangulation systems is an essential requirement and hence that verification procedures are given. Signals and towers of various types to be adopted in the triangulation system are briefly described and illustrated. ******

SURVEYING UNIT NO.14 BASE LINE MEASUREMENT

CE401/17

146

CONTENTS: 1. Aims and Objectives. 2. Introduction 3. Equipment and use 4. Correction to base line measurement 5. Extension of a base 6. Types of triangulation stations 7. Satellite Computations 8. Heights of stations 9. Summary. 14.1.Aims and Objectives: Base line must be very accurately measured. Corrections are applied for the field measurement. Triangulation stations and satellite computations are also dealt here. 14.2.Introduction: Base line measurement is the basis on which computations are made in a triangulation system. Hence in alignment and measuring the base, first importance is given to the accuracy. Depending upon the order of triangle, base line may be upto a length of 10 to 15 km. Site selection:

Base line is taken on a fairly level, smooth and firm ground, free from obstructions. Uniform or

gentle slope may be considered but undulating surfaces are avoided. whole length of base must be laid so as to form well shaped triangles in the system. Terminal stations must be intervisible. There should always be a possibility for future expansion and extension to triangulation systems. 14.3.Equipment and Use : Rigid bars offer more precision. Ends of bars are brought into successive contacts for measuring total length.

Marks engraved on the effective lengths of bars may be observed through optical

instruments. Bars may be monometallic or bimetallic. They may be of compensating type to minimise temperature corrections. By using steel or inver tapes, brass or steel wires longer lengths may be measured at a time. Accurate measurement is possible inspite of undulations on the ground. This method is flexible. quick and economical. Bars or tapes must be standardised before and after measurement . They should be carefully handled. 14.3.1 Colby’s apparatus :

CE401/17

147

Fig.14.1. Colby’s apparatus This is a compensating type of equipment for measuring base line. The rigid bar eliminates the effects due to temperature variations in the field. Due to temperature increase, steel bar increases its length to ‘b1’ position and simultaneously brass bar expands upto ‘c1’ position, on one side. Similarly on the otherside. These variations do not have an effect on distance a a’ as the ratios of lengths ab to ac is according to the ratios of coefficient of expansion of meterials (3:5) and due to the similar triangles formation in the figure. 14.3.2. Wheeler’s apparatus (Hunter’s short base) : All supports must be kept in one line using a theodolite. Tape or chain is supported by intermediate stakes, (with a nail or strip at the top supporting stake) one end of tape is fixed to a straining pole and the other end to a spring balance of a counter weight. Tension of the tape is kept constant so as to adjust to the standard length of tape.

Fig. 14.2. Wheelers apparatus The Equipment set up consists of standard tapes, straining devices, tripods, supports, spring balance, thermometers etc. In field work, line alignment is done by one party and the distance measurement is done by another group. Field work includes, clearing the site, making sections, alignment of line, fixing up of tripods, measuring distances, packing the equipment for further extension of work etc.

CE401/17

148

In undulating grounds, tacheometric observations are useful. Accurate observations may be carried out using electronics distance measurement devices. 14.4.Corrections to base line measurement: i) Abosolute length (±) measured length is not equal to the designated length. Correction for absolute length =

measured length x designated length (correction for one tape length).

ii) Temperature (±) Field temperature is different from the temperature at which tape is standardised. Correction = length x difference in temperature x coefficient of expansion iii) Pull or tension (±) pull applied in the field is more than that at which tape length is standardised. Correction =

Difference in pull x length Area of section of tap x E

iv) Sag ( - ) tape due to self weight, between supports may take the shape of catenary, making the measured length more.

Standard length x ( weight of tape )2 Correction = 24 ( pull applied )2 Normal tension may be aplied so as to minimise sag and balance with the pull correction. v) Slope ( - ) Distance measured along a sloping ground is more than the horizontal. Correction = Length x [ 1 - cos ( angle of slope ) ]

CE401/17

149

( difference in level )2 or = 2 x length vi) Wrong alignment ( - ) If the tape is out of line, correction is applied for alignment mistake. vii) M.S.L ( - ) Horizontal distances are reduced to geodetic distances at mean sea level. Difference between measured distance and equivalent length at M.S.L. =

measured length x Difference in level radius of earth' s curvature

14.5.Extension of a base line:

14.3 (a) Extension of base Base line AB is extended by means of well conditioned triangles as shown in the figure 14.3 (a). In the triangle ABE, all the angles and side AB are measured. Other sides are calculated. Length of BC is calculated using the data of

AEC and AFC.

ii)

14.3 (b) From the known angles at A and B, sides AC, BC, AD, BD and later DC are obtained as in fig. 14.3 (b). From the measured angles at C and D towards F and E, sides FC, EC, FD, DE and later EF is calculated. The new base line forms a larger base for a greater triangle. 14.6.Types of triangulation stations:

CE401/17

150

There are four categories i) Main stations These are the stations used to carry forward the net work of main triangulation. Observations at all these stations must be done with sufficient care. ii) Subsidiary stations: These stations provide additional intersecting rays to triangulation stations. iii) Satellite stations: These are stations selected close to main triangulation stations to avoid intervening obstructions. Measurements are taken with high degree of accuracy. Data from these satellite stations are reduced to the values from triangulation stations. iv) Pivot stations Angles measured at these stations are used for making the continuity in the series of triangulation. 14.7.Satellite Computations: Some times observations are taken to satellite stations from one or more triangulation stations. Distance of the satellite station from the main triangulation station may be measured. Data is used to fixing the main triangle. When one of the main stations is not visible or obstructed from the view, from other station, observations are taken from a satellite station and later the data is used to compute and fix up the main station. In the figure 14.4, A, B, C are three triangulation stations and S is the satellite station, at a distance of ‘d’ from station C.

Fig.14.4 Satellite Observations. A S B,

B S C are observed at S:

C A B,

C B A are observed from A and B.

CE401/17

151

A B is the known side

Sin ABC and Sin ACB Sin CAB B C = AB Sin ACB Applying sine rule to A C S, A C = AB

Sin

d Sin ( + ) Sin ASC = b AC

= SC

Similarly Sin

as

d Sin SC sin CSB = a BC Knowing the values of , , =

ACB = 1 +

and

,

A C B is computed from the figure.

d sin d sin ( + ) a sin 1 ' ' b sin 1 ' '

Knowing all the three sides and three angles of a triangle, Lalitudes and departures of all sides are computed, starting from a known bearing value. Example 1: (Using the same figure and notation) d = 14 m = 320 45’ 48’’ = 680 26’ 36’’ BC = 5678 m and AC = 4441 m. Find

A C B.

Ans:

sin

=

14 sin ( 32 0 45' 48'' + 68 0 26' 36'' ) 4441

= 0

Sin

0

32' 45.7''

14 sin 32 0 45' 48' ' = 5678 0 = 0 04' 35'' A C B = 680 45' 48' + 0 0 32' 46'' - 0 0 04' 35' ' = 690 13' 59'' .

CE401/17

152

14.8.Heights of triangulation stations: Heights and distances are computed as in cases of trignometrical levelling, considering all aspects of applications of corrections to observed angles. 14.9.Summary: In this Chapter, how base line is measured in triangulation systems and the various corrections to be applied are described. Some details on stations and extension of a base line and computations regarding satellite stations are given. (Chapter 13 and 14 are giving the fundamentals of geodetic surveying. Care has been taken to avoid astronomical observations and their applications, as this is the first courses intended at the graduate level). *******

SURVEYING UNIT NO.15.

THEORY OF ERRORS AND ADJUSTMENTS CONTENTS: 1. Aims and Objectives 2. Introduction 3. Terminology 4. Law of weights 5. Probable errors 6. Most probable value 7. Method of least squares 8. Additional examples 9. Summary. 15.1.Aims and Objectives: i) To know the different methods of evaluating the probable values from the observed values with the knowledge of theory of errors. ii) Application of different methods have been explained through illustrative examples. 15.2.Introduction: At the time of taking linear and angular measurements in field surveying, errors may come due to imperfect instruments, atmospheric conditions and individual limitations. Principles explained in this chapter are intended to arrive at the most probable value of an observed quantity that is nearer to its true value. Computations can be carried from the knowledge of trust worthiness of the measured value. Errors may be broadly categorised into four groups:

CE401/17

153

i. Cumulative or Systematic Errors: If measurements are taken with a chain of 29.9 m, only instead of 30 m, for each unit of length measured, a. the error is positive (as observed value is in excess of true value). b. it is cumulative. c. it is of constant value during total measurement. Such type of errors are systematic errors. They are of known value and corrections are applied for the observed quantity. ii. Compensating errors: If measurements are taken with a theodolite, having incorrect graduations on the horizontal circle, a. Some times the error is positive and negative for some observations. They may tend to compensate each other. b. the magnitude and sign of error may vary during a set of observations. c. corrections are applied only if the extent of error is known completely. iii. Accidental errors: Some times accidental errors are unavoidable. During the observations in a day, the atmospheric variations may affect the measured quantities. Results may be increased or decreased. Errors caused beyond the observer’s control are termed as Accidental errors. As they cannot be fully accounted for and taken care of, all observations are combined to arrive at a most probable value of measurement which is likely to be close to the real value. In this chapter, such type of computations are dealt in detail. iv. Personal mistakes: Sighting a wrong signal, manipulating wrong screw, reading or recording a wrong value are classfied as personal or observers mistakes. Sufficient care must be taken to avoid such type of errors which may be difficult even to detect and rectify. 15.3.Terminology: 1.a. Direct observation:- Reading angle on a theodolite is a direct observation. b. Indirect observation:- In trignometrical levelling, heights and distances are computed from direct observations, indirectly. 2.a. Independent quantity:- Observed measurements such as length of a line, elevation of a point, which do not have any relationship with or not based on anyother data, are known as independent quantities. b. conditioned Quantity:- If two angles in a triangle are observed first, then the third angle observation is conditioned quantity as the sum of angles must be 1800. 3.a. Observed Value:- Standard corrections are applied to the measured value, (due to known and possible sources of error) for getting the observed value. b. Most probable value:- It is considered as the most likely nearer value to the true or real value and free from all possible errors. c. True value:- It is observed value corrected for all true errors. If the true error is unknown, true value is always indeterminate. (True value of a quantity is the weighted arithmetic mean of number of observations).

CE401/17

154

4. Weight of an observation:- Two measurements may not be made under truly identical conditions. If angles are measured with a compass and a theodolite, or length is measured by steel tape or chain, accuracy or reliability of observation is not going to be same. Weight of a quantity is expressed as a measure of precision of measurement. It is a number and used to compare different quantities. Suppose two values are measured as 10 and 9 with equal consideration for precision. The average is taken as

measurement is twice

10 + 9 = 9.5. But if the first 2

as accurate than the second measurement, the better mean value would be

10 + 10 + 9 = 9.67. That is the weight of first observation is 2 and the weight of the second 3 observation is 1 only. Weight of quantity is inversely proportional to square of its probable error. 5.a. True error:- It is the difference between observed value and true value. b. Probable error:- From this quantity, one can understand the limiting values of a true values from observed quantity. c. Residual error:- It is difference between the most probable value and the observed value (as arithmetic mean is influenced by errors). d. Average error:- It is arithmetic mean of separate errors. If three separate set of observations cause errors of +20’’ , +30’’ and -20’’ , the average error =

+20 + 30 + 20 3

e. Mean square error =

2

[20 + 30

2

2

+ + 20 ]/

Probable error = 0.845 x average error = 0.674 mean square error. 6.a. Observation equation:- It is the relationship between the observed quantities such as 50 = K0 (S) + C in techeometry. b. Conditioned equation: - It is a equation stipulating relationship between several independent quantities, such as

A

+

B

+

C

0

= 180 in a triangle.

c. Normal equation:- For each unknown, one normal equation is framed. Those equations are solved to obtain the probable values of quantities. To find a normal equation of A, multiply each observation equation by the coefficient of A and add all such equations. 15.4.Law of Weights: When weights of individual observations are known, these rules help in finding the weight of the result. They are illustrated with suitable examples below. a. If the length is measured three times as 10.1 m, 10.2 m, 10.3 m, with equal weights (1, 1, 1), then weight of arithmatic mean

[

10.1 + 10.2 + 10.3 ] is 1 + 1 + 1 i.e.3 only. 3

CE401/17

155

b. In the case above, if the respective weights are 2,3 and 2, then weight of arithmatic mean

[

10.1 x 2 + 10.2 x 3 + 10.3 x 2 ] is 2 + 3 + 2 i.e. 7 only. 2 + 3 + 2

c. If the weight of A is 2, wt. of 3A will [2/(3)2]. d. If the weight of B is 2, wt. of [A/3] will be 2 x (3)2. e. If weight of A + B + C is [3/4], wt. of [3/4]ths of (A + B + C) is [4/3]. f. If weight of A + B is 2, there is no change in the weight of quantity for (A + B) ± a constant quantity. g. If the weight of A is 4 and weight of B is 3, then weight of A ± B is [(1)/(

1 1 + )] i.e. [ 12/7] i.e. sum of 4 3

reciprocal of individual weights. 15.5.Determination of probable Errors: Probable errors of observation may be determined in the following three ways: r = residual error n = total number of observations w = weight Es = Probable error of a single observation of unit weight. Ew = Probable error of any observation of weight w. a. Direct observations of equal weights: 2

[ r ] / [n - 1]

i. Es = ± 0.6745 ii. Ew = [ Es/

w]

iii. Probable error of arithmetic mean of observations = [Es/ b. Direct observations of unequal weights: i. Es ± 0.6745

[

2

(wr )/n - 1]

ii. Ew = [Es/ w ]

c. Error of computed quantities Y = Computed quantity, ey = Probable error of y. x = observed quantity,

ex = Probable error of x.

c = a constant. If Y = ± x + c, ey = ex Y = ± x.c, then ey = c. ex

n]

CE401/17

156

2

2

e x + e z + .....

Y = ± x ± z, they ey =

Y = f(x) , then ey = ex [dy/dx] Y = f(x,Z) then ey =

dy 2 dy 2 ) + (e z ) + . . . . dz dx

(e x

Example.15.5.1: i. Calculate the probable value of

A , if

A +

B = 900 and

B = 500 30' 30' ' and probable error of

B =

± 5'' .

A = 90 - 50 0 30' 30' ' ± 5'' .

Ans : Probable value of

ii. Calculate the probable error in the sum of 1800 in a triangle from the data

A = 250 ± 2'' ,

B = 750 ± 3' ' and

Ans : Probable error of sum =

±

C = 800 ± 1' ' .

22 + 32 + 12

iii. Find the probable error of the angle, from three measurements of an angle 500 30’ 10’’ (weight 2) and 500 30’ 12’’ (weight 3).and 50” 30’ 9” (weight 1) Ans: Probable error = ± 0.6745 Where n = 3,

2

[ wr / (n - )]

w = 2 + 3 + 1 = 6.

Weighted arithmatic mean of seconds value of angle =

( 10' ' x 2 ) + ( 12' ' x 3 ) + ( 9' ' x 1 ) 2 + 3 + 1

= 10.83.

r1 = 500 30’ 10’’ ------- 500 30’ 10’’ 83 = - 0.83 r2 = 500 30’ 12’’ -------- 500 30’ 10’’ 83 = + 1.17 r3 = 500 30’ 9’’ ---------- 500 30’ 10’’ 83 = - 1.83 2

Probable error = ± 0.6745

2

2

2 x ( 0.83 ) + 3 ( 1.17 ) + 1 ( 1.83 ) 6(3 - 1)

15.6.Determination of most probable values: There are few cases in general for obtaining probable values: i. from direct observations of equal or unequal weights. ii. from indirectly observed quantities involving unknowns of equal or unequal weights, and iii. from observation equations accompained by conditioned equations.

CE401/17

157

These are illustrated in the following methods: Example.15.6.1: i.

Determine the probable value of angle A from the following observations: Case (i )

Case (ii)

A = 100 10’

weight 1

weight 2

2A = 200 22’

weight 1

weight 2

4A = 400 41’

weight 1

weight 3

Ans: in case (i)

A =

10 010' + 200 22' + 40 0 41 = 10010' 43 1 + 2 + 4

in case (ii)

A =

( 10 010' x 1 ) + ( 200 22' x 2 ) + ( 400 41' x 3 ) (1x1) + (2x2) + (4x4)

(multiply equation by weight and algebraic coefficient of angle) ii. Determine the probable value of angles A and B from the following data.

A = 360

.................(1)

B

= 98 0

.................(2)

A

+

B

= 13401' ........(3)

Normal equation in A is

2A + B = 17001’

2

(by adding (1) + (3)) Normal equation in B is (by adding (2) + (3))

2

2

A + 2B = 23201’

2

Hence, solving normal equations,

A

0

= 36 0' . 33

B

= 98 0 0' . 33.

iii. Determine the probable values of angles a,b,c from the data, a = 300 12’

weight 1

- - - - - - Eq. (1)

b = 420 11’

weight 1

- - - - - - -Eq. (2)

= 10 0 10' . 33.

CE401/17

158

a + b = 720 24’

weight 2

- - - - - - -Eq. (3)

b + c = 118032’

weight 2

- - - - - - -Eq. (4)

b+c+a = 148044’

weight 1

- - - - - - Eq. (5)

Ans: i. to obtain normal equation in a, a = 30012’

eq. (1) x 1

2a + 2b = 2 x 72024’ a + b Add

eq. (3) x 2

+ c = 148044’ eq (5) x 1

4a + 3b + c = 323044’

eq. (6)

ii. to obtain normal equation in b, b = 42011’

eq. (2) x 1

2a + 2b = 144048’

eq (3) x 2

2b + 2c = 2 x 118032’ a + b + c

eq (4) x 2

= 148044’

eq (5) x 1

Add 3a + 6b + 3c = 5720

eq. (7)

iii. to obtain normal equation in c, a + b + c = 1480 44’

eq. (5) x 1

0

2b + 2c = 237 04’

eq. (4) x 2

___________________ Add a + 3b + 3c = 3850

eq. (8)

By solving equations 6, 7 and 8 a = 300 12’ , b = 420 11’ and c = 760 21’. iv. Find the probable values of angles of a quadrilateral from the following data, a = 1220 5’

weight (2)

----- Equation 1

b = 860 45’

weight (1)

------Equation 2

c = 720 50’

weight (3)

------ Equation 3

d = 780 18’

weight (1)

------ Equation 4

Note: d is eliminated by using the equation 4 as a + b + c = 3600 - 780 18’ and framing normal equations in a, b, c and further solved for obtaining values. 15.7.Method of Least Squares: Suppose, observed equations consisting of two unknowns are as follows: 1.5

=

0.4a + b

CE401/17

159

4.9

=

1.1a + b

6.83

=

1.6a + b

8.98

=

2.2a + b

In each case, observational errors are 1.5 - 0.4a - b = r1 4.9 - 1.1a - b = r2 6.83 - 1.6a - b = r3 and 8.98 - 2.2a - b = r4 The condition to be satisfied in the least squares method would be, sum of squares of errors should be equal to minimum i.e., r12 + r22 + r32 + r42 = minimum. Differentiating this equation with respect to ‘a’ and equating the sum to zero. 8.77a + 5.3b - 36.674 = 0. This is the normal equation in similarly, differentiating the same equation with respect to ‘b’ and equating the sum to zero. 5.3a + 4b - 22.21 = 0. This is the normal equation in b - (2). Solving the two equations 1 and 2, a and b are obtained.. Hence the equation in general form is y = 4.146x + 0.058. This principle, involved in the least squares method is given as follows: Rule: Most probable values of the observed quantities are those that render the sum of squares of residual errors a minimum. In observations of equal weights, errors may be +ve or -ve and the probable value must be symmetrical to all observed quantities. i.e., the arithemetic mean of observed quantities must be equal to:

observed quantities x number of observations n Let p be the most probable value of a quantity, x1 , x2 , x3 . . . . . . be the n observations of a quantity and r1 , r2 , r3 . . . . . . be the residual error. then

r2 must be a minimum

or (P - x1)2 + (P - x2)2 + . . . . must be a minimum or differentiating the equation, (P - x1) + (P - x2) + . . . . . . . = 0 and P =

x1 + x 2 + . . . . n

. . . . = (1)

This is the case for direct observations of equal weight. In the case of direct observations of unequal weight, the most probable value of the observed quantity is equal to the weighted arithmetic mean of the observed values. If w1, w2, w3.

are the weights of n observations x1, x2, x3 etc., then

CE401/17

160

w1(P - x1)2 + w2(P - x2)2 + . . . . must be a minimum. i.e.

w1(P - x1) + w2(P - x2) . . . . . = 0 or P =

w1x 2 + w2 x2 + . . . . . . . . (2) w1 + w2 + . . . .

In all the cases explained earlier, normal equations are

framed and solved for the most probable values.

Conditioned equations are avoided or eliminated. 15.8.Additional Examples: Example 15.8.1.By the method of differences: Determine the most probable values of angles A, B and C from the following observations of triangulation. A = 700 30’ 20’’

wt 2

B = 550 20’ 40’’

wt 4

C = 100o 40’ 30’’

wt 2

A + B = 1250 50’ 30’’ wt 3 B + C = 1560 2’ 10’’

wt 2

0

A + B + C = 226 30’ 30’’wt 1 Ans: Let e1, e2 e3 be the most probable corrections to angles A, B, C, then Probable value of A = 700 30’ 20’’ + e1 B = 550 20’ 40’’ + e2 C = 1000 40’ 30’’ + e3 and A + B = 1250 50’ 30’’ + e1 + e2 B + C = 1560

2’ 10’’ + e2 + e3

A + B + C = 2260 30’ 30’’ + e1 + e2 + e3 Hence reduced observation equations are, e1 = 0

wt 2

e2 = 0

wt 4

e3 = 0

wt 2

e1 + e2 = 30’’

wt 3

e2 + e3 = 60’’

wt 2

e1 + e2 + e3 = 60’’

wt 1

normal equations for the above set of observed equations are: 2e1 =

0

CE401/17

161

3e1 + 3e2 =

3 x 30’’

e1 + e2 + e3 = 60’’ ______________________________ Adding 6e1 + 4e2 + e3 = 150’’ Similarly normal equation for e2 and e3 are formed and simultaneous equations are solved to obtain the ultimately the probable values of A, B and C. Example 15.8.2.By the method of correlates: In this method, one extra condition is taken as that the sum of the squares of residual errors should be minimum. If c1, c2, c3 are the corrections to be observed angles A, B, C with weights W1, W2, W3 .....then c

=

wc2

=

c1 + C2 + c3 = 180 - (A + B + C)

(1)

w1c12 + w2c22 + w3c32

(2)

Equation (2) must be a minimum differentiating these two equations, c1 +

c2 + c3 = 0

(3)

w1c1 c1 + w2c2 c2 + w3c3 c3 = 0

(4)

multiply equation (3) by a correlative (-’K’) add that resultant equation to equation (4). Hence K = c1w1 = c2w2 = c3w3 and K (

(5)

1 1 1 ) = 180 - (A + B + C ) + + w1 w2 w3

(6)

Corrections are known and probable values of angles are thus calculated. Determine the probable values of A, B, C from the following data. A = 700 10’

wt 2

B = 500 20’

wt 3

C = 590 20’

wt 2

A + B + C = 180 - 1790 50’ = 10’ Ans:

c1 + c2 + c3 = +10’

(1)

2c12 + 3c22 + 2c32 must be a minimum

(2)

c1 + c2 + c3 = 0 2c1 c1 + 3c2 c2 + 2c3 c3 = 0 multiplying (3) by -K and adding it to (4), c1(2c1 - K) + c2(3c2 - K) + c3(2c3 - K) = 0 Since the coefficients of c1, c2, c3 must independently vanish

(3) (4)

CE401/17

162

K K K , c2 = , c3 = 2 3 2 K K K 10 x 3 from equation (1) - - - ( + + ) = 10' or K = = 7.5' 4 2 3 2

c1 =

and c1 = 3.’ 75 ;

c2 = 2.’ 5 ;

c3 = 3.’75

Hence the corrected angles are, A = 700 10’ + 3,’ 75 B = 500 20’ + 2.’ 5 C = 590 20’ + 3.’ 75. 15.9. Summary : 1..Errors are classified as systematic, compensating, accidental and mistakes. 2. Observed values are corrected. Based on the weight or strength of observation probable values of errors are determined.

3. In different cases, most probable values are determined from the principles of law of weights, Least squares, through a number of illustrative worked examples. *********

SURVEYING UNIT : 16 TRIANGULATION ADJUSTMENT Contents 1.

Aims and Objectives

2.

Introduction

3.

4.

5.

Chain of Triangles

6.

Summary.

16.1 Aims and Objectives: In this chapter, triangle adjustment has been illustrated with suitable principles of use. Method of adjusting chain of triangles has also been explained. Student should understand the procedures and method of adjusting the triangulation system. 16.2 Introduction: Geodetic triangle should satisfy a number of geometric and trignometric conditions in the set up. Some relate to a station, whereas other requirements pertain to the figure and framework of system. Probable values should be estimated from the methods illustrated in chapter 15. Few examples have been shown in this chapter, on adjustment of triangles. 16.3 Principles of Adjustment: Angles are measured in Geodetic triangulation surveys. They have to satisfy a number conditions.

CE401/17

163

i) First step is to adjust individual observed angles most probable value of an angle may be obtained by calculating the weighted arithmetic mean, or corrections may be applied inversely proportional to the respective weights. ii) Second step is to determine the most probable values of two or more angles measured at each and every station. The geometric condition to be satisfied is that at one station sum of angles around a station must be 3600. As the horizon is closed, discrepancy or error is distributed among all angles at that station, inversely as their respective weights. In case of equal weights, the error is distributed equally among all measured angles at that station. When individual and combined angles are measured, most probable values of angles are obtained by framing normal equations. iii) Third step is to satisfy the geometric conditions of a triangle or a quadrilateral or a polygon regarding the sum of angles in the figure. The following are the general rules for making necessary corrections to observed angles in figure adjustment of a triangle, if the sum of angles is not 1800. Let

n1, n2, n3 - number of observations of angles A, B, C respectively w1, w2, w3 - weights of observations of angles A, B, C respectively e1, e2, e3 - error of each angle A, B, C (e = e1 + w2 + e3) E1, E2, E3 - Probable errors of angles A, B, C respectively rule I - For angles measured with equal weights e1 = e2 = e3 =

1 e 3

rule II - For angles measured with unequal weight e1 = e2 = e3 =

1 1 = w1 w2

=

1 or w3

1 ( % & w1 # e, Similarly e2 and e3 e1 = 1 1 1 & # + + &' w1 w2 w3 #\$ rule III - If weights of angles are not known, but number of observations are taken e1 = e2 = e3 =

1 1 = n1 n2

=

1 n3

rule IV - rule III may be also taken as e1 = e2 = e3 =

1 n1

2

1 = n2

2

1 = n3

2

rule V - From the probable errors of angles, e1 = e2 = e3 = E12 = E22 = E32 i.e., e1 =

E 12

E1 2 , e Similarly e2 and e 3 . + E 22 + E32

CE401/17

164

rule VI - From the residual error of each observation, weights can be calculated by Gauss rule as 1 2 n 2 w= ( residual error )2

knowing the weights of angles, rule II may be used for correcting the angles. 16.4 Triangle adjustment Examples 16.4.1: Find the probable values of the following angles measured at station P.

/ A P B = 1580 56' 42'' - -- - - - - - - - - weight 1 0

/ B P C = 144 21' 26' '

/A P C =

0

56 43' 06' ' 0

- - - - - - - - - - weight 2 - - - - - - -- - - weight 3

total error of closure = 360 - (158 56’ 42’’ + 1440 21’ 26’’ + 560 43’ 06’’) = 00 1’ 14’’

Ans:

0

error is distributed in proportion to weights as

1 1 1 : : or 6 : 3 : 2 1 2 3

Hence corrections are

6 x 74' ' 11 3 x 74' ' 11 2 x 74' ' 11

=

- 40'' .36 to angle APB

=

- 20'' .18 to angle BPC

=

- 13'' .45 to angle APC

The probable values of angles are (1580 56’ 42’’ ________ 40’’ . 36) (1440 21’ 26’’ _______ 20’’ . 18) and (560 43’ 06’’ ______ 13’’ . 45) Example 16.4.2: Find the most probable values of angles A, B and C, from the following observations at a stations. A = 350 22’ 26’’ 0

E = 38 20’ 8’’ 0

A + B = 73 44’ 32’’ 0

B + C = 112 44’ 30’’ 0

A + B + C = 148 Ans:

6’ 45’’

Assuming the value of C as (B + C) - B from the above data i.e. C = 1120 44’ 30’’ - 380 20’ 8’’ = 740 24’ 22’’

Let C1, C2, C3 be the corrections for angles A, B and C respectively. A = 350 22’ 26’’ + C1 B = 380 20’ 8’’ + C2 C = 740 24’ 22’’ + C3 The reduced observation equations are

--- wt 1 --- wt 1 ---- wt 2 ---- wt 2 ---- wt 1

CE401/17

165

C1 = 0

---- wt 1

C2 = 0

---- wt 1

C1 + C2 = -2’’

---- wt 2

C2 + C 3 = 0

---- wt 2

C1 + C2 + C3 = - 11’’ -

---- wt 1

Normal equations are 1 C1 + 2 C1 + 2 C2 + 1 C1 + 1 C2 + 1 C3 - (2 x 2 + 1 x 11) --

(1)

1 C2 + 2 C1 + 2 C2 + 2 C3 + 2 C3 + 1 C1 + 1 C2 + 1 C3 = - (2 x 2 + 11) ---

(2)

2 C2 + 2 C3 + C1 + C2 + C3 = -11

---

(3)

Solving these three normal equations, values of C1, C2, C3 are obtained and from them, most probable values of angles are calculated. Example 16.4.3: Adjust the following angles of a triangle. A = 550 14’ 26’’

B = 640 32’ 50’’

C = 600 12’ 56’’

= 550 14’ 24’’

= 640 32’ 49’’

= 600 12’ 54’’

= 550 14’ 28’’

= 640 32’ 48’’

Ans: mean value A = 550 14’ 26’’

n1 = 3

0

B = 64 32’ 49’’

n2 = 3

0

n3 = 2

C = 60 12’ 54’’ total correction = - 9’’ As per Gauss’s law

weight of Angle A =

wA

1 2 n1 2 ( mean value - observed value ) 2

1 × 9 9 2 = = 2 2 2 16 (0 + 2 + 2 )

wB =

wC =

1 × 32 9 2 = 2 2 2 4 (1 + 0 + 1 ) 1 × 22 2 2 2 (2 + 2 )

=

1 4

CE401/17

166

1 wA

Correction to angle A =

1 1 1 + + wB wC wA =

4 9 4 Correction to angle C = 1 Correction to angle B =

× ( - 9' ' )

16 9 × × ( - 9' ' ) 9 56 9 × × ( - 9'' ) 56 9 × × ( - 0.9'' ) 56

Hence the corrected angles are obtained. 16.5 Adjustment of a chain of triangles

Adjustment is carried as in the previous case in two steps for station and figure adjustments at stations A, 1 + 2 = 3600 B, 3 + 4 + 5 = 3600 C, 6 + 7 + 8 + 9 = 3600

etc.

and In triangle ABC, 1 + 3 + 6

= 1800

BDC, 5 + 7 + 10 = 1800

etc.

Then, error at each station or triangular error is distributed inversely propositional to their weights. Example 16.5.1: In the above figure taken as ABCD only, the following angles are observed. 1 = 630 42’ 12’’ 6 = 670 28’ 40’’ 7 = 710 05’ 58’’ 3 = 480 49’ 18’’ 5 = 50o 29’ 50’’ 10 = 580 24’ 18’’ 6 + 7 = 1380 34’ 28’’ 3 + 5 = 990 19’ 08’’ Adjust the angles

CE401/17

167

The four unknown Angles

or and

B,

C and

D of the figure are expressed in terms of independent

1 = 630 42’ 12’’

unknown angles 3, 5 and 6, 7 as, = 180 - ( 3 +

A,

6)

6 = 180 - 630 42’ 12’’ = 1160 17’ 48’’

3+

7 = 180 - 580 24’ 18’’ = 1210 35’ 42’’

5 +

3 +

5 = 990 19’ 08’’

6+

7 = 1380 34’ 28’’

Let C1, C2, C3 C4 be the required corrections to angles 6,

7,

3,

5, then

0

6 = 67 , 28’ 40’’ + C1 7 = 710, 05’ 58’’ + C2 From the Right observed 0

3 = 48 , 49’ 18’’ + C3

equations only.

0

5 = 50 , 29, 50’’ + C4 3 +

6 = 1160 17’ 58’’ + C1 + C2

5 +

7 = 1210 35’ 42’’ + C2 + C4

6 +

7 = 1380 34’ 38’’ + C1 + C2

3 +

5 = 990 19’ 08’’ + C3 + C4

C1 = 0 C2 = 0 C3 = 0 C4 = 0 C1 + C3 = 0 C2 + C4 = 0 C1 + C2 = - 10’’ C3 + C4 = 0’’ Hence normal equations are 3 C1 + C2 + C3 = - 10’’ C1 + 3 C2 + C4 = - 10 C1 + 3 C3 + C4 = 0 C2 + C3 + 3 C4 = 0 values of corrections are obtained and angles are corrected 16.6.SUMMARY : In adjusting the angles at a station or in a triangle geometric condition of the figure is observed as the required condition. Most probable values of a triangle or a chain of triangles have been calculated in a set of examples.

CE401/17

168

(NOTE: Problems on triangulation adjustment in the few chapters mentioned here, have been taken from standard text books on surveying II by Shahani, Punmia, Kanetkar and Agor. In the contact programme, they will be discussed in detail. In normal theory examinations, it is very difficult for a student to answer these problems in a limited time. However, the student should understand the procedures involved so that in a practical and field situation, he may be in a position to solve the computations).

****

CE401/17

169

1. Rigorous method of least squares. 2. Approximate method. 3. Method of equal shifts. 3. Least squares method: Angles

1,

3,

5,

7,

are known as left angles ( L) and

2,

4,

6,

8,

are right angles ( R)

considering the sides from the intersection point of diagonals. Conditions to be satisfied are (from the figure) (i)

3+

4+

5+

6+

7+

= 3600

1+

2+

1+

2

=

5+

6

(2)

3+

4

=

7+

8

(3)

8

(1)

(ii) Even if the angle conditions are satisfied, the quadrilateral may have a closing error i.e., figure may not close at the same starting point. From the various triangles in the quadrilateral, applying sine rule,

AB = BC.

DC = AB

sin \5 sin \8

sin \1 sin \ 4

CE401/17

170

sin \ 7 sin \ 2

BC = DC.

sin \ 3 sin \ 6

1.

3.

Hence sin or

sin

log sin

sin

L=

5.

sin

log sin

7.

= sin

2.

sin

4.

sin

6.

sin

8

(4)

R

(iii) Let E1, E2, E3, E4, are the discrepancies for the four condition equations, then E1 = 360 – (

1+

2

+ ….. 8)

E2 = (

5+

6)

–(

1+

2)

E3 = (

7+

8)

=(

3+

4)

E4 =

log sin

R-

log sin

If e1, e2 . . . .e8 are corrections to

1,

L 2

....

8,

then

e1 + e2 + e3 + . . . . .= E1 (e1 + e2) – (e5 + e6) = E2 (e3 + e4) – (e7 + e8) = E3 (e1 f1 + e3 f3 + e5 f5 + e7 f7 ) – (e2 f2 + e4 f4 + e6 f6 + e8 f8 ) = E4 Where f1, f2 . . . .f8 are log sine differences for 1” in the values of eight measured angles respectively from the log tables. Another condition to be satisfied from the least squares rule is.

e12 + e 22 + e33 + . . . . . = a minimum. (iv) Differentiate the above 5 conditional equations and multiply the first four equations by –k1, k2, -k3 and –k4 respectively. Then the correction equations may be written as, e1 = k1 + k2 + f1k4 e2 = k1 + k2 - f2k4 e3 = k1 + k3 + f3k4

CE401/17

171

e4 = k1 + k3 - f4k4 e5 = k1 - k2 + f5k4 e6 = k1 - k2 - f6k4 e7 = k1 - k3 + f7k4 e8 = k1 - k3 - f8k4 Hence E1 = 8k1 + [(f1 + f 3 + f 5 + f 7 )

(f 2 + f 4 + f6 + f8 )] k 4

E 2 = 4k 2 + [(f1 f 2 )

(f5

f 6 )] k 4

E 3 = 4k 3 + [(f 3

(f 7

f8 )] k 4

f4 )

E 4 = [(f1 + f 3 + f 5 + f 7 ) + [(f1 f 3 ) (f 5

(f 2 + f 4 + f 6 + f8 )] k1

f 6 )]k2 + [(f 3

(

f4 )

(f 7

f8 )] k 3

)

+ f12 + f 22 + f 32 + . . . . . .f 82 k 4

As the values of E1, E2, E3, E4 are known k1, k2, k3, k4 are calculated. Then e1, e2 . . . .e8 are known and later angles are corrected. Example 4.1: Adjust the quadrilateral from the given data of angles of equal weights. 1

= 710 261 511

2

= 530 391 5511

3

= 310 181 1011

4

= 230 351 5011

5

= 890 401 1011

6

= 350 251 5011

7

= 140 181 511

8

= 400 351 4011

CE401/17 Ans.

172

=3590 5914511 = 1511

E1 = 360 E2 = (

5

+

6)

-(

1

+

2)

=0

E3 = (

7

+

8)

-(

3

+

4)

= 0.00833

E4 =

log sin

R

-

log sin

L

= 0.0002

Correction equations are: e1 + e2 + . . . . . . . .e8

= 1511

(1)

(e1 + e2) – (e5 + e6)

=0

(2)

(e3 + e4) – (e7 + e8)

= 0.008

(3)

(e1 f1+ e3 f3 +e5 f5 + e7 f7) – (e2f2 + e4 f4 + e6 f6 + e8 f8) = 0.0002

(4)

i.e., e12 + e22 + . . . . . . . . e82 = minimum

(5)

Multiply equation (1) by – k1, eq. (2) by –k2, eq. (3) –k3, eq. (4) by –k4, add to equation (5), equating the coefficients e1, e2 . . . . to zero, the equations are e1 = k1 + k2 + (-0.0232) k4 e2 = k1 + k2 – (- 0.0938) k4 e3 = k1 + k3 + (- 0.2843) k4 e4 = k1 + k3 – (- 0.3976) k4 e5 = k1 - k2 + (0)

k4

e6 = k1 - k2 – (- .2367) k4 e7 = k1 - k3 + (- 0.6070) k4 e8 = k1 - k3 – (- 1866) k4 Substituting these values of e1 to e8, 8k1 – 0.0002 k4 = 1511 4k2 + .1661

k4 = 0

4k3 - .3071

k4 = 0.008

.0002 k1 + .1661 k2 - .3071 k4 + 0.71 = 0.0002 Solving these equations for k1, k2, k3, k4; e1 =

CE401/17

173

e2 = e3 = e4 =

are obtained

e5 = e6 = e7 = e8 = Hence the corrected angles are 1

=

2

=

3

=

4

=

5

=

6

=

7

=

8

=

(The student may completely solve the problem. He is expected to know the procedure only) 4. Approximate method:

This gives fairly reasonable values. Except the least squares

conditions, all the other four conditioned equations are to be satisfied. i)

360 – (

) = correction total or discrepancy

Each angle is corrected by 1/8 of error. ii)

(

1

+

2)

–(

5

+

6)

= correction

Each angle is corrected by ¼ of discrepancy. iii)

(

3

+

4)

–(

7

+

8)

= correction

Each angle is corrected by ¼ of error. iv)

1,

2

....

8

are tested for the condition

log sin

L

=

log sin

R.

CE401/17

174

If there is any discrepancy now, angles are corrected again by an amount of f1 ( log sn _f 2

L

log sin

R

) to angle

1 … 2.etc.

5. Method of equal shifts: As in the other method, conditions to be satisfied in the case of quadrilateral with a central station are a) Sum of angles of a triangle = 1800 figure equation b) Sum of angles at a station = 3600 station equation c)

log sin

L

=

log sin

R

side equation.

Any shift made in the angles to satisfy local station equation, should also be equally applied to each of the triangle of the quadrilateral, and any shift made to satisfy the side equation should be equally applied for each triangle. The following problem illustrates the method.

Example 2: Following are the measured angles of a quadrilateral ABCD with a central station E, Adjust the quadrilateral.

Central angle

AEB

590 031 1011

BFC

1180 231 5011

CED

600 321 0511

DEA

1220 001 5311

L.H. angle

R.H. angle

610 001 5411

2

= 590 561 0611

3

= 320 031 5411

4

= 290 321 0611

5

= 560 281 0111

6

= 620 591 4911

7

= 280 421 0011

8

= 290 171 0011

1=

CE401/17

Ans.

175

9

= 590 031 1011

10

= 1180 231 5011

11 =

600 321 0511

12 =

1120 001 5511

Fig 17.2 Quadrilateral with a central station

CE401/17

9

LEFT Triangle

Observed angles

Observed

Log sine

RIGHT Differen 11

angles

ce for 1

Correc

Correc-

Observed

tion

tion diff.

angles

Log sine

(2)

Correct

Correc

Sum

ence

ion

tion

observed

ion

11

secs (1)

Differ-

in

of

Correct

for 1

secs

Diff.

angles

secs.

(3)

(4)

(5)

(6)

(7)

(8)

(9)

(10)

(11)

(12)

(13)

(14)

AEB

5900311011

6100015411

9.9418823

11.67

+ 1.23

+ 14

5905610611

9.9372458

12.68

- 7.07

- 90

18000011011

- 1011

BEC

11802315011

3200315411

9.7249972

33.63

+ 7.80

+ 263

2903210611

9.6928074

37.15

- 0.40

- 15

17905915011

+ 10

CED

6003210511

5602810111

9.9209407

13.93

+ 6.23

+ 87

6205914911

9.9498691

10.73

- 2.06

- 22

17905915511

+ 5

DEA

12200015511

2804210011

9.6814434

38.45

+ 6.23

+ 240

2901710011

9.6894232

37.57

- 2.06

- 78

17905915511

+ 5

39.269236

97.68

36000010011

+ 604 39.2693240

+604

39.2693455 - 205 39.2693250 39.2693240 10

-205

in

CE401/17

10

Centre Triangle

Ist

Trial

correction (15)

Face Left

II

Trial

correction (16)

Final

Corrected

Remaining

Trial

Trial

Final

Trial

Corrected

correction

angles

correction

correction

correction

correction

correction

angles

for angles

for angle

difference

(19)

(20)

(21)

(22)

(23)

(24)

-.2.92

-34

+1.23

+14

6100155.2311

(17)

(18) 0 1

11

11

difference

AEB

- 3.33

- 4.17

- 4.16

59 3 5.84

- 5.84

BEC

+ 3.33

-2.49

+ 2.50

118023152.511

+7.5011

+ 3.75

+136

+7.80

+263

320411.8011

CED

+ 1.67

+ 0.83

+ 0.83

6003215.831

+4.1711

+ 2.08

+29

+6.23

+87

5602817.2311

DEA

+ 1.67

+ 0.83

+ 0.83

122000155.8311

+ 4.1711

+ 2.08

+80

+6.23

+240

2804216.2311

+ 3.34

- 0.02

+ 0.00

3600

+ 211

+ 604

Face Right Triangle correction for

Trial correction Difference

Final correction

Final correction difference

Corrected angles

angles

Sum angles

(25) - 2.92

(26) - 37

(27) - 7.07

(28) - 9011

(29) 59055158.9311

(30) 1800

+ 3.75

+ 140

- 0.30

- 1511

2903215.711

1800

+ 2.09

+ 22

- 2.06

- 2211

62059146.8411

1800

+ 2.09

+ 78

- 2.06

- 7811

29016157.9411

1800

+ 203

- 205

Note Face Right observation are given under the same table columns corresponding to Face left due to lack of space in single page. Please observe it.

of

corrected

CE401/3A

i.

11

Observed angles at centre, left, right, sum of observed angles are filled up in the table. Then correction values are obtained to each triangle in column 14.

ii.

In column (15), 1/3 of corrections obtained in column (14) are entered as first trial corrections. Sum of four central angles must be 3600. 3.34 = 0.84 to each of the first trial corrections. 4

Second trial correction must be

Find corrections are made in column (17). To make the sump zero. This satisfies the station equation. Corrected angles at the centre are entered. iii.

Remaining correction in column (19) is the difference of values in column (14) and column (17). This is distributed equally to left and right angles of each triangles. These values are shown in column (20) and column (25).

iv.

Log sine values and difference for 111 are noted from log tables. Colum 20 and 25 gives the actual shift after applying the correction to the angles. Calculatate the difference for correction (using the value of difference of 111) for column 20 and col. 25 and enter in column 21 and column 26. (Note: 11.67 x – 2.9211 = -34).

v.

Total shift = (sum of column 21 and column 4) % (sum of colmn 26 and colmn 9) = 39.2693658 – 39.2692847) = 811 from right to left. Increase the left hand angles and decrease the right hand angles.

vi.

log sin difference for 111 for all right angles = 97.68 + 98.13 = 195.81 i.e., for a shift of 111. For shifting 811 value,

811 = 411.15 is the required correction value for 195.81

angles. Column 22 and column 27 are obtained by changing the angles. Correction are applied to angles and noted in column 24 and column 29. vii.

Check:

column 23 and 28 are noted, by multiplying the corrections by the

corresponding difference of 111 in log sine. Add

(column 23) to

column (4),

CE401/3A

12

column 28 to

column 9.

Comparing the above two values, the required correction is 10 which is negligible as it causes an error of

10 i.e., 0.00511 in each angles. 195.81

Column (6 and 7) are corresponding column (22 and 23) Column (11 and 12) corresponding column (27 and 28). viii.

Column (30) is the (column 18 + column 24 + column 29)

Adjustment of Geodetic triangle with central station (by method least squares):

Fig 17.3 Triangle with central station i.

In this case, the conditioned equations are: e1 + e2 + e7 = E1 = 180 – (

1

+

2+

7)

1

e3 + e4 + e8 = E2 = 180 – (

3

+

4+

8)

2

e5 + e6 + e9 = E3 = 180 – (

5

+

6+

9)

3

e7 + e8 + e9 = E4 = 360 – (

7

+

8+

9)

4

e1 f1 – e2f2 + e3f3 + e4f4+ e5f5 – e6f6 = E5 =

log sin

R

-

log sin

e12 + e22 +. . . e92 = minimum according to least squares methods ii.

L

5 6

Differentiating the above six equations and multiplying each of the first five equations thus obtained by –k1, - k2, - k3, - k4, - k5 respectively and adding to the last equation, the following equations are obtained. (Coefficients of e1, e2, . . . .are equated to zero) e1 = k1 + f1 k5 e2 = k1 - f2 k5 e3 = k2 + f3 k5

CE401/3A

13

e4 = k2 - f4 k5 e5 = k3 + f5 k5 e6 = k3 - f6 k5 e7 = k1 + k4 e8 = k2 + k4 e9 = k3 + k4 iii.

e1, e2, e3 . . . .values are substituted in equations 1, 2, 3, 4 and 5. Five equations are obtained as given below. 3k1 + k4 + k5 (f1 – f2) = E1 3k2 + k4 + k5 (f3 – f5) = E1 3k3 + k4 + k5 (f5 – f6) = E1 k1 + k2 + k3 + 3k4 = E4 k1 (f1 – f2) + k2 (f3 – f4) + k3 (f5 – f6) + k5 ( f)2 = E5

First k1, k2, k3, k4, k5 values are obtained by solving the above simultaneous equations. Later on the correction values e1, e2, . . .e9 are obtained and the angles

1,

2.

...

9

are corrected.

17.7 Summary: Procedures for adjusting a quadrilateral and triangle with central stations have been worked out in brief, by various methods. In minor triangulation problems, these computations are quite useful in field practice.

*****

SURVEYING UNIT - 18 HYDROGRAPHIC SURVEYING SOUNDINGS METHODS OF LOCATING SOUNDINGS THREE POINT PROBLEMS 18.0 Hydrographic Surveying: Hydrographic surveying so far as civil engineer is concerned covers the survey work for project in or adjoining bays, harbours, lakes or rivers.

CE401/3A

14

18.1 AIM: Generally speaking the types and purpose of the various branches of hydrographic surveying may be summarised as

a) Measurement of tides for sea-coast work. Ex: construction of sea defence works jetties harbours etc. For the establishment of a levelling datum and for reducing soundings. b) Determination of bed depths (under water) by soundings (explained later). (i) For navigation including the location of rocks, sand bars, navigation lights buoys etc. (ii) For the location of under water works, volumes of under water excavation etc. (iii) In connection with irrigation and land-drainage schemes. c) Determination of direction of current in connection with (i) Location of sewer outfalls and similar works. (ii) Determination of areas subject to scour and silt. (iii) For navigational purpose. d) Measurement of quantity of water and flow of water in connection with (i) Water schemes (ii) Power schemes and (iii) Flood control.

18.2. OBJECTIVES: After going through this chapter the students is expected to know thoroughly the different methods of measurement of depth of water (or depth of bed below the surface of water) under various conditions and using different instruments suitable for a particular condition. He must also be able to locate the points where depth measurements are taken (soundings) wherever required. 18.3. INTRODUCTION:

CE401/3A

15

Hydrographic surveying as the name implies is concerned with water-determination of topography of the land under water. For this the depth of the bed below the surface of water is to be determined and the position where these measurement are taken should be recorded with reference to some permanent marks on the ground. For locating these details the survey should have both horizontal and vertical control. 18.4.

LESSON:

18.4.1 Soundings: The measurement of depths below the water surface are called soundings. It is to determine the configuration of the bottom of the body of water.

This is done by

measurement from a boat, the depths of water at various points (similar to levelling). 18.4.2 Horizontal Control: When making sounding of the depth of a river bed or a sea bed the location of the sounding vessel (sounding boat) is made by reference to fixed control points on shore and the accurate establishment of this shore frame work is of utmost importance. Primary horizontal control is established by triangulation. By running transit and tape traverse between triangulation station very close to shore line secondary horizontal control is established. 18.4.3 Vertical Control: A chain of bench marks must be established near the shore line and these serve for setting and checking tide gauges etc to which the soundings are referred. Since the elevation of the water surface which is taken as level surface of reference is continuously varying in tidal water it is necessary to ascertain the water level at the time each sounding is made by taking the gauge readings at regular intervals of time during the period of soundings so that the observed sounding can be reduced to the datum. 18.4.3 18.4.4 Gauges: (a) Non self registering (observer to note the reading) (b) Self registering (automatic) (a) Non registering - Staff gauge: It is about 15 to 25 cm wide 2

1 cm thick board of suitable 2

height having graduations to a least count of 5 to 10 cm Zero of the gauge should be connected to a permanent Bench Mark on shore by levelling. (b) Self registering gauge: It automatically registers the variation of water level with time. It essentially consists of float protected from wind, waves etc. The float has attached to it a wire or cord which passes over a wheel (called the float wheel) and is maintained at constant tension by some suitable arrangement. The movement of the float is transferred to the wheel which

CE401/3A

16

reduces it through some gear system and is finally communicated to a pencil attached to a level and the reading recorded on a graph paper attached to a rotating drum. 18.5 EQUIPMENT FOR TAKING SOUNDINGS: I.

Sounding boat: Flat bottomed boat for quiet water and round bottomed boat for rough water are suitable. Steam or motor launch is suitable even under winds and stray currents of water.

II.

Sounding rods: For shallow and smooth waters (4 to 6 m depth) a timber pole (or rod) of well seasoned tough timber about 5 to 6 cm diameter and length 4 to 7.5 meters can be used. At the lower end of the wooden fole an iron or lead shoe of sufficient weight is attached. The purpose of the weight is to hold the rod upright in water and to facilitate plunging. The shoe should have sufficient area at the base to prevent the pole from sinking into the mud or sand.

III.

Lead lines: Also called sounding lines. These are used for depths over 6 m. They are made up of hemp (a fibre from plants for making rope) cotton rope or brass chain having at its end a weight called lead. Because of possible change in length the lines are stretched and subjected to alternate wetting and drying continuously till there is no appreciable change in length and then graduated with zero at the bottom of the lead. The line should be kept dry when not in use but should be soaked in water for about an hour before use in order to bring it to the tested length. For regular work brass chain is preferable. Each meter marked with cloth or leather tag (cloth or leather to avoid injury to the leads man). Each one meter marked with a tag of different colour and each 5 cm interval with a leather tag similar to the brass tag of a measuring chain. The weight attached to a lead line is conical in shape and varies from 2.5 kg to shallow still water to about 10 kg for greater depths and stormy current.

IV.

Sounding Machine (for depth greater than 30 m): This is used when large scale sounding operations are carried out. It consists of a piano wire carrying 7 kg of lead weight at the end and wound round a drum. Two dials are provided on the drum the outer one, indicating the depth in meters and the inner one tenth of meter.

V.

Fathometer: For depths greater than 30 m and for ocean sounding an instrument known as fathometer is used. It is an electric device and measures the time required for the sound to travel to the bottom of the water and back.

CE401/3A

VI.

17

Signals: They are required to mark ranges and range lines along which soundings are to be taken and the reference points to which angular observations are to be taken from a boat. For angular observations objected such as church spires, wind mills light houses and chimneys etc are used as signals. Triangulation tripod signal may also be used. For identification the signals should be distinguished from each other by flags of different colours at their top, or by nailing figures of different geometrical forms etc.

Fig 18.1 VII.

Buoy: Sometimes the signals may have to be placed in water. For shallow water pole signals may be used. For deep waters BUOYS are used. A buoy is a float made of light wood or hollow air tight vessel properly weighted at the bottom and anchored in vertical position by means of guy ropes on the top of buoy in a hole in which is inserted a short flag pole.

VIII. Sextant: It is a portable and very accurate hand instrument for measuring angles from a boat in hydrographic surveying. (It is also used for astronomical observations and for measuring vertical angles. It can also measure oblique angles when the observed objects are at different altitudes. It consists of an index mirror or glass I rigidly fixed to a movable arm called an index arm. The index arm rotates about a pivot placed at the centre of the graduated arc and carries a vernier reading to 10” at the end and is fitted with a clamp and tangent screw.

CE401/3A

18

Fig 18.2 Sextant H is horizon glass, the lower half of it is silvered and the upper half un-silvered. Both index glass and the horizon glass are perpendicular to the plane of the instrument and are parallel to each other when the index of the venier is at the zero of the graduated arc. T, the telescope is rigidly attached to the frame pointing towards the horizon glass. Graduated arc ABC called the limb is 1/6 of the arc of the circle (600). The arc is divided into degree and 10 minutes and measures angles upto 1200. It is read by the vernier to 1 minute or 1011. Measurement of Horizontal angles with sextant: Move the index arm until the image of the right hand object seen in the silvered portion of the horizon glass is coincident with the object sighted directly. Clamp the index arm and bring the two images into exact coincidence by means of the tangent screw. Vernier reading is the required angle. Vertical angles: On land artificial horizon is required. On sea visible horizon is taken as the reference point in which case correction for dip is to be applied.

Fig 18.3 Arc vernier Shallow vessel filled with mercury, water or oil. = angle of incidence = angle of reflection = altitude. Adjustments of sextant:

CE401/3A

(i)

19

To make index glass (I) and horizon glass perpendicular to the plane of the graduated arc.

(ii)

To make horizon glass H parallel to the index glass when the vernier reads zero.

(iii)

To make the line of sight of the telescope parallel to the plane of the graduated arc.

(ix) Range and Range lines: The lines on which soundings are taken are called ranges or range lines. These are laid on the shore parallel to each other and at right angles to the shore line or radiating from a prominent natural object. Spacing of range lines vary from 6 to 30 m.

(a) Parallel range lines

(b) Radiating range lines Fig 18.4

Each range line should be marked by means of signals erected at two points on it at considerable distance apart. 18.6 Methods of locating soundings: i)

By cross rope

ii)

By range and line interval

iii)

By range and angle from shore

iv)

By range and angle from boat

v)

By intersecting ranges

vi)

By two angles from shore

vii)

By two angles from boat

viii)

By tacheometry.

CE401/3A

(i)

20

By cross rope: This is the most accurate method of locating soundings. It involves stretching across the line of sounding a rope marked off by equidistant tags, the soundings being taken opposite the tags. The method is suitable for sounding in harbour, and across rivers if the section does not exceed about 500 meters in length.

(ii)

Range and time interval: The sounding boat is kept in range with two shore poles defining the section line and is rowed at a uniform rate. The sounding being taken at regular time intervals. This is useful in conjunction with other methods. While proceeding along the section the boatsman must maintain a steady stroke so that equal time intervals may correspond to equal distances.

(iii)

Range and angle from shore:

Fig 18.5 BP = d tan measured ‘d’ known It is customary to locate every 10th sounding by angle and the intermediate soundings by time interval. (iv)

Range and angle from boat:

In this method the instrument man will be in the boat itself and therefore there will be better control over the work.

CE401/3A

21

Fig 18.6 AP = cot AB

measured AP = AB cot (v)

= d cot .

By intersecting ranges: In this method angular observations are avoided if the position of each sounding is defined by the intersection of two ranges as shown in fig 18.7. The method is suitable for the location of a few isolated soundings and proves highly accurate if the intersections are good.

Fig 18.7 (vi)

Two angles from shore: In this system the fix is made independent of a range by having simultaneous observations to the boat taken by theodolite from two shore stations the sounding being located by the intersection of the sight lines. This method is used when it is not possible to keep the boat exactly on a range.

CE401/3A

22

Fig 18.8 Co-ordinates of P (x, y)

(vii)

x=

d tan ` tan \ + tan `

y=

d tan \ tan ` tan \ + tan `

By two angles from boat: By observation of the two angles subtended at the boat by three suitable shore objects of known position the boat can be located by solution of the three point problem explained in article 18.7. This method is largely used particularly when periodical soundings are not required. It not only possesses the merit of concentrating the party as in the range and one boat angle method but if a sufficient number of land marks are exhibited on an existing map no preliminary shore work is required.

18.9

CE401/3A

23

(viii) By Tacheometry: A tacheometric observation on a staff held in the boat affords a simple method of location since it gives the bearing and distance from a shore station. This method however can be used only in smooth water. 18.7 Three point problem: The problem may be stated as: Given three points A, B and C representing the shore signals and the values

and

of the angles APB and BPC subtended

by them at the boat P it is required to plot P. The problem may be solved mechanically, graphically and analytically. 18.7.1 Mechanical solution: (a) By station pointer: The station pointer or three arm protractor consists of a graduated circle with one fixed and two movable arms. The fiducial edges of the arms pivot about the centre of graduation. The edge of the fixed arm passes through the zero division and the movable arms can be set so that their fiducial edges subtend the observed angles with that of the fixed arm.

Fig 18.10 Station pointer (b) Tracing paper method: On a piece of tracing paper protract

and

between three

radiating lines from any point. Apply the tracing paper to the plan and move it about until A, B and C simultaneously lie under the lines, then prick through the point P at the apex of the angles. 18.7.2 Graphical method:

CE401/3A

24

(a) A, B, C three known points. Join AC. At A make DAC =

. At C make DCA =

.

Draw a circle passing through A, D and C, join DB and produce it to cut the circle in P. Join AP, CP

APD = DCA = BPC = DAC = p is the position of the boat.

Fig 18.11 (b) Join AB, BC. Draw AO1, BO1, making 90 Similarly at B, and C make 90 -

each with AB on the side towards P.

intersecting at O2 towards P. With center O2 describe a

circle passing through A and B. With center O2 describe a circle passing through B and C. The point of intersection P of the two circles is the required boat position P. Proof APB = 1 / 2 AO1 B = BPC = 1/2 BO2 C = a

CE401/3A

25

Fig 18.12 (3) Join AB and BC and from B set off BD and BE making angles of 90 -

and 90 -

with

BA and BC respectively and on the side of P.

Fig 18.13 From A erect a perpendicular to AB to cut BD at D. And from C a perpendicular to CB to cut BE at E. Join DE from B, drop a perpendicular BP on DE. This will intersect DE at the required pt P, for the quadrilateral BADP and BCEP can be circumscribed by circles so that APB = ADB = BPC = BEC = 18.7.3 Analytical Method:

Fig 18.14 Given the lengths AB and BC and angles. , ,

the values

and

of angles BAP

and BCP and the distances AP, BP and CP may be computed trigonometrically. By sine ratio. BP =

AB sin \ BC sin b = sin c sin a

CE401/3A

But

26

+ = 360 – ( + + )= say S

So that by substitution AB sin sin c

=

BC sin(S - ) sin a

BC(sin S cos \ - cos S sin \ ) sin a

=

sin S cos

- cos S sin

=

AB sin \ sin a BC sin c

Divide through out by sin S sin Cot

- cot S =

AB sin a BC sin sin

Cot

= cot S +

AB sin a BC sin S sin

Knowing

other angles can be determined and hence the sides PA, PB, PE.

18.8 Problems: A, B and C are three shore stations on a coast line and P is sounding point at sea AB = 400 m, BC = 381 m, APB = 480.361 BPC = 450241 A and C are respectively East and West of BP, B and P are respectively North and South of AC. Calculate the distance AP, BP and CP. Solution: BCP = \ BAP = b

+ = 3600 - 1220301 - 480361 - 450241 = 1430301 = 1430301 -

CE401/3A

27

Fig 18.15 BP 381 = sin \ sin 450 241 BP 400 = sin b sin 480361 BP =

381 sin \ 400 sin b = 0 1 sin 45 24 sin 480361

sin

=

400 sin 450 241 × sin 381 sin 480361

on simplifying = 720221

(

= sin 1430301

)

= 710281

CBP = 180 450 241 710 281 = 630 281 ABP = 1800

480361 720021 = 590 221

AP 400 = 0 1 sin 59 22 sin 480361 (0.86) (0.75) BP 381 = 0 1 sin 71 28 sin 450 241 (0.948) (0.712) CP 381 = 0 1 sin 63 08 sin 450 241 (0.892) (0.712)

AP = 459 m

BP = 507.28 m

CP = 477 m

CE401/3A

28

18.9 Problem for Practice: During a hydrographic survey three shore stations A, B and C were established such that AB = 792 m, BC = 870 m the three stations lying in a straight line. Angles APB and BPC were measured simultaneously by sextent as 480361 and 460241 respectively from a float P which was then due East of B. Determine the reduced bearing of ABC given that A lies southwards of B. (Ans. N 040291).

*****

SURVEYING UNIT – 19 PHOTOGRAPHIC SURVEYING CONTENTS: 1. Terrestrial photogrammetry 2. Photo theodolite 3. Definitions 4. Calculation of horizontal and vertical angles from terrestrial photographs. 5. Elevation of a point by photographic measurements, stereo photogrammetry. 6. Problems. 19. Photogrammetry: Photogrammetry can be defined as the method of determining the shapes, sizes and positions of objects using photographs and therefore, it is, in the main , an indirect method measurement. Some linear or angular measurements in the object ‘object space’ need to be obtained or to be known for control purposes. 19.1 Aim: To study about phototheodolite method of taking photographs with the same and then preparation of plans etc with the help of the photographs. 19.2 Objectives: Given the data connected with a photograph the student must be able to prepare the plane with details making use of various coordinates of the points taken from the photographs. 19.3 Introduction: Photogrammetry is classified into two groups. (a) Terrestrial photogrammetry or ground photogrammetry: where in the photographs are taken from ground only and maps prepared with the help of the same.

CE401/3A

29

(b) Aerial photogrammetry: Where in maps are prepared with the help of photographs taken from air. 19.4 Lesson: Terrestrial photogrammetry is a further development of plane table survey. It is not suitable for wooded country. A modern development of the method consists in taking stereoscopic views of the surface features in pairs at the end of a base line and this method is known as stereo-photogrammetry. 19.4.1 Photo Theodolite (and definitions): The phototheodolite is a combination of a camera and theodolite (of least count 111) and is used for taking photographs and for measuring the angles which the vertical plane of collimation makes with the base line. The camera is mounted on a conventional theodolite. This enables. i)

The determination of the orientation of camera axis and

ii)

The angular measurements needed to fix the camera stations.

The image plane of the camera is represented by the surface of a glass plate upon which are engraved for fiducial marks which appear on the negative or photograph obtained after an exposure has been made when joined by lines these produce horizontal and vertical principle axes for coordinate measurement, their intersection lactating the principal point which is also defined on the plate in the camera. The collimation line or optical axis of the camera contain the optical center of the lens and the intersection of the principal axis mentioned above. This line is to be perpendicular to the glass plate and hence to the photographic material (film or plate). Also this line or axis has to be parallel to the line of collimation of the theodolite telescope and the line joining the fiducial marks on the short sides of the format must be horizontal when the theodolite axis is vertical. 19.4.2 (a) Principle involved: The principle of survey is same as that of plane table for locating points (method of intersection). Reconnaissance should be carried out first to cover the required area with minimum number of photographs. The following points should be taken care off while fixing the camera stations.

CE401/3A

30

Fig 19.1 Wild P32 Terrestrial Camera Mounted on Theodolite (Courtesy of Wild Heerbrugg UK Ltd) i)

Every object to be plotted should be visible at least from two photgraphs and photographs taken in pairs from the end of a base line.

ii)

Intersection angles should not be too acute or too obtuse.

iii)

Camera stations should be fixed on higher ground to command the area.

iv)

Control should be established by triangulation stations. Sometimes triangulation stations can be camera stations.

v)

Each photograph should contain at least one triangulation station or a point which has already been fixed previously.

vi)

No. of photographs of an area depends on the size of the area to be covered. In plane table survey most of the work is done in the field whereas it is done in the office here. 19.4.2 (b) Principle explained:

CE401/3A

31

Fig 19.2 AB is the base line = D A and B are camera stations from which photographs were taken AC and BC are the directions of camera axis from stations A and B respectively. AC and BC are the vertical planes containing the vertical axis. P is the point to be located from the photographs. It is shown as P on the prints (fig 19.2.b (i) (ii)) XA and YA the co-ordinates of the point P in the photograph at A. XB and YB the co-ordinates of the point P in the photograph at B. F = Focal length of the camera lens. To obtain the position of the point P photographed with reference to the base line. First draw the base line to a convenient scale. Then draw AC making an angle AB, and BC with angle

with

with BA. On AC mark a point d at a distance equal to f in front of

A and mark e at a distance of f from B on BC. Through the points d and e draw perpendiculars XA and XB equal to dd1 and ee1 respectively. Join Ad1 and Be1 and produce them to meet at P which gives the required position of P on plan. AP can be measured on plan. Knowing the distance AP and the co-ordinate of P (YA) from the print the elevation of the P can be determined with reference to the level of the axis of the camera.

CE401/3A

32

Fig 19.3 h = AP

YA f +X 2

2 A

h=

YA + A P f 2 + X 2A

AP is the distance of the point P from camera station from the plan plotted. YA co-ordinate of P from the print of the photograph at A. Knowing the R.L of camera axis R.L of P can be determined either by adding ‘h’ suitably.

Analytical method (fig 19.2) CAB =

CBA =

CAP =

CBP =

tan bc =

XA X tan ba = B f f

PAB = c + bc APB = 180

PBA = a bB

(c + bc ) (a

ba )

AP AP = sin (a ba ) sin 180 - (c + bc ) AP =

(a

D sin (a - ba ) sin (180 (c + bc ) (a ba ))

Similarly BP can be determined.

ba )

CE401/3A

h=

33

YA × AP f 2 + X 2A

19.5 Stereo photogrammetry: In this method pairs of photographs are taken from stations at each end of a line with the instrument set up normally so that the image planes are in the same vertical plane and parallel to the line joining the stations. Thus the camera axis when set up at each station in turn must be made parallel by setting them at right angles to the line. Which is achieved by initially sighting the theodolite along the line and then rotating through 900 in the horizontal plane. The length of the base line horizontal distance between parallel principle planes is usually between 30 to 120 m. It is measured either with a tape or by transit and stadia method. Alternatively a device such as the “wild stereo-metric camera” can be used. It essentially consists of two similar cameras mounted one at each end of a bar (either 1200 mm or 400 mm long) such that their optical axes are parallel and perpendicular to the bar. This in turn is supported on a tripod and it is possible for the camera axis to be tilted through a range of angles from the horizontal plane into the vertical. These instruments are used in close range photogrammetry. The pairs of photographs or negatives can be viewed stereoscopically subject to certain conditions so that plotting can be effected by machine (“Wild A 40 Autograph) or by measurement of co-ordinates in a stereo-comparator. 19.5.1 Principle explained:

CE401/3A

34

Fig 19.4 A and B optical centers of the lenses at camera stations A and B f = focal length of the lens. XA and YA the distances (co-ordinates) of the point P on the plate exposed at A from the vertical and horizontal hairs respectively. XB and YB distance (co-ordinates) of the point P on the plate exposed at B. X and YA co-ordinate of the point P ( on the ground) with reference to station A. X and YB co-ordinate of the point P ( on the ground) with reference to station B. D = length of the stereoscopic base. hA height of the point P above the horizontal plane of collimation at A. hB height of the point P above the horizontal plane of collimation at B. YA =

XA ×X f

YB =

XB ×X f

D = YA

X=

YB =

X (X A f

XB )

fD (f, D, X A , X B Known ) XA XB

YA =

YB =

XA fD XAD × = f XA XB XA XB

XB D XA XB

knowing X, YA and YB the co-ordinates of the point P on the ground known (with reference to base AB) hA AP X = = YA APA f hA =

X × YA f

CE401/3A

35

similarly h B =

X × YB f

The difference in level of the two collimation planes of the camera hA

hB =

YA

YB f

×X

Note: If the point P appears on the point at B to the left of the vertical hair D = YA + YB 19.6 Problems: 19.6.1 The horizontal angle between two points P and Q was measured directly at a station R and found to be 240 481. A phototheodolite was set up at R and on the photograph P was found to be 3 cm to the left of the vertical hair and 1.25 cm above the horizontal hair. While Q was found to be 3.5 cm to the right of the horizontal hair and 2.0 cm below the horizontal hair. Determine the focal length of the camera lens and the difference in level between P and Q. If PR = 85 m and QR = 67 m

Fig 19.5 tan (c + a ) =

+

tan c + tana 1 tan c tan a

= 240.481

tan c =

3 f

tan a =

3.5 f

CE401/3A

36

(

)

(

)

3 3.5 + f f 3 3.5 1- × f f

tan 240 481 =

tan 240.481 =

3 + 3.5 f

f 2 - 3 + 3.5 f

6.5 f2 0.461 = × 2 f f 10.5 0.461 (f2 – 10.5) = 6.5f 0.461 f2 – 4.84 = 6.5f 0.461 f2 – 6.5 f – 4.84 = 0 f2 – 14.1 f – 10.5 = 0

14.1 ± 198.81 + 42 14.1 ± 155 = = f = 14.8 cm 2 2 hP =

hQ =

y P × RP X 2P + f 2

yQ × R Q X 2R + f 2

=

=

1.25 × 85 32 + 14.82

=

2 × 67 3.52 + 14.82

106.25 = 7.036 m 15.1

=

134 = 8.82 m 15.2

P is higher than Q by 7.036 + 8.82 = 15.856 m 19.6.2 Two photographs were taken with a photo theodolite from stations A and B 100 m

apart the line of collination being 900 to AB in each case. A point P appears on the photograph from A, 5 cm to the right of the vertical line and 2.5 cm above the horizontal line while on the photograph from B it appears 7.5 cm to the left of the vertical line and 2 cm above the horizontal line. B is east of A. Calculate the co-ordinates of P and the difference in level of the two collimation planes if the focal length of the camera lens is 15 mm. A1P AA1 = 5 15

CE401/3A

37

B1P BB1 AA1 A1P = = = 7.5 15 15 5 B1P =

7.5 × A1P 5

A1P + PB1 = 100 A1P +

7.5 × A1P = 100 = 2.5 A1P = 100 5

A1P =

100 = 40 2.5

A1P AA1 40 = = 5 15 5 AA1 =

40 × 153 = 120 m 5

AP = AA12 + A1P 2 = 1202 + 402 = 14400 + 1600 = 126.49 BP = 1202 + 602 = 14400 + 3600 = 134.16 Difference of level between A and P hA 2.5 2.5 × 120 , hA = = 20 m 120 15 15 hB 2 2 × 120 = , hB = = 16 m. 120 15 15

From both the photographs at A and B, P is above the horizontal. It is 20 m above A and 16 m above B i.e., A is lower than B by 4 m.

CE401/3A

38

Fig 19.6 19.7 Problems for practice: 19.7.1 Two photographs are taken with a photo theodolite from stations A and B 160 m apart, the lines of collinations being at right angles to AB in each case. A point C appears on the photograph from A as 32 mm to the right of the vertical hair and 16 mm below the horizontal hair and on the photograph from B as 35 mm to the left of the vertical hair and 11 mm below the horizontal hair. B is to the right of A and the focal length of the instrument is 165 mm. Calculate the co-ordinates of C from A as origin and the difference in levels of the two collimation planes. 19.7.2 From two stations A and B 100 m apart photographs of an area where taken. The focal length of the camera is 160 mm. The camera axis makes an angle of 600 and 400 with the base line at stations A and B respectively. The image of a point P appears 2 cm to the right and 1.5 cm above the hair line on the photograph taken at A and 3.5 cm to the left on the photograph taken at B. Calculate the distance AP and BP and elevation of point P if the elevation of the axis of the instrument at A is 125.00 m.

*****

SURVEYING UNIT – 20

CE401/3A

39

AERIAL PHOTOGRAMMETRY 20.1 Aerial photogrammetry: Definition scale of vertical photograph, Flying height Determination of no. of photographs to cover a specified area. Problems, Mosaics and Relief displacement. 20.2

Aim:

The method of aerial photogrammetry, Determination of scale of the

photographs, flight altitude no. of photography’s required to cover the given area, preparation of Mosaics etc., explained. 20.3 Objectives: Given the data connected with the aerial photography the student is expected to know how to calculate from the available data scale of the photograph, or the flight altitude or, the length of a base line or the focal length of a camera lens. He must be able to distinguish the different types of Mosaics and their uses. 20.4 Introduction: Photographs from the air may be used for compilation of topographical maps, but for the photography to give a TRUE PLAN certain conditions must be fulfilled namely. (i)

The ground on the photograph should be horizontal

(ii)

The camera must not be tilted from the vertical when the exposure is made

(iii)

The camera lens and photographic material should be as perfect as possible and there should be no atmospheric refraction. In addition when flying at high altitude the curvature of the earth is of some account.

20.5 Lesson: As the photographs are taken from air and to cover a large area aerial camera is mounted at the bottom of a fast moving aeroplane. 20.5.1 Requirements of a aerial camera a) Fast lens b) High speed efficient shutter c) High speed emulsion for the film d) Provision to hold (Magazine) large rolls of film. Since the air craft is at a considerable distance from the terrain to be photographed all the points on the ground can be considered to be at an infinite distance from the lens and hence the focal plane of the camera can be fixed at one location. Thus the aerial camera is always of a fixed focus type the focus being set for infinity.

CE401/3A

40

20.5.2 Definitions: a) Vertical Photograph: A vertical photograph is an aerial photograph with the axis of the camera truly vertical coinciding with the direction of gravity. b) Tilted photograph: In a photograph made with the axis of the camera tilted unintentionally from the vertical axis by a small amount not exceeding 30 (for tilts of 20 to 30 )the photograph is treated as a ‘near vertical photograph’. c) Oblique photograph: For tilts greater than 30 the photographs are classified as (i) low oblique photograph and (ii) high oblique photograph. (i) If the apparent horizon is not included in the photograph it is called a low oblique photograph. (ii) If the horizon is included in the photograph it is called a high oblique photograph. d) Flight height: It is the elevation of the camera station is space (air space above mean sea level or another datum. d) Focal length: It is the distance from the front nodal point of the lens to the plane of the photograph. e) Principal point: In a point where the perpendicular dropped from the front nodal point strikes the photograph. f) Nadir point: Nadir point is a point where a plumb line dropped from the front Nodal point pieces the photograph. g) Ground Nadir point: Ground Nadir point or the ground plumb point is the datum intersection with the plumb line through the front nodal point. h) Iso Center: Iso center is the point in which the bisector of the angle of tilt meets the photograph. 20.5.3 Scale of a Vertical Photograph

CE401/3A

41

Fig 20.1 l3 oda and OA ‘D’ are similar. da A'

f H

=

D'

Scale for points A and B are

f H ha

and

f H hb f ha + hb 2

Scale for the line AB = H

=

f H

h ave

H is the flying height or flight height. f is the focal length of camera lens. (ha + hb)/2= Ave height of ground photographed. 20.5.4 Flight height: Flight height depends on: (i)

The scale of map or plan

(ii)

Contour interval in the map

(iii)

Type of Terrain – flat or mountainous

(iv)

Characteristics of camera – focal length etc.,

CE401/3A

(v)

Type of plotting equipment

(vi)

Type of aircraft available.

42

When sophisticated plotting machines are used it is economical to increase the flying height upto 6000 meters and thereby ground coverage per photograph can be increased. Lower flying heights are required for close contouring over flat terrain. 20.5.5 Number of photographs required to cover a given area. Notation: LP = length of photograph in cm in the direction of flight Wp = Width of photograph in cm at right angles to the direction of flight O1 = % of longitudinal over lap (usually 60% adopted minimum 50% desirable) Ow = % of side overlap. (30% desirable) L = net ground distance corresponding to Lp in m or km. Wg = net ground distance corresponding to Wp in m or km. S = Scale of photograph 1 cm = S m or S km. N = No. of photographs required Ap = net area of each photograph Ag = area of the track to be photographed in sq. m or sq.km. Lg = SLp (1 – 0L) Wg = SWp (1 – 0w) Net area of each photograph = AP = Lg. Wg. No of photographs required =

Aq Ap

Alternatively: Theoretical spacing of flight strips Net width of a single photograph = Wg.

CE401/3A

43

Theoretical No. of strips =

Width of Area =K Wg

Actual no. of strips = K + 1 ( one strip being added to cover the sides) Theoretical No. of photographs per strip =

Length of the area =M Lg

Actual No. of photographs per strip = M + 1 One photograph being added to cover the ends. There would be K + 1 strips with M + 1 photographs in each actual No.of photographs = (K + 1) (M + 1). 20.6 Problems: 20.6.1: The scale of a photograph is 1 cm = 100 m. the photograph size is 20 x 20 cm. Determine the no.of photographs required to cover an area of 160 sq. km if the longitudinal overlap is 60% and side overlap = 30%. Lp= 20 cm

OL = 0.6

Wp = 20 cm

Ow = 0.3 S = 100

Lg = ground length covered = S Lp (1 – 0L) = 100 x 20 (1 – 0.6) = 800 m = 0.8 km. W = ground width covered = S Wp (1 – 0w) = 100 x 20 (1 – 0.3) = 1400 m = 1.4 km. Ap = Net ground area covered = Lg Wg = 0.8 x 1.4 = 1.12 sq.km. No. of photographs =

Ag 160 = A p 1.12

CE401/3A

44

= 142. = 143. 20.6.2: A line A B measures 10.92 cm on a photograph taken with a camera having a focal length of 20 cm. The same line measures 2.85 cm on a map drawn to a scale of

1 . 50,000

Calculate the flying height of the air craft if the average altitude of the ground is 320 m. Solution: Photo scale photo distance of line AB = Map scale Map distance of line AB

S = Photo scale S 10.92 = 1/50,000 2.85 Simplifying S =

1 13500

S = Photo scale =

1 H h

=

0.2 1 = H 320 13500

0.2 1 = H 320 13500

Simplifying H = 3020 meters 20.7.1 Calculate the aeroplane flying height to obtain the average scale of the photograph equal to

1 7500

. Ground surface elevation vary from 150 m to 450 m. Focal length of camera

lens = 15 cm. 20.7.2: The scale of an aerial photograph is 1 cm = 150 m and the size of the photograph is 22 x 22 cm. If the longitudinal lap is 62% and the side lap = 30% determine the number of photographs required to cover an area of 230 sq. km. 20.8 Mosaics: It is an assembly of vertical aerial photographs into a single composite photographic representation of the terrain. It is made from overlapping vertical photographs and is sometimes spoken of as an aerial map. 20.8.1 Comparison with Maps: a) Topographic details are shown pictorially rather than by conventional signs.

CE401/3A

45

b) Mosaic produces a greater amount of detail than can be shown on a topographic map. c) As compared to maps making of Mosaics is easy, particularly to photograph inaccessible points such as extensity swamps etc. Disadvantages: No elevation of ground points furnished. Because of showing of too much details sometimes some important points may be obscured. 20.8.2 Classification of Mosaics: Mosaics are classified as 1) Uncontrolled Mosaic 2) Controlled Mosaic 3) And Semi-controlled Mosaic. 1) Uncontrolled Mosaic: If a set of overlapping vertical photographs is joined together on a backing board a rough map or aerial view of the are is obtained, which has no uniformity of scale. This is called an uncontrolled Mosaic, which allows preliminary inspection of conditions in the area and ensures full photographic coverage. 2) Controlled Mosaics: If the photographs are rectified and brought to a common scale the Mosaic is said to be controlled and is superior to the uncontrolled mosaic. Controlled mosaics are constructed so that points identified on the photograph will be super imposed over the plotted positions which may be geographical control points. The process is done by means of a rectifying projector. 3) Semi-controlled mosaic: If some of the points of original photographs are adjusted to their corresponding ground survey plotted positions and the details surrounding those points matched the resulting mosaic is called semi-controlled mosaic. In semi-controlled mosaics directions and distance are mentioned. 20.9. Relief displacement on a vertical photograph: If the photograph is truly vertical and the ground is horizontal and if other sources of error are neglected the scale of the photograph will be uniform. Such a photograph represents a true orthographic projection and hence the true map of the terrain. In actual practice, however, such conditions are never fulfilled. When the ground is not horizontal the scale of the photograph varies from point to point and is not constant. Since the photograph is the perspective view the ground relief is shown in perspective view on the photograph. Every

CE401/3A

46

point on the photograph is therefore displaced from their true orthographic position. This displacement is called relief displacement. A, B and K (fig 20.2) are the ground points having elevations ha,. hb, hk above datum. A0, B0, K0 are their datum position respectively when projected vertically downwards on the datum planes. On the photograph their positions are a, b and k respectively the point k being chosen vertically below the principal point. If the datum points A0, B0, K0 are imagined to be photographed along with the ground points their positions will be a0, b0 and k respectively. As is clear from the figure the points a and b are displaced outwards from their datum photographic positions, the displacement being along the corresponding radial lines from the principal point. The radial distance aa0 is the relief displacement of A while bb0 is the relief displacement of B. The point K has not been displaced since it coincides with the principal point of the photograph.

CE401/3A

47

Fig 20.2 References: 1) ‘PLANE AND GEODETIC SURVERYING FOR ENGINEERS’ By the Late David Clark M.A., B.Sc. fifth Edition revised and Enlarged. By James Cleoinning O.B.E. B.Sc. (Engg). 2) “SURVEYING AND LEVELLING (Part II)” by Late T.P.Kanetkar, B.A (Hons), B.E., A.M.I.E (Ind), B.E.S and Prof S.V.Kulkarni, B.E. (Hons), M.Sc(Engg). F.I.E (Ind). 3) ‘TEXT BOOK OF SURVEYING’ Volume II BY B.P.Shahani, Ph.D., B.E (Civil), M.I.E (Ind), MIS (Ind). 4) ‘SURVEYING’ by the Late W.Norman Thomas. C.B.E., M.A., D.Phil. (Oxom) M.Sc (B.hon), M.I.C.E., F.R.I.B.A. 5) ‘SURVEYING’ by A.Bannister, M.C., M.Sc., F.I.C.E and S. Raymond M.Sc., Ph.D., Dip.T.P (Manchester) F.I.C.E., M.RT.P.I., M.Inst. H.C.

CE401/3A

48

6) ‘’SURVEYING’ (Volumn I, II, III) by B.C.Punmia, B.E (Hons), M.E(Hons), M.I.G.S., M.I.E. 7) ‘ADVANCE SURVEYING’ by V. Natarajan, B.E(Hons), M.Sc., M.A.C.I., M.I.E (Ind).

*****