PLANE TABLE SURVEY AND ITS ACCESSORIES

PLANE TABLE SURVEY AND ITS ACCESSORIES

In plane table surveying a table top, similar to drawing board fitted on to a tripod is the main instrument. A drawing sheet is fixed on to the table top, the observations are made to the objects, distances are scaled down and the objects are plotted in the field itself. Since the plotting is made in the field itself, there is no chance of omitting any necessary measurement in this surveying. However the accuracy achieved in this type ofsurveying is less. Hence this type of surveying is used for filling up details between the survey stations previously fixed by other methods.

The most commonly used plane table is shown in Fig. 1. It consists of a well seasoned wooden table top mounted on a tripod. The table top can rotate about vertical axis freely. Whenever necessary table can be clamped in the desired orientation. The table can be levelled by adjusting tripod legs.

Fig: Plane table with tripod stand

The following accessories are required to carry out plane table survey:

1. Alidade

2. Plumbing fork with plumb bob.

3. Spirit level

4. Trough compass

5. Drawing sheets and accessories for drawing.

1. Alidade

It is a straight edge ruler having some form of sighting device. One edge of the ruler is bevelled and is graduated. Always this edge is used for drawing line of sight. Depending on the type of line of sight there are two types of alidade:

(a) Plain alidade

(b) Telescopic alidade

Plain Alidade: Figure 2 shows a typical plain alidade. A sight vane is provided at each end of the ruler. The vane with narrow slit serves as eye vane and the other with wide slit and having a thin wire at its centre serves as object vane. The two vanes are provided with hinges at the ends of ruler so that when not in use they can be folded on the ruler. Plain alidade is not suitable in surveying hilly areas as the inclination of line of sight in this case is limited.

Fig: 2 – Plane Alidade

Telescopic Alidade: It consists of a telescope mounted on a column fixed to the ruler [Fig. 3]. The line of sight through the telescope is kept parallel to the bevelled edge of the ruler. The telescope is provided with a level tube and vertical graduation arc. If horizontal sight is required bubble in the level tube is kept at the centre. If inclined sights are required vertical graduation helps in noting the inclination of the line of sight. By providing telescope the range and the accuracy of line of sight is increased.

Fig. 3: Telescopic alidade

2. Plumbing Fork and Plumb Bob

Figure 4 shows a typical plumbing fork with a plum bob. Plumbing fork is a U-shaped metal frame with a upper horizontal arm and a lower inclined arm. The upper arm is provided with a pointer at the end while the lower arm is provided with a hook to suspend plumb bob. When the plumbing fork is kept on the plane table the vertical line (line of plumb bob) passes through the pointed edge of upper arm. The plumb bob helps in transferring the ground point to the drawing sheet and vice versa also.

Fig. 4: Plumbing fork and plumb bob

3. Spirit Level

A flat based spirit level is used to level the plane table during surveying (Fig.5). To get perfect level, spirit level should show central position for bubble tube when checked with its positions in any two mutually perpendicular directions.

Fig. 5: Spirit level

4. Trough Compass

It consists of a 80 to 150 mm long and 30 mm wide box carrying a freely suspended needle at its centre (Ref. Fig. 6). At the ends of the needle graduations are marked on the box to indicate zero to five degrees on either side of the centre. The box is provided with glass top to prevent oscillation of the needle by wind. When needle is centred (reading 0–0), the line of needle is parallel to the edge of the box. Hence marking on the edges in this state indicates magnetic north–south direction.

Fig. 6: Trough compass

5. Drawing Sheet and Accessories for Drawing

A good quality, seasoned drawing sheet should be used for plane table surveying. The drawing sheet may be rolled when not in use, but should never is folded. For important works fibre glass sheets or paper backed with thin aluminium sheets are used.

Clips clamps, adhesive tapes may be used for fixing drawing sheet to the plane table. Sharp hard pencil, good quality eraser, pencil cutter and sand paper to keep pencil point sharp are other accessories required for the drawing work. If necessary, plastic sheet should be carried to cover the drawing sheet from rain and dust.

LESSON Note on Computation of area and volume

LESSON Note on Computation of area and volume

The main objective of the surveying is to compute the areas and volumes.

Generally, the lands will be of irregular shaped polygons.

There are formulae readily available for regular polygons like, triangle, rectangle, square and other polygons.

But for determining the areas of irregular polygons, different methods are used.

Earthwork computation is involved in the excavation of channels, digging of trenches for laying underground pipelines, formation of bunds, earthen embankments, digging farm ponds, land levelling and smoothening. In most of the computation the cross sectional areas at different interval along the length of the channels and embankments are first calculated and the volume of the prismoids are obtained between successive cross section either by trapezoidal or prismoidal formula.

Calculation of area is carried out by any one of the following methods:

a) Mid-ordinate method

b) Average ordinate method

c) Trapezoidal rule

d) Simpson’s rule

The mid-ordinate rule

Consider figure.

Let  O₁, O₂, O₃, O₄……….On= ordinates at equal intervals

l=length of base line

d= common distance between ordinates

h₁,h₂,……..h_n=mid-ordinates

Area = common distance* sum of mid-ordinates

Average ordinate method

Let O₁, O₂, …..O_n=ordinates or offsets at regular intervals

l= length of base line

n= number of divisions

n+1= number of ordinates

THE TRAPEZOIDAL RULE

While applying the trapezoidal rule, boundaries between the ends of ordinates are assumed to be straight. Thus the areas enclosed between the base line and the irregular boundary line are considered as trapezoids.

Let O1, O2, …..On=ordinate at equal intervals, and   d= common distance between two ordinates

Total area=d/2{ O₁+2O₂+2O₃+…….+2O_n-1+O_n}

Thus the trapezoidal rule may be stated as follows:

To the sum of the first and last ordinate, twice the sum of intermediate ordinates is added. This total sum is multiplied by the common distance. Half of this product is the required area.

Limitation: There is no limitation for this rule. This rule can be applied for any number of ordinates

SIMPSON’S RULE

In this rule, the boundaries between the ends of ordinates are assumed to form an arc of parabola. Hence simpson’s rule is some times called as parabolic rule. Refer to figure:

Let

O₁, O₂, O₃= three consecutive ordinates

d= common distance between the ordinates

area AFeDC= area of trapezium AFDC+ area of segment FeDEF

Here,

Area of segment= 2/3* area of parallelogram FfdD

                            = 2/3* eE*2d

                             = 2/3 *{ O₂- O₁+O₃  /2 }*2d

So, the area between the first two divisions,

     = d/3(O₁+4O₂+O₃)

Similarly, the area of next two divisions

            ∆₂ =  d/3(O₁+4O₂+O₃) and so on

Total area = d/3[O₁+O_n+4(O₂+O_4+……) + 2(O₃+O₅)]

Thus the rule may be stated as the follows

To the sum of the first and the last ordinate, four times the sum of even ordinates and twice the sum of the remaining odd ordinates are added. This total sum is multiplied by the common distance. One third of this product is the required area.

Trapezoidal rule	Simpson’s rule
1.       The boundary between the ordinates is considered to be straight 1.       There is no limitation. It can be applied for any number of ordinates 1.      It gives an approximate result	The boundary between the ordinates is considered to be an arc of a parabola To apply this rule, the number of ordinates must be odd It gives a more accurate result.

Limitation: This rule is applicable only when the number divisions is even i.e. the number of ordinates  is odd.

The trapezoidal rule may be compared in the following manner:

Note: sometimes one or both the end of the ordinates may be zero. However they must be taken into account while applying these rules.

Worked- out problems

Problem 1: The following offsets were taken from a chain line to an irregular boundary line at an interval of 10 m:

0, 2.50, 3.50, 5.00, 4.60, 3.20, 0 m

Compute the area between the chain line, the irregular boundary line and the end of offsets by:

a) mid ordinate rule

b) the average –ordinate rule

c) the trapezoidal rule

d) Simpson’s rule

Solution: (Refer fig)

Mid-ordinate rule:

Required area= 10(1.25+3.00+4.25+3.90+1.60)

                         = 10*18.80=188 m²

By average-ordinate rule:

Here d=10 m and n=6(no of devices)

Base length= 10*6=60 m

Number of ordinates= 7

Required area=10((1.25+3.00+5.00+4.60+3.20+0)/7)

   

By trapezoidal rule:

Here d=10m

Required area=10/2{0+0+2(2.50+3.50+5.00+4.60+3.20+)}

                         = 5*37.60=188 m²

By Simpson’s rule:

d=10m

required area=10/3{0+0+4(2.50+5.00+3.20)+2(3.50+4.60)}

                         = 10/3{ 42.80+16.20}=10/3*59.00

                        10/3*59= 196.66m²

Problem 2: The following offsets were taken at 15 m intervals from a survey line to an irregular boundary line

3.50,4.30, 6.75, 5.25, 7.50, 8.80, 7.90, 6.40, 4.40, 3.25 m

Calculate the area enclosed between the survey line, the irregular boundary line, and the offsets, by:

a) the trapezoidal rule

b) simpson’s rule

solution:

a) the trapezoidal rule

required area=15/2{3.50+3.25+2(4.30+6.75+5.25+7.50+8.80+7.90+6.40+4.40)}

                        = 15/2{6.75+102.60} = 820.125 m²

c) simpson’s rule

if this rule is to be applied, the number of ordinates must be odd. But here the number of ordinates must be odd. But here the number of  ordinate is even(ten).

So, simpson’s rule is applied from O₁ to O₉  and the area between O₉ and O₁₀  is found out by the trapezoidal rule.

A₁= 15/3{ 3.50+4.40+4( 4.30+5.25+8.80+6.40)}+2(6.75+7.50+7.90)

    = 15/3( 7.90+99.00+44.30)= 756.00 m²

A₂= 15/2(4.40+3.25)=  57.38 m²

Total area= A₁+ A₂ =756.00+57.38 = 813.38  m²

Problem 3: the following offsets are taken from a survey line to a curves boundary line, and the first and the last offsets by:

a) the trapezoidal rule

b) simpson’s rule

solution:

here the intervals between the offsets are not reglar through out the length.

So, the section is divided into three compartments

Let

 ∆_I= area of the first section

 ∆_II= area of 2^nd section

∆_III= area of 3^rd section

Here

d1= 5 m

d2=10 m

d3=20 m

a) by trapezoidal rule

∆_I= 5/2{2.50+6.10+2(3.80+4.60+5.20)} = 89.50 m²

∆_II= 10/2{6.10+5.80+2(4.70)} =106.50 m²

∆_III= 20/2{5.80+2.20+2(3.90)} = 158.00 m²

Total area = 89.50+106.50+158.00 = 354.00 m²

b) by simpson’s rule

∆_I= 5/3{2.50+6.10+4(3.8+5.20) + 2(4.60)} = 89.66 m²

∆_II= 10/3{6.10+5.80+4(4.70)} =102.33 m²

∆_III= 20/3{5.80+2.20+4(3.90)} = 157.33 m²

Total area= 89.66+102.33+157.33 = 349.32 m²

FORMULA FOR CALCULATION OF VOLUME:

D= common distance between the sections

A. trapezoidal rule

volume (cutting or filling), V=D/2(A1+An+2(A2+A3+….+An-1))

1.     Prismoidal formula

Volume( cutting or filling), V= D/3{A₁+ A_n +4(A₂+ A₄+ A_n-1)+ 2(A₃+ A₅+….+ An_n-1)}

i.e. V=common distance  {area of 1^st section+ area of last section+ 4(sum of areas of even sections)

                       3                                                 +2(sum of areas of odd sections)

Note: the prismoidal formula is applicable whrn there is an odd number of sections. If the number of sections is even, the end strip is treated separately and the area is calculated according to the trapezoidal rule. The volume of the remaining strips is calculated in the usual manner by the prismoidal formula. Then both the results are added to obtain the total volume.

Works out problems

Problem 1: an embankment of width 10 m and side slopes 1 ½:1 is required to be made on a ground which is level in a direction transverseto the centre line. The central heights at 40 m intervals are as follows:

0.90,1.25,2.15,2.50,1.85,1.35, and 0.85

Calculate the volume of earth work according to

i) Trapezoidal formula

ii) Prismoidal formula

Solution:  the c/s areas are calculated by

∆= (b+sh)*h

∆₁= (10+1.5*0.90)*0.90 = 10.22 m²

∆₂= (10+1.5*1.25)*0.90 = 14.84 m²

∆₃= (10+1.5*1.25)*2.15 = 28.43 m²

∆₄= (10+1.5*2.50)*2.50 = 34.38 m²

∆₅= (10+1.5*1.85)*1.85 = 23.63 m²

∆₆=(10+1.5*1.35)*1.35 = 16.23 m²

∆₇=(10+1.5*0.85)*0.85= 9.58 m²

(a) Volume according to trapezoidal formula

V= 40/2{10.22+ 9.58+2(14.84+28.43+34.38+23.63+16.23)}

  = 20{19.80+235.02} = 5096.4 m²

(b) Volume calculated in prismoidal formula:

V = 40/3 {10.22+9.58+4(14.84+34.38+16.23)+2(28.43+23.63)}

    = 40/3 (19.80+ 261.80+104.12) = 5142.9 m²

Problem the areas enclosed by the contours in the lake are as follows: 

Contour (m)	270	275	280	285	290
Area (m2)	2050	8400	16300	24600	31500

Calculate the volume of water between the contours 270 m and 290 m by:

i) Trapezoidal formula

ii) Prismoidal formula

Volume according to trapezoidal formula:

=5/2{2050+31500+2(8400+16300+24600)}

=330,250 m³