Vertical Controls Surveying & Its Methods

VERTICAL CONTROL & ITS METHODS:

The vertical control consists of establishment of reference marks of known height relative to some special datum. All levels at the site are normally reduced to the nearby bench mark, usually known as master bench mark.

The setting of points in the vertical direction is usually done with the help of following rods:

1. Boning rods and travelers

2. Sight Rails

3. Slope rails or batter boards

4. Profile boards

A boning rod consist of an upright pole having a horizontal board at its top, forming a 'T 'shaped rod. Boning rods are made in set of three, and many consist of three 'T' shaped rods, each of equal size and shape, or two rods identical to each other and a third one consisting of longer rod with a detachable or movable 'T' piece. The third one is called traveling rod or traveler.

Sight Rails:

A sight rail consist of horizontal cross piece nailed to a single upright or pair of uprights driven into the ground. The upper edge of the cross piece is set to a convenient height above the required plane of the structure, and should be above the ground to enable a man to conveniently align his eyes with the upper edge. A stepped sight rail or double sight rail is used in highly undulating or falling ground. Slope rails or Batter boards:

These are used for controlling the side slopes in embankment and in cuttings. These consist of two vertical poles with a sloping board nailed near their top. The slope rails define a plane parallel to the proposed slope of the embankment, but at suitable vertical distance above it. Travelers are used to control the slope during filling operation.

Profile boards:

These are similar to sight rails, but are used to define the corners, or sides of a building. A profile board is erected near each corner peg. Each unit of profile board consists of two verticals, one horizontal board and two cross boards. Nails or saw cuts are placed at the top of the profile boards to define the width of foundation and the line of the outside of the wall.

An instrument was set up at P and the angle of elevation to a vane 4 m above the foot of the staff held at Q was 9 o 30?. The horizontal distance between P and Q was known to be 2000 metres. Determine the R.L. of the staff station Q given that the R.L. of the instrument axis was 2650.38.

Solution:

Height of vane above the instrument axis

= D tan ? = 2000 tan 9 o 30?

= 334.68 m

Correction for curvature and refraction

C = 0.06735 D² m, when D is in km

= 0.2694 ? 0.27 m (+ ve)

Height of vane above the instrument axis

= 334.68 + 0.27 = 334.95

R.L. fo vane = 334.95 + 2650.38 = 2985.33 m

R.L. of Q = 2985.33 - 4 = 2981.33 m

An     instrument  was  set up at       P  and         the     angle of  depression  to  a  vane  2  m above  the foot     of  the  staff  held at Q  was  5 o 36?. The horizontal  distance between P and Q was known to be 3000 metres. Determine the R.L. of the staff station Q given that staff reading on a B.M. of elevation 436.050 was 2.865 metres.

Solution:

The difference in elevation between the vane and the instrument axis

=   D tan ?

= 3000 tan 5 o 36? = 294.153

Combined correction due to cuvature and refraction

C = 0.06735 D² metres , when D is in km

= 0.606 m.

Since  the   observed   angle   is   negative,   the   combined   correction   due   to

curvature and refraction is subtractive.

Difference in elevation between the vane and the instrument  axis

= 294.153 - 0.606 = 293.547 = h.

R.L. of instrument axis  = 436.050 + 2.865 = 438.915

             R.L. of the vane = R.L. of instrument aixs - h

= 438.915 - 293.547 = 145.368

R.L. of Q = 145.368 - 2

= 143.368 m.

In order to ascertain the elevation of the top (Q) of the signal on a hill, observations were made from two instrument stations P and R at a horizontal distance 100 metres apart, the station P and R being in the line with Q. The angles of elevation of Q at P and R were 28 o 42? and 18 o 6? respectively. The staff reading upon the bench mark of elevation 287.28 were respectively 2.870 and 3.750 when the instrument was at P and at R, the telescope being horizontal. Determine the elevation of the foot of the signal if the height of the signal above its base is 3 metres.

Solution:

Elevation of instrument axis at P = R.L. of B.M. + Staff reading

= 287.28 + 2.870 = 290.15 m

Elevation of instrument axis at R = R.L. of B.M. + staff reading

= 287.28 + 3.750 = 291.03 m

Difference in level of the instrument axes at the two stations

S      =291.03 - 290.15 = 0.88 m.

?  -- = 28 o 42 and ? ---- = 18 o 6?

s cot ?--- = 0.88 cot 18 o 6? = 2.69 m

= 152.1 m.

h-- = D tan ?-- = 152.1 tan 28 o 42? = 83.272 m

R.L. of foot of signal = R.L. of inst. aixs at P + h-- - ht. of signal

= 290.15 + 83.272 - 3 = 370.422 m.

Check   :   (b + D)     = 100 + 152.1 m = 252.1 m

h-- = (b + D) tan ?-- = 252.1 x tan 18 o 6?

= 82.399 m

R.L. of foot of signal = R.L. of inst. axis at R + h--+ ht. of signal = 291.03 + 82.399 - 3 = 370.429 m.