The sensitiveness of a bubble tube in a level would decrease if
Hydrographic surveys deal with the mapping of
If "L" is the length of the chain , "W" is the weight of the chain and 'T' is the tension , the Sag Correction for the chain line is
If the whole circle bearing is 315°20′, its quadrantal bearing would be
The radius of a 1 degree curve is __________ m
A level when set up 25 m from peg A and 50 m from peg B reads 2.847 on a staff held on A and 3.462 on a staff held on B, keeping bubble at its centre while reading. If the reduced levels of A and B are 283.665 m and 284.295 m respectively, the collimation error per 100 m is
Actual height difference between A and B = 284.295 - 283.665 = 0.63.
Height difference between A and B based on staff readings = 3.462 - 2.847 = 0.615.
Distance between A and B = 50-25 = 25 m.
Therefore collimation error per 100 m.
= (0.63 - 0.615)*(100/25).
= 0.060 m.
An invar tape, 50 m in length, standardized at 20oC temperature and 10kg pull, is used to measure a baseline. The Correction per tape length, if at the time of measurement the temperature was 30o and the coefficient of linear expansion of the tape was 1x10-6 per oC , will be
Correction for temperature
Ct = L ∝ ( Tm - To)
∝ = Co-efficient of thermal expansion
Tm = Mean temperature in the field during measurement
To = the standard temperature in the field during
L = measured length
Ct = 1 x 10-6 (30 - 20) x 50
= 0.0005 m
A Rectangular plot of 16km2 in area is shown on a map by a similar rectangular area of 1 cm2 , R.F of the scale to measure a distance of 40km will be
R.F is ratio of two measurements with common unit
Following observations were taken with a transit fitted with stadia wires.
The line of sight was horizontal and the staff was held vertical.
The tacheometric constants k and C are 100 and 0.4 m, respectively. The horizontal distance between staff and instrument is
D = KS + C
S = 2.830 - 1.726 = 1.104
D = 100 x 1.104 + 0.4 = 110.8m
Following observations were taken during a reciprocal leveling:
If reduced level of P is 140.815 m, the reduced level of Q is
True difference = 1.824-0.928/ 2.748-1.606 = 0.7845
Reduced level of Q is = 140.815-0.7845 = 140.030m