Question

Question l (a) A level survey was conducted on site between two temporary benchmark and the staff reading were collected in Table 1. Calculate the survey accuracy and make adjustments to reduced levels where deemed appropriate. (60%) Table 1. Staff Readings for Levelling Survey Intermediate Foresight Backsight Sight Remarks 0.951 1.565 TBM1 10.00mOD Point A Point B Point C Point D Point E TBM2 9.85mOD 1.255 1.422 1510 0.751 0.923 0.812 0.658 (b) From a station at point I with reduced level 101.00mOD, a theodolite with height 1.50m is sighted onto two staffs at points A and B, respectively. The staff and angle readings are recorded in Table 2. Table 2. Staff and Angle Readings Stadia ReadingsHrizontaVertical Staff Location Upper/Middle/Lower Angle Angle 952/1.852/1.752 0 15' 20 91 35' 10" 0.846/0.646/0.446 124° 40' 50" 87 50' 20" Given K-100 and C-0 for the instrument calculate: the reduced level at point A and the horizontal distance from the station to point B (40%)

Answer 1

a)

Station	BS	IS	FS	HI	Elevation	Formula used
TBM1	0.951			10.951	10	HI = Elev. + BS = 10 + 0.951 = 10.951
A	1.565		1.255	11.261	9.696	Elev. = HI - FS = 10.951 - 1.255 = 9.696 ; HI = Elev. + BS = 9.696 + 1.565 = 11.261
B		1.422			9.839	Elev. = HI - IS = 11.261 - 1.422 = 9.839
C	0.751		1.51	10.502	9.751	Elev. = HI - FS = 11.261 - 1.51 = 9.751 ; HI = Elev. + BS = 9.751 + 0.751 = 10.502
D		0.923			9.579	Elev. = HI - IS = 10.502 - 0.923 = 9.579
E		0.812			9.69	Elev. = HI - IS = 10.502 - 0.812 = 9.69
TBM2			0.658		9.844	Elev. = Hi - FS = 10.502 - 0.658 = 9.844

Given elevation of TBM2 = 9.85 m

But obtained elevation of TBM2 = 9.844 m

Therefore, Error in elevation = 9.844 - 9.85 = - 0.006 m

Therefore total correction = + 0.006

Average correction = Total correction / Number of instrument stations

= 0.006 / 3 = + 0.006

Correction for each elevation and final corrected elevations are as shown below:

Station	BS	IS	FS	Obtained Elevation	Correction	Corrected elevation
TBM1	0.951			10	-	10
					0.002 * 1 = 0.002
A	1.565		1.255	9.696		9.696 + 0.002 = 9.698
					0.002 * 2 = 0.004
B		1.422		9.839		9.843
					0.002 * 2 = 0.004
C	0.751		1.51	9.751		9.755
					0.002 * 3 = 0.006
D		0.923		9.579		9.585
					0.002 * 3 = 0.006
E		0.812		9.69		9.696
					0.002 * 3 = 0.006
TBM2			0.658	9.844		9.85

b)

Reduced level of A = Reduced level of I + Height of instrument at I - Middle hair reading at A

= 101 + 1.5 - 1.852

= 100.648 m

Distance from I to B = K * S + C

Where,

S = Staff intercept = Top reading - Bottom reading

= 0.846 - 0.446

= 0.4

Therefore, Distance = 100 * 0.4 + 0

= 40 m