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TACHYMETRY TACHYMETRY LESSON 4

Incline Sights With

the Staff NormalStaff Normal

ContentsContents

PRINCIPLES1

2 CALCULATION3

s = the staff intercept ABh = the length of the centre hair reading from the staff baseV = the vertical component XY, the height of the centre hair reading above (or below) the instrument axisD = the length of the line of sight IXH = the horizontal distance required.hi = instrument height

PrinciplesPrinciples

dH

hi

Incline Sight With The Staff NormalIncline Sight With The Staff Normal

Formula staff normal Formula staff normal

Horizontal Distance;Horizontal Distance; H = (Ks + C) cos θ ± h sin θ

Vertical DistanceVertical Distance ; V = (Ks + C) sin θ

First Reduced Level Station First Reduced Level Station ; RLRL1 1 = RL= RLTBMTBM – hi – hi V – V – h cos θ

Difference HeightDifference Height ; dH = hi . ± V – h cos θ

Cycle Diagram Cycle Diagram

Where

DD = Distance

K & CK & C = constant (if not given assume K = 100 & C = 0)

S = staff intercept

HH = horizontal distance

VV = vertical distance

= zenith angle (positive for angles of the elevation, negative for angles depression)

hihi = the height of instrument (always positive)

hh = the centre hair reading (always negative)

EXAMPLE CALCULATION EXAMPLE CALCULATION Inst

. Stn. and Ht. of

inst.

axis

Staff

stn.

Horizontal angle

Vertical angle

Zenith angle,

Upper stadia

Middle

stadia

Lower stadia

Horizontal

distanceH (m)

Vertical distanc

eV (m)

Difference height,dH (m)

R.L. at

stn.

R.L. at

staff

Remarks

A(1.64m

)

B 97 30’ 00” 81 30’ 00” +8 30’ 00”2.025 1.515 1.000

Take the first RL = 42.353 m

HH = (Ks + C) cos θ ± h sin θ = ((((100 . (2.025 – 1.000)100 . (2.025 – 1.000))) + 0.65 + 0.65)) . (Cos . (Cos +8+800 30’ ))) + (1.515 . Cos + (1.515 . Cos +8+800 30’ )) = (103.150. (Cos . (Cos +8+800 30’ ))) ) + + 0.2240.224 = 102.240 m= 102.240 m

VV = (Ks + C) sin θ = = ((100 . (2.025 – 1.000) + 0.65100 . (2.025 – 1.000) + 0.65) ) . (Sin . (Sin +8+800 30’ )) = = (103.150. (Sin . (Sin +8+800 30’ ))) ) = 15.250 m= 15.250 m

dH = hi dH = hi V – h kos V – h kos θ = 1.64 + 15.25 - 1.515 kos= 1.64 + 15.25 - 1.515 kos +8 +800 30’ = 15.387 m= 15.387 m

RLRLA A = RL= RLTBMTBM – hi – hi V – h V – h = RL= RLTBMTBM – – dHdH = 42.353 + 15.387= 42.353 + 15.387 = 57.740 m= 57.740 m

Given K = 100 and C = 0.65

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