LOGO TACHYMETRY

LESSON 5 Variable angleTangential systemTangential system

s

1

2

3

WORK PROCEDUREWORK PROCEDURE4

PRINCIPLESPRINCIPLES

CALCULATIONCALCULATION

PRACTICAL 3PRACTICAL 3

ContentsContents

PrinciplesPrinciplesThe Tangential System of TachymetryThe Tangential System of Tachymetry

s = the staff intercept ABV = the vertical component XY, the height of the centre hair reading above (or below) the instrument axis ,,= vertical angle (variable)= vertical angle (variable)H = the horizontal distance required.hi = instrument height

Publication formulaAY = H tan θBY = H tan AY –BY = s

= H ( tan θ – tanH = s

tan θ – tanHorizontal distance

If used θ, If used ,Vertical distance

Vθ = H tan θ V = H tan

Difference height

dH = hi ± Vθ – hθ dH = hi ± V – h

Reduced level

RLRL1 1 = RL= RLTBMTBM + + hi ± Vθ – hθ RLRL1 1 = RL= RLTBMTBM + + hi ± V – h

Where

S = staff intercept

HH = horizontal distance

VV = vertical distance

= zenith angle ( > )

hihi = the height of instrument (always positive)

hh = the centre hair reading ( @ )

RLRL= Reduced level

Work procedure Work procedure EXAMPLE RESULT PRACTICAL 3 - TACHEOMETRY BOOKING FORM

(The Tangential System )

Inst. Stn. and

Ht. of inst. axis

Staff

stn

Horizontal angle, HL

Horizontal angle, HR

Average horizontal

angle

Vertical angle, V

Stadia Remarks

A BSETTING

105015’30”

SETTING285015’30

”

95011’25” h1.205

GIVEN REFERENCE BERING AB = 105015’30”

92010’20” h1.525

(1.455 m) C 155010’3

6”335012’3

0”335011’33

”

90012’20” h1.050

90010’10” h1.450

C ASETTING

335011’33”

SETTING155011’33

”

83020’25” h1.345

85016’25” h1.205

(1.305 m) D 45020’26

”225032’3

0”225026’28

”

88018’10” h1.250

89028’00” h1.205

Practical 3Practical 3

• GIVEN THE TBM (Temporary bench mark) – Stn A & BStn A & B• DETERMINING THE NUMBER OF STATIONSDETERMINING THE NUMBER OF STATIONS• READING VERTICAL ANGLE BELONG THE SITUATION• TRAVERSE METHOD• TAKE TOPOGRAPHY ITEMS (Tree, Building, Pedestrian)

Sketches topographySketches topographyB Tree 1B Tree 1H =V =S=V =S=

A b1A b1H =V =S=V =S=

B b2B b2H =V =S=V =S=

Pedestrian

Pedestrian

E Tree 2E Tree 2H =V =S=V =S=

C B3C B3H =V =S=V =S=

C P1C P1H =V =S=V =S=

C P5C P5H =V =S=V =S=

C P3C P3H =V =S=V =S=

A P2A P2H =V =S=V =S=

A P4A P4H =V =S=V =S=

A P6A P6H =V =S=V =S=

Inst. Inst. Stn. Stn.

and Ht. and Ht. of inst. of inst.

axisaxis

Staff Staff stn.stn.

Horizontal Horizontal angle, HLangle, HL

HorizontaHorizontal angle, l angle,

HRHR

Average Average horizontal horizontal

angleangle

Vertical Vertical angle, Vangle, V

StadiaStadia RemarksRemarks

AA(155

)

BB

SS

GIVEN REFERENCE BERING AB =

b1b1

SS Building 1

P2P2

SS

Pedestrian 2

P4P4

SS

Pedestrian 4

P6P6

SS

Pedestrian 6

CC

SS

BB(155

)

AA

SS

T1T1

SS Tree 1

b2b2

SS Building 2

EE

SS

Example for field book - topography

EXAMPLE CALCULATION EXAMPLE CALCULATION Inst. Stn. and

Ht. of inst. axis

Staff

stn.

Horizontal angle

Vertical angle

Zenith angle,

Stadia Horizontal distance

H (m)

Vertical distance

V (m)

Difference height,dH (m)

R.L. at

stn.

R.L. at

staff

Remarks

A(1.5m) B 27 30’

00”

77 00’ 00”

+13 00’ 00” 3.50 Take

the first RL = 50 m85 00’

00” +5 00’ 00” 1.50

H = s = (3.50 – 1.50) = 13.94 m

tan θ – tan tan +13 – tan+5

If used θ, If used ,

Vθ = H tan θ = 13.94 . Tan +13 = 3.22 m

V = H tan = 13.94 . Tan +5 = 1.22 m

dH = hi ± Vθ – hθ = 1.5 +3.22 – 3.50 = 1.22 m

dH = hi ± V – h= 1.5 + 1.22 -1.50 = 1.22 m

RLRL11 = RL = RLTBMTBM + + hi ± Vθ – hθ = 50 + 1.5 +3.22 – 3.50 = 51.22 m

RLRL11 = RL = RLTBMTBM + + hi ± V – h = 50 + 1.5 + 1.22 -1.50 = 51.22 m

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JIKA ADA MASALAH AMALI,