TBM 9.09 m A.A.D.
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TBM 9.09 m A.A.D. TBM 10.00 m A.A.D. Main Gate Burnaby Building Approximate North a TBM outside the main entrance of Burnaby Building and obtain the s of three points before closing onto another TBM near the main gat Point 1 Ground level at entrance to structures laboratory Top of door level at entrance to structures laboratory
description
Approximate North. Main Gate. TBM 9.09 m A.A.D. Burnaby Building. Top of door level at entrance to structures laboratory. Ground level at entrance to structures laboratory. TBM 10.00 m A.A.D. Point 1. Start at a TBM outside the main entrance of Burnaby Building and obtain the - PowerPoint PPT Presentation
Transcript of TBM 9.09 m A.A.D.
TBM9.09 m A.A.D.
TBM10.00 m A.A.D.
Main Gate
Burnaby Building
Approximate North
Start at a TBM outside the main entrance of Burnaby Building and obtain the RL values of three points before closing onto another TBM near the main gate.
Point 1
Ground level at entrance to structures laboratory
Top of door level at entrance to structures laboratory
Site: …………………………………. Instrument: ………………………………….
Date: …………………………………. Observer: ………………………………….
Weather: …………………………………. Booker: ………………………………….
BS IS FS HPC RL Corr Corr RL Remarks
Burnaby Building L 52
07/10/98 M.A.R.
Good M.A.R.
TBM 10.00m AAD
It is important to complete details at the top of booking forms or on every page of field books.
TBM Level Posn. BS
Key
Site: …………………………………. Instrument: ………………………………….
Date: …………………………………. Observer: ………………………………….
Weather: …………………………………. Booker: ………………………………….
BS IS FS HPC RL Corr Corr RL Remarks
Burnaby Building L 52
07/10/98 M.A.R.
Good M.A.R.
TBM 10.00m AAD10.0001.546
HPC = RL + BS HPC = 10.000 + 1.546 = 11.546
11.546
We now signal to the staff person to move to the next point.
As the next required point is too far away (it is also round a corner) we will eventually need to move the instrument.
So, we must move the staff to a change point (CP), to allow us to move theinstrument to a better position later on.
TBM CP Level Posn. BS FS
Key
Site: …………………………………. Instrument: ………………………………….
Date: …………………………………. Observer: ………………………………….
Weather: …………………………………. Booker: ………………………………….
BS IS FS HPC RL Corr Corr RL Remarks
Burnaby Building L 52
07/10/98 M.A.R.
Good M.A.R.
TBM 10.00m AAD1.546 10.00011.546
C.P.
New staff position therefore a new row.
Each rowrepresentsa staffposition.
1.562
RL = HPC - FS RL = 11.546 - 1.562 = 9.984
9.984
After we make a FS and we have calculated the new RL we are finished with that instrument position.
Move the Instrument (about the CP) to a new position where we can see the CPand also the next point we want the RL value of.
TBM CP Level Posn. BS FS ISKey
Site: …………………………………. Instrument: ………………………………….
Date: …………………………………. Observer: ………………………………….
Weather: …………………………………. Booker: ………………………………….
BS IS FS HPC RL Corr Corr RL Remarks
Burnaby Building L 52
07/10/98 M.A.R.
Good M.A.R.
TBM 10.00m AAD1.546 10.00011.546
C.P.1.562 9.9841.418
Same staff position as last reading therefore the same row
HPC = RL + BS HPC = 9.984 + 1.418 = 11.402
11.402
TBM CP Level Posn. BS FS ISKey
This reading is not the first so it is not a BSIt is not the last from this position (we can see the next points) so it is not a FSSo it is known as an INTERMEDIATE SIGHT (IS)
Site: …………………………………. Instrument: ………………………………….
Date: …………………………………. Observer: ………………………………….
Weather: …………………………………. Booker: ………………………………….
BS IS FS HPC RL Corr Corr RL Remarks
Burnaby Building L 52
07/10/98 M.A.R.
Good M.A.R.
TBM 10.00m AAD1.546 10.00011.546
C.P.1.562 9.9841.418 11.402
Point 11.390
RL = HPC - IS RL = 11.402 - 1.390 = 10.012
10.012
New staff position therefore a new row
TBM CP Level Posn. BS FS ISKey
Site: …………………………………. Instrument: ………………………………….
Date: …………………………………. Observer: ………………………………….
Weather: …………………………………. Booker: ………………………………….
BS IS FS HPC RL Corr Corr RL Remarks
Burnaby Building L 52
07/10/98 M.A.R.
Good M.A.R.
TBM 10.00m AAD1.546 10.00011.546
C.P.1.562 9.9841.418 11.402
Point 11.390 10.012
GL Struct. Lab Door
New staff position therefore a new row
1.281
RL = HPC - IS RL = 11.402 - 1.281 = 10.121
10.121
TBM CP Level Posn. BS FS ISKey
Site: …………………………………. Instrument: ………………………………….
Date: …………………………………. Observer: ………………………………….
Weather: …………………………………. Booker: ………………………………….
BS IS FS HPC RL Corr Corr RL Remarks
Burnaby Building L 52
07/10/98 M.A.R.
Good M.A.R.
TBM 10.00m AAD1.546 10.00011.546
C.P.1.562 9.9841.418 11.402
Point 11.390 10.012
GL Struct. Lab Door1.281 10.121
Top Struct. Lab Door
New staff position therefore a new row
Requires an inverted staff i.e turn the staff upside down
Read and then book the staff with a negative sign
-2.420
The negative sign will keep all the calculations correct
RL = HPC - IS RL = 11.402 - (-2.420) = 11.402 + 2.420 = 13.822
13.822
TBM CP Level Posn. BS FS ISKey
The last point required is the TBM. However it is too long a sight.So we need a CP. This will be the last sighting from this positionTherefore a FS
Site: …………………………………. Instrument: ………………………………….
Date: …………………………………. Observer: ………………………………….
Weather: …………………………………. Booker: ………………………………….
Burnaby Building L 52
07/10/98 M.A.R.
Good M.A.R.
BS IS FS HPC RL Corr Corr RL Remarks
Top Struct. Lab Door-2.420 13.822
TBM 10.00m AAD1.546 10.00011.546
C.P.1.562 9.9841.418 11.402
Point 11.390 10.012
GL Struct. Lab Door1.281 10.121
CP New staff position therefore a new row
1.321
RL = HPC - FS RL = 11.402 - 1.321 = 10.081
10.081
Last Reading -- FS -- Move the instrument
TBM CP Level Posn. BS FS ISKey
Site: …………………………………. Instrument: ………………………………….
Date: …………………………………. Observer: ………………………………….
Weather: …………………………………. Booker: ………………………………….
Burnaby Building L 52
07/10/98 M.A.R.
Good M.A.R.
BS IS FS HPC RL Corr Corr RL Remarks
Top Struct. Lab Door-2.420 13.822
TBM 10.00m AAD1.546 10.00011.546
C.P.1.562 9.9841.418 11.402
Point 11.390 10.012
GL Struct. Lab Door1.281 10.121
CP1.321 10.0811.011 Same staff position as last reading therefore the same row
HPC = RL + BS HPC = 10.081 + 1.011 = 11.092
11.092
TBM CP Level Posn. BS FS ISKey
Site: …………………………………. Instrument: ………………………………….
Date: …………………………………. Observer: ………………………………….
Weather: …………………………………. Booker: ………………………………….
Burnaby Building L 52
07/10/98 M.A.R.
Good M.A.R.
BS IS FS HPC RL Corr Corr RL Remarks
Top Struct. Lab Door-2.420 13.822
TBM 10.00m AAD1.546 10.00011.546
C.P.1.562 9.9841.418 11.402
Point 11.390 10.012
GL Struct. Lab Door1.281 10.121
CP1.321 10.0811.011 11.092
TBM 9.09m AAD
New staff position therefore a new row
2.007
RL = HPC - FS RL = 11.092 - 2.007 = 9.085
9.085
TBM CP Level Posn. BS FS ISKey
Before we look more fully at the results we will complete the second half of the levelling exercise
TBM CP Level Posn. BS FS ISKey
Top of door level at entrance to structures laboratory
Ground level at entrance to structures laboratory
Point 2
TBM CP Level Posn. BS FS ISKey
TBM CP Level Posn. BS FS ISKey
TBM CP Level Posn. BS FS ISKey
TBM CP Level Posn. BS FS ISKey
TBM CP Level Posn. BS FS ISKey
TBM CP Level Posn. BS FS ISKey
TBM CP Level Posn. BS FS ISKey
TBM CP Level Posn. BS FS ISKey
TBM CP Level Posn. BS FS ISKey
