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7/30/2019 6 Differential Level
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Differential Leveling
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Introduction
Differential surveying is used to
determine the difference in
elevation between two or morepoints.
It is commonly used to establish
the elevation of a benchmark
referenced to an existing
benchmark.
It is also useful for comparing the
elevation of several points or
objects.
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Differential Leveling Example
An example of
comparing theelevation of multiple
points is setting the
top of the forms
before placing
concrete.
In common practice, abacksight would be
recorded from the
bench mark and the
target would be set for
the desired elevation
of the forms.
The rod holder would then place the rod at
several point along the forms to determine if
they were at the correct height.
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Establishing A Benchmark
Another use of differential leveling
is establishing the elevation of abenchmark.
When the existing benchmark and
the location of the new benchmark
can be seen from one instrument
position, the procedure is very
simple. The instrument is set up halfway
between the points and leveled.
A rod reading is taken on the
existing benchmark, this iscalled a backsight.
The backsight reading is added
to the elevation of the
benchmark to establish the
instrument height (reference
line).
HI = Elevation + Backsight
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Benchmark Example-cont.
In this example the benchmark
elevation is 850.47 feet and the
backsight is 3.56 ft.
The height of the instrument is:
HI = 850.47 ft + 3.56 ft = 854.03 ft
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Benchmark Example-cont.
The instrument isrotated until it is alignedwith the second
benchmark. A rod reading
(foresight) is recordedfor the secondbenchmark.
In this example the
foresight is 5.21 ft.
The rod reading is subtracted from the height of
the instrument to find the elevation of the second
benchmark.
The elevation is:
Elev = HI - FS
= 854.03 ft - 5.21 ft
= 848.82 ft
BM1 is 1.65 feet higher
than BM2.
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Benchmark Example TP
When both
benchmarks cannotbe reached from one
instrument position,
turning points are
used.
Because a turning
point is a temporarybenchmark, it must
be a stable structure.
A backsight is taken on BM1. The 4.31 is added to the elevation of the bench
mark to find the height of the instrument (104.31).
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Benchmark Example TP-cont.
A turning point is establishedand a foresight is recorded
(4.92).
The foresight is subtracted from
the height of instrument to
determine the elevation of the
turning point (99.39) .
Then the instrument is moved toa point between the turning
point and the next station.
In this example the next station
is BM2.
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Benchmark Example TP-cont.
A backsight is taken on theturning point (4.22).
The backsight is added to theelevation of the turning point tofind the new instrument height(103.61).
The instrument is rotated and aforesight is recorded on BM2.
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Benchmark Example TP-cont.
The foresight on BM2 (2.35) is
subtracted from the instrument
height to determine the
elevation of BM2 (101.08)
Tables are an excellent way oforganizing numbers.
Surveyors have developed a
standard table for differential
leveling.
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Differential Leveling Table
STA BS HI FS ELEV
BM1 4.31 104.31 100.0TP 4.22 103.61 4.92 99.39
BM2 2.53 101.08
Five columns are used.
STA = Station IdentificationBS = Backsight
HI = Instrument Height
FS = Foresight
ELEV = Elevation
The table for this example:101.08 - 100.0 = 1.08BM2 is 1.08 feet higher than BM1
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Benchmark Example TP-cont.
Assuming no errors occurred during the survey, BM2 is 1.08 feet
higher than BM1.
This is not a good assumption.
Differential leveling uses three checks for errors.
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Three Checks For Error
1. Closing the loop
2. Note check
3. Allowable error check
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1. Closing the Loop
To close the loop the survey is continued back to the beginning.
In the previous example, surveying from BM1 to BM2 resulted in adifference in elevation between the two benchmarks of 1.08 feet.
Surveying from BM2 to BM1 should result in the same difference in
elevation.
Any difference in elevation for BM1 between the initial elevation of
BM1 and the closing elevation of BM1 is error.
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Closing the Loop Example
The steps are the same.
The instrument is moved and a backsight is recorded for BM2 (3.27).
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Closing the Loop Example-cont.
The instrument is rotated.
A foresight is recorded on TP2 (2.21) .
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Closing the Loop Example-cont.
The instrument is moved between TP2 and BM1
A BS is recorded on TP2 (3.29).
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Closing the Loop Example-cont.
The instrument is rotated.
The loop is closed by recording a foresight on BM1 (5.42).
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Differential Table
STA BS HI FS Elev
BM1 4.31 104.31 100
TP1 4.22 103.61 4.92 99.39
BM2 3.27 104.35 2.53 101.08
TP2 3.29 105.43 2.21 102.14
BM1 5.42 100.01
When the closing data is entered into the table thefirst error check is completed.
The second check for error is called the note check.
The note check uses an equation:
| BS - FS |=| BM1i - BM1c |
If the equation is true, there is no math error in the
notes.
If the equation is not true, the notes have a math error.
What should you do if the note check is not true?
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2. Note Check
The note check statement is true.
The 0.01 difference in the elevation of BM1i and BM1c is not
caused by a math error in the notes
BM1i
BM1c
STA BS HI FS Elev
BM1 4.31 104.31 100.00
TP1 4.22 103.61 4.92 99.39
BM2 3.27 104.35 2.53 101.08
TP2 3.29 105.43 2.21 102.14
BM1 5.42 100.01
15.09 - 15.08
0.01 = 0.01
OK
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3. Allowable Error of Closure
The third check for error is called the allowable error.
Early surveyors realized that the sources of error were so large that itwould be impossible to control for all of them.
It is common practice for the agency/individual contracting the work to
specify the acceptable level of error.
Professional standards may also specify allowable error.
A simple one is called the allowable error and it is based on an
equation:
AE = k M
k = 1.0 to 0.01
M = Distance surveyed (miles)
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Allowable Error of Closure-cont.
For the differential example, the distance between BM1 and BM2 waspaced and a distance of 1.100 feet was recorded.
A k value of 0.1 is acceptable for general work.
AE = k M = 0.1 1,100 x 25280
= 0.1 x 0.417 = 0.04
Is pacing an appropriate method for measuring distance?
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Allowable Error of Closure-cont.
The actual error was 0.01 and the allowable error is 0.04, thereforethe survey is acceptable.
0.01 < 0.0
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The Complete Data Table
STA BS HI FS Elev
BM1 4.31 104.31 100.00
TP1 4.22 103.61 4.92 99.39
BM2 3.27 104.35 2.53 101.08
TP2 3.29 105.43 2.21 102.14
BM1 5.42 100.01
15.09 - 15.08
0.01 = 0.01
OK
AE = k M = 0.1 1100 x 2
5280= 0.06
0.01 < 0.06
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Allowable Error-cont.
In this example the actual error was less than the allowable error.
What should happen if the actual error is greater than the allowable error?
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Allowable Error-cont.
What would be the conclusion about the error in the data if a higher
standard was used, k = 0.01.
AE = 0.01 x 0.417 = 0.004
0.01 > 0.004
The data would be unacceptable.
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