Transcript Leveling

Differential Leveling
Introduction
 Differential surveying is used to
determine the difference in
elevation between two or more
points.
 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.
Differential Leveling Example
 An example of
comparing the
elevation of multiple
points is setting the
top of the forms
before placing
concrete.
 In common practice, a
backsight 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.
Establishing A Benchmark
 Another use of differential leveling
is establishing the elevation of a
benchmark.
 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 is
called a backsight.
 The backsight reading is added
to the elevation of the
benchmark to establish the
instrument height (reference
line).
HI = Elevation + Backsight

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
Benchmark Example-cont.
 The instrument is
rotated until it is aligned
with the second
benchmark.
 A rod reading
(foresight) is recorded
for the second
benchmark.
 The rod reading is subtracted from the height of
 In this example the
the instrument to find the elevation of the second
foresight is 5.21 ft.
benchmark.
 The elevation is:
BM1 is 1.65 feet higher
than BM2.
Elev = HI - FS
= 854.03 ft - 5.21 ft
= 848.82 ft

Benchmark Example TP
 When both
benchmarks cannot
be reached from one
instrument position,
turning points are
used.
 Because a turning
point is a temporary  A backsight is taken on BM1.
benchmark, it must  The 4.31 is added to the elevation of the bench
be a stable structure.
mark to find the height of the instrument (104.31).
Benchmark Example TP-cont.
 A turning point is established
and 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 to
a point between the turning
point and the next station.
 In this example the next station
is BM2.
Benchmark Example TP-cont.
 A backsight is taken on the
turning point (4.22).
 The backsight is added to the
elevation of the turning point to
find the new instrument height
(103.61).
 The instrument is rotated and a
foresight is recorded on BM2.
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 of
organizing numbers.
 Surveyors have developed a
standard table for differential
leveling.
Differential Leveling Table
Five columns are used.
STA
BS
HI
FS
ELEV
STA = Station Identification
BM1
4.31
104.31
100.0
BS = Backsight
TP
4.22
103.61 4.92
99.39
HI = Instrument Height
BM2
2.53
FS = Foresight
ELEV = Elevation
The table for this example:
101.08 - 100.0 = 1.08
BM2 is 1.08 feet higher than BM1
101.08
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.
Three Checks For Error
1.
Closing the loop
2.
Note check
3.
Allowable error check
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 a
difference 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.
Closing the Loop Example
 The steps are the same.
 The instrument is moved and a backsight is recorded for BM2 (3.27).
Closing the Loop Example-cont.
 The instrument is rotated.
 A foresight is recorded on TP2 (2.21) .
Closing the Loop Example-cont.
 The instrument is moved between TP2 and BM1
 A BS is recorded on TP2 (3.29).
Closing the Loop Example-cont.
 The instrument is rotated.
 The loop is closed by recording a foresight on BM1 (5.42).
Differential Table
ST A
BM 1
TP1
BM 2
TP2
BM 1
BS
4 .3 1
4 .2 2
3 .2 7
3 .2 9
HI
1 0 4 .3 1
1 0 3 .6 1
1 0 4 .3 5
1 0 5 .4 3
FS
4 .9 2
2 .5 3
2 .2 1
5 .4 2
E lev
100
9 9 .3 9
1 0 1 .0 8
1 0 2 .1 4
1 0 0 .0 1
 When the closing data is entered into the table the
first 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?
2. Note Check
ST A
BM 1
TP1
BM 2
TP2
BM 1
BS
4 .3 1
4 .2 2
3 .2 7
3 .2 9
HI
1 0 4 .3 1
1 0 3 .6 1
1 0 4 .3 5
1 0 5 .4 3
1 5 .0 9
0 .0 1
FS
4 .9 2
2 .5 3
2 .2 1
5 .4 2
E lev
1 0 0 .0 0
9 9 .3 9
1 0 1 .0 8
1 0 2 .1 4
1 0 0 .0 1
1 5 .0 8
=
0 .0 1
BM1i
BM1c
OK
 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
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 it
would 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)

Allowable Error of Closure-cont.
 For the differential example, the distance between BM1 and BM2 was
paced 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 2
5280
= 0.1 x 0.417 = 0.04
Is pacing an appropriate method for measuring distance?

Allowable Error of Closure-cont.
The actual error was 0.01 and the allowable error is 0.04, therefore
the survey is acceptable.
0.01 < 0.04

The Complete Data Table
ST A
BM 1
TP1
BM 2
TP2
BM 1
BS
4 .3 1
4 .2 2
3 .2 7
3 .2 9
HI
1 0 4 .3 1
1 0 3 .6 1
1 0 4 .3 5
1 0 5 .4 3

1 5 .0 9
0 .0 1
FS
4 .9 2
2 .5 3
2 .2 1
5 .4 2
E lev
1 0 0 .0 0
9 9 .3 9
1 0 1 .0 8
1 0 2 .1 4
1 0 0 .0 1
1 5 .0 8
=
0 .0 1
OK

AE = k M = 0.1 1100 x 2
= 0.06
5280
0.01< 0.06

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?

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.