Taping Errors: (Section 6.14-6.16) Taping equipment and methods have been covered in
the field school. However, it is useful to consider taping errors which consist of:
1) Incorrect Tape Length Correction
2) Temperature Correction
3) Pull Correction
4) Sag Correction
5) Alignment Error
These sources of error tend to be systematic and can be eliminated with careful planning
and technique. Accuracy ratio 1/10,000 can be obtained using taping.
ENGI 3703
Surveying and Geomatics
Topic
Taping Distance Errors
Quantified
Instructor: Prof. Ken Snelgrove
Lect 7 - Sept 24/07 Slide 1 of 7
Incorrect Tape Length Correction: (Section 6.14.1) Steel tapes may become damaged
over time through “kinking”, “stetching” or through breakage and repair. A kinked tape will
have its length shorted somewhat and rather then discarding the instrument a tape
correction can be applied as:
l l CL = L
l where: CL
l
l’
L
is the correction applied to the recorded measurement
is the actual tape length
is the “nominal” tape length
is the recorded measurement
Example: A measurement of 171.278 m was recorded with a 30-m tape that was only
29.996 m long under standard conditions. What is the corrected measurement?
ANS:
l l CL = L
l 29.996 30.000 CL = 171.278
30.000
CL = (133.33 106 )171.278
CL = 0.022836m
ENGI 3703
Surveying and Geomatics
Actual Length = Measured Length + CL
= 171.278 - 0.022836 = 171.255 m
Topic
Taping Distance Errors
Quantified
Instructor: Prof. Ken Snelgrove
Lect 7 - Sept 24/07 Slide 2 of 7
Temperature Correction: (Section 6.14.2) Steel tapes provide direct measurement at a
“standard” temperature of 20oC. Above this temperature, the tape elongates and provides
distances that are shorter then actual. Below 20oC, the tape contracts to a shorter length and
produces longer than actual distances. The coefficient of thermal expansion / contraction of
steel is 11.6 x 10-6/oC.
Because tape temperatures can vary from air temperature due to moisture, “micro-climate” and
radiation, very accurate tapes made from a iron-nickel alloy (64%-36%) called Invar are
sometimes used. Invented in 1896 by Charles Guillaume it won him a Nobel Prize in physics in
1920. Invar has ~1/10th the coefficient of expansion (1.1 x 10-6 / oC) making it ideal for precise
instruments, watches and survey tapes. The tape temperature correction formula is give as:
CT = k(T1 T)L
where : C T is the tape correction (m)
k is the coefficient of thermal expansion
T1 is the tape temperature during measurement
T is the standard temperature (m)
Example: You must lay out two points that are
exactly 100.000 m apart. Field conditions
indicate that standard conditions apply except
the measured temperature is 27oC. Determine
the distance to be laid out.
ANS: CT = 11.6x10-6 /oC (27-20 oC) 100.000 m
= +0.008 m
The corrected length if 100.000m was taped
would be 100.008 m. As a result, the correct
layout tape distance should be 99.992m
ENGI 3703
Surveying and Geomatics
Topic
Taping Distance Errors
Quantified
Instructor: Prof. Ken Snelgrove
Lect 7 - Sept 24/07 Slide 3 of 7
Pull Correction: (Section 6.14.3) Along with temperature, steel tapes provide direct
measurement at a “standard” pull tension of 50 N (~11 lbs). Above this pull, the tape elongates
and provides measured distances that are shorter then actual. Below the standard pull, the
tape contracts to a shorter length and produces measurements longer than actual distances.
The pull correction depends on the both the x-section area of the tape and its modulus of
elasticity. The average modulus (E) for steel tapes is 200x109 N/m2 (1 N/m2 = 1 Pascel). The
x-section areas may differ but are generally i) 8mm x 0.45mm for Heavy Duty tapes (3.6 mm2)
and ii) 6mm x 0.30mm for Normal usage tapes (1.8 mm2). The tape pull correction formula is
give as:
Example: A 30-m tape is used with a 100N
force instead of the standard tension of 50N. If
the x-section area of the tape is 1.8 mm2, what
is the tension error per tape length?
L
CP = (P1 P)
AE
where : C P is the tape correction (m)
E is the Young's modulus (Pa)
ANS: CP =
A is the tape area (m2 )
P1 is the tape tension during measurement (N)
P is the standard pull (N)
(100-50 N) 30.000 m
(1.8x10-6 m2) 200x109 N/m2
= +0.0042 m
The corrected tape length would be 30.004m.
ENGI 3703
Surveying and Geomatics
Topic
Taping Distance Errors
Quantified
Instructor: Prof. Ken Snelgrove
Lect 7 - Sept 24/07 Slide 4 of 7
Sag Correction: (Section 6.14.4) Steel tapes provide direct measurement only when they are
“fully supported” along their length. As a result, when only end supports are provided, the tape
sag in the form of a catenary and always produces a shorter tape length (hence the negitive).
This shorting produces measurements longer than actual distances.
The sag error (B’B) depends on the pull tension (P1), the weight of the tape per unit length (w)
and the length of the sag tape (Ls), given as:
w 2 L3s
CS = 24P12
where : C s is the sag tape correction (m)
P1 is the tape tension during measurement (N)
w is the weight of tape per unit length (N/m)
L s is the tape reading including sag (m)
Example: A 30-m tape is used with standard
tension of 50N. However, the tape is not fully
supported. If the unit weight of tape is 0.14N/m,
determine the sag correction?
ANS: CS = - (0.14 N/m)**2 (30.000 m)3
24 (50 N)2
= -0.0088 m
A
B’
B
The corrected tape length would be 30.004m.
Note: Sag errors depend on taping distance are therefore
accumulated for each taping sub-distance.
ENGI 3703
Surveying and Geomatics
Topic
Taping Distance Errors
Quantified
Instructor: Prof. Ken Snelgrove
Lect 7 - Sept 24/07 Slide 5 of 7
Normal Tension: Since sag tends to shorten the tape and pull tends to extend it, we can
determine a pull amount that will exactly balance the sag amount. If we add the pull and sag
formula and set to zero, we can solve for P to get:
L w 2 L3s (P1 P ) + =0
AE 24P12 (P1 P )
W 2L
L
=
AE 24P12
W 2 AE
P =
24 ( P1 P )
2
1
P1 =
0.204W AE
P1 P
if w = W/L where W is the total weight of the tape (N)
Example: Determine the normal pull tension of 30-m tape if the unit
weight of tape is 0.14 N/m. The tape is supported at only two points. The
properties of the tape are: standard tension of 50N, x-section area is 1.8
mm2 and the Young’s modulus is 200 x 109 Pa.
ANS: i) Determine the total tape weight, W = 0.14 N/m x 30 m = 4.2 N
ii) Solve for P1 by trial and error
P1 =
P1 =
0.204 ( 4.2) 200 10 9 (1.8 106 )
P1 50
514.08
P1 50
P1 = 85.8 N
Note: If total tape weight given then we need to know
the density of steel to solve = 7850 kg/m3
ENGI 3703
Surveying and Geomatics
Topic
Taping Distance Errors
Quantified
Instructor: Prof. Ken Snelgrove
Lect 7 - Sept 24/07 Slide 6 of 7
Type and Class of Error
Error
Source
Systematic or
Random
Departure from Normal to
Produce 3 mm Error for
30 m tape (1:10,000)
Tape Length
I
S
3 mm
Temperature
N
S or R
8.6 oC
Pull
P
S or R
86 N
Sag
N,P
S
200 mm at centre
Alignment
P
S
420 mm at one end
210 mm at mid point
Tape not Level
P
S
420 mm difference in
elevation
Plumbing
P
R
3 mm
Marking
P
R
3 mm
Interpolation
P
R
3 mm
ENGI 3703
Surveying and Geomatics
Topic
Taping Distance Errors
Quantified
Instructor: Prof. Ken Snelgrove
Lect 7 - Sept 24/07 Slide 7 of 7