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Corrections
in
Taping
1st RULE : When a line is measured with a tape
that is “too long”, the corrections are
applied to the observed length by adding.
2nd RULE : When a specified or required length is
to be laid out with a tape that is “too
long”, the corrections are subtracted from
the known length to determine the
corrected length to be laid out.
3rd RULE : When measuring or laying out lengths
with a tape that is “too short”, the
corrections are applied opposite to those
stated in the first two rules.
Measuring :
“too long” → add
“too short”→ subtract
Laying out :
“too long” → subtract
“too short”→ add
1
Correction
Due to
Incorrect Tape
Length
Correction Due to Incorrect Tape Length
--
Corr = TL – NL
Cl = Corr
CL = ML ± Cl
note:
+Corr = too long
- Corr = too short
Ex.1:
The length of line AB measured with a
50-m tape is 465.285m. When the
tape is compared with a standardized
invar tape it is found to be 0.016m too
long in almost the same conditions of
support, tension, and temperature that
existed during measurement of the
line. Determine the correct length of
AB.
GIVEN:
NL = 50m
ML = 465.285m
Corr = 0.016m
Where:
Corr = correction per tape length
TL = true or actual length of tape
NL = nominal length of tape
Cl = total correction
ML = measured length or length
to be laid out
CL = correct length
GIVEN:
NL = 50m
ML = 465.285m
Corr = 0.016m (too long)
SOLUTION:
Cl = Corr
= 0.016 m
.
Cl = 0.149 m
CLAB= ML ± Cl
= 465.285 m ± 0.149 m
= 465.285 m + 0.149 m
CLAB= 465.434 m
2
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Ex.2:
A building 38m x 45m is to be laid out
with a 50-m long metallic tape. If
during standardization the tape is
found to be only 49.950m, determine
the dimensions to be laid out, using
this tape, in order that the building
shall have the desired dimensions.
Corr = TL – NL
= 49.950m – 50.000m
Corr = - 0.050m (too short)
CL = Corr
CW = Corr
= 0.050 m
GIVEN:
NL = 50 m
TL = 49.950 m
W = 38 m
L = 45 m
= 0.050 m
CW = 0.038 m
CL = 0.045 m
CLW= W ± CW
= 38 m ± 0.038 m
= 38 m + 0.038 m
CLW= 38.038 m
CLL= L ± CL
= 45 m ± 0.045 m
= 45 m + 0.045 m
CLL= 45.045 m
38.038m x 45.045m
Correction Due to Temperature
Correction
Due to
Temperature
SOLUTION:
GIVEN:
NL = 50 m
TL = 49.950 m
W = 38 m
L = 45 m
Ct= α L (T-Ts)
steel α = 0.0000116/°C
L’= L ± Ct
L’= L ± CT
Where:
α = coefficient of linear expansion
L = length of tape or length of line
measured
T = observed temperature of tape
at time of measurement
Ts= standard temperature
L’ = corrected length of line
3
Correction
Due to Sag
Cp=
steel E = 2.00x106 kg/cm2
L’= L ± Cp
L’= L ± CP
A=
Where:
Cp= total elongation in tape length
due to pull
Pm= pull applied to the tape
during measurement
Ps= standard pull
L = length of tape
A= cross-sectional area of the tape
E= modulus of elasticity of the tape
material
L’ = corrected length of line
Correction Due to Sag
Correction
Due to
Tension
Correction Due to Tension
--
Cs=
CS=Cs1+Cs2+…+ Csn
CP = Cp
C s=
W = wL
W2= w2L2
Cs1
Cs2
Cs3
L1
L2
L3
L’= ML ± CS
note:
Where:
Cs is always “too short”
CS= correction due to sag
w = weight of tape per unit length
W = total weight of tape between supports
L = interval between supports or
unsupported length of tape
P = tension or pull applied on the tape
L’ = corrected length of line
4
Combined
Corrections
Combined Corrections
--
L’= L ± CT ± CP ± CS
5