Systematic Errors in Taping
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Systematic errors in taping are caused
by:
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Incorrect tape length or standardization error
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Variations in temperature
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Variations in tension
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Tape Sag
Tape is not of Standard Length
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The true length of the tape typically
differs from the nominal tape length
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To determine the true length, we compare
it to a known standard length under given conditions of temp., tension
& support (standardization)
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Therefore,
Cd / tape length =
true length - nominal length
Standardization Error (cont.)
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Standardization error is directly proportional
to the number and fractional portions of the tape used
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Total correction = Cd/tape
length x no. of tape lengths
= Cd/tape length x (Distance/tape
length)
Sign of the Correction
Too Long
Too Short
Measuring
+ve
-ve
Layout
-ve
+ve
Example
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A distance is measured with a 100-ft steel
tape and is found to be 896.24 ft. Later that tape was standardized and
found to be actually 100.04 ft. What is the correct distance?
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It is desired to layout a distance of
360 m. using a 30-m tape that actually has a length of 29.98 ms. What field
measurement should be made?
Variation in Temperature
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The tape expands & contracts with
temp.
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Ct = a
.L.(T - T0) (Equation 4.6)
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a = coeff. of thermal expansion
0.0000065/°
F OR 0.0000116/°
C
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L = measured length
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T = temp. during measurements
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T0 = temp. at standarization
Sign for Temperature Correction
-
If standardized at a given temp.,
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measurements at a higher temp Þ
too long
-
measurements at a lower temp. Þ
too short
Example
-
A steel tape having a standard length
of 29.990 m at 20°
C was used to measure a distance at 30°
C. The measured distance was 915.258. Determine the correct distance.
Variation in Tension
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Cp = (P - P0).L
/ (a.E) (Eqn. 4.7)
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P = tension during measurements (lb or kg)
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P0 = tension at standardization
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L = length (ft or m)
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a = cross section in in2 or cm2
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E = elasticity modulus
30
x 106 lb/in2 OR 2.1 x 106 kg/cm2
Sign for Tension Correction
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Standardized at a given tension:
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Tension is greater Þ
too long
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Tension is smaller Þ
too short
Tape Sag
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When held bet. points of support, tape
sags
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Cs = W 2.L / (24.P2)
(Eqn. 4.8)
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Cs = correction bet. points of support (ft
or m)
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W = weight of tape bet. pts of support (lb or kg)
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L = length bet. pts. of support (ft or m)
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P = applied tension (lb or kg)
Sign for Sag Correction
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Depending on the difference bet. methods
of support at standardization and measurements
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Case 1: fully supported
at standardization & supported at 2 ends while measuring Þ
too short
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Case 2: supported
at 2 ends at standardization & fully supported while measuring Þ
too long
Example
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A steel tape weighing 2 lbs is 100.00
ft when supported continuously on a floor and pulled with a tensile force
of 10 lbs. If the tape is used for measurement with only 2 ends supported.
The measured distance was 570.52 ft. What is the correct distance?
Normal Tension
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The amount of tension at which the elongation
due to tension is equal to the shortening due to sag.
Combined Corrections - Example
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A 30-m tape weighing 0.55 kg and with a cross-sectional
area of 0.02 cm2 was standardized and found to be 30.005 m at
20 C, with 5 kg tension and supported at the 0- and 30- points. The tape
was used to measure a distance of about 89 m over terrain of a uniform
5% slope. The temp. was constant at 30 C, the tape was fully supported
throughout, and a tension of 5 kg was applied to each tape length. The
observed distances were 30 m, 30 m and 29.5 m. Calculate the horizontal
distance between the 2 points.
Homework Problems - Due 09/18/98
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The textbook, page159 & 161
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Problem 4.5
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Problem 4.20
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Problem 4.23