The correction for sag, Cs, unsupported between two given points L apart, and subject to a pull (tension) P, is given by
Cs=ω²L³/(24P²)
where ω is the weight of the tape/unit length.
In the given case,
L=48.88 m
ω=0.32N/m
P=100N
Cs=0.32²*48.88³/(24*100²)
=0.0498m
Check:
percentage correction
=0.0498/48.88=0.1%
This is a rather substantial correction, but in consideration of the span of 49m, it is understandable.
The two other problems are similar, using different units. The percentages correction are about 1/4 of that of problem (1).
Notes:
(1) It is important that the tension applied is the specified tension for the tape, or else an additional correction for tension has to be applied.
(2) This kind of measurement is becoming less and less day-to-day work, as the new instruments in total stations can measure to 2ppm±2mm in a matter of a second, up to 500 m or more.
Hey, I need help working these problems. Thanks you.
1) A 50-m tape is used to measure between two points. The average weight of the tape per meter is 0.320 N. If the measured distance is 48.888 m, with the tape supported at the ends only and with a tension of 100 N, find the corrected distance.
2) A100-ft steel tape weighing 1.8 lb and supported only at the ends with a tension of 24 lb is used to measure a distance of 471.16 ft. What is the distance corrected for sag?
3) A distance of 72.55 ft is recorded using a steel tape supported only at the ends with a tension of 15 lb and weighing 0.016 lb per foot. Find the distance corrected for sag.
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