![]() W: weight of tape per unit length = total weight / length Mathematical: Equation E-3 is the error for an unsupported length of tape. ![]() ![]() This can be be accomplished by using a support at mid-tape. However, it can be substantially reduced by using shorter unsupported sections. Procedural: Sufficient pull cannot be applied in the field to completely remove sag nor the error it causes for an entire tape length. Sag is a function of the tape weight and the pull applied to it. The error for a 100 ft tape segment is more than the total error of two consecutive 50 ft segments, Figure E-9. The weight of a tape will cause it to sag when it isn't supported along its length and pull the ends together.The sag approximates a parabolic curve which causes the error to be nonlinear. L M: Measured distance (1) Sag (a) Principle and Behavior Equation E-1 is the error as a ratio (ft/ft, m/m, etc) and Equation E-2 is the total correction for a measured distance. Use the differnce between calibration and use pulls to determine the correction. Mathematical: Apply a specific pull during measurement. Procedural: Apply the pull indicated by calibration results to stretch the tape to 100.00 ft. We would subtract the error from the observed measurement since our measurement would be long.īecause the material and cross-sectional area of a tape is constant, the error is uniform along the length of the tape. That amounts to 0.030 ft of error per full tape length. If we apply 20 pounds in the field, that means every time we measure 100.00 ft with the tape, we're actually measuring 99.970 ft. At a more comfortable 20 pound pull if may be only 99.970 feet long. For example, calibratiion may determine that 27.2 pounds is needed to stretch the tape to 100.000 ft. Calibration may indicate a tape is 100.00 ft long at a particular pull which may not be easy to achieve in the field. Instrumental (1) Tape length (a) Principle and BehaviorĪ tape is 100.00 feet long under specific physical conditions. Many of these require the tape be calibrated and that differences between calibration and measurement conditions be compared. Systematic taping errors can be eliminated by procedure or computations. Unlike many other survey measurements, taping errors cannot be compensated with reversion. Therefore, it is important to consider the sample size while estimating the standard error in measurement.A tape is a simple device, but measurements with it are subject to a number of errors. The standard error in measuring a longer length with the same precision decreases as the sample size increases. Therefore, the standard error of measuring 1.08 km with a precision of ± 0.01 m is 0.000833 m. Standard error = Standard deviation / √Sample size Standard deviation = 0.01 m / 2 = 0.005 m Standard deviation can be calculated as half of the precision as it is given as ± 0.01 m. ![]() The standard error is the standard deviation of the sample divided by the square root of the sample size. Sample size = Total length / Length that can be measured with precision We can determine the sample size by dividing the total length to be measured by the length that can be measured with precision. We need to convert 1.08 km to meters as the precision of measurement is given in meters. We need to find the standard error in measuring 1.08 km with the same precision. Given, the length that can be taped with a precision of ± 0.01 m is 30 m.
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