Take α = 6.2 Ã- 10^(6) /✯.Įxample 12.5: To measure a base line, a steel tape 30 m long, standardised at 15º C with a pull of 80 N was used. If the temperature at which tape was standardised is 80º F, find the true length of the line. Measured distances on the ground and slope of the ground are as given below: ![]() Solution: Initial length of the chain = 20 mĮxample 12.4: A 20 m tape was used for measuring a line at an average temperature of 65º F. Assuming the length was exactly 20.0 m in the beginning of survey work, determine the true area of the field. ![]() However when the chain was tested at the end of work, it was wound to be 20.10 m. The area of the plan was found to be 62.8 cm2. Solution: Average correction per chain lengthĮxample 12.3: A survey was conducted with a 20 m chain and plan of the field was drawn to a scale of 1 cm = 5 m. If the length of chain was perfectly correct while starting measurement, what is the true length of the line measured? After the measurement chain was found to be 80 mm longer. Hence normal tension Pn may be found as,Įxample 12.2: A distance of 1500 m was measured with a 20 m chain. The pull for which these two corrections neutralise each other is called normal tension. It may be noted that if pull is more than standard pull, the correction for pull is +ve, while correction for sag is always ve. P = the pull applied and L = measured length W = the weight of the tape of span length The correction, which is difference between the length of catenary and true length is given by Hence, measured length is more than actual length. This technique eliminates errors due to measurement along slopes, but necessitates correction for sag. While measuring on unevenly sloping ground, tapes are suspended at shorter length and horizontalĭistances are measured. If the length measured is L and the difference in the levels of first and the last point is h, then slope correction Then the temperature correction Ct is given by, α = Coefficient of thermal expansion of the material of the tape and Let T0 = Temperature at which tape is standardised The above expression takes care of sign of the correction also. If pull applied while standardising the length of tape and pull applied in the field are different, thisĮ = Youngs modulus of the material of tape, then Therefore, it is important to consider the sample size while estimating the standard error in measurement.The following five corrections may be found for the measured lengths of tape: 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.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |