Probable Error-a.b.Example 2PLANE SURVEYINGSU1200ASSIGNMENT 11. The sides of a closed figure have been measured and found to have a total length of 1018.910 m.If the error in the measurement is estimated to be 0.073 m, what is the accuracy ratio of the work?2. A random error of ± 0.004 m is estimated for each of 22 length measurements (which are to beadded together to get the total length).what is the estimated total error?what is the probable total error if the random error is ± 0.006 m per length?3. A surveying crew or party is capable of taping distances with an estimated probable error of ± 0.005for each 30.000 m distance. What estimated total error should be expected if a distance of 1400 mis measured?4.It is desired to tape a distance of 360 m with a total estimated probable error of not more than ±0.006 m.a. How accurately should 30.000 m distance be measured so that the permissible value is notexceeded?b. How accurately should each 30.000 m distance have be measured so that the estimated

Note : Question 2 is circled, so solving only circled question. - Calculate the estimated total…

2. A random error of ± 0.004 m is estimated for each of 22 length measurements (which are to be added together to get the total length). a. what is the estimated total error? b. what is the probable total error if the random error is ± 0.006 m per length?

2. A random error of ± 0.004 m is estimated for each of 22 length measurements (which are to be added together to get the total length). a. what is the estimated total error? b. what is the probable total error if the random error is ± 0.006 m per length?

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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PLANE SURVEYING SU1200 ASSIGNMENT 1 1. The sides of a closed figure have been measured and found to have a total length of 1018.910 m. If the error in the measurement is estimated to be 0.073 m, what is the accuracy ratio of the work? 2. A random error of ± 0.004 m is estimated for each of 22 length measurements (which are to be added together to get the total length). a. what is the estimated total error? b. what is the probable total error if the random error is ± 0.006 m per length? 3. A surveying crew or party is capable of taping distances with an estimated probable error of ± 0.005 for each 30.000 m distance. What estimated total error should be expected if a distance of 1400 m is measured? 4. It is desired to tape a distance of 360 m with a total estimated probable error of not more than ± 0.006 m. a. How accurately should 30.000 m distance be measured so that the permissible value is not exceeded? b. How accurately should each 30.000 m distance have be measured so that the…
can you pls help me understand how this is solve.. Topic: Tape corrections (fundamentals surveying) •Pls answer with Complete Solution • 3 decimal places ALL answer 2. A 50m tape of standardized length has a weight of 0.05kg/m, with a cross sectional area of 0.04 sq.cm. It has an E of 2.10x10^6 kg/sq.cm. The tape is of standardized length under a pull of 5.5kg when supported throughout its length and a temp. of 20degC. This tape was to measure a distance between A and B and was recorded to be 458.65m. At the time of measurement, the pull applied was 8kg with the tape supported only at its end points and the temp. observed was 18degC. Assuming a (in greek alpha) of the tape is 0.0000116/degC. Compute the true length of the measured line AB due to combined effects of temperature, tension and sag. Thank you. I'll give your answer a LIKE for helping me.
1. A line is measured on a windy day as 338.65 m. The same line measured 338.37 m on a calm day. If the latter measurement is given four times the reliability of the first, determine the most probable value of the measured line 2. In this problem the weight an angle is assumed to be proportional to the number of times it has been measured by repetition. Five angles in a pentagon were measured with the of following results: 134°44’35’’, 167°02'05'’, 86°15’20’’, 75°48’50’’, and 76°08’50’’. If the number of repetitions for each measurement and were 2, 6, 6, 8, and 4, respectively, determine the adjusted values of the angles.
1. The surveying team at UTM can make taping measurements with an error of ±0.03 ft per 100-ft tape length. What is the error associated with taped distance of 700 ft? 2. Since the mean is computed from individual observed values, each of which contains an error, the mean is also subject to error. We can assume that the error or the mean is =Em= Eseries/n u What will be the 90% error of the mean for a 25 observations having a standard deviation of +0.0495 ft 3. Determine the weighted mean for the following angles 88° 54'21", wt 3, 88° 54'55", wt 5, 88° 54'11", wt 6, 88° 53'58", wt 4,
Hello, please solve the problem attached image. Thank you! Using 4 decimal places at final answer. Hint: Ah=33.5...kn, Av=150kn,Bh=-33.5...kn.
SURVEYING A 30 m tape is 1.2 cm wide and 0.080 cm thick. If the tape is correct under a pull of 54 N. Compute the error made by using a pull of 66 N. E = 2.1 GPa.
situation 4] In the two-peg test method performed by a group of surveying students, the following observations were taken: Rod reading Instrument Instrument set up near set up near K 1.851 On A On B 1.318 1.741 1.048 Point J is equidistant from A and B, and K is 12.14 m from A and 74.25 m from B. The difference in elevation between A and B. The correct rod reading on B with the instrument set up at K. The correct rod reading on A with the instrument set up at K. iii.
1. Calculate the linear error of closure. 2. Calculate the relative precision.
Plugging when performing the longitudinal section leveling of a road, the readings were recorded according to the figure below. Note that the real level of the first leveling marker (BMI) is 70.00 meters and the second leveling figure (BM2) is 67.85 meters. Complete the table and verify the correctness of the calculations and the correctness of work (with adjusting the levels (). R is the ruler reading (7 degrees)) R= 2.00 R= 1.30 R= 1.60 R= 1.90 R= 2.02 R= 2,50 0+00 1+00 2+00 3+00 4+00 5-00 R= 0.50 R= 2.90 R= 1.10 R= 2.30 R= 2.40 R= 1.75 В.М.1 B.M.2 Elevation = 67.85 m Elevation = 70.00 m Т.Р.2 L3 T.P.1 L2 L1
With your pencil, make a dot on the position of your best visual estimate of the centroid of the triangular area. 17.0 13.6 10.2 6.8 3.4 0 0 3.4 (x, y) = (i 6.8 10.2 13.6 Now refer to Sample Problem 5/2 and to Table D/3 and find the exact location of the centroid. 17.0 i )
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In the figure, the width of a river being observed is 80m, assuming the transit being used has a K value of 99 and C value of 0.20. Determine the distance DL and DR , and stadia intercept of DL. (Assume stadia intercept of DR is 1.25) Topic: Tacheometric Surveying Pls answer in COMPLETE solution. DIAGRAM/ILLUSTRATION of the solution is a must, for better understanding. 3 dec places all answers.