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- A horizontal collimation test is performed on an automatic level. With the instrument setup at point. 1, the rod reading at A was 5,221 ft, and to Bit was 5.533 ft. After moving and leveling the instrument at point 2, the rod reading to A was 4.950 ft and to B was 5.116 ft. The distance between the points in the test was 100 ft. Y What is the collimation error of the instrument? Express your answer to three significant figures and include the appropriate units. 6= Submit Part B HA Value Request Answer What is the corrected reading to A from point 2? Express your answer to four significant figures and include the appropriate units. Correct reading to A Units d μA Value UnitsAn angle was carefully measured 10 times with an optical theodolite by observers A and B on two separate days. The calculated results are as follows: Observer A Mean: 42°16'25" Probable Error: +3.2" Observer B Mean: 42°16'20 Probable Error: +1.6" Compute the most probable value of the angle.This question has 3 parts. Please answer all 3. Thank you! a.) What is the most efficient method to reduce the overall error in the function? b.) A line is observed as 156.58 ft, 156.60 ft, and 156.61 ft. How many redundant observations were observed? c.) Why do surveyors measure angles using both faces of a total station (i.e., direct and reversed)?
- 1) 2 points B & C were observed from point A where the instrument was set up at a height of 1.25m. Points B & C are ahead of point A. The surveyor used an internally focusing telescope with a stadia interval factor of 100. The table below shows the data recorded as observed for points B & C. Answer the following. (Show complete figures and solutions) a) What is the stadia constant of the instrument? b) What is the slope distance between the telescope at point A and the rod reading at point B? c) What is the difference in elevation between point B & C? d) What is the slope distance between the ground surface at point B and the ground surface at C? e) What is the elevation of point C if point A is at an elevation of +0316.15m above mean sea level? - 1 attachment Vertical Stadia Hair Readings ROD/Point Angle Upper Middle Lower В -25°15' 2.316 1.877 1.438 -45°57' 3.333 2.011 0.689An elevation must be established on a benchmark on an island that is 2536.98 ft from the nearest benchmark on the island’s shore. The surveyor decides to use a total station that has a stated distance measuring accuracy of (3 mm +3 ppm) and a vertical compensator accurate to within 0.4”. The height of instrument was 5.37 ft with an estimated error of 0.05 ft. The prism height was 6.00 ft with an estimated error of 0.02 ft. The single zenith angle is read as The estimated errors in instrument and target centering are 0.003 ft. If the elevation of the occupied benchmark is 632.27 ft, what is the corrected benchmark elevation on the island? (Assume that the instrument does not correct for Earth curvature and refraction.) What is the estimated error in the computed benchmark elevation if the instrument has a DIN 18723 stated accuracy of +-3"?An elevation must be established on a benchmark on an island that is 2536.98 ft from the nearest benchmark on the island’s shore. The surveyor decides to use a total station that has a stated distance measuring accuracy of (3 mm +3 ppm) and a vertical compensator accurate to within 0.4”. The height of instrument was 5.37 ft with an estimated error of 0.05 ft. The prism height was 6.00 ft with an estimated error of 0.02 ft. The single zenith angle is read as The estimated errors in instrument and target centering are 0.003 ft. If the elevation of the occupied benchmark is 632.27 ft, what is the corrected benchmark elevation on the island? (Assume that the instrument does not correct for Earth curvature and refraction.) After completing the job, the surveyor discovered that the instrument had a vertical indexing error that caused the sight line to be inclined by 15. (a) How much error would be created in the elevation of the island benchmark if the indexing error were ignored? (b) What…
- Surveying Question: A differential leveling circuit starts at benchmark Concrete Benchmark (Elevation=232.051m) goes out to an unknown benchmark and then continues to a known control point (Elevation=227.975m). Two turning points are used with the measurements included in the attached table. Determine the elevation of the unknown benchmark Determine the order and class of the survey. For First order Class 1: Allowable Error= +/- 4(K)1/2 Range of Constant= m less than/equal to 4 For First Order Class 2: Allowable error= +/- 5(K)1/2 Range of constant= m greater than/equal to 4, and m less than equal to 5 (between 4 and 5) For Second Order Class 1: Allowable error = +/- 6(K)1/2 Range of Constant = m greater than/equal to 5, and m less than equal to 6 (between 5 and 6) For Second Order Class 2: Allowable error = +/- 8(K)1/2 Range of Constant = m greater than/equal to 6, and m less than equal to 8 (between 6 and 8) For Third Order, Any Class: Allowable Error = +/-12(K)1/2 Range…Problem 1 To determine the distance between two points A and B, a tacheometer was set up at P and the following observations were recorded. (a) Staff at A Staff readings = 2.225, 2.605, 2.985 (b) Staff at B Staff readings 1.640, 1.920, 2.200 Vertical angle = -1°46' (c) Horizontal angle APB = +68°32′30" k = 100 m c = 0.00 m Determine the distance ABA horizontal collimation test is performed on an automatic level and the collimation error of the instrument is found to be 0.001 m on the distance between the points of 100 m. Immediately before the test, the instrument was used in a survey where the observed elevation difference between two benchmarks was -13.038 m. The sum of the plus sight distances between the benchmarks was 1100 m and the sum of the minus sight distances was 580 m. Part A What is the corrected elevation difference between the two benchmarks? Express your answer to five significant figures and include the appropriate units. Aelev = Submit O μA - 12.518 m Previous Answers Request Answer X Incorrect; Try Again ?
- Determine: (a) The probable error of the mean in Taping Operation 1. (b) The probable error of the mean in Taping Operation 2. (c) The combined most probable measurement of the frontage, in meters. Answer these questions using this taping scenario. Two surveying tapes were used in measuring the frontage of an open field. The resulting measurements are as follows: Taping Operation 1 Taping Operation 2 300.977m 300.761m 300.415m 300.772m 299.743m 300.776m 299.994m 300.713m3. An angle a is observed using various types of theodolites and observing techniques. The results of these observations and their variances are iT = [121° 32' 14", 121° 33' 9", 121° 32' 46", 121° 32' 50"], C = diagonal [12, 10, 5, 15] (units = arcseconds squared). What is the weighted mean value of the angles? What is the standard deviation of the weighted mean?A distance AB is observed repeatedly using the same equipment and procedures, and the results, in meters are listed. 36.) 65.401 65.402 65.406 65.396 65.405 65.400 65.396 65.401 65.401 65.404 Calculate (a) the line's most probable length, (b) the standard deviation, and (c) the standard deviation of the mean for each set of results.