ITCM 310 - Plane Surveying - Fall 2015 Assignment 3
Name __________________________Show formulas and work on this sheet.
90% ErrorA distance was taped six times with the following results: 85.67, 86.33, 85.90, 85.95, 86.16, and 85.92 feet. Compute the 90 percent error and give the maximum and minimum range of accuracy for the survey.
Relative AccuracyA distance of 345.75 feet is measured by a survey crew. The true distance is known to be 345.85 feet. What is the relative accuracy of the measurement?
Maximum Error of ClosureWhat is the maximum error of closure in a measurement of 2500 feet if the relative accuracy is 1:5000?
Trigonometric Review Problems
The hypotenuse of a right triangle is 106.32 ft long, and one of the other sides has a length equal to 69.66 ft. Find the angle opposite to the 69.66 ft side.
The three sides of a triangle are 60, 80, and 100 ft. Determine the magnitude of the interior angles.
A surveyor measures an inclined distance and finds it to be 1642.5 ft. The angle between the horizontal plane and the inclined line is measured at 2°56'30". Determine the horizontal distance and the difference in elevation between the two ends of the line.
A section of a road with a constant 3% slope or grade (i.e., 3 ft vertically for every 100 ft horizontally) is to be paved. If the road is 24.000 ft wide and its total horizontal length is 900.0 ft, compute the road area to be paved.
Exploring Key Concepts in Surveying
In this blog post, we will begin working through this assignment and break down two problems to showcase both methodology and conceptual understanding.
Problem:Six distance measurements were taken with the results:85.67, 86.33, 85.90, 85.95, 86.16, and 85.92 feet.
The task requires calculating the 90% error and determining the accuracy range.
Step-by-Step Solution:
To compute the 90% error, we first need to calculate the mean of the measurements:
Where:
The 90% error is typically calculated by multiplying the standard deviation by a factor that corresponds to the 90% confidence level (approximately 1.645 for normal distributions). Once obtained, this value can help in determining the range of accuracy for the measurements.
We can also calculate the maximum and minimum ranges by applying this error margin to the mean distance, yielding a high and low boundary of possible true distances.
Problem:A distance of 345.75 feet was measured, while the known true distance is 345.85 feet. We are tasked with determining the relative accuracy.
Relative accuracy (RA) is a measure of how close a measurement is to the true value, often expressed as a ratio. The formula for relative accuracy is:
In surveying, a higher relative accuracy ratio indicates a more precise measurement. For example, 1:3459 means that for every 3459 feet measured, there is likely to be an error of 1 foot. This level of accuracy can be critical depending on the project's requirements.
By demonstrating such calculations, students gain insights into the precision needed for tasks like error analysis and trigonometric applications. Stay tuned for the continuation, where we will address the remaining problems, including trigonometry and road area computations.
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