Solution: Least Squares 5
Problem (1)
Determine the best-fit straight line equation for the data in the following table:
| X, m | Y, m |
| 10 | 300 |
| 14 | 190 |
| 20 | 100 |
Solution
Using observation equations least squares solution.
Equation of a straight line is y = mx + b
The observation equations for the three points are:
| 300 + v1 = 10m + b | |
| 190+ v2 = 14m + b | |
| 100 + v3 = 20m + b |
Initial matrices:
Intermediate
Solution
Equation: Y = -19.605X + 484.210
Problem (2)
The shape of a freely hanging powerline between supports is a complex curve called a catenary. The catenary equation uses hyperbolic trigonometery to define the entire curve. A section of the curve can be closely approximated using a parabola, same as a vertical curve.
A powerline crossing a waterway has targets for monitiring purposes at approximately 74 ft spacing along the wire. The utility company wants to know the clearance between the line and the river at its Oridinary High Water Mark elevation of 1896.5 ft.
A surveyor established a baseline from which she was able to make sufficient measurements to determine the target poisitons. Using X for horizontal position and Y for vertical, she assumed X = 200.0 ft and Y = 1000.0 ft for point A. All target coordinates are listed in the table.
| Target | X | Y |
| A | 200.0 | 1000.0 |
| B | 274.3 | 966.7 |
| C | 347.7 | 950.2 |
| D | 422.3 | 949.6 |
| E | 496.5 | 965.6 |
| F | 570.1 | 1000.2 |
Part (a) What is the observation equation for target E?
Solution:
The equation for a parabola is Y = aX2 + bx + c
At point E: 965.6 + vE = a(496.5)2 + b(496.5) + c
Part (b) A least squares adjustment yielded these results:
| a | 0.001531 | |
| b | -1.18053 | |
| c | 1175.1 |
Using the information on the diagram and adjustment results, determine the minimum distance between the wire and river at OHWM.
Solution
Because the wire endpoints are at (essentially) the same elevation, the lowest part will be in the middle.
The coordinate of the midpoint is 200 plus the difference of the X coordinates at A and F, divided by two.
XM =200+(570.1-200)/2 = 385.05
The Y coordinate at M: YM =0.001531(385.05)2 +(-1.18053)(385.05) + 1175.1 = 947.6
The Y coordinate difference from A to M is dY = 947.6 - 1000 = -52.4
To compute M's elevation starting at the tower base elevation
ElevM = 1923.5+75-52.4 = 1946.1
The clearance is the elevation difference between the elevation of M and the OHWM
T = 1946.1-1896.5 = 50.5 ft
Clearance is 50.5 ft.