1. Adjustment of Level Nets

2. Introduction In this chapter we will deal with differential leveling only In SUR2101 you learned how to close and adjust a level loop – that is actually a least squares solution In complex situations where an interconnected level net is involved, we need least squares Method of observation equations is most common

3. Leveling Observation Equation The observed elevation difference between stations I and J The ΔElev may be simply a Backsight minus Foresight, or the result of many Backsights and Foresights along a leveling path

4. Unweighted Example

5. Example - Continued

10. Compute Residuals

11. Weighted Level Adjustment Recall that weights are inversely proportional to line length Weights are also inversely proportional to number of setups Weights can also be estimated a priori based on standard deviations of readings, propagated as error of a sum

12. Weighted Level Example Same as previous example, but consider line lengths for weighting

13. Example - Continued

16. Reference Standard Deviation Standard deviation of an observation of unit weight

17. Unweighted Example – S 0

18. Weighted Example – S 0

19. Another Weighted Adjustment From:To:ΔE (m)σ (m) AB10.5090.006 BC5.3600.004 CD-8.5230.005 DA-7.3480.003 BD-3.1670.004 AC15.8810.012 Benchmark A has an elevation of 437.596 m. What are the most probable values for B, C, D?

20. Observation Equations +B= A + 10.509 + v 1 = 448.105 + v 1 -B +C= 5.360 + v 2 -C +D= -8.523 + v 3 -D= -A – 7.348 + v 4 = -444.944 + v 4 -B +D= -3.167 + v 5 +C= A + 15.881 + v 6 = 453.477 + v 6 See spreadsheet for solution.

21. Units for Standard Deviation of Unit Weight Unweighted (all unit weight), first example: feet Weighted by line length, second example: Weighted by standard deviation, third example: Unitless

22. S 0 – Weights by A-Priori σ When weights are based on a priori standard deviations, the computed reference variance should be the input value (typically 1) We can do a Chi-square test to see if it is significantly different from 1 If a Chi-square test fails, it may be due to one or more blunders or incorrect a priori standard deviations Blunders can cause the reference variance to be much greater than 1

23. Chi-Square Test for Example H 0 : σ 2 = 1 H a : σ 2 ≠ 1(two-tail test at 0.05 significance) Degrees of freedom = 6-3 = 3 χ 2 0.975,3 = 0.216 0.65 2 = 0.42 0.42 > 0.216, therefore do not reject the null hypothesis In other words, we have no statistical evidence that the a priori standard deviations were incorrect.

24. Reference Variance Test We can make the reference variance equal 1 by adjusting the a priori standard deviations Usually, if the computed value is < 1, nothing further is done If the value fails the Chi-Square test by being much greater than 1, the first thing to do is look for blunders Large residuals often indicate blunders