A line of three-wire differential levels goes from benchmark Gloria to benchmark Carey. The length of the line was determined to be 2199 m. The instrument had a stated compensator accuracy of ±0.3″. The instrument-rod combination had an estimated reading error of ±1 mm per km. The instrument is found have a inclined collimation error of 1.0 mm per km so the sight distances were kept to approximately 50 ± 5 m. If the observed difference in elevation is 15.601 m, what is the estimated error in the final elevation difference?

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Asked Dec 2, 2019
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A line of three-wire differential levels goes from benchmark Gloria to benchmark Carey. The length of the line was determined to be 2199 m. The instrument had a stated compensator accuracy of ±0.3″. The instrument-rod combination had an estimated reading error of ±1 mm per km. The instrument is found have a inclined collimation error of 1.0 mm per km so the sight distances were kept to approximately 50 ± 5 m. If the observed difference in elevation is 15.601 m, what is the estimated error in the final elevation difference?

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Expert Answer

Step 1

Given information:

D 50 m.o ±5
length of line 2199 m
0.001
a 1.0 mm per km
106
o0.3 0.3x4.848 x 10 radian
D 1mm/km
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D 50 m.o ±5 length of line 2199 m 0.001 a 1.0 mm per km 106 o0.3 0.3x4.848 x 10 radian D 1mm/km

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Step 2

Number of instrument setups

2199
N =
2x50
21.99
=22
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2199 N = 2x50 21.99 =22

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Step 3

formula us...

CR(D)
2ND2 (o ,D+2Noa+
500, 000
Where
Total estimated error in an elevation difference
N = Number of instrument setups
D Sight distance
ODEstimated error in a rod reading
oEstimated instrument mis levelment
Op-Estimated standard errors in the sight distances
a Actual collimation error
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CR(D) 2ND2 (o ,D+2Noa+ 500, 000 Where Total estimated error in an elevation difference N = Number of instrument setups D Sight distance ODEstimated error in a rod reading oEstimated instrument mis levelment Op-Estimated standard errors in the sight distances a Actual collimation error

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