Most Probable Value - surveying material PDF

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ILLUSTRATIVE PROBLEMS (Most Probable Value) 1. A surveying instructor sent out six group of students to measure a distance between two points marked on the ground. The students came up with the following six different values: 250.25, 250.15, 249.90, 251.04, 250.50 and 251.22 meters. Assuming these values are equally reliable and that variations result from accidental errors, determine the most probable value of the distance measured. Solution: mpv or  X

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= (250.25 + 250.15 + 249.90 + 251.04 + 250.50 + 251.22) / 6 = 250.51 m (the most probable value of the distance measured) 2. The angles about a point Q have the following observed values: 130°15’20”, 142°37’30”, and 87°07’40”. Determine the most probable value of each angle.

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Θ3=87°07’40”

Θ1=130°15’20”

    Θ2=142°37’30”

Solution: a.) Determining the Correction to be Applied.

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Sum

= Θ1 + Θ2 + Θ3 = 130°15’20” + 142°37’30” + 87°07’40” = 360°00’30” (sum of the angles observed about point Q)

Disc

= 360° - 360°00’30” = -30”

(discrepancy in the observation)

Note: The sum of the angles observed about point Q is subtracted from 360° (the expected correct sum) to determine the discrepancy.

Correction

= Disc / n = -30” / 3 = -10”

(correction to be subtracted from each observed angle)

b.) Determining the Most Probable Values Θ1 ’

= Θ1 ± Correction = 130°15’20” – 10” = 130°15’10” (most probable value of Θ1)

Θ2 ’

= Θ2 ± Correction = 142°37’30” – 10” = 142°37’20” (most probable value of Θ2)

Θ3 ’

= Θ3 ± Correction = 87°07’40” – 10” = 87°07’30” (most probable value of Θ3)

c.) Solution Check: Θ1’ + Θ3’ + Θ3’ = 360°00’00” 130°15’10” + 142°37’20” + 87°07’30” = 360°00’00” 360°00’00” = 360°00’00”

Note: Since the two quantities are equal, the above solution is assumed to be correct.  

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Reference: Putt,J.P.(2010).ElementarySurveying(3rdEdition).MandaluyongCity:BaguioResearch& PublishingCenter.  ...