Compute the corrected distance and its error with data given with attached file. Given the following tape calibration data: C=(I - LC,. = k(T —T)L 2 T 3 L C =w P, —P) A c E 24P2 A = 0.006 in2, 1 =...

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Compute the corrected distance and its error with data given with attached file.
Given the following tape calibration data:
C=(I -
LC,. = k(T —T)L
2 T 3 L C =w P, —P) A c E 24P2
A = 0.006 in2, 1 = 99.994 ft., l'= 100 ft., w= 0.02 lb. k= 0.00000645/°F, E = 29,000,000 lb/in2, P = 10 lb, T = 68°F
Compute the corrected distance and its expected error if the measured distance is 145.67 ft. Assume that Tf = 50°F ± 5°F, Pf= 20 lb ± 4 lb, a reading error of 0.005 ft., and that the distance was measured as two end-support distances of 80.00 ft. and 65.67 ft.


Answered Same DayDec 22, 2021

Answer To: Compute the corrected distance and its error with data given with attached file. Given the following...

Robert answered on Dec 22 2021
124 Votes
Solution
Correction for length can be calculated as
CL = 0.006 ft x 145.67 ft / 100ft = -0.008
7 ft ( Corrected Distance)
Error in Length Measurement (By std deviation Method) = √(


)

(


)

(


)
= √( ) ( ) (

)
On...
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