Taping Corrections Incorrect length n Slope n Temperature

Taping Corrections Incorrect length n Slope n Temperature n Sag n Stretch n

Incorrect Length n Sources of Error Bad repair n Poor standardization n CL = (Ltrue-L) n Determining - add correction n Establishing - subtract the correction n

Incorrect Length - Determining n n Your Task: Measure distance A to B You measure 500. 00’ n n n Your tape is actually 100. 02’ long Actual length = 500. 10’ CL = (100. 02’ – 100’) = 0. 02’/pull Dtrue = Dmeasured + CL Dtrue = 500. 00 + 5*(. 02’) = 500. 10’

Incorrect Length - Establishing n Your task: Establish point B exactly 500. 00 feet from point A n n Your tape is actually 100. 02’ long B is set 5(100. 02’) = 500. 10’ from A CL = (100. 02’ – 100’) = 0. 02’/pull Correct by subtracting 5(. 02’) = 0. 10’

Slope n Trigonometry Horizontal: h = s*cos( ) n Calculation s h v

Slope Example n If s = 300. 00’ = 5° n n n s h h = 300 cos(5) = 298. 86’ v = 300 sin(5) = 26. 15’ If you had measured v = 26. 15’ n n CS = v 2/2 S = 26. 152/600. 00 = 1. 14’ h = v – CS = 300. 00 – 1. 14 = 298. 86’ v

Temperature

Temperature Example n n n Tape calibrated to 100. 00’ at 68°F Determine Dist AB = 368. 50’ at 22°F Calculate true distance n n CT =. 0000065(22 -68)(368. 50) = -0. 11’ True Dist AB = 368. 50 - 0. 11 = 368. 39’

Sag n Example: n 2. 8 -lb chain calibrated at 100. 00’ when supported throughout n Established B 348. 75’ from A n n n P = 12 lb, Supported at the ends only Measured 3 pulls of 100 ft, one of 48. 75’ CS=-[3(2. 82*100)+0. 0282*48. 753]/(24*122)=-. 71’ n n If P = 18 lb, CS = -0. 31’ (Pull hard!) If pulls are 6 at 50’ plus 48. 75’, P = 12 lb, CS = -0. 20

Sag and Tension W = 2. 8, A = 0. 015, PS = 12 Trial and error -> P = 31 lb If P = 18 -lb, PS = 12 -lb, L = 100’, A = 0. 015 in 2, CP = (18– 12)100/(0. 015*29, 000) = 0. 0014’

Taping Precision 1/2500 - Poor n 1/5000 - Average n 1/10, 000 - Good n