Taping Corrections Incorrect length Slope Temperature Sag Stretch

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

Incorrect Length - Determining Your Task: Measure distance A to B You measure 500.00’ 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 Your task: Establish point B exactly 500.00 feet from point A 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 Trigonometry Horizontal: h = s*cos() Calculation s h v a

Slope Example If s = 300.00’  = 5° If you had measured v = 26.15’ h = 300 cos(5) = 298.86’ v = 300 sin(5) = 26.15’ If you had measured v = 26.15’ CS = v2/2S = 26.152/600.00 = 1.14’ h = v – CS = 300.00 – 1.14 = 298.86’

Temperature

Temperature Example Tape calibrated to 100.00’ at 68°F Determine Dist AB = 368.50’ at 22°F Calculate true distance CT = .0000065(22-68)(368.50) = -0.11’ True Dist AB = 368.50 - 0.11 = 368.39’

Sag Example: 2.8-lb chain calibrated at 100.00’ when supported throughout Established B 348.75’ from A 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’ 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 in2, CP = (18–12)100/(0.015*29,000,000) = 0.0014’

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