Taping Corrections Incorrect

length Slope Temperature Sag Stretch

Incorrect Length Sources of Error

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

v

hs

vC

hshshsv

hsC

2

))((22

222

s

h

v

Slope Example If s = 300.00’

= 5° 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’

s

h

v

Temperature

))((0000065.

/0000116.

/0000065.

))((

LTTC

C

Fk

LTTkC

st

st

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

2

2

2

32

2424 P

LW

P

LwCs

Sag and Tension

2

2

2

32

2424 P

LW

P

LwCs

psiEAE

LPPC

Steel

sp

000,000,29

)(

sPP

AEWP

204.0

If P = 18-lb, PS = 12-lb, L = 100’, A = 0.015 in2,

CP = (18–12)100/(0.015*29,000,000) = 0.0014’

W = 2.8, A = 0.015, PS = 12 Trial and error -> P = 31 lb

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