1 / 11

Taping Corrections

Taping Corrections. Incorrect length Slope Temperature Sag Stretch. Incorrect Length. Sources of Error Bad repair Poor standardization C L = (L true -L) Determining - add correction Establishing - subtract the correction. Incorrect Length - Determining.

leiko
Download Presentation

Taping Corrections

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Taping Corrections • Incorrect length • Slope • Temperature • Sag • Stretch

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

  3. 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’

  4. 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’

  5. s v a h Slope • Trigonometry Horizontal: h = s*cos() • Calculation

  6. s v a h 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’

  7. Temperature

  8. 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’

  9. 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

  10. 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’

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

More Related