Exam Notes - Tape Corrections

Corrections in Taping

• Corrections to consider:

  • Temperature

  • Pull or tension

  • Sag

  • Slope

• Rules:

  Measuring

  • Tape too long: Subtract correction

  • Tape too short: Add correction

    Laying out

  • Tape too long: Add Correction

  • Tape too short: Subtract Correction

• 
  

Correction Due to Temperature

• Formula: $CT = αL(T - Ts)$

  • Where:

    • $α$ is the coefficient of thermal expansion (11.6 x $10^{-6}$ per degree Celsius for steel)

Correction Due to Pull

• Formula: $CP = \frac{(P - P_0)L}{AE}$

  • Where:

    • $P$ is the applied pull

    • $P_0$ is the standard pull

    • $A$ is the cross-sectional area

    • $E$ is the modulus of elasticity

• Negative correction if initial lead is subtracted.

• Ensure units are consistent (e.g., kg/m).

Corrected Length

• Formula: Corrected Length = Measured Length + Total Correction

• Total Correction = Sum of all corrections (temperature, pull, sag, slope)

Correction Due to Sag

• Formula: $CS = \frac{W^2 L^3}{24P^2}$

  • Where:

    • $W$ is the weight of the tape per unit length

    • $L$ is the length of the tape

    • $P$ is the applied tension

Compute True Distance of Line

• $L<em>{corrected} = L</em>{measured} + Total Correction$

• $Total Correction= \frac{CT}{L<em>{tape}} * L</em>{measured}$

• Where:

  • $CT$ is the length of tape

  • $L_{tape}$ is the total tape

  • $L_{measured}$ is the measured line

Laying Out/Layout

• Measurements from plans to execute on site.

Correction Due to Slope

• Formula: $CH = \frac{h^2}{2S}$

  • Where:

    • $h$ is the vertical distance (height difference)

    • $S$ is the slope distance

• $D = S - CH$ (Horizontal Distance = Slope Distance - Slope Correction)

Okay, here's a sample problem demonstrating the application of these surveying corrections:

Problem:
A steel tape is used to measure a distance. The tape is 30m long at 20°C and has a cross-sectional area of $5 * 10^{-6} m^2$. The modulus of elasticity (E) of the steel is $200 * 10^9 N/m^2$. During measurement, the tape is supported at both ends with an applied tension of 50N. The measured length is 25m, the average temperature during the measurement is 28°C, and the tape sags between supports. The tape weighs 0.02 N/m. Also, there is a vertical height difference of 1m over measured 25m. Calculate the corrected length of the line.

Solution:

1. Correction Due to Temperature

   • Formula: $CT = αL(T - Ts)$

   • Where: $α = 11.6 * 10^{-6} /°C$, $L = 30m$, $T = 28°C$, $Ts = 20°C$

   • $CT = (11.6 * 10^{-6})(30)(28 - 20) = 0.002784 m$

2. Correction Due to Pull

   • Formula: $CP = \frac{(P - P_0)L}{AE}$

   • Where: $P = 50N$, Assume $P_0 = 20N$ (standard pull), $L = 30m$, $A = 5 * 10^{-6} m^2$, $E = 200 * 10^9 N/m^2$

   • $CP = \frac{(50 - 20)(30)}{(5 * 10^{-6})(200 * 10^9)} = 0.0009 m$

3. Correction Due to Sag

   • Formula: $CS = \frac{W^2 L^3}{24P^2}$

   • Where: $W = 0.02 N/m$, $L = 30 m$, $P = 50 N$

   • $CS = \frac{(0.02)^2 (30)^3}{24(50)^2} = 0.00009 m$

4. Correction Due to Slope

   • Formula: $CH = \frac{h^2}{2S}$

   • Where: $h = 1 m$, $S = 25 m$

   • $CH = \frac{(1)^2}{2(25)} = 0.02 m$