Question

1. compute the required layout distance under the following conditions.

· temperature: 100 °f
· tape length: 100.04 ft
· design distance: 88.92 ft

Explanation:

1. First, we need to convert the temperature from Fahrenheit to Celsius for the tape - correction formula. The conversion formula is $T_C=\frac{5}{9}(T_F - 32)$.

• Given $T_F = 100^{\circ}F$, then $T_C=\frac{5}{9}(100 - 32)=\frac{5}{9}\times68\approx37.78^{\circ}C$.

1. Assume the standard temperature for the tape is $T_{0}=68^{\circ}F$ (or $T_{0C}=\frac{5}{9}(68 - 32)=20^{\circ}C$), and the coefficient of thermal expansion of the tape $\alpha = 6.5\times10^{-6}/^{\circ}C$.

• The formula for the thermal - correction of the tape length $\Delta l_{t}=\alpha(T - T_{0})L_{0}$, where $L_{0}$ is the nominal length of the tape. Here, $L_{0}=100$ ft (nominal length, since the actual length is $L = 100.04$ ft), $T = 37.78^{\circ}C$, and $T_{0}=20^{\circ}C$.
• $\Delta l_{t}=6.5\times10^{-6}/^{\circ}C\times(37.78 - 20)^{\circ}C\times100$ ft $\approx0.011667$ ft.

1. The actual length of the tape $L = 100+ \Delta l_{t}+ \Delta l_{n}$, where $\Delta l_{n}$ is the nominal - length correction. Here, $\Delta l_{n}=100.04 - 100 = 0.04$ ft. So, $L\approx100+0.011667 + 0.04=100.051667$ ft.
2. The measured distance $l_{m}$ corresponding to the design distance $l_{d}=88.92$ ft can be found using the proportion:

• $\frac{l_{m}}{l_{d}}=\frac{L}{L_{0}}$.
• $l_{m}=\frac{L}{L_{0}}\times l_{d}=\frac{100.051667}{100}\times88.92\approx88.967$ ft.

Step1: Convert temperature to Celsius

$T_C=\frac{5}{9}(100 - 32)\approx37.78^{\circ}C$

Step2: Calculate thermal - correction of tape length

$\Delta l_{t}=6.5\times10^{-6}/^{\circ}C\times(37.78 - 20)^{\circ}C\times100$ ft $\approx0.011667$ ft

Step3: Find actual tape length

$L\approx100+0.011667 + 0.04 = 100.051667$ ft

Step4: Calculate measured distance

$l_{m}=\frac{100.051667}{100}\times88.92\approx88.967$ ft

Answer:

$88.967$ ft