Electronic Distance Measuring Equipments - Problem 2

A distance of 10ft. was measured by an EDM with ±(5mm. + 5ppm). What is the error in this measurement?

Solution:

\begin{equation} e = \pm A + (B \times C) \end{equation}

\begin{equation} e = \pm 5mm. + [(10ft. \times \frac{12in.}{1ft.} \times \frac {25.40mm.}{1in.}) \times (5 \times 10^{-6})] \end{equation}

\begin{equation} e = \pm 5.015mm. \times \frac{1in.}{25.40mm.} \times \frac{1ft.}{12in.} \end{equation}

\begin{equation} e = \pm 0.016ft. \end{equation}

Answer: \begin{equation} e = \pm 0.016ft. \end{equation}