Electronic Distance Measuring Equipments - Problem 1

A distance of 1000ft. 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. + [(1000ft. \times \frac{12in.}{1ft.} \times \frac{25.40mm.}{1in.}) \times (5\times 10^{-6}) ] \end{equation}

\begin{equation} e = \pm 6.524mm. \times \frac{1in.}{25.40mm.} \times \frac{1ft.}{12in.} \end{equation}

\begin{equation} e = \pm 0.021ft. \end{equation}

Answer: \begin{equation} e = \pm 0.021ft. \end{equation}