Question

a drain has a run of 52 ft at a grade of 1/8 in/in. the high end of the drain has an elevation of 126.70 ft. what is the elevation at the low end?

Explanation:

Step1: Recall grade formula

Grade = $\frac{\text{Vertical change}}{\text{Horizontal change}}$. Here, the horizontal change is the run of the drain which is 52 ft. The grade is $\frac{1}{8}$ in/in. First convert the grade to ft/ft. Since 1 ft = 12 in, $\frac{1}{8}$ in/in = $\frac{1/8}{12}$ ft/ft=$\frac{1}{96}$ ft/ft.

Step2: Calculate vertical change

Let the vertical change be $h$. Using the grade formula $\text{Grade}=\frac{h}{\text{Horizontal change}}$, we have $\frac{1}{96}=\frac{h}{52}$. Solving for $h$, we get $h = \frac{52}{96}\text{ ft}\approx0.5417\text{ ft}$.

Step3: Find low - end elevation

The high - end elevation is 126.70 ft. The low - end elevation $E_{low}=E_{high}-h$. So $E_{low}=126.70 - 0.5417=126.1583\approx126.16\text{ ft}$.

Answer:

126.16 ft