ure 6.18). If the absorption field is to be used for irrigation and, by necessity, is larger than required for disposal only, the distance between the trenches should be increased. The total trench length is found by dividing the total absorption area by 5 ft. For example, a two-bedroom house would have a 300 gal per day design flow. In sandy clay loam having a percolation rate of 50 min per in., the application rate would be 0.25 gal per day per square foot (from Figure 6.17). Therefore, the total absorption area needed 300 gpd 0.25 gpd/ft = 1200 ft The distribution pipe and trench must have a total length of 1200 ft2 5 ft = 240 ft CONVENTIONAL SYSTEM MODIFICATIONS One foot contour line Absorption field area Dosing chamber Septic tank Figure 6.18 Shallow subsurface plot and distribution system. The number of distribution laterals needed is found by the following formula: laterals needed = field width (ft) 5 ft In our example, the number of laterals needed if we make our plot 30 ft wide and 40 ft long (1200 ft2 is 30 ft 5 ft = 6 Six laterals, each 40 ft long, is a total lateral length of 240 ft. Surprise! Remember that because of friction losses the far end of any lateral should be no more than 70 ft from the manifold. There- fore the maximum length for a trench is 140 ft and then only if the manifold intersects the lateral midway between the two ends. Trench width is normally 4 to 6 in. and trench depth is normally 18 146 ONSITE WASTEWATER DISPOSAL in. If there is a potential for freezing below 18 in. in the winter and the system is to be used year-round, the trenches must extend below frost-line depth. Trenches can be as shallow as 6 in. into natural soil if 1 ft of imported fill is placed in the absorption field area, and the trenches dug through the fill and 6 in. into the natural soil below. As shown in Figure 6.19, there must be a minimum of 3 in. of gravel below the distribution laterals, and preferably 6 in. Above the lateral, there must be a minimum of 2 in. of gravel, and preferably 4 Next, the hydrauli