to figure the dose as 60% to 75% of the volume of the manifold and distribution pipe. This rule works best for 18-in. to 2-ft wide trenches. Example 1 For example, for a two-bedroom house where: (a) silty clay loam has a percolation rate of 60 min per in.; (b) the drain field area will be 744 ft2 (see the next section for method of calculation); (c) with 2-ft wide trenches, the distribution pipe will be 744 2 = 372 ft long; (d) the distribution pipe volume will be (4 in.)2 X 0.0408* X 372 ft = 243 gal; (e) the 4-in. manifold pipe will be 20 ft long and have a volume of (4 in.)2 X 0.0408 X 20 = 13 gal; WASTE PROCESS (f) the whole distribution system will have a volume of 243 gal + 13 gal = 256 gal; (g) 75% of 256 gal = 192 gal per dose; then (h) the (maximum) dosing frequency will be 300 gpd 192 gal per dose = 1.6 per day. *The volume in gallons of a pipe is the square of the pipe's inside diameter in inches times the constant 0.0408, times the pipe's length in feet. See the appendix of this book for this and other formulas. Example 2 For a two-bedroom house where: (a) sand has a percolation rate of 10 min per in.; (b) the drain field will be 304 ft2; (c) with 2-ft wide distribution trenches, the distribution pipe will be 304 2 = 152 ft long; (d) the volume of the distribution pipe will be 42 X 0.0408 X 152 = 99 gal; (e) the manifold will have a 13-gal volume as above; the total distribution system will have a volume of 99 gal + 13 gal = 112 gal; (g) 60% of 112 gal = 67 gal per dose; and (h) the (maximum) dosing frequency will be 300 gpd 67 gal per dose = 4.5 doses per day. Note that we used 60% of the pipe volume for the rapidly perco- lating soil and 75% of the pipe volume for the slower percolating soil. The discharge rate per dose is important to effluent distribution. As a rule of thumb, a reasonable minimum rate can be calculated by dividing the dosing volume in gallons by the percolation rate of the soil in minutes per inch. For example, if: (a) dosing volume is 192 gal; and (b) percolation rate is 60 min p