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, <br /> Otto Associates, Engineers and Land Surveyors, Inc. <br /> Mound Design Worksheet f or flows up to 1200 gpd) <br /> All boxed rectangles must be entered,the rest will'be calculated. <br /> A. FLOW 141:Esfinaled Smage Rows in Calors per Day <br /> Estimated 750 1 gpd(see figure A-1) rimmot <br /> or measured 0 x 1.5(safety factor)= 0 gpd bedwm Class I Cku 0 C Ims 01 ckw IV <br /> 2 300 225 180 60% <br /> B. SEPTIC TANK LIQUID VOLUMES 3 450 300 218 ofihe <br /> Septic tank capacity 2-1250 gallons(see figure C-1) 4 600 375 256 vatres <br /> 5 750 450 244 inihe <br /> C. SOILS(Site evaluation data) 6 9M 525 332 Grist <br /> 1. Depth to restricting layer- 1.67 ]feet 7 1050 600 310 I,orW <br /> 2. Depth of percolation tests= 12 inches <br /> 8 1 1200 675 408 1 cohurrra <br /> 3. Texture I Clay Loam <br /> 4, Soil loading rate(see Figure D-33) 0.45 gpd/fe <br /> Percolation rate 60 MPI <br /> 5. %Land Slope 7 % n-3& Aeaorptim WNMslzWg'ftW <br /> Pxam Lolam xme <br /> C-1; a Tank C itle$fin ort z sones p day <br /> � � <br /> 1 tY Caqwdkcapacity' aa>a�.,mms c mmsmw 1.20 1.00 <br /> Number d Mnimwn-liquid U capacity with ,with di asst& Mediamsand <br /> Bedrooms capacity garbattdisposal h io om �' <br /> 2 or lest 750 1125 16'°3° 60 <br /> 1 71 a Loam SO 4 <br /> 5 ori 1000 1500 2000 �m eo 0.43 3Ar <br /> 5 or 6 1500 2250 3000 a�Lo.m <br /> ?,8 or 9 20DQ 3040 61�izo suty 5-9'12.1 <br /> os4 sea <br /> i <br /> : daolgaee flOrltrasmn.mut d amerarprtonaam. <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flow(A)by 1.0 to obtaih required area of rock layer.Item A x 1.0= <br /> 750 . gpd x 1.0 ft-gpo= 750.0 fe <br /> 2. Determine rock layer width =1.0 fe/gpd x Line: r Loafing Rate(LLR)(see LLR chart) <br /> 1 fe/gpd, X 10 10.0 ft <br /> LLVhart <br /> Peat R to LLR. r. <br /> <120 MPI <=12 <br /> >=120 MPI <=6 <br /> 3. Length of rock layer=area divided by width= <br /> 750 fe / 10 feet= 75.0 feet <br /> E. ROCK VOLUME <br /> 1. Multiply rock area by rock depth to get cubic feet of rock <br /> 750 X 1 ft= 750.0 ft' <br /> 2. Divide ft3 by 27 fe/yd'to got <br /> cubic yards <br /> 750.0 ft3 / 27 = 27.8 yd,3 <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> 27.8 yd' X 1.4 ton/yd' = 38.9 tons <br /> P <br />