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-a��r�w,.s � rodcbed#2 Job#� <br /> Tw�wTrn�rr <br /> r�eosww�r <br /> Universi of Minnesota Mound Design Worksheet <br /> Greater than 196 Slopes <br /> a F�ow <br /> Estimated ; 750 9pd(see figure A-1),' <br /> or measured I x 1.5(safety factof)= 0 9Pd <br /> B. SEPTIC TANK LIQUID VOLUMES <br /> ��k��y �p gallons(see figure G1) <br /> Number of tanksloompartments 0 <br /> Effluent FRer (Yes/no) ye.s <br /> C-1 Septic Tank Cap�dty in Galbns <br /> Number of Minimum Capacity with Capaaty with <br /> Bedroort�s Capaaty Garb.Disp. Disp.and Lift <br /> 2 or less 11 5 <br /> 3 a 4 1500 <br /> 5 or 6 2250 <br /> 7,8 or 9 3000 <br /> C. SOILS(Site evaluatior►data) <br /> 1. Depth to reshiding layer- 1.5 feet <br /> 2. D�th of peroolation tests= 12 inches <br /> 3. Texture clay loam <br /> 4. Sal lo�ng rate(see Figure D-33) 0.45 9Pd�� <br /> Pe�ol�ion rate 6 MPI <br /> 5. %Land Sbpe 4.0 % <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flow(A)by 0.83 to�tain required area of rodc layer:Item A x 0.83= <br /> 750 gpd x 0.83 ft/gpd= 630 ft <br /> 2. Detertnine rodc layer width =0.83 ft`/gpd x Linear Loading Rate(LLR)(see LLR chart <br /> 0.83 fl�/gpd x 12.00 = 10.0 ft <br /> LLR Chart <br /> Perk Rate LLR <br /> <120 MPI <=12 <br /> >=120 MPI <=6 <br /> 3. Length of rodc layer=area divided by width= <br /> 630.0 ft2 / 10.0 feet= 32.0 ft <br /> E. ROCK VOLUME <br /> 1. Multiply rodc area by rodc depth to get cubic feet of rodc <br /> 630.0 X 1.0 ft= 630.0 ft3 <br /> 2. Divide ft3 by 27 ft3/yd3 to get cubic yards <br /> 630.0 ft3 I 27 = 23.3 yd3 <br /> 3. Mulfiply cubic yards by 1.4 to get weight of rock in tons; <br /> 23.3 yd3 X 1.4 ton/yd3 = 32.7 tons <br /> Page 1 of 5 <br />