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a� Uusboe rockbed Job#� <br /> Tw�w�sNr <br /> 'Pwoorew�r <br /> Universi of Minnesota Mound Design Worksheet <br /> Greater than 196 Slopes <br /> a F�ow <br /> Estim�ed � 900 9Pd(see figure A-1) <br /> � or measured I � x 1.5(safety factor)_ � 9Pd ; <br /> B. SEP'TIC TANK LIQUID VOLUMES <br /> Se�C tank c�ac�ty 225p galtons(see figure G1) <br /> Number of tanks/compartments 0 <br /> Eifluent Flter (Yes/no) Yes <br /> C-1 Septic Tank Capacily in Galbns <br /> Number of Minimum Capaat�r with C�aaty with <br /> Bedrooms C�aaty Garb.Disp. Disp.and Lift <br /> 2 or 1125 <br /> 3 or 4 1500 <br /> 5 or 6 2250 <br /> 7,8 or 9 3000 } <br /> C. SOILS(Site evalu�ia►data) <br /> 1. De�h to r�eshic�ng laye�= 1.8 feet <br /> 2. Depth of peroola6on tests= 12 inches <br /> 3. Textune bam <br /> 4. Shc bading r�e(see F'�gure D-33) 0.60 9Pd�� <br /> Peroda6on rate 14 MPI <br /> 5. 96 Land Sbpe 5.0 96 <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flow(A)by 0.83 to obtain required area of rock layer:Item A x 0.83= <br /> 900 gpd x 0.83 ft�/gpd= 390 ft2 <br /> 2. Detertmne rock layer width =0.83 ft`/gpd x Linear Loading Rate(LLR)(see LLR chart <br /> 0.83 ft/gpd x 12.00 = 10.0 ft <br /> w�cnart <br /> P�k Rate LLR <br /> <120 MPI <=12 <br /> >=120 MPI <� <br /> 3. Length d rock layer=area divided by width= <br /> 390.0 ft! 10.0 feet= 39.0 ft <br /> E. ROCK VOLUME <br /> 1. Muitiply rodc area by rock depth to get cubic feet of rock <br /> 390.0 X 1.0 ft= 390.0 ft3 <br /> 2. Divide ft3 by 27 ft3lyd3to get cubic yards <br /> 390.0 ft3 1 27 = 14.4 yd3 <br /> 3. Multiply cubic yards by 1.4 to get weight of rock in tons; <br /> 14.4 yd3 X 1.4 ton/yd3 = 20.2 tons <br /> Page 1 of 5 <br />