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a w.wa Downslooe rockbed Job#� <br /> Twsw��rr <br /> �Pwosre�w <br /> Universi of Minnesota Mound Design Worksheet <br /> Greater than 1X Slopes <br /> a F�� <br /> Estimated ! g0p gpd(see figure A-1) <br /> or measured ' x 1.5(safety factor)_ � 0 gpd <br /> B. SEPTIC TANK UQUID VOLUMIES <br /> Septic t�k capaaty 2250 gallons(see figu�Gl) <br /> Number of tankslcortipartrr�ents 0 <br /> Effluent Rlter (yes/no) yes <br /> G1 Sepfk Tank Ca�adt�/in Galbns <br /> Number of Minimum C�acity with Capaaty with <br /> Bedrooms Capacity Garb.Disp. Disp.and Lift <br /> 2 ur less 1125 <br /> 3 or 4 1500 ;� <br /> 5 or 6 2250 <br /> 7,8 or 9 3000 <br /> C. SOILS(Site evelua6on d�a) <br /> 1. D�h to restricting layer= 1.3 feet <br /> 2. Depth of peroolation tests= 12 inches <br /> 3. Textu�e lo�n <br /> 4. Sal loading rate(see F'x�ure D-33) 0.60 gpd�ft� <br /> Per�da6on rate 14 MPI <br /> 5. %Land Slope 5.0 % <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flaw(A)by 0.83 to obt�n required�ea of rodc layer Item A x 0.83= <br /> �0 gpd x 0.83 it�/gpd= 390 ftZ <br /> 2. Determine rock layer width =0.83 ft`/gpd x Unear Loading Rate(LLR)(see LLR chaR <br /> 0.83 tt�lgpd x 12.00 = 10.0 ft <br /> LLR Chart <br /> Pe.rk Rate LLR <br /> <120 MPI <=12 <br /> >=120 MPI <=6 <br /> 3. Length of rodc layer=area divided by width= <br /> 390.0 ft/ 10.0 feet= 39.0 ft <br /> E. ROCK VOLUME <br /> 1. Multiply rodc area by rodc depth to get cubic feet of rock <br /> 390.0 X 1.0 ft= 390.0 ft' <br /> 2. Divide ft3 by 27 ft31yd3 to get cubic yards <br /> 390.0 ft3 / 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 />