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. o.,...�.� <br /> s.v,u„�s� Job#� <br /> -r.�..�.� <br /> P.eoo.e.... <br /> Universi of Minnesota Mound Design Worksheet Site B <br /> Gre�than 1%Slopes <br /> a �ow <br /> Estimated 750 9Pd(see figure A-1) <br /> a�� x 1.5(safety factor)= 0 gpd <br /> B. SEPTIC TANK LIQtJID VOLUMES <br /> ��� 2�p gallons(see figure Gi) <br /> Nurt�r d tanksloompartn'�ents � <br /> Etfluent F�r (Yesha) Y� <br /> G1 Septic Tnik C�in Gallons <br /> Nurr�r of Minimum Capacity with Capaaty with <br /> gedropms Capacity Garb.Disp. Disp.and l.ift <br /> Of 12S3 .1 � �' '�`- <br /> ?J^r �1 zF ,.��y�:: <br /> 3 or 4 ;�' � � 150U ��;� �'��, <br /> 5 or 6 � �� 2250 _�°�'�� '��.< <br /> � �� � <br /> 7,8 or 9 +�. { * 3000 ��:' �� <br /> C. SOILS(Sife sv�uatror►data) <br /> 1. Depih to resUic6ng layer- 1.6 feet <br /> 2. Depth of pe�daUion tests= 12 ir�hes <br /> 3. Texlure � <br /> 4. Sal IoadN►g rabe(see Figws D-33) 0.60 �� <br /> p���� 7 MPI <br /> 5. 96 l.and Sbpe 9.0 96 <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply avera�ge design flaw(A)by 0.83 to obtain required a�ea�rock layer:Item A x 0.83= <br /> 750 gpd x 0.83 it�/gpd= 630 (t� <br /> 2. Oe�er►rune rodc layer width =0.83 ft`Igpd x Linear Loading Rate(LLR)(see LLR ch <br /> 0.83 ft2lgpd x 12.00 = 10.0 ft <br /> LLR Chart <br /> Perk Rate lLR <br /> <120 MPI <=12 <br /> >=120 MPt <� <br /> 3. Length of rock�ayer=area divided by width= <br /> 630.0 ft! 10.0 feet= 63.0 ft <br /> E. ROCK VOLUME <br /> 1. Muitiply rock area by rodc depth to get cubic feet of rodc <br /> 630.0 X 1.0 ft= 630.0 ft <br /> 2. Divide ft Dy 27 ft/yd3 to get cubic yards <br /> 630.0 ft3 / 27 = 23.3 yd3 <br /> 3. Muttiply 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 />