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� r <br /> Otto Associates, Engineers and Land Surveyors, Inc. <br /> Mound Design Worksheet (For flows up to 1200 gpd) <br /> All boxed rectangles must be entered,the rest will be calculated. <br /> A. FLOW AI: <br /> ls�rrated Setaoge R=In galans PB Do! <br /> Estimated 750 . gpd(see figure A-1) <br /> or measured 0 x 1.5(safety factor)= 0 gpd bedam Choi Clnss 1 Odes til Odes N2 300 225 180 60% <br /> B. SEPTIC TANK LIQUID VOLUMES 3 4% 3M 218 of$re <br /> Septic tank capacity 2-1250 gallons(see figure C-1) 4 600 375 256 <br /> vdms <br /> 5 730 450 244 Inihe <br /> C. SOILS(Site evaluation data) 6 900 525 332 Ckal, <br /> 1. Depth to restricting layer- 1.5 feet 7 1050 600 370 I,drll <br /> 2. Depth of percolation tests= 12 inches <br /> B 1200 1 675 408 cd urrrs. <br /> 3. Texture I Clay Loam <br /> 4. Soil loading rate(see Figure D-33) 0.6 gpd/fe <br /> Percolation rate 30 JMPI <br /> 5. %Land Slope 5 % n-M Ah0n7 lon wsam Sizing 9latue <br /> pPaaolnllaaReaie Lo�tmrAoe <br /> In Mi.pa San Testme Gelfeus <br /> C-i; c rang C.a �(� 61W) pen <br /> } � � � <br /> 4CI 1?OW*Ws CWW s..d 1.20 r oa <br /> Number of Minimum Liquid 0*id c 9j4L ty jaith & MedlmnSaad <br /> sjBeftom- Capacity liffh, a dispo§ I lifliaside ysa <br /> 6' 90 I.om 60 <br /> 2 0r 1CSS 740 j I, t 1:►0p W 4s Oso 40 <br /> 3 ar 4 1000 15002pOp ;WW-so M Mt 0.45 2.0 <br /> Sorb 1500 M% 300p I z <br /> 7'$ 9 2DDO 3000 61 w 120 =7 C- 024 4.00 <br /> t -SyEw wem+�soumnt6�o0rorpuramam <br /> " D. ROCK LAYER DIMENSIONS <br /> 1. Multiply average design flow(A)by 1.0 to obtai�r requii*area p rock layer.Item A x 1.0= <br /> 750 gpd x 1.0 ft"gp� f 750.0 fe <br /> 2. Determine rock layer width =1.0 ftz/gpd x Liner Loading Rate'(LLR)(sem LLR chart) <br /> 1 fe/gpd. X j 10 = <10.0 ft <br /> LLR Ckart A. <br /> w, 1 <br /> Perk Rate <br /> <10 MPI <br /> <=12'' <br /> >=120 MPI <- ' . <br /> 3 .Length of rock layer=area•ivided by width= ., <br /> 750" / ' .. :10 ''v i ,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 ft3 <br /> 2. Divide 1`3 by 27 fe/yd3 to get-cubic yards <br /> 750.0 ft3 / 27 ;, 27.8 Yd' <br /> :1 <br /> 3. Multiply cubic yards by 1.4 to get weight of"in tons; <br /> 27.8 yds X 1:4 ton/yd3 _ 38.9 tons <br />