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2006-P10164 - septic
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4455 Bayside Road - 06-117-23-21-0002
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2006-P10164 - septic
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Last modified
8/22/2023 5:24:18 PM
Creation date
4/1/2016 2:22:39 PM
Metadata
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Template:
x Address Old
House Number
4455
Street Name
Bayside
Street Type
Road
Address
4455 Bayside Rd
Document Type
Septic
PIN
0611723210002
Supplemental fields
ProcessedPID
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3. TRENCH OR BED BOTTOM AREA , <br /> H. For trenches with 6 inches of rock below the pipe: <br /> A x F= gpd x fUgpd = ftz � <br /> I. For trenches with 12 inches of rock below the pipe: <br /> A x F x 0.8= 600 gpd x 2.2 ftlgpd x 0.8 = 1056.0 ft2 ` <br /> J. For trenches with 18 inches of rock below the pipe: <br /> A x F x 0.66= gpd x fUgpd x 0.66= ft2 � <br /> K. For trenches with 24 inches of rock below the pipe: <br /> A x F x 0.6= gpd x fUgpd x 0.6= ft2 ' <br /> L. For gravity beds with 6 or 12 inches of rock below the pipe; <br /> 1.5xAxF= 1.5x gpdx fUgpd = ft2 � <br /> M. For pressure beds with 6 or 12 inches of rock below the pipe; <br /> A x F= gpd x ff/gpd = ft2 <br /> 4. DISTRIBUTION (Check all that apply) � <br /> Bed(<6%slope) X Drop Boxes(any slope) Rock <br /> X Trenches Distribution Box(<3%) X Chamber <br /> Pressure Gravity Gravelless , <br /> 5. SYSTEM WIDTH,LENGTH AND VOLUME <br /> M. Select width= ft <br /> N. If using rock,divide bottom area b width:(H, I,J or K)divided by P=lineal feet <br /> ft2 � ft = lineal feet ! <br /> Rock depth below distribution pipe plus 0.5 foot times bottom area: <br /> (Rock depth+0.5 foot x Area H, I,J, K, L) <br /> � ft+0.5 ft)x �� ft2 = ft3 , <br /> Volume in cubic yards=volume in cubic feet divided by 27 <br /> / 27= yd3 <br /> Weight of rock in tons=cubic yards times 1.4 � <br /> x 1.4= tons <br /> 0. If using 10"Gravelless Pipe,length=Flow(A)x Gravelless SSF(see figure D-9) ' <br /> gpd x � ft/gpd = lineal feet <br /> P. If using a Chamber H, I,J,K[based on height of chamber slats divided by width of chamber in ft) � <br /> 1056.0 ft2 / 3.0 ft = 352.0 lineal feet <br /> 7. LAWN AREA � ' <br /> Q. Select trench spacing,center to center= 6 feet <br /> R. Multiply trench spacing by lineal feet R x Q=sq.ft.of lawn area <br /> 6 x 352.0 = 2112 ft2 , <br /> , <br /> � <br /> ' <br /> Page 2 of 2 ' <br />
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