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2005-P09293 - new septic system
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0160 Old Crystal Bay Road North - 33-118-23-43-0006
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2005-P09293 - new septic system
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Last modified
8/22/2023 4:52:01 PM
Creation date
3/7/2018 9:39:33 AM
Metadata
Fields
Template:
x Address Old
House Number
160
Street Name
Old Crystal Bay
Street Type
Road
Street Direction
North
Address
160 Old Crystal Bay Road North
Document Type
Septic
PIN
3311823430006
Supplemental fields
ProcessedPID
Updated
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1 • • <br /> SII�{��1• Y <br /> G. MOUND SLOPE WIDTH &LENGTH <br /> Gandslope 1% or more) �s <Rock 1• 7 • <br /> . Subtract rock layer width from absorption width `"em�"'" s ro�,ou <br /> to obtain minimum downslope width '�' S '' <br /> umtnry,L"ar <br /> Z7 ft- 16 ft= feet t �.z am Don, an <br /> 2. Calculate minimum mound size M%7,dtn <br /> a. Determine depth of clean sand fill at Y <br /> upslope edge of rock layer: t.lb9 abs W '' ��� sept+l"�' " "°""�''� 1 <br /> Separation 3'- 1,!5" ft=! feet�� <br /> b. Add depth of clean sand for separation(2a) <br /> at upslope edge, depth of rock layer (1 foot) to SLOPE MULTIPLIER TABLE <br /> depth of cover(1 foot) to find the mound height s;o�, MotNpuliars°;v„i„u mul up � CUB <br /> at the upslope edge of rock layer; '"'' 'tope ratios sVPe ntlw <br /> )'S- ft+ 1ft+ 1ft= 3.s feet <br /> c. Enter table with landslope and upsle ratio. 0 3.0 4.0 5.0 6.0 7o e.o 3.0 4.0 5.0 6.0 7.0 <br /> op <br /> 3.57 1 2.91 3.85 4.76 S." 16.54 7.41 3.09 4.17 5.26 6.38 7.53 <br /> Select berm multiplier of <br /> d. Multiply berm multiplier by upslope mound 2 2'83 3'70 454 536 6•x4 6.90 3.19 435 556 6 82 eai <br /> 3 2 <br /> height to find upslope width: .75 q3574.35 5.08 5.79 6.45 3.30 4.54 S.88 7.32 8.864 2.68 4.17 4.84 5.46 6.06 3.41 4.76 6.25 7.89 9.72 <br /> 3.5- X 3.57= J 3 feet 5 2.61 3.33 4.00 4.62 L.9 5.71 3.53 5.00 6.67 837 10.77 <br /> e. Multiply rock layer width by 6 2.54 3.23 3.0 4.41 4.93 5.41 3.66 S2 7.14 938 M07 <br /> landslope to determine drop in elevation; i 2.48 ,.,: 3.70 423 4. 5.13 360 5.56 7.69 1034 13.73 <br /> �% <br /> l_x + 100= a.46 fee sr �a'�'ovU� <br /> b� �..;d.E8 2.42 3.3 337 4.5 4.1 4.68 3.95 5.88 8.33 11.51 15.91 <br /> f.Add depth of clean sand for slope difference 9 2.36 2.94 3.45 3.90 4.30 4.65 <br /> -'I P 4.11 6.25 9.09 13.01 18.92 <br /> le)at downslope rock edge,to the mound 10 2.31 2.86 3.33 3.75 4.12 4.N 4.29 6.67 10.00 15.0 2333 <br /> -height at the upslope edge of rock layer(2b) 11 L26 2.78 3.23 3.61 3.95 4.26 4.46 7.14 11.11 19.ss no <br /> to find the downslope mound height; 12 2.21 2.70 3.12 3.49 3.80 4.08 4.69 7.69 12.50 21.43 43.75 <br /> •Sb ft+0•do ft= 3.40 feet <br /> g. Enter table with landslope and downslope ratio. Select <br /> downslope multiplier of 5'.'L r- <br /> h. Multiply downslope multiplier by downslope mound height to <br /> get downslope width: <br /> _x 5===feet <br /> i. Compare the values of step G.1-- 1-7 <br /> and Step G.2h <br /> Select the greater of the two values as the UP313 aVVIdth <br /> downslope.width:- mg feet <br /> j. Total mound width is the sum of J' Up"slope Width , P R gip¢ .Ups1 width <br /> upslope(G.2d) width plus rock layer 3 H►id ° ? 'Ups <br /> lope <br /> �.' <br /> width(D-2) �• t.en .potia• � �''�;� ` , <br /> plus downslope width(G.2i); <br /> ft+=ft+_Z,'0 ft= 43 feet a Downslope warn zo , <br /> k. Total mound length is the sum of upslope H Absorption Width z�� <br /> width(G.2d)plus rock layer length (D.3) <br /> plus upslope width(G.2d); <br /> 13" ft+_ft+ 13 ft= 89 feet <br /> Total Length_ 89 <br /> Final Dimensions: <br /> 43-0 X 89.0 <br />
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