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1996-008339 - new septic system
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Old Crystal Bay Road North
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0325 Old Crystal Bay Road North - 33-118-23-31-0003
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1996-008339 - new septic system
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Last modified
8/22/2023 4:49:13 PM
Creation date
3/7/2018 12:36:29 PM
Metadata
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Template:
x Address Old
House Number
325
Street Name
Old Crystal Bay
Street Type
Road
Street Direction
North
Address
325 Old Crystal Bay Road North
Document Type
Septic
PIN
3311823310003
Supplemental fields
ProcessedPID
Updated
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G. DOWNSLOPE DIKE WIDTH <br /> 1. If landslope is 2.9 % or less adsorption width inculdes both pslope and <br /> downslope widths <br /> 2. Calculate Minimum mound size bsaed on geometery: <br /> a. Determine depth of clean sand fill at upslope edge of r ck layer: <br /> Separation Z.a S feet <br /> b. Multiply rock layer width by landslope -toot c.�.r <br /> to determine drop in elevation; _ C' ' foot R° 6' <br /> Slope Difference s.o6r ue -. r..t. <br /> x�%+100 = o oC feet s1op. Ofit•rene• =^�' et <br /> uoslooe veldt <br /> c. Add depth of clean sand for separation (2a) '= 'feet <br /> P P Rotk Bed Width <br /> at upslope edge,depth of rock layer(1 foot) to L feet Downslope Width <br /> depth of cover(1 foot) to find the mound height at the :rest <br /> upslope edge of rock layer;2. ;s ft+Ift+ Ift= S-d Slfe t <br /> d. Enter table with landslope and upslope dike ratio. <br /> Select dike multiplier of 3.70 <br /> e. Multiply dike multiplier by upslope mound height to fi d <br /> upslope dike width: .?:7 x�S- /-5,7 feet <br /> f. Add depth of clean sand for slope difference (2b) at do lope edge, to <br /> the mound height at the upslope edge of rock layer(20 t find the <br /> downslope height; . 2. ft+_:LR S-ft= y!%'Zeet <br /> g. Enter table with landslope and downslope dike ratio. <br /> Select dike multiplier of 5! 2 S <br /> h. Multiply dike multiplier by downslope mound height to get <br /> downslope dike width:_OSx JYL3Y= /?,/feet <br /> i. Minimum mound width is the sum of upslope dike widt i plus rock <br /> layer width plus downslope dike width; <br /> 9- - <br /> + y + <br /> ft ft /r� f - <br /> 5. t <br /> feet <br /> j. 1� <br /> slo a rile <br /> D <br /> t <br /> Subtract the minimum width 2i <br /> from the ` <br /> f ea t <br /> .`> <br /> adsorption area <br /> (F 4 tofin <br /> d the additional <br /> onal <br /> X. Roe:e.e wletn <br /> downslope area for adsorption v! UP5e � YUPS:I.:0:9:•.•• <br /> Mleln <br /> ,2 oft- --I' S- ft= meet 3 1 feet teat <br /> 0 <br /> additional - <br /> k Add thewidth <br /> 2.' <br /> to the <br /> e <br /> r <br /> downsolP a width and recalculate to <br /> :,".Downslope Wl <br /> e <br /> fes v{5' % <br /> the total width <br /> t <br /> f + f + ft _ <br /> t t <br /> -feet <br /> 1. To un <br /> tel mound length is the <br /> um of <br /> upslope dike width plus rock layer length <br /> plus upslope dike width: <br /> /`/ ft+ y5-ft+�ft= '73 feet T•�3 Length <br /> ^ <br /> Downslope O upslope <br /> 3:1 LI 5:1 6:1 7:1 3:1 4:1 5:1 6:1 7:1 8:1 <br /> !L elope <br /> 0 311 LO U 6.0 7.0 3.0 4.0 S.0 6.0 7.0 611 S26 638 lz 7M 191 4.76 5,66 <br /> 6.90 <br /> 330 435 SIAS 732 6.86 � 435 5.06 5796A5 <br /> 4 3A1 4.76 6.25 7.89 972 2.68 3.45 4.17 464 5.46 6.06 <br /> S 353 5.00 6.67 857 1077 2.61 317 4.00 4.62 5.19 S21 <br /> 6 366 S.26 7.14 938 1207 254 323 3SS 4.41 L93 SAI t <br /> 7 330 S56 7.69 1034 1373 248 3.12 3.70 4.23 1.70 533 <br /> 6 3.95 5.68 837 1154 15.91 142 3.03 357 4.05 4.49 4A <br /> 72 9 4.11 6.25 9.09 13.04 16.92 2.36 294 3.45 3.90 430 465 <br /> 10 429 6:67 10.0 IS.00 2333 231 286 333 375 4.12 4A4 <br /> 11 4A8 7.14 lilt 1765 30.43226 278 3.23 161. 195 426 <br /> 12 469 7.69 1250 21.43 4375 211 2.70 3.12 3.49 3.80 4118 <br />
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