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Septic design - 1999
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4375 Bayside Road - 06-117-23-12-0009
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4375 Bayside Rd - PID: 06-117-23-12-0008 - Old PID
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Septic design - 1999
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Last modified
8/22/2023 5:23:15 PM
Creation date
9/2/2015 2:36:38 PM
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x Address Old
House Number
4375
Street Name
Bayside
Street Type
Road
Address
4375 Bayside Rd
Document Type
Land Use
PIN
0611723120009
Supplemental fields
ProcessedPID
Updated
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, . • � t-ly <br /> ' N10UND DESIGi�1 WORKSHEET <br /> (For Flows up to 1200 gpd) <br /> A. FLOW D-7 . <br /> Estimated ��� gpd (see pages D-7 or I-3, 4,5) ES��W1[p SCM�G[RO'MS�M G41Dy1�OI�.= <br /> 1ar1EEA lrrf pf IKS�OEMCE� <br /> �f <br /> or measured gpd. i °`°"'°'6 ` � � s <br /> : �oo ae ��o son <br /> 3 •!O 700 i�� or <br /> • f00 37] t3i <br /> B. SEPTIC TANK LIQCTID VOLUMES � � �n � `'�` <br /> T i 050 .t� 3�0 <br /> Z_� �C`.� gallons (see pages C-3 or C-5) ° '_°° °„ '°' �� <br /> C-3 <br /> C. SOILS (refer to site evaluation) sflanc T�r+K c�v�cmes, iN GALlONS <br /> / „�o�.,.�m <br /> 1. Depth to restricting layer = 7C9 in es ""`"" "'�"'" '""°.".°. <br /> �.._ �FAOOd LIOUA W�QT� 0�]IOl�I <br /> 2. Depth of percolarion tests = %'L in es ,a,,,. ,�� ,,,, <br /> 3. Percolation rate i .-�' mpi '°"' ,••° ,�•° <br /> ��� ��as »�o <br /> 4. Land slope ..� °'o �.�a� <br /> ,.o. ..eo <br /> D. ROCK LAYER DIMENSIONS <br /> 1. Multiply flow rate by 0.83 to obtain required area of rock <br /> layer: A x 0.83 = _ <br /> -;���%� gpd x 0.83 sq. ft./gpd =/r�s . ft. <br /> ?„---- <br /> 2. Select width of rock layer (10 feet or less) = l0 ft. <br /> 3. Length of rock layer = area y width = '.:-,� <br /> � � <br /> ,, ��e_ sq. ft. i _l!�' ft. _ (:-:%? ft Rock Bed <br /> _ f'f•f.f.f.f.f•f.f.,.f•f•r.f�f T I <br /> 1•�•ti•�•�•ti•ti•ti•ti•ti•ti•ti•�•ti•`.• � <br /> f.l.J•t•f•I•f•t l•l.r•l.r•f•f � <br /> ti.ti.�.ti.ti.ti.ti•`•ti•�•ti•ti.ti•�•ti• idth <_1(' <br /> � � 1•f•t•f•f.l•f•l l•I./.f•r•1•f <br /> ti�ti•t•ti•ti•ti•ti•�•�•ti�ti•ti•ti�ti•ti• <br /> .f•f.�•f•r•I•r•J•r•f•f•!•,r•f•f <br /> E. ROCK VOLUME �- Length -� <br /> , 1. Multiply rock area by rock depth to get cu 'c feet of rock; <br /> ;!-�,? sq. ft. x � ft. =�'��cu. ft. <br /> 2. Divide cu. ft. by 27 cu. ft./cu. yd. to get cub c yards; <br /> ;1-?� cu. ft. 127 = �,�cu. yd. <br /> 3. Multiply cubic yards by 1.4 to get weight o rock in tons; I <br /> �=<cu. yd. x 1.4 ton/cu. yd. _ �to s. <br /> F. ADSORPTION WIDTH I <br /> 1. Percolation rate in top 12 inches of soil is mpi E-16 <br /> 2. Select allowable soi! loading rate from table o page E-16; �-�����d����u��•�� I <br /> �. �'"-' gpd/ft2 _ ��� ' �... I <br /> 3. Calculate adsorption width ratio by dividing ro layer �~"" "M� �"� �� �� � ' <br /> .,. . ,,� .., ,.a ,�o <br /> laadin� rate of 1.20 �pd/ft2 by allowable soil lo ding rate; ' " "' "' '" "' <br /> �� ��o a w a.•� t... t.eo I <br /> ..1�--r�___ ."p�_._.QN".—�.0]__"—.T.O <br /> 1.2� gpd/ft2�- �y,���Pd/ft2 = � �- �� .. ..e e.. o.,, ,.� ,., <br /> .. .,:< a t. ,,, ... ,:a � <br /> Che�K this t�a(ue on page E-16. I <br /> 4. Multiply adsorption width ratio by rock layer idth to get , <br /> required adsorprion width; � <br /> � !� x <=% ft = ��ft <br />
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