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On-Site Wastewater Treatment and Disposal Systems - Forced ...

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1.0 + 1.4<br />

= - t 0.75 + 1.5<br />

1<br />

[<br />

= (3.5) x (3)<br />

= 10.5 ft, or 11 ft<br />

Basal Area Required = (B) x t(I) + (AlI<br />

(I) + (A) = ,e<br />

= m-%&$ = 1,800 ft2<br />

.<br />

(I) = # - (A)<br />

1,800<br />

= ---6<br />

= 21.7 ft, or 22 ft<br />

x (3)<br />

Check to see that the downslope setback (I) is great<br />

enough so as not to exceed a 3:l slope:<br />

(mound height at downslope edge of bed) x (3:l slope)<br />

= C(E) +<br />

= (1.4 +<br />

= 9.5 ft<br />

(F) + (G)] x (3)<br />

0.75 + 1.0) x (3)<br />

Since<br />

less<br />

the<br />

than<br />

distance needed to maintain a 3:l slope is<br />

the distance needed to provide sufficient<br />

basal area, (I) = 22 ft<br />

Mound Length (L) = (8) + 2(K)<br />

= 65 + 2 (11)<br />

= 87 ft<br />

Mound Width (W) = (J) + (A) + (I)<br />

= 8 + 6 + 22<br />

= 36 ft<br />

252

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