# Sec. 22-161. Appendix C

Codified through Bill No. CB-018-2026, adopted May 12, 2026, effective July 24, 2026. Source: Municode, Supplement 2026 Update 2. Editorial notes and legislative history omitted.

## Sec. 22-161. Appendix C

Prince George's County, Md., Code § 22-161 (2026)

(a) State Standard Test.

  (1) Area for absorption.

    Bottom = 1 ft 2

    4 sides at 1/2 ft 2

    TOTAL = 1 + 4 (.5) = 3 ft 2

  (2) Absorptive Rate in gallons / ft 2 / 30 minutes.

    | .62333 gallons | = | .2077 gallons |
    | --- | --- | --- |
    | 3 ft 2 / 30 mins. | ft 2 / 30 mins. |

(b) Conversion to State equivalent.

  (1) Theory. State standard passing percolation test is .2077 gallons/ft/30 minutes. Any diameter hole can be used for a test hole provided that the equivalent State percolation test rate is achieved in terms of gallons per square foot in 30 minutes, that being .2077 gallons for every square foot of absorptive area (i.e., a hole with ten (10) square feet of wetted area would have to absorb 2.077 gallons of water). This volume can be converted into an inch drop as a function of test hole diameter.

    (A) Shallow Test Conversion: (Test hole with six (6) inches of water)

      (i) Volume as a function of test hole diameter.

        Formula H × Pi R 2 × 7.48 H = Height = 6 inches = .5 feet R = Radius = D/2 Pi = 3.14159 7.48 converts cubic feet to gallons

        Volume = (.5) (Pi) (D/2) 2 (7.48)

        D in inches = (.5) (Pi) (D/12) 2 (1/4) (7.48)

        Volume absorbed = Pi (X/12) (D/12) 2 (1/4) (7.48) X = inches dropped

      (ii) Available Absorptive Area (Based on six (6) inches of water).

        Includes sidewall and bottom

        Area = Pi D H + Pi (D/2)2

        H = Height = 6 inches = .5 feet D = Diameter (inches)

        Converting D into feet = D/12

        Area = Pi(D/12)(.5) + Pi (D/12) 2 (1/4)

      (iii) Volume/Area as a function of diameter and inch drop.

        Pi (X/12) (D/12) 2 (1/4) (7.48) Pi (D/12) (.5) + Pi (D/12)2(1/4)

        Simplifying:

        (X/12) (D/12) (1/4) (7.48) (.5 + D/48)

      (iv) Set equal to .2077 and solve for X.

        .2077 = .0129861 (D) (X) (.5 + D/48)

        Varying D from 4 to 12 inches gives minimum inch drop in 30 minutes being equal to State standard test in terms of gallons per square feet in 30 minutes. This was reported in Table 2.

      (v) To convert to equivalent State test rate, multiply the rate in minutes per inch by the correction factor found in Table 2.

    (B) Deep Test Conversion: (Bottom area not included, with four (4) feet of water)

      (i) Volume formula: V = Pi (H) (D/12) 2 (1/4) (7.48)

        Volume absorbed = Pi (D/12) 2 (1/4) (7.48) (X/12)

        Pi = 3.14159 D = Diameter (inches) X = Equivalent drop inches = H in general formula

      (ii) Area formula: Pi (D/12)

        Pi = 3.14159 D = Diameter (inches) H = Height = 4 feet

        Simplifying: Pi (D/12) (4) = Area

      (iii) Volume (gallons) / Area (ft 2 )

        | Pi (X/12)(D/12)(D/12)(1/4)(7.48) | = | (X/12)(D/12)(7.48)(1/4) = |
        | --- | --- | --- |
        | Pi (D/12)4) |  | 4 |
        |  |
        | .0129861 (D)(X) | = | .0032465 (D)(X) |
        | 4 |

      (iv) Set this formula equal to the minimum State standard rate.

        .0032465 (D) (X) = .2077 gals/ ft 2 /30 minutes

        Varying D and solving for X gives minimum rates found in Table 3.

(v) To convert to State equivalent rate, 30 minutes/inch is divided by the ratio of percolation rate gallons/ft/minute to the State maximum rate of .2077 gallons/ ft 2 /30 minutes by the following formula:

  Where D = Diameter (inches) X = Drop recorded (inches) T = Time to measure drop (minutes)

  Simplifying equation.

  | 30 minutes/inch .469 (D)(X) | = | ____ minutes/inch |
  | --- | --- | --- |
  | T |

  | 64 = (D) (X) | = | ____ minutes/inch |
  | --- | --- | --- |
  | T |

  i.e., 8 inch diameter hole drops 8 inches in 15 minutes. Find equivalent State standard rate.

  |  | 64 |
  | --- | --- |
  | = Equivalent rate = | (D) (X) | = 15 minutes/inch |
  |  | T |