10.4 Topography, Contour Lines, Slope Calculations, and Plan Amendments

Reading Topographic Maps and the Seven Fundamental Contour Rules

A topographic map depicts three-dimensional terrain on a flat two-dimensional sheet using contour lines (isolines connecting points of equal elevation). As detailed in the CESSWI General Principles Review Manual (Chapter 11, Sections 11.6–11.8), seven fundamental rules govern contour interpretation:

1. Rule 1 — Equal Elevation Along a Contour: Every point along a single continuous contour line sits at the exact same vertical elevation above the benchmark datum. Therefore, a BMP installed strictly along a single contour line (such as a level perimeter silt fence or contour wattling) has a $0\%$ longitudinal grade along its length, preventing runoff from flowing laterally along the barrier.
2. Rule 2 — Contour Interval (CI) and Index Contours: The Contour Interval (CI) is the constant vertical elevation difference between any two adjacent contour lines on a given map (typically $1\text{ ft}$ or $2\text{ ft}$ on civil site grading plans, and $5\text{ ft}, 10\text{ ft},\text{ or }20\text{ ft}$ on regional USGS maps). Every fifth contour line is printed bolder and labeled with its numerical elevation as an index contour. To find an unlabeled contour interval between two adjacent index contours (for example, index contours labeled 200 and 210 with $4$ intermediate lines—which form $5$ vertical spaces—between them), divide the elevation difference by the number of spaces: $$\text{Contour Interval (CI)} = \frac{210\text{ ft} - 200\text{ ft}}{5\text{ spaces}} = 2\text{ ft per contour}$$
3. Rule 3 — Spacing Indicates Slope Steepness and Profile Geometry:
   • Closely Spaced Contours: Indicate a steep slope where elevation changes rapidly over a short horizontal distance, producing high runoff velocities and severe rill/gully erosion risk.
   • Widely Spaced Contours: Indicate a gentle or flat slope where runoff moves slowly and sediment settles.
   • Uniformly (Evenly) Spaced Contours: Indicate a uniform planar slope of constant gradient.
   • Closer at the Top (Crest) and Wider at the Bottom (Toe): Indicate a concave slope. Because the gradient flattens near the bottom, runoff decelerates at the toe—making concave slopes the most stable profile against erosion.
   • Wider at the Top (Crest) and Closer at the Bottom (Toe): Indicate a convex slope. Because the slope steepens sharply toward the bottom, runoff accelerates as it approaches the toe, creating high shear stress and severe toe scour.
4. Rule 4 — The Rule of V's (Valleys/Swales vs. Ridges/Noses):
   • Valleys, Swales, Gullies, and Streams (V Points UPSTREAM / UPSLOPE): Whenever contour lines cross a drainage swale, ditch, valley, or stream channel, they bend into V or U shapes whose apex points upstream / upslope toward higher elevation. Because gravity pulls surface water downhill perpendicular ($90^\circ$) to contour lines, overland runoff converges into the center of the swale and flows downstream out of the open mouth of the V.
   • Ridges, Spurs, Noses, and Divides (V Points DOWNSLOPE): Whenever contour lines cross a ridge crest or nose, they form V or U shapes whose apex points downslope toward lower elevation. Surface water diverges (splits away) from the ridge centerline toward adjacent watersheds.
5. Rule 5 — Closed Concentric Loops (Hills/Stockpiles vs. Depressions/Basins):
   • Concentric Closed Loops with Increasing Elevation Toward the Center: Represent a hill, knoll, summit, or soil stockpile; runoff sheds radially outward ($360^\circ$) away from the center.
   • Concentric Closed Loops with Short Inward Tick Marks (Hachures / Depression Contours): Represent an enclosed depression, sinkhole, sump, or excavated sediment basin/trap; runoff flows radially inward toward the bottom.
6. Rule 6 — Continuity and Non-Intersection: Contour lines never cross, split, or fork (except at a vertical retaining wall or overhanging cliff) and always form closed loops—either closing on the plan sheet or extending beyond the neatline border of the drawing.
7. Rule 7 — Underground Pipe Invert Elevations (INV): When stormwater enters an underground storm sewer pipe or culvert, surface contours no longer govern flow direction; instead, gravity flow inside the pipe always travels from the higher pipe invert elevation ($\text{INV}_{\text{up}}$) to the lower pipe invert elevation ($\text{INV}_{\text{down}}$), with pipe slope calculated as: $$\text{Pipe Slope } (S) = \frac{\text{INV}_{\text{up}} - \text{INV}_{\text{down}}}{\text{Pipe Length } (L)}$$

---

Core CESSWI Exam Calculations and Step-by-Step Worked Examples

1. Slope Calculations: Decimal, Percentage, and Horizontal-to-Vertical ($H:V$) Ratio

Slope measures the vertical change in elevation (Rise, $V$) divided by the horizontal map distance (Run, $H$) measured perpendicular to the contour lines:

$$\text{Decimal Slope } (S) = \frac{\text{Rise}}{\text{Run}} = \frac{\text{Elev}_2 - \text{Elev}_1}{\text{Horizontal Distance } (L)}\text{ (ft/ft)}$$ $$\text{Percent Slope } (\%\text{Slope}) = \frac{\text{Rise}}{\text{Run}} \times 100$$ $$\text{Horizontal-to-Vertical Ratio } (H:V) = \frac{\text{Run}}{\text{Rise}}:1 = \frac{100}{\%\text{Slope}}:1$$

Conversely, if you know the horizontal distance ($\text{Run}$) and decimal slope ($S$), or if you know the vertical change ($\text{Rise}$) and decimal slope ($S$), rearrange the equation:

$$\text{Rise} = \text{Run} \times S \qquad \text{and} \qquad \text{Run} = \frac{\text{Rise}}{S}$$

Worked Example 1: Calculating Slope from Map Contours and Scale

Problem: On a grading plan with a scale of 1" = 50' and a contour interval of $\text{CI} = 2\text{ ft}$, a fill embankment drops across $8$ contour intervals from the top of the pad at elevation $426.0\text{ ft}$ to the toe at elevation $410.0\text{ ft}$. Using an engineer's scale perpendicular to the contours, the horizontal width of the slope on the paper is $1.28\text{ inches}$. Calculate the Rise, Run, Decimal Slope, Percent Slope, and $H:V$ ratio.

Step-by-Step Solution:

1. Calculate Vertical Rise: $$\text{Rise} = 426.0\text{ ft} - 410.0\text{ ft} = 16.0\text{ ft}\quad (\text{or } 8\text{ intervals} \times 2\text{ ft/interval} = 16.0\text{ ft})$$
2. Calculate Horizontal Run from Map Scale (1" = 50'): $$\text{Run} = 1.28\text{ in} \times 50\text{ ft/in} = 64.0\text{ ft}$$
3. Calculate Decimal Slope, Percent Slope, and $H:V$ Ratio: $$S = \frac{16.0\text{ ft}}{64.0\text{ ft}} = \mathbf{0.25\text{ ft/ft}} \implies \%\text{Slope} = 0.25 \times 100 = \mathbf{25.0\%} \implies H:V = \frac{64.0}{16.0}:1 = \mathbf{4\text{H}:1\text{V}}$$

---

2. Check Dam Spacing Formula ($L = H / S$)

When rock, wattle, or aggregate-bag check dams are installed in a conveyance swale or roadside ditch, they must be spaced close enough so that the toe of the upstream check dam sits at the exact same elevation as the center overflow weir crest of the downstream check dam. This creates a series of low-velocity ponding steps that prevent supercritical channel bed scour between structures.

$$L = \frac{H}{S}$$

where:

• $L$ = maximum center-to-center spacing between check dams along the channel ($\text{ft}$);
• $H$ = effective center weir height of the downstream check dam ($\text{ft}$) (measured at the depressed center notch, not the higher outer wing edges keyed into the channel banks!);
• $S$ = longitudinal channel slope expressed as a decimal ($\text{ft/ft}$).

Worked Example 2: Check Dam Spacing in a Graded Swale

Problem: A $300\text{-foot}$ long trapezoidal drainage swale is graded at a longitudinal slope of $4\%$ ($S = 0.04\text{ ft/ft}$). The standard detail sheet specifies rock check dams with a $2.0\text{-foot}$ high center weir crest ($H = 2.0\text{ ft}$) and $3.0\text{-foot}$ outer bank wings. What is the maximum center-to-center spacing ($L$) between check dams, and how many check dams are required for the $300\text{-foot}$ swale?

Step-by-Step Solution:

1. Use the Center Weir Height ($H = 2.0\text{ ft}$, NOT the $3.0\text{-ft}$ wing height): $$L = \frac{H}{S} = \frac{2.0\text{ ft}}{0.04\text{ ft/ft}} = \mathbf{50\text{ feet on center}}$$
2. Determine Total Check Dams Required: $$\text{Number of Check Dams} = \frac{300\text{ ft}}{50\text{ ft/dam}} = \mathbf{6\text{ check dams}}$$

---

3. Pure Live Seed ($\text{PLS}$) and Bulk Seeding / Fertilizer Application Math

Commercial seed bags contain inert chaff, stems, and a fraction of dormant or non-germinating seeds. Therefore, SWPPP revegetation specifications state seeding rates in pounds of Pure Live Seed ($\text{PLS}$) per acre. Before hydroseeding or drill-seeding begins, the CESSWI inspects the certified seed bag tags to verify that the contractor applies enough bulk seed to deliver the specified $\text{PLS}$ weight:

$$\%\text{PLS} = \frac{\%\text{Purity} \times \%\text{Germination}}{100} \qquad (\text{or Decimal PLS} = \text{Decimal Purity} \times \text{Decimal Germination})$$ $$\text{Bulk Seed Rate (lb/acre)} = \frac{\text{Specified PLS Rate (lb/acre)}}{\%\text{PLS} / 100}$$

Worked Example 3: Pure Live Seed ($\text{PLS}$) and Bulk Fertilizer N-P-K Math

Problem A (Bulk Seed): The SWPPP requires applying $40\text{ lb PLS/acre}$ of perennial tall fescue across a $3.5\text{-acre}$ disturbed slope. The certified seed tag on the contractor's pallets lists $90\%\text{ Purity}$ and $80\%\text{ Germination}$. How many total pounds of bulk seed must the crew apply across the $3.5\text{-acre}$ slope?

1. Calculate $\%\text{PLS}$: $$\%\text{PLS} = \frac{90 \times 80}{100} = 72\%\quad (\text{Decimal PLS} = 0.72)$$
2. Calculate Required Bulk Seed Rate per Acre: $$\text{Bulk Seed Rate} = \frac{40\text{ lb PLS/acre}}{0.72} = 55.56\text{ lb bulk seed/acre}$$
3. Multiply by Disturbed Acreage ($3.5\text{ acres}$): $$\text{Total Bulk Seed Required} = 55.56\text{ lb/acre} \times 3.5\text{ acres} = \mathbf{194.4\text{ lb of bulk seed}}$$

Problem B (N-P-K Fertilizer): Commercial fertilizer bags display a three-number guarantee (N-P-K, such as $10\text{-}20\text{-}10$), representing the weight percentages of elemental Nitrogen ($\text{N}$), available Phosphate ($\text{P}_2\text{O}_5$), and soluble Potash ($\text{K}_2\text{O}$). If the SWPPP specifies applying $40\text{ lb}$ of actual Nitrogen per acre using $10\text{-}20\text{-}10$ bulk fertilizer ($10\%\text{ N} = 0.10$), the required bulk fertilizer rate per acre is:

$$\text{Bulk Fertilizer Rate} = \frac{40\text{ lb N/acre}}{0.10} = \mathbf{400\text{ lb/acre of } 10\text{-}20\text{-}10\text{ bulk fertilizer}}$$

---

4. Area, Volume, RECP Blanket Rolls, and Sediment Basin Storage Math

CESSWI inspectors routinely verify surface areas for Rolled Erosion Control Products (RECPs) and excavation volumes for temporary sediment traps and basins using standard civil unit conversions:

• $1\text{ acre} = 43,560\text{ ft}^2$; $1\text{ square yard (yd}^2\text{)} = 9\text{ ft}^2$
• $1\text{ cubic yard (yd}^3\text{)} = 27\text{ ft}^3$; $1\text{ acre-foot} = 43,560\text{ ft}^3$

Worked Example 4A: Rolled Erosion Control Blanket (RECP) Roll Quantity

Problem: A graded embankment measures $150\text{ ft}$ long by $60\text{ ft}$ wide. The contractor is installing double-net straw erosion control blankets where each roll measures $8\text{ ft} \times 112.5\text{ ft}$ ($900\text{ ft}^2 = 100\text{ yd}^2$ per roll). Including a standard $10\%$ waste/overlap allowance for crest anchor trenching and adjacent roll overlaps, how many rolls must be staged on site?

$$\text{Slope Area} = 150\text{ ft} \times 60\text{ ft} = 9,000\text{ ft}^2 \implies \text{Area with } 10\%\text{ Overlap} = 9,000\text{ ft}^2 \times 1.10 = 9,900\text{ ft}^2$$ $$\text{Required Rolls} = \frac{9,900\text{ ft}^2}{900\text{ ft}^2/\text{roll}} = \mathbf{11\text{ rolls}}$$

Worked Example 4B: Temporary Sediment Basin Storage Volume

Problem: Under the standard NPDES CGP volumetric sizing rule requiring $3,600\text{ ft}^3$ of storage per acre drained ($134\text{ yd}^3/\text{acre}$), what minimum storage volume (in both cubic feet and cubic yards) must be excavated for a temporary sediment basin serving a $10\text{-acre}$ disturbed watershed?

$$\text{Required Storage Volume} = 10\text{ acres} \times 3,600\text{ ft}^3/\text{acre} = \mathbf{36,000\text{ ft}^3} = \frac{36,000\text{ ft}^3}{27\text{ ft}^3/\text{yd}^3} = \mathbf{1,333.3\text{ yd}^3}$$

---

SWPPP Plan Amendments vs. Minor Field Red-Lines

As construction progresses, field conditions inevitably diverge from the original baseline drawings. Under SAOP 10 Test Objective 8, a CESSWI must distinguish between major engineering changes that trigger a formal SWPPP Plan Amendment (within 7 calendar days) and routine operational adjustments that are documented as minor field red-line markups on the on-site living map: