9.3 Ratios, Averages, Area & Volume Measurements

9.3 Ratios, Averages, Area & Volume Measurements

The final section of the IO Solutions NFSI Numeric Facilitation (Number Facility) subtest evaluates candidates on applied geometric, statistical, and proportional reasoning. Firefighters utilize these principles daily: setting up hydraulic proportions for extended hose stretches, calculating weighted average response times across fire stations, determining roof ventilation cut areas, calculating containment pool dimensions for chemical spills, and computing the total weight of water stored in apparatus tanks. Because non-calculator testing demands rapid setup and computation, understanding core formulas and physical constants is essential.

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Ratios & Proportions

A ratio compares two quantities by division, expressed as $A:B$, $\text{A to B}$, or $\frac{A}{B}$. A proportion states that two ratios are equal:

$\frac{A}{B} = \frac{C}{D}$

Proportions are solved by cross-multiplication ($A \times D = B \times C$).

Hydraulic Application: Friction Loss Proportions

Hose friction loss ($FL$) varies directly with length. If 100 feet of 2.5-inch hose carrying 250 GPM creates 12 psi of friction loss, what is the friction loss for a 350-foot hose stretch operating at the same flow rate?

$\text{Set up proportion: } \frac{12\text{ psi}}{100\text{ ft}} = \frac{X\text{ psi}}{350\text{ ft}}$ $\text{Cross-multiply: } 100 \times X = 12 \times 350$ $\text{Mental Math: } 12 \times 350 = 12 \times (300 + 50) = 3,600 + 600 = 4,200$ $100X = 4,200 \rightarrow X = 42\text{ psi}$

Alternatively, use the Hose Length Factor Method: $\frac{350}{100} = 3.5$. Multiply baseline loss by 3.5: $3.5 \times 12 = 42\text{ psi}$.

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Averages: Simple Arithmetic Mean & Weighted Averages

1. Simple Arithmetic Mean

The arithmetic mean sums all individual values and divides by the total number of items ($N$):

$\text{Mean } (\bar{X}) = \frac{\sum X}{N}$

2. Weighted Average

When averaging groups of different sizes or calls of varying durations, a simple average introduces significant error. You must calculate a weighted average:

$\text{Weighted Average} = \frac{\sum (w_i \cdot x_i)}{\sum w_i}$ Where $w_i$ represents the weight (count or volume) of each subgroup $x_i$.

Worked Example: Department Response Times

Calculate the department's overall average response time across three stations:

• Station 1: 4.0 minute average across 100 calls
• Station 2: 5.0 minute average across 200 calls
• Station 3: 6.0 minute average across 100 calls

1. Calculate Call-Minutes per Station:
   • Station 1: $100 \times 4.0 = 400\text{ call-minutes}$
   • Station 2: $200 \times 5.0 = 1,000\text{ call-minutes}$
   • Station 3: $100 \times 6.0 = 600\text{ call-minutes}$
2. Sum Total Call-Minutes: $400 + 1,000 + 600 = 2,000\text{ call-minutes}$
3. Sum Total Calls: $100 + 200 + 100 = 400\text{ calls}$
4. Divide: $\frac{2,000}{400} = 5.0\text{ minutes}$

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2D Geometry: Perimeter & Area

Firefighters calculate surface areas to determine foam blanket application rates, roof ventilation cut dimensions, and building collapse safety zones.

1. Rectangles & Squares

• Perimeter: $P = 2L + 2W$
• Area: $A = L \times W$
• Unit Conversion: $1\text{ sq ft} = 144\text{ sq in}$

2. Triangles

• Area: $A = \frac{1}{2} \times B \times H = 0.5 \times B \times H$ (Used for calculating gable wall dimensions and triangular collapse zones)

3. Circles

• Circumference: $C = \pi d = 2\pi r$
• Area: $A = \pi r^2$ (Use $\pi \approx 3.14$ or $\frac{22}{7}$ for manual non-calculator math)

Worked Example: Circular Storage Tank Footprint

A circular petroleum storage tank has a diameter of 20 feet ($radius = 10\text{ ft}$). What is the ground footprint area covered by the tank base?

$A = \pi r^2 = 3.14 \times (10)^2 = 3.14 \times 100 = 314\text{ square feet}$

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3D Geometry: Volume & Water Weight

Volume measures three-dimensional space ($L \times W \times H$). In fire hydraulics, converting volumetric measurements (cubic feet) into fluid capacity (gallons) and total water weight (pounds) is a core operational requirement.

Fundamental Fire Service Hydraulic Constants

• 1 Cubic Foot of Space $= 1,728\text{ cubic inches}$
• 1 Cubic Foot of Water $\approx 7.48\text{ gallons}$
• 1 Gallon of Fresh Water $\approx 8.34\text{ pounds}$
• 1 Cubic Foot of Fresh Water $\approx 62.4\text{ pounds}$ ($7.48 \times 8.34 \approx 62.4\text{ lbs}$)

1. Rectangular Water Tanks & Rooms

$\text{Volume} = L \times W \times H$ $\text{Capacity (Gallons)} = V \text{ (cu ft)} \times 7.48$ $\text{Water Weight (lbs)} = \text{Gallons} \times 8.34$

Worked Example: Tender Tank Weight

A wildland water tender carries a rectangular tank measuring 10 feet long, 6 feet wide, and 4 feet high.

1. Calculate Volume: $V = 10 \times 6 \times 4 = 240\text{ cubic feet}$
2. Calculate Water Capacity: $240 \text{ cu ft} \times 7.48 \text{ gal/cu ft} = 1,795.2\text{ gallons}$
3. Calculate Total Water Weight: $1,795.2 \text{ gal} \times 8.34 \text{ lbs/gal} = 14,972\text{ pounds}$

Understanding water weight is critical for structural stability: adding 1,500 gallons of water to a building roof or upper floor adds over 12,500 pounds of dead weight, increasing collapse risk.

2. Cylindrical Water Tanks & Standpipes

$\text{Volume} = \pi r^2 H$

If a vertical cylindrical water tower has a radius of 5 feet and a height of 20 feet: $V = 3.14 \times 5^2 \times 20 = 3.14 \times 25 \times 20 = 3.14 \times 500 = 1,570\text{ cubic feet}$ $\text{Gallons} = 1,570 \times 7.48 = 11,743.6\text{ gallons}$

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Reference Table: Formulas, Constants & Fireground Geometry