5.3 Perimeter, Area & Spatial Geometry at Crime Scenes

5.3 Perimeter, Area & Spatial Geometry at Crime Scenes

When patrol officers arrive at major crime scenes, fatal traffic collisions, or hazardous industrial incidents, establishing physical control of the environment is the foremost operational priority. Officers must secure evidence, establish secure cordons to preserve chain of custody under the Fourth Amendment, coordinate multi-officer evidence searches across open terrain, and document roadway skid marks before weather or vehicle traffic destroys perishable physical clues.

All of these tactical functions rely on applied geometry. FPSI describes PACT math as basic addition, subtraction, multiplication, and division, and its published samples do not include geometry formulas. Perimeter and area problems are still excellent practice, because they are the same four operations applied in a few steps, and they must be solved on scratch paper without a calculator. The skid-mark formula at the end of this section goes beyond FPSI's published samples; treat it as enrichment, and expect any formula to be supplied if a test uses one.

---

Perimeter Fundamentals & Crime Scene Barrier Taping

The perimeter of a geometric figure is the total linear distance around its outer boundary. In field policing, perimeter calculations determine the exact yardage or footage of yellow barrier tape required to secure an incident zone.

The Rectangular Perimeter Formula

Most buildings, residential parcels, and cordoned roadway intersections are rectangular. The perimeter of a rectangle is calculated by summing all four boundary sides:

$\text{Perimeter} = \text{Length} + \text{Width} + \text{Length} + \text{Width} = 2L + 2W = 2(L + W)$

Where:

• $L$ is the length of the rectangular boundary.
• $W$ is the width of the rectangular boundary.

Cordoning Two-Tiered Crime Scenes: Inner vs. Outer Perimeters

Crime-scene guidance commonly calls for two perimeters at a major scene:

1. Inner Perimeter: Encloses the immediate incident area, physical evidence, victims, weapon drop zones, and trace materials. Access is strictly restricted to authorized primary investigators and crime scene technicians. A designated log officer records every entry and exit by name, badge number, and time.
2. Outer Perimeter: Encloses the inner perimeter and creates a protective buffer zone. It encompasses the mobile incident command post, equipment staging, patrol vehicle parking, witness holding areas, and a media briefing point.

Loading diagram...

The Concentric Offset Rule (Adding Setbacks to Both Sides)

Perimeter word problems often describe an outer barrier placed a fixed setback distance ($d$) away from a building or inner perimeter in all four directions.

$\text{Length}_{\text{outer}} = L + 2d$ $\text{Width}_{\text{outer}} = W + 2d$ $\text{Perimeter}_{\text{outer}} = 2(L + 2d) + 2(W + 2d) = 2L + 2W + 8d = \text{Perimeter}_{\text{inner}} + 8d$

Step-by-Step Cordoning Calculation

Scenario: Officers respond to a commercial warehouse burglary where physical evidence is scattered outside the building. The warehouse footprint measures 100 feet long by 60 feet wide. The incident commander directs officers to:

1. Run inner barrier tape tightly around the warehouse exterior walls.
2. Run outer barrier tape 20 feet beyond the building walls in all four directions.
3. Determine how many 500-foot rolls of barrier tape are needed to complete both perimeters.

Step 1: Calculate Inner Perimeter Tape: $\text{Perimeter}_{\text{inner}} = 2(L) + 2(W) = 2(100) + 2(60) = 200 + 120 = 320\text{ feet}$

Step 2: Calculate Outer Dimensions with 20-Foot Setback:

• Length: $L_{\text{outer}} = 100 + 20 + 20 = 140\text{ feet}$
• Width: $W_{\text{outer}} = 60 + 20 + 20 = 100\text{ feet}$

Step 3: Calculate Outer Perimeter Tape: $\text{Perimeter}_{\text{outer}} = 2(140) + 2(100) = 280 + 200 = 480\text{ feet}$ (Note: Using the shortcut formula $\text{Perimeter}_{\text{inner}} + 8d = 320 + 8(20) = 320 + 160 = 480\text{ feet}$.)

Step 4: Total Tape Required and Roll Procurement: $\text{Total Tape} = 320 + 480 = 800\text{ feet}$

• Since barrier tape is packaged in 500-foot rolls: $\frac{800}{500} = 1.6\text{ rolls}$
• Because supply rolls cannot be purchased as partial units, the officers must retrieve 2 full 500-foot rolls (1,000 feet total) to secure the scene.

---

Area Calculations for Premises & Evidence Search Sectors

While perimeter measures linear boundary distance, area measures the two-dimensional surface space enclosed within a boundary, expressed in square units (such as square feet, square yards, or acres).

Primary Geometric Area Formulas

1. Rectangular Zones: $\text{Area} = \text{Length} \times \text{Width} = L \times W$
2. Triangular Sectors (used when searching triangular park quadrants, highway off-ramp medians, or alley sightlines): $\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{b \times h}{2}$
3. Composite (L-Shaped) Zones: Break the irregular polygon into non-overlapping sub-rectangles, compute the area of each sub-rectangle, and sum the results.

+-----------------------+
|      Sector A         |
|   (40 ft x 30 ft)     |
|     1,200 sq ft       |
+-----------+-----------+
            | Sector B  |
            | (20 x 50) |
            | 1,000 sq  |
            +-----------+
Total Composite Area = 1,200 + 1,000 = 2,200 sq ft

Personnel Workload Allocation by Square Footage

When commanding large search operations, supervisory personnel must distribute search zones fairly among available officers based on square footage.

Scenario: An evidence search team must canvass an open rectangular field measuring 120 feet wide by 250 feet long for a discarded firearm. The team consists of 6 search officers. How many square feet of search area is each officer responsible for covering?

1. Calculate total field area: $\text{Total Area} = L \times W = 250\text{ ft} \times 120\text{ ft} = 30{,}000\text{ sq ft}$
2. Divide total area by the number of officers: $\text{Area per Officer} = \frac{30{,}000\text{ sq ft}}{6\text{ officers}} = 5{,}000\text{ square feet per officer}$

---

Search Logistics: Strip / Lane Searches and Swath Spacing

In field evidence recovery, officers utilize structured search patterns to ensure $100\%$ ground coverage. The most common tactical pattern is the strip or lane search (also called a line search).

Strip Search Mechanics

• A line of officers stands abreast along the starting boundary line.
• Officers walk forward in lockstep, scanning the ground immediately in front of them within a defined search lane.
• Lane Width (Swath): The lateral distance assigned to each searcher. In heavy vegetation, tall grass, or when searching for tiny physical evidence (such as 9mm cartridge casings or teeth), lane width is narrowed to 3 to 4 feet. In open pavement or short grass, lane width expands to 6 to 10 feet.

$\text{Total Lanes Required} = \frac{\text{Total Width of Search Zone}}{\text{Lane Width per Officer}}$ $\text{Number of Team Passes} = \frac{\text{Total Lanes Required}}{\text{Number of Officers Walking Abreast}}$

Step-by-Step Line Search Problem

Scenario: A search coordinator must systematically search a 160-foot wide by 300-foot long open rectangular parcel for discarded narcotics packaging. Because the packaging is small, the coordinator narrows the lane width to 4 feet per officer even though the grass is mowed short. The department provides an 8-officer search team.

1. Calculate total search lanes across the 160-foot width: $\text{Total Lanes} = \frac{160\text{ feet}}{4\text{ feet/lane}} = 40\text{ lanes}$
2. Calculate the lateral width cleared during each pass by the 8-officer team: $\text{Width cleared per pass} = 8\text{ officers} \times 4\text{ feet/officer} = 32\text{ feet per pass}$
3. Calculate the number of full passes needed to cover the 160-foot zone: $\text{Number of Passes} = \frac{40\text{ lanes}}{8\text{ officers}} = \frac{160\text{ feet}}{32\text{ feet/pass}} = 5\text{ passes}$
4. Result: The team must complete exactly 5 passes across the 300-foot length to sweep the entire zone.

---

Skid Mark Distance & Collision Speed Reconstruction

At vehicular crash scenes, tire skid marks represent irreplaceable physical evidence of emergency braking. When vehicle brakes lock wheels, kinetic energy is converted into thermal friction as tires slide across the road surface. By measuring the length of these skid marks and determining the surface friction, investigators calculate the vehicle's minimum speed prior to braking.

Averaging Four-Wheel Skid Distances

Because vehicle weight shifts forward during hard braking, each of the four tires leaves skid marks of varying lengths. To establish the representative skid distance ($d$), investigators measure all four individual marks and compute their arithmetic average:

$d_{\text{avg}} = \frac{d_{\text{Front Left}} + d_{\text{Front Right}} + d_{\text{Rear Left}} + d_{\text{Rear Right}}}{4}$

Example: Investigators measure four locked-wheel tire marks on asphalt:

• Front Left: $82\text{ feet}$
• Front Right: $86\text{ feet}$
• Rear Left: $76\text{ feet}$
• Rear Right: $80\text{ feet}$

$d_{\text{avg}} = \frac{82 + 86 + 76 + 80}{4} = \frac{324}{4} = 81\text{ feet}$

The Standard Minimum Speed Formula

In traffic crash investigation, the simplified minimum speed formula (assuming level road and 100% braking efficiency) is expressed as:

$S = \sqrt{30 \times d \times f}$

Where:

• $S$ is the minimum pre-braking vehicle speed in miles per hour (mph).
• $30$ is the gravitational conversion constant derived from physics ($2g \approx 64.4\text{ ft/s}^2$ converted to mph).
• $d$ is the average skid mark distance in feet.
• $f$ is the drag factor (coefficient of friction) of the roadway surface.

Drag Factor ($f$) Reference Values

Published tables vary with the age of the surface, temperature, and tires, so a problem will always state the drag factor to use.

Step-by-Step Speed Calculation Walkthrough

Scenario: A vehicle skids to a complete halt on dry asphalt with a verified drag factor of $f = 0.75$. The average skid mark length across all four wheels is measured at 90 feet. What was the minimum speed of the vehicle when braking commenced?

1. Multiply the constant, distance, and drag factor inside the radical: $30 \times d \times f = 30 \times 90 \times 0.75$
2. Perform intermediate multiplication on scratch paper: $30 \times 90 = 2{,}700$ $2{,}700 \times 0.75 = 2{,}700 \times \frac{3}{4} = \frac{8{,}100}{4} = 2{,}025$
3. Extract the square root of 2,025: $S = \sqrt{2{,}025}$
4. Find the square root using standard benchmark testing squares:
   • $40^2 = 1{,}600$
   • $50^2 = 2{,}500$
   • Because the last digit of 2,025 is 5, test $45$: $45 \times 45 = 2{,}025$
5. Result: The vehicle was traveling at a minimum speed of 45 mph.

Perimeter, area, and search-logistics problems build the same multi-step, four-operation fluency that the PACT's Mathematical Reasoning items reward.