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newSchoolWorks.com › VoTechWorks › Transportation › Small Engine & Power Equipment › Universal Core: Safety, Math, Communication & Career for Small Engine & Power Equipment › Lesson 4
TRN-SMAL74-01 · Lesson 4 · 18 clock hours · practice

Trade Math: Ratio, Proportion, Area, Volume, and Electrical Math

Objective. Students apply ratio and proportion, area and volume formulas, and Ohm's law calculations to solve service problems representative of fuel mixing, engine displacement, cutting area, and basic circuit analysis encountered in small engine repair.

⚠ Pending professional review

This lesson, its worked examples and its practice questions has not yet been verified by a certified examiner or master technician in the trade or by a certification or approval body. It is a study draft and may contain errors. Do not use it as instructions for work on a job site or on live equipment. Always follow your instructor, your supervising professional, and current official references and regulations.

VoTechWorks program approval and accreditation are still in progress. Credentials are issued only by the certifying bodies named in each program, after their own exams.

Why Math Matters in Small Engine Work

Math errors in the shop cost money and can cost lives. A displacement figure that is off by several cubic centimeters leads to the wrong replacement parts. An Ohm's law mistake sends you chasing an ignition fault that does not exist. A fuel-mix calculation that is even slightly wrong can seize a two-stroke engine on the first pull or, in the other direction, leave unburned oil fouling the plug and the air. Treat every calculation as a safety task, not a paperwork task.

Ratio and Proportion

A ratio expresses the relationship between two quantities. In small engine work the most common ratio you will see is the fuel-to-oil ratio for two-stroke engines, written as, for example, 50:1. This means 50 parts gasoline to 1 part two-stroke oil by volume.

A proportion states that two ratios are equal: a/b = c/d. You use proportions to scale a known ratio to any volume you need. To find the amount of oil for a given volume of gasoline, rearrange the proportion:

> Oil needed = Total fuel volume ÷ (ratio number + 1)

For a 50:1 mix and 1 gallon of gasoline, the total mixture volume is approximately 1 gallon, so:

> Oil = 128 fluid ounces ÷ 51 ≈ 2.51 fl oz

Most technicians round to the nearest 0.1 fl oz, or they use a pre-calculated reference table. Always write the complete calculation on your work order and on the mixing label. Errors have safety implications: too little oil causes seizure; too much oil causes carbon fouling, rough running, and excessive emissions.

Label safety note: Any fuel-mixing label you create must be legible, permanently attached, and must state the ratio clearly (e.g., "50:1 — add 2.6 fl oz two-stroke oil per 1 gallon gasoline"). Follow the engine or equipment manufacturer's specified ratio — never substitute your own preference.

Area and Volume Formulas

Area of a circle: A = π × r², where r is the radius (half the diameter). You use this when thinking about the bore area of a cylinder — the bore is the diameter of the cylinder.

Engine displacement is the swept volume of all cylinders combined. The standard formula used in the trade is:

> Displacement = Bore² × Stroke × Number of cylinders × 0.7854

All measurements must be in the same unit. If bore and stroke are in inches, the result is in cubic inches (ci). Convert to cubic centimeters (cc) by multiplying by 16.387. The constant 0.7854 is π ÷ 4, the area factor for a circle with a diameter of 1 unit.

Displacement determines engine class for certification purposes, replacement part selection, emissions compliance, and competition or regulatory eligibility. Getting it right means comparing your calculated result to the manufacturer's published spec within 0.1 cc — if you are further off than that, recheck your unit conversions before you recheck your arithmetic.

Volume of a rectangular area: Length × Width × Height. You may use this when estimating oil capacity in a sump or fuel in a tank with a known shape.

Ohm's Law and Basic Electrical Math

Three formulas cover nearly every basic circuit check you will perform on a small engine:

| Formula | Use |

|---|---|

| V = I × R | Find voltage when you know current and resistance |

| I = V ÷ R | Find current when you know voltage and resistance |

| R = V ÷ I | Find resistance when you know voltage and current |

| P = I × V | Find power (watts) when you know current and voltage |

In small engine ignition and charging circuits, typical values are low voltages (DC charging systems often operate at 12 V or under) and resistances measured in ohms. A primary coil winding might measure a few ohms; a kill-switch wire with a poor connection might show unexpected resistance that drops available voltage below threshold.

Voltage drop across a component equals I × R for that component. If you know the rated current through a circuit and you measure the resistance of a suspect wire or connector, you can calculate how much voltage is being lost before it reaches the coil, solenoid, or sensor. Excessive voltage drop explains many no-start and weak-spark complaints without requiring expensive part swaps.

Electrical safety note: Before probing any circuit, identify the system voltage and confirm the circuit is de-energized where required. On equipment with capacitor-discharge ignition (CDI), treat the secondary (high-voltage) side as dangerous even when the engine is stopped — residual charge can remain. Always wear appropriate PPE and follow the equipment manufacturer's service manual for discharge procedures before handling CDI components.

Working Accurately Under Shop Conditions

Use a scientific calculator for every multi-step problem; mental math is for estimates only. Record all given values, your formula, your substituted values, and your answer with units before you move to the next step. Unit labels are not optional — "2.5" means nothing without "fl oz" or "cc" after it.

When your answer does not match the published spec, work backward step by step. The most common errors are: wrong unit conversion, using diameter instead of radius (or vice versa), and mis-keying a decimal point. A 10× error almost always means a decimal mistake.

Key terms8

RatioA comparison of two quantities expressed as a/b or a:b; in small engine work, commonly the proportion of gasoline to two-stroke oil, such as 50:1.
ProportionAn equation stating that two ratios are equal, used to scale a known ratio to a different total volume when mixing fuel or diluting fluids.
Engine DisplacementThe total swept volume of all cylinders, calculated as Bore² × Stroke × Number of cylinders × 0.7854; expressed in cubic inches or cubic centimeters.
BoreThe internal diameter of an engine cylinder, measured in inches or millimeters; used in displacement and compression calculations.
StrokeThe distance the piston travels from bottom dead center to top dead center inside the cylinder; used together with bore to calculate displacement.
Ohm's LawThe relationship V = I × R, where V is voltage in volts, I is current in amperes, and R is resistance in ohms; the foundation of basic circuit analysis.
Voltage DropThe reduction in voltage across a component or conductor due to its resistance, calculated as I × R; excessive voltage drop indicates a wiring or connection fault.
Power (Electrical)The rate of energy use in a circuit, calculated as P = I × V and expressed in watts; used to check whether components are operating within their rated limits.

Worked examples4

1. Calculate the displacement of a single-cylinder engine with a bore of 2.76 inches and a stroke of 2.44 inches. Express the answer in cubic inches and cubic centimeters.
  1. Write the formula: Displacement = Bore² × Stroke × Cylinders × 0.7854
  2. Substitute values: Displacement = (2.76)² × 2.44 × 1 × 0.7854
  3. Square the bore: 2.76 × 2.76 = 7.6176 in²
  4. Multiply by stroke: 7.6176 × 2.44 = 18.587 in³
  5. Multiply by 0.7854: 18.587 × 0.7854 = 14.598 in³ ≈ 14.60 ci
  6. Convert to cc: 14.60 × 16.387 = 239.2 cc
Answer: Approximately 14.60 cubic inches (239.2 cc)
2. A two-stroke chainsaw requires a 50:1 fuel-to-oil ratio. How many fluid ounces of two-stroke oil are needed to mix with exactly 2 gallons of gasoline?
  1. Convert 2 gallons to fluid ounces: 2 × 128 = 256 fl oz of gasoline
  2. At 50:1, there are 50 parts gasoline to 1 part oil, so the total parts = 51
  3. Oil needed = Total gasoline volume ÷ ratio = 256 ÷ 50 = 5.12 fl oz
  4. Note: dividing by 50 (not 51) is the standard field method used in the trade when the gasoline volume is already measured; this is acceptable practice and matches most manufacturer mixing charts
Answer: 5.12 fl oz of two-stroke oil (approximately 5.1 fl oz)
3. A small engine ignition kill-switch circuit operates on 12 V. The kill-switch wire has a measured resistance of 3.5 ohms and carries 0.8 amperes when the circuit is active. What is the voltage drop across that wire, and how many watts are dissipated?
  1. Voltage drop: V = I × R = 0.8 A × 3.5 Ω = 2.8 V
  2. This means only 12 − 2.8 = 9.2 V reaches the load on the other side of this wire — potentially a fault condition
  3. Power dissipated: P = I × V = 0.8 A × 2.8 V = 2.24 W
Answer: Voltage drop = 2.8 V; Power dissipated = 2.24 W
4. A technician needs to find the current flowing through a stator charging circuit component that has a resistance of 6 ohms and is connected to a 12 V source. What is the current?
  1. Use the Ohm's law rearrangement: I = V ÷ R
  2. Substitute: I = 12 V ÷ 6 Ω
  3. Calculate: I = 2.0 A
Answer: Current = 2.0 amperes

The math in this lesson2

Learners convert engine displacement from 14.60 cubic inches to about 239 cc.
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Brush up in the math halls at newSchoolWorks.com:
In class: Calculate engine displacement (bore × stroke × number of cylinders × 0.7854) for three different small engines using given bore and stroke data, then verify by comparing to published manufacturer displacement figures. · Solve a set of 15 Ohm's law problems (V = IR, power P = IV) drawn from small engine ignition and charging circuit scenarios, such as determining expected voltage drop across a resistive component at rated current. · Design a fuel-mixing label for a hypothetical two-stroke product line specifying the correct amount of two-stroke oil for 1, 2, and 5 gallons of gasoline at three different ratios (32:1, 40:1, 50:1), showing all math work.
Materials: Scientific calculator, trade math workbook or instructor-prepared worksheets, engine spec sheets (bore and stroke data), Ohm's law reference card, blank label template, colored markers for label design
Check: Displacement calculations: within 0.1 cc of published spec; Ohm's law problems: 80% correct; fuel-mixing label: all nine values mathematically correct (100% required, as errors have safety implications) and legible.

Practice check8

Answer each question, then check it. Nothing is recorded; this is your own practice.

Trade Math / Engine Theory
1. A technician uses the formula Displacement = Bore² × Stroke × Cylinders × 0.7854. What does the constant 0.7854 represent?
Trade Math / Fuel Systems
2. A two-stroke trimmer specifies a 40:1 fuel-to-oil ratio. How many fluid ounces of two-stroke oil should be added to 1 gallon (128 fl oz) of gasoline?
Electrical Systems / Ohm's Law
3. Using Ohm's law, what is the resistance of a component if 12 V is applied and 2.4 A flows through it?
Electrical Systems / Circuit Analysis
4. A technician measures 0.5 Ω of resistance in a ground wire that carries 3 A. What voltage is dropped across that wire?
Safety / Trade Math
5. Why is 100% accuracy required when calculating fuel-mixing values for a two-stroke label, rather than the 80% threshold used for Ohm's law problems?
Trade Math / Engine Theory
6. A single-cylinder engine has a bore of 2.50 inches and a stroke of 1.88 inches. Which calculation correctly begins the displacement formula?
Electrical Systems / Charging Circuits
7. A charging stator output measures 14.5 V at rated RPM and drives 3.5 A into the battery circuit. What is the power output in watts?
Safety / Electrical Systems
8. When working near capacitor-discharge ignition (CDI) components, a technician should:
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