BIZOMAT MAKEOS · Mechanics
Flywheel Rim Design (Inertia for Speed Fluctuation, Size and Stress)
What inertia does this flywheel need to hold the speed fluctuation within limits, what disc size gives it, and is the rim speed and stress safe?
Design of a flywheel as a plain annular disc: inertia needed for the energy fluctuation at the mean speed and coefficient of speed fluctuation, the inner diameter that gives that inertia at the outer diameter and width chosen, mass, the energy check by the flywheel energy method, the speed range, rim speed at the top speed, and the largest stress of a rotating disc at the top speed (as the rotating disc tool) against your safety factor.
Open the calculatorRuns in your browser. No account needed.
How this tool works
Inputs, units and where to find them
| Input | Unit | Typical source |
|---|---|---|
| Energy fluctuation per cycle | kJ | Largest difference in kinetic energy over the cycle (from the torque–angle diagram). |
| Mean speed | rpm | Mean running speed. |
| Coefficient of speed fluctuation | Allowed (n_max − n_min) ÷ n_mean (about 0.002 generators, 0.02–0.05 machine tools, 0.1–0.2 punches and crushers). | |
| Outer diameter | mm | Outside diameter of the flywheel. |
| Width (axial) | mm | Width of the disc or rim along the shaft. |
| Density | kg/m³ | Density (about 7200 cast iron, 7850 steel). |
| Poisson’s ratio | Poisson’s ratio. | |
| Strength (yield or tensile for cast iron) | MPa | Strength of the material for the stress check. |
| Safety factor required | Safety factor you require on the rotating stress (higher for cast iron). | |
| Largest rim speed | m/s | Largest rim speed for the material and construction (your value). |
What each input means
Open an input to see what it is, its units, the worked example and the common mistakes, with a small diagram.
What this means: Energy fluctuation per cycle
- What it is
- The largest swing in kinetic energy the flywheel must absorb and release in one cycle, from the area of the torque against crank angle diagram.
- Units and format
- Enter a number in kJ: kilojoules. The form accepts 0.000001 kJ to 100000 kJ. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 5 kJ as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- The inertia needed rises in direct proportion.
- Get it right
- Enter kJ, not J. Use the largest excess energy over the cycle, not the total work per cycle.
What this means: Mean speed
- What it is
- The mean running speed of the flywheel, halfway between its highest and lowest speed in the cycle.
- Units and format
- Enter a number in rpm: revolutions per minute. The form accepts 0.000001 rpm to 100000 rpm. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 600 rpm as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- The inertia needed falls with the square of speed, so a faster flywheel can be much smaller.
- Get it right
- Use the speed of the flywheel shaft itself, not the motor if there is a reduction between them.
What this means: Coefficient of speed fluctuation
- What it is
- The speed variation you allow: highest minus lowest speed, divided by the mean speed.
- Units and format
- Enter a plain number with no unit. The form accepts 0.002 to 0.5. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 0.05 as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- A tighter (smaller) coefficient needs proportionally more inertia.
- Get it right
- Type it as a fraction, not a percentage.
What this means: Outer diameter
- What it is
- The outside diameter of the flywheel disc. The tool works out the inner diameter that gives the inertia needed.
- Units and format
- Enter a number in mm: millimetres. The form accepts 0.000001 mm to 10000 mm. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 800 mm as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- A larger diameter stores inertia more effectively but raises rim speed and stress.
- Get it right
- Choose it within the space available; the rim speed check limits it.
What this means: Width (axial)
- What it is
- The axial width of the disc or rim, measured along the shaft.
- Units and format
- Enter a number in mm: millimetres. The form accepts 0.000001 mm to 2000 mm. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 100 mm as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- A wider disc reaches the inertia with a larger bore and less depth.
- Get it right
- The whole disc is taken as this width; a rim with a thinner web stores less.
What this means: Density
- What it is
- The density of the flywheel material, which sets both the inertia per unit volume and the rotating stress.
- Units and format
- Enter a number in kg/m³: kilograms per cubic metre. The form accepts 0.000001 kg/m³ to 20000 kg/m³. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 7200 kg/m³ as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Get it right
- Take it from the material data sheet for the grade used.
What this means: Poisson’s ratio
- What it is
- The Poisson’s ratio of the flywheel material, used in the rotating disc stress formula.
- Units and format
- Enter a plain number with no unit. The form accepts 0.1 to 0.5. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 0.26 as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Get it right
- Take it from material data; it is a pure number, not a percentage.
What this means: Strength (yield or tensile for cast iron)
- What it is
- The strength of the flywheel material used for the stress check: the yield strength for steel, or the tensile strength for brittle cast iron.
- Units and format
- Enter a number in MPa: megapascals (1 MPa = 1 N/mm²). The form accepts 0.000001 MPa to 3000 MPa. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 150 MPa as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Effect on the result
- The safety factor rises in direct proportion.
- Get it right
- Use the strength for the actual casting or forging, which can be lower in thick sections.
What this means: Safety factor required
- What it is
- The smallest safety factor you accept between the material strength and the largest rotating stress.
- Units and format
- Enter a plain number with no unit. The form accepts 1 to 20. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 5 as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Get it right
- Use the factor your design rules require; brittle materials and burst risk call for a higher factor.
What this means: Largest rim speed
- What it is
- The highest rim surface speed you accept for the material and construction, compared with the rim speed at top speed.
- Units and format
- Enter a number in m/s: metres per second. The form accepts 0.000001 m/s to 500 m/s. Required: the tool does not calculate until every input is valid.
- Example
- The form opens with 30 m/s as a worked example. It is an illustration, not a recommended value: replace it with your own figure.
- Get it right
- Take it from your material and design rules for flywheels.
Method and formulas
- Inertia I = ΔE ÷ (ω² C_s), ω the mean angular speed.
- Annular disc I = ρ π b (R_o⁴ − R_i⁴) ÷ 2, so R_i = [R_o⁴ − 2I ÷ (ρ π b)]^¼; mass = ρ π b (R_o² − R_i²).
- Speed range n (1 ± C_s ÷ 2); energy check ΔE = I (ω_max² − ω_min²) ÷ 2 (as the flywheel energy tool).
- Rotating disc stress at the bore at top speed: σ = (3 + ν) ÷ 4 × ρ ω² [R_o² + (1 − ν) ÷ (3 + ν) R_i²] (as the rotating disc tool).
- Example: 5 kJ at 600 rpm needs 25.33 kg·m²: a 800 mm disc 100 mm wide with a 475.8 mm bore, 233.89 kg.
Limitations
- Plain annular disc; spoked and rimmed flywheels have different inertia and stress.
- Hub, keys and bolts need their own checks.
- Example values are illustrations. Replace them with values for your actual task.
Typical uses
- Size a flywheel for a compressor, press or engine
- Check a flywheel for a higher speed
- Trade diameter against width and mass
Who it is for
- Design engineers
- Mechanical engineers
- Maintenance engineers
What you get
- Inertia needed
- Inner diameter, mass and speed range
- Rim speed, rotating stress and safety factor
Interpreting the result
- Inertia rises with the fourth power of diameter: a slightly larger diameter saves a lot of mass, until rim speed limits it.
References
- Machinery’s Handbook, 29th edition (Industrial Press, 2012), printed pp. 3000–3010, 179–190. Topic reference; no source tables or recommended settings are reproduced.
Learn the Concepts Behind This Tool
These BIZOMAT Library readings explain the method this tool uses, with worked examples.
Method explained here · BIZOMAT original book
Designing a flywheel
In Inertia and Flywheels by the Numbers
What it covers: How to size a flywheel rim from an energy fluctuation and speed variation: inertia, bore, mass, rim speed and rotating stress, and the effect of putting it on the fast or slow shaft.
BIZOMAT tools are screening and decision-support utilities. Verify important financial, engineering, safety, legal or operational decisions using appropriate professional standards and your organisation’s approved data.
Open the Flywheel Rim Design (Inertia for Speed Fluctuation, Size and Stress) calculator