Library / Automotive and vehicle dynamics / VEH-03
Automotive and vehicle dynamics
Equivalent mass with rotating inertia
The effective mass that resists acceleration once the spinning wheels and engine are counted.
- Ref
- VEH-03
- Rev
- v1
- Field
- Automotive and vehicle dynamics
- Result
- meq [kg]
m is the vehicle mass, Iw the total moment of inertia of all the wheels, tyres and rotating brake parts, Ie the inertia of the engine and flywheel, G the overall reduction from engine to wheels and Re the rolling radius. It is the kinetic energy of the spinning parts, ½Iω², written as an extra mass at the wheel using ωwheel = U/Re and ωengine = GU/Re. Use the engine-side term only in the gear you are analysing, and do not count an inertia twice.
The effective mass that resists acceleration once the spinning wheels and engine are counted. Rotating parts must be spun up as well as the vehicle moved, so the effective mass is greater than the weight divided by g.
Calculator
- Equivalent mass factor ε = meq / m1.19
- Wheel inertia as extra mass, Iw / Re²44.44 kg
- Engine inertia as extra mass, Ie G² / Re²240 kg
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- At the defaults the rotating parts add 284.4 kg to a 1,500 kg car (19 %). In a tall top gear (G = 4) the extra mass falls to 71.11 kg, because the engine term scales with G².
- This is why a car accelerates noticeably harder in a high gear than the drive force alone would suggest, and why a heavy flywheel slows acceleration most in the low gears.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (kg) | Local step | Change in result |
|---|---|---|---|---|---|---|
| m Vehicle mass | 500 to 5,000 kg | 784.4 | 5,284 | 4,500 | +10% | +8.41 % |
| Ie Engine and flywheel inertia | 0.05 to 2 kg·m² | 1,624 | 4,744 | 3,120 | +10% | +1.34 % |
| G Total reduction ratio between engine and wheels | 1 to 20 | 1,546 | 2,211 | 665 | +10% | +2.82 % |
| Re Effective rolling radius | 0.2 to 0.5 m | 1,602 | 2,140 | 537.6 | +10% | −2.77 % |
| Iw Total inertia of wheels and tyres | 1 to 20 kg·m² | 1,751 | 1,962 | 211.1 | +10% | +0.249 % |
Sweep of Total reduction ratio between engine and wheels
| Total reduction ratio between engine and wheels | Equivalent mass (kg) |
|---|---|
| 1 | 1,546 |
| 3.375 | 1,563 |
| 5.75 | 1,600 |
| 8.125 | 1,654 |
| 10.5 | 1,728 |
| 12.88 | 1,821 |
| 15.25 | 1,932 |
| 17.63 | 2,062 |
| 20 | 2,211 |
Sweep one variable
Variables and how each one contributes
Equivalent mass
What the equation gives you. Everything else feeds it.
Vehicle mass
The mass of the whole vehicle, in kg. Adds one-for-one to the equivalent mass, so a heavier vehicle makes the rotating-inertia share smaller. Raising it by 10 % (to 1,650) moves the result by +8.41 %, from 1,784 to 1,934 kg.
Total inertia of wheels and tyres
The total moment of inertia of all wheels, tyres and rotating brake parts about their axes, in kg·m². A passenger car with four wheels is typically 3 to 6. Adds a term proportional to Iw, independent of the gear. Raising it by 10 % (to 4.4) moves the result by +0.249 %, from 1,784 to 1,789 kg.
Engine and flywheel inertia
The moment of inertia of the engine crankshaft, flywheel and clutch, in kg·m². Typical values are 0.1 to 0.3. Adds a term proportional to Ie that grows with the square of the gear ratio, so it dominates in first gear. Raising it by 10 % (to 0.165) moves the result by +1.34 %, from 1,784 to 1,808 kg.
Total reduction ratio between engine and wheels
The overall reduction between the engine and the wheels in the gear being analysed. The engine term grows with G², so a 10 % bigger ratio raises the engine contribution by about 21 %. Raising it by 10 % (to 13.2) moves the result by +2.82 %, from 1,784 to 1,835 kg.
Effective rolling radius
The effective rolling radius of the tyre, in m. Both rotating terms scale as 1/Re², so a smaller radius makes the spinning parts count for more. Raising it by 10 % (to 0.33) moves the result by −2.77 %, from 1,784 to 1,735 kg.
Worked example with the values above
- Wheel term: Iw / Re² = 4 / 0.3² = 44.44 kg.
- Engine term: Ie G² / Re² = 0.15 × 12² / 0.3² = 240 kg.
- meq = m + wheel term + engine term = 1,500 + 44.44 + 240 = 1,784 kg.
Assumptions and limits
- The wheels roll without slip, and the gear ratio stays fixed over the interval analysed.
- Rotating inertias are lumped into one wheel term and one engine term. Counting driveshaft, gearbox and clutch inertia separately is optional but must not duplicate them.
Common mistakes
- Leaving out the G² factor on the engine term, which makes the effect look the same in every gear.
- Using the equivalent mass for weight or tyre loads. Weight is always m g; the equivalent mass only governs acceleration.
Statistics and uncertainty
meq is a sum, so the uncertainty of the larger term dominates: the vehicle mass. The inertia terms are often known only to 10 to 20 %, but they carry only a few per cent of the total, so a 20 % error in the engine inertia changes meq by about 3 % at the defaults. The gear ratio matters more than its inertia error, because it enters squared.
Where it is used
- Acceleration and coast-down calculations, where ignoring rotating inertia overstates the acceleration.
- Comparing flywheel and wheel-weight choices in low gears.
Maths you need
- Kinetic energy: ½ m U² and ½ I ω²
- Rolling without slip: ω = U / R