Machine Dynamics, Mechanical Press Flywheels & Rotational Inertia
Flywheel Speed Fluctuation (Cs) & Moment of Inertia Formula
Calculate Mean Angular Velocity omega (rad/s), Required Mass Moment of Inertia I (kg·m²) for Target Speed Fluctuation Cs, Equivalent Thin-Rim Mass (kg), and Rim Tip Speed (m/s).
Delta_E equals I * omega^2 * Cs—WITHOUT the 1/2 factor—because (omega_max^2 - omega_min^2) factors into 2*omega*d_omega!
During a punch-press shearing stroke or single-cylinder compressor compression stroke, cyclic torque demand exceeds motor torque by Maximum Fluctuation of Energy Delta_E (J), momentarily slowing the shaft from omega_max to omega_min. Students frequently write 0.5 * I * omega_mean^2 * C_s by analogy with total kinetic energy 0.5 * I * omega^2—under-sizing inertia or miscalculating speed droop by 2x! Algebraically, 0.5 * I * (omega_max^2 - omega_min^2) = 0.5 * I * (omega_max + omega_min) * (omega_max - omega_min) = I * omega_mean^2 * C_s, where Coefficient of Speed Fluctuation C_s = (omega_max - omega_min) / omega_mean (typically 0.02 for punch presses, 0.003 for synchronous generators)—the 0.5 cancels cleanly against 2 * omega_mean!
Verify on a 600 RPM Mechanical Stamping Press—Stroke_Energy_Delta_E = 12,000 J (12.0 kJ, B2), Mean_Shaft_RPM = 600.0 RPM (B3 -> omega_mean = 20 * PI() = 62.83 rad/s in E2), Target_Fluctuation_Coeff C_s = 0.02 (±1.0% speed band, B4), and Mean_Rim_Radius R = 0.50 m (B5): Required Flywheel Inertia I = 12,000 / (62.83185^2 * 0.02) = 151.98 kg·m² (E3), Equivalent Concentrated Rim Mass m = I / R^2 equals 607.93 kg (E4), and Rim Peripheral Velocity omega * R stays safely at 31.42 m/s (E5, below the ~35–40 m/s cast-iron tensile hoop-burst limit)!
| Press Flywheel Sizing KPI (dE = 12,000 J, 600 RPM, Cs = 0.02, R = 0.50 m) | Formula | Verified Output |
|---|---|---|
| Mean Angular Speed omega (rad/s) | (2 * PI() * Mean_RPM) / 60 | 62.83 rad/s |
| Required Inertia I (kg·m²) | Delta_E_J / ((omega^2) * Cs) | 151.98 kg·m² (303.96 @ Cs = 0.01) |
| Equivalent Thin-Rim Mass (kg) | Required_Inertia_I / (Rim_Radius_m^2) | 607.93 kg (1,215.85 @ Cs = 0.01) |
| Rim Peripheral Velocity (m/s) | omega_rad_s * Rim_Radius_m | 31.42 m/s |
Exact Flywheel Moment of Inertia Formula for Target Speed Droop Cs
Validate Energy_Delta_J > 0, Mean_RPM > 0, Speed_Fluctuation_Cs in (0.001, 0.25], and Rim_Radius_m > 0!
=LET(de_j, B2, rp_m, B3, cs_f, B4, rr_m, B5, IF(OR(de_j <= 0, rp_m <= 0, cs_f <= 0.001, cs_f > 0.25, rr_m <= 0), "Invalid Flywheel Input", LET(om_g, 2 * PI() * rp_m / 60, ROUND(de_j / ((om_g ^ 2) * cs_f), 2))))Tighter ±0.5% Generator-Grade Speed Regulation Check (Set Cs = 0.01)
Halve allowable Speed_Fluctuation_Cs B4 from 0.02 down to 0.01: Required Inertia E3 doubles from 151.98 to 303.96 kg·m² and Equivalent Rim Mass E4 doubles from 607.93 to 1,215.85 kg!
=IF(ISTEXT(E3), E3, LET(om_g, 2 * PI() * B3 / 60, in_r, B2 / ((om_g ^ 2) * B4), ROUND(in_r / (B5 ^ 2), 2)))Try the example
Use the original values or change them to test your understanding.