Designing a conveyor system involves more than just bolting a motor to a frame. If you undersize the motor, it won't start under load due to breakaway torque . If you oversize it, you waste thousands on electricity and oversized VFDs. In this guide, we will walk through the engineering math required to size a conveyor motor and gearbox correctly, specifically focusing on the critical "Dynamic Tension" resulting from inertia. Table of Contents 1. The Physics: Effective Pull (Te) 2. Calculating Motor Power (Worked Example) 3. The Inertia Problem: VFD vs DOL 4. Gearbox Ratio Selection 5. Frequently Asked Questions Advertisement 1. The Physics: Effective Pull (Te) The first step in any sizing calculation is determining the Effective Pull ( T e ) . This is the sum of all forces resisting the motion of the belt. The Basic Formula: T e = F friction + F gravity + F material...
Figure 1: Modern flywheels act as mechanical batteries, storing kinetic energy (0.5 Jω²) for rapid release.
The energy-storage capacity of a flywheel is determined essentially by two factors: its Polar Moment of Inertia (J) and its Rotational Speed (ω).
While modern "Grid-Scale" flywheels are used as massive batteries, the principal use in machine design is to smooth out ripples in shaft speed. By absorbing torque spikes (excess energy) and releasing it during dead spots, a flywheel acts as a mechanical reservoir.
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Step 1: Determining the Allowable Fluctuation (Cs)
No machine runs at a perfectly constant speed. The goal is to keep the variation within harmless limits. We define this limit as the Coefficient of Speed Fluctuation (Cs):
Cs = ( ωmax - ωmin ) / ωavg
Where ω (omega) is the angular velocity in rad/s.
| Machine Application | Recommended Cs |
|---|---|
| Stone Crushers / Hammers | 0.200 |
| Punch Presses / Shears | 0.100 - 0.150 |
| Pumps & Compressors | 0.030 - 0.050 |
| Precision Machine Tools | 0.020 - 0.030 |
| Electric Generators | 0.003 - 0.005 |
Step 2: Calculating Inertia (J)
Once Cs is selected, we calculate the required Mass Moment of Inertia (J). This tells us "how heavy" or "how large" the flywheel needs to be.
Figure 2: The flywheel must absorb the "Excess Energy" (Green Area) to prevent the shaft from speeding up.
The fundamental energy equation is the same for both systems, but the units differ.
J = E / ( Cs · ωavg2 )
SI Units (Metric)
- J: Inertia in kg·m²
- E: Energy in Joules (J)
- ω: Speed in rad/s
Note: If using RPM, convert to rad/s:
ω = (2π · RPM) / 60
Imperial Units
- J: Inertia in lb·ft·s² (slug·ft²)
- E: Energy in ft·lb
- ω: Speed in rad/s
Caution: Ensure weight (lbs) is converted to mass (slugs) by dividing by gravity (g=32.2).
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Step 3: Energy Estimation for Engines
For internal combustion engines, we often estimate the Energy Variation (E) based on the engine's power if the exact torque curve is unknown.
E ≈ ( K · Power ) / N
|
For SI Units: K ≈ 60 (Constant) Power in Watts N in RPM Result E in Joules |
For Imperial Units: K ≈ 33,000 (Constant) Power in Horsepower (hp) N in RPM Result E in ft·lb |
Step 4: Safety Check (Hoop Stress)
Knowing J is only half the battle. You must design a shape that provides that inertia without bursting. The limiting factor is Hoop Stress (σ) caused by centrifugal force.
Figure 3: A "Rim" type flywheel is more efficient because mass is concentrated further from the center.
SI Formula:
σ = ρ · v²
σ = Stress (Pa or N/m²)
ρ = Density (kg/m³)
v = Rim Velocity (m/s)
σ = ρ · v²
σ = Stress (Pa or N/m²)
ρ = Density (kg/m³)
v = Rim Velocity (m/s)
Imperial Formula:
σ = (w · v²) / g
σ = Stress (lb/ft²)
w = Specific Weight (lb/ft³)
v = Rim Velocity (ft/s)
g = Gravity (32.2 ft/s²)
σ = (w · v²) / g
σ = Stress (lb/ft²)
w = Specific Weight (lb/ft³)
v = Rim Velocity (ft/s)
g = Gravity (32.2 ft/s²)
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