Calculation
TK = JL/(JA+JL) · (TA−TL) + TL
A large load inertia JL relative to JA pushes the clutch torque toward the drive torque TA; a small JL pushes it toward the load torque TL.
TK = JL/(JA+JL)·(TA−TL)+TL
Clutch engagement torque during start-up (two-mass system)
A large load inertia JL relative to JA pushes the clutch torque toward the drive torque TA; a small JL pushes it toward the load torque TL.
MINTSI
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Result
Select a target and calculate.
Understand the inputs
- Clutch torque TK — Torque transmitted during the clutch engagement.
- Driver-side inertia JA — Inertia of all driver-side masses, referred to the clutch shaft.
- Load-side inertia JL — Inertia of all load-side masses, referred to the clutch shaft.
- Drive torque TA — Torque supplied by the driver during engagement.
- Load torque TL — Assumed constant opposing torque on the load side; 0 for the no-load start-up case.
Example
JA = 0.02 kg·m², JL = 0.08 kg·m², TA = 50 N·m and TL = 10 N·m give TK = 0.8·40+10 = 42 N·m.
Assumptions and limits
Ideal two-mass system with constant drive and load torque during slip; shaft elasticity and damping are excluded.
Understand Clutch engagement torque during start-up (two-mass system)
This calculator determines the torque transmitted while engaging a clutch in a two-mass system, from the driver- and load-side inertias plus drive and load torque.
What does this quantity describe?
When starting up through a slipping clutch, drive torque TA and load torque TL split across the actually transmitted clutch torque TK in proportion to the ratio of the two inertias JA and JL: TK = JL/(JA+JL)·(TA−TL) + TL. For the no-load special case (TL = 0), this simplifies to TK = JL/(JA+JL)·TA — a pure inertia split.
Picture two flywheels of different weight connected through a slipping clutch: a very light flywheel on the load side (small JL) can be spun up with almost no resistance, so the transmitted torque approaches the load torque; a very heavy flywheel (large JL) demands nearly the full drive torque to be accelerated.
Formula and variables
TK = JL/(JA+JL) · (TA−TL) + TL
TK = JL/(JA+JL) · (TA−TL) + TLTA = (TK−TL)·(JA+JL)/JL + TLTL = 0 → TK = JL/(JA+JL) · TA
| Symbol / input | Meaning |
|---|---|
| Clutch torque TK | Torque transmitted during the clutch engagement. |
| Driver-side inertia JA | Inertia of all driver-side masses, referred to the clutch shaft. |
| Load-side inertia JL | Inertia of all load-side masses, referred to the clutch shaft. |
| Drive torque TA | Torque supplied by the driver during engagement. |
| Load torque TL | Assumed constant opposing torque on the load side; 0 for the no-load start-up case. |
Choose the inputs correctly
JA is the inertia of all driver-side masses referred to the clutch shaft, JL the corresponding load-side inertia, TA the torque supplied by the driver during engagement, and TL the assumed constant opposing torque on the load side (0 for the no-load start-up case).
How to use the calculator
Enter JA, JL, TA and TL to get TK. For the no-load start-up case, set TL = 0.
Worked example
JA = 0.02 kg·m², JL = 0.08 kg·m², TA = 50 N·m and TL = 10 N·m give TK = 0.08/0.10·(50−10)+10 = 42 N·m.
Understand the result and units
TK = 42 N·m is the torque actually transmitted through the clutch during slip, which governs the clutch's rating (nominal torque, switching work) — not the full drive torque TA and not the pure load torque TL.
JA and JL are given in kg·m², TA, TL and TK in N·m.
The no-load special case TL = 0
Setting TL = 0 simplifies the formula to TK = JL/(JA+JL)·TA. This expression shows clearly that the transmitted torque is a pure split of the drive torque proportional to the load inertia's share of total inertia: if JL is very small relative to JA, only a small fraction of TA is transmitted (the load spins up easily); if JL is very large, nearly the full drive torque is transmitted because the heavy load can only be accelerated slowly.
Typical applications
The formula is used when sizing slipping friction clutches for start-up events, e.g. on conveyors, presses, machine tools and any drive with significant load inertia.
Assumptions, limits and common mistakes
The formula assumes an ideal two-mass system with constant drive and load torque throughout the slip event. Shaft elasticity, damping and a time-varying motor torque (e.g. an induction-motor curve) are not captured.
Common mistake: A common mistake is getting the sign of TL wrong — a load torque that opposes acceleration is entered positive, as shown in the formula; a driving (assisting) load torque would need to be entered negative.
Frequently asked questions
What does TL = 0 mean in practice?
It corresponds to starting up with no external load, e.g. spinning up an unloaded driven machine — then TK is determined purely by the inertia split.
How do I choose the sign of TL?
Positive if TL opposes acceleration (the typical load case); negative if the load assists acceleration.
What is TK used for specifically?
As the sizing quantity for the friction clutch's nominal torque and switching work during start-up.
What happens with a very large JA?
A large driver-side inertia relative to JL reduces the share transmitted to the load, per the formula.
Does this formula apply to braking too?
No, braking uses its own formulas for required brake torque and braking time with the opposite direction of action.