Calculation
Ff = (m/L) · v²
Centrifugal tension scales with belt mass per length and with speed squared, reducing the tensile-force reserve available for useful pull.
Ingenieurmäßige Vorbemessung
Centrifugal tension in a running belt
Centrifugal tension scales with belt mass per length and with speed squared, reducing the tensile-force reserve available for useful pull.
MINTSI
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Result
Select a target and calculate.
Understand the inputs
- Belt centrifugal tension Ff — Speed-induced tensile-force component acting throughout the running belt.
- Mass of measured section m — Mass of a known belt section of length L.
- Measured section length L — Length to which the entered belt mass refers.
- Belt speed v — Linear speed along the belt's neutral layer.
Example
A 1 kg belt section over 2 m gives q = 0.5 kg/m; at 20 m/s the centrifugal tension is 200 N.
Assumptions and limits
Uniform running and constant mass per length; imbalance, splice, vibration, bending and product-specific maximum speed are excluded.
Understand Centrifugal tension in a running belt
This calculator finds the centrifugal component in a fast-running belt. Instead of requiring an unfamiliar mass per length directly, it accepts mass m of a known belt section and its matching length L, giving q = m/L.
What does this quantity describe?
Every mass element of the circulating belt needs radial acceleration around a pulley. The required force appears as a uniform tensile component Ff = q·v² = (m/L)·v² throughout the belt and grows with speed squared.
A light rope can be swung slowly with little perceptible centrifugal force. At twice the speed, the same mass needs four times the radial force. That quadratic rise adds load to a fast-running belt.
Formula and variables
Ff = (m/L) · v²
q = m/LFf = q·v²v = √(Ff/q)m = Ff·L/v²
| Symbol / input | Meaning |
|---|---|
| Belt centrifugal tension Ff | Speed-induced tensile-force component acting throughout the running belt. |
| Mass of measured section m | Mass of a known belt section of length L. |
| Measured section length L | Length to which the entered belt mass refers. |
| Belt speed v | Linear speed along the belt's neutral layer. |
Choose the inputs correctly
m and L must describe the same belt section. v is linear belt speed at the neutral layer, not shaft rpm. If a datasheet supplies kg/m, choose any convenient L and enter m = q·L.
How to use the calculator
Enter section mass, matching length and operating speed, then solve for Ff. The equation can be inverted for a speed at a selected force, but that result never overrides a lower product-specific manufacturer speed limit.
Worked example
A 2 m section weighing 1 kg has q = 0.5 kg/m. At v = 20 m/s, Ff = 0.5·20² = 200 N. At 40 m/s it is already 800 N.
Understand the result and units
The 200 N acts throughout the circulating belt. It transmits no useful power, yet loads the tensile member and consumes part of the margin to maximum allowable span tension.
Mass per length must reduce to kg/m and v to m/s for a result in newtons. Selectable units are converted through SI. Divide g/m by 1,000 to obtain kg/m.
Why is high belt speed critical?
Because Ff is proportional to v², a 10% speed increase raises centrifugal tension by 21%. Flex frequency, windage and imbalance sensitivity also tend to rise, so manufacturer limits depend on belt profile, pulley diameter and construction.
Typical applications
Use this estimate for fast flat and V-belts to quantify a speed increase, reserve a centrifugal-force allowance, and convert a manufacturer belt-mass value into an operating tension.
Assumptions, limits and common mistakes
The model assumes uniform motion and mass per length. It excludes splice peaks, imbalance, transverse vibration, bending stress, heating and product-specific speed ratings.
Common mistake: Do not enter rpm as v; calculate circumferential speed first. Use total belt mass only with total belt length. Do not confuse Ff with useful effective pull Fᵤ.
Frequently asked questions
What is centrifugal belt tension?
It is the tensile component required to keep belt mass moving around the curved pulley path; Dubbel gives Ff = q·v².
Why is pulley diameter absent?
For a differential belt element, radius cancels between radial acceleration and the tension force acting over the arc.
How do I find belt mass per length?
Use a manufacturer kg/m value or divide the measured mass of a known belt section by its length.
Does centrifugal tension transmit power?
No. It is a speed-induced load that reduces the tensile reserve available for useful pull.
Can this set maximum rpm?
Only as one partial check. Manufacturer limits for speed, flex rate, minimum diameter, heat and splice construction also apply.