Ft = P / v
At the same power, a lower chain speed requires a higher pull; this is the useful circumferential force, not the chain's full operating tension.
At the same power, a lower chain speed requires a higher pull; this is the useful circumferential force, not the chain's full operating tension.
Select a target and calculate.
At the same power, a lower chain speed requires a higher pull; this is the useful circumferential force, not the chain's full operating tension.
5.5 kW at a 5.3975 m/s chain speed gives a chain pull of about 1,019 N.
Steady, uniform force transmission; centrifugal tension, catenary tension, starting shocks and dynamic load peaks are not captured separately here.
This calculator finds the useful tangential chain pull from transmitted power and mean chain speed. It shows why, at the same power, a slower chain carries a higher pull.
Chain pull (circumferential force) Ft at the sprocket follows from the power relation P = Ft·v, so Ft = P/v. Equivalently, Ft can be found from drive torque T1 and the driving sprocket's pitch radius r1: Ft = T1/r1.
This works like a bicycle: at the same pedaling power, a lower gear with lower chain speed at the sprocket produces a higher chain pull than a higher gear at greater speed.
Ft = P / v
Ft = P / vP = Ft · vv = P / Ft| Symbol / input | Meaning |
|---|---|
| Chain pull Ft | Tangential circumferential force available at the sprocket. |
| Transmitted power P | Mechanical power transmitted through the chain drive. |
| Chain speed v | Mean chain travel speed. |
Transmitted power P and mean chain speed v set the pull. Chain speed can first be found with the chain speed calculator from pitch, tooth count and rotational speed if needed.
Enter the power to be transmitted and the mean chain speed to compute chain pull. For a sizing question — such as what speed is needed to stay under an allowable maximum pull — choose v as the target.
5.5 kW at a chain speed of 5.3975 m/s (19.05 mm pitch, 17 teeth, 1,000 rpm) gives a chain pull of Ft = 5,500 W / 5.3975 m/s ≈ 1,019 N.
A chain pull of about 1,019 N is the useful circumferential force that transmits torque. It is markedly smaller than the actual operating tension in the loaded chain span, which additionally includes centrifugal tension, catenary tension and a service factor for shock loading.
Power is given in kW, speed in m/s and the resulting force in N.
The chain pull Ft computed here is only the share that actually transmits torque. The real operating tension in the loaded chain span additionally consists of the centrifugal tension Fz = q·v² (significant at higher speeds), the catenary tension Fs of the sagging slack span, and a service factor KA for shock and starting loads: Fges ≈ Ft·KA + Fz + Fs. Ft alone is often enough for a first-pass estimate; a complete verification needs these additional force contributions from manufacturer literature.
The relationship is used for first-pass chain drive sizing, for estimating shaft load together with the other force contributions, and to sanity-check whether a chosen chain size can transmit the required force at all.
This is the pure useful force derived from power and speed, not a complete chain design. For the actual operating tension in the loaded span, centrifugal tension at higher speeds, catenary tension in the slack span, and a service factor for shock loading must additionally be considered — see the linked service-factor calculator.
Common mistake: A common mistake is treating this useful force as the chain's complete design tension without adding centrifugal tension, catenary tension and a service factor. It is also easy to substitute an instantaneous speed instead of the speed averaged over one revolution.
Because P = Ft·v: if P stays fixed and v falls, Ft must rise correspondingly to keep the product constant.
No. It is only the torque-transmitting share; the full operating tension additionally includes centrifugal tension, catenary tension and a shock-load service factor.
Via Ft = T1/r1, where r1 is the driving sprocket's pitch radius.
From the chain speed calculator, which derives it from pitch, tooth count and rotational speed.
Only indirectly: at a fixed rotational speed, a different tooth count changes chain speed and, via P = Ft·v, therefore the pull.