A Chinese windlass uses two different drum diameters on one shaft. One end of a continuous rope winds in while the other pays out; the moving load block rises by half the difference in rope taken up.
Chinese Windlass Interactive Calculator
Adjust crank radius, both effective drum diameters and load. The animation follows one lifting turn and then reverses, while the calculator reports ideal force, torque and rise per turn.
Equation Used
- Constant effective drum diameters and rigid common shaft.
- One continuous rope, wound in opposing senses on the drums.
- Vertical supporting legs, freely rotating load sheave and negligible losses.
- D1 must exceed D2 for the depicted lifting direction. Equal diameters give zero rise.
Ideal differential windlass with a continuous rope and two supporting load-block legs. Effective diameters are measured at the rope centerline. Results exclude losses and structural ratings.
One shaft, two winding diameters
Hiscox figure 40 shows two barrels on the same shaft, a hand crank and a sheave suspended in the rope loop between them. Turning the shaft winds rope onto the larger barrel while releasing rope from the smaller one.
The free rope shortens by the difference in barrel circumferences. Two rope legs support the moving block, so the block rises by half that amount. The drums turn together; they are not two independently counter-rotating wheels.
The animation shows the common shaft, bearings, opposed winding, moving sheave and attached load. Drum diameters retain their entered ratio; shaft spacing and vertical travel are scaled for readability.
Studying differential lifting
This mechanism illustrates how a small difference between winding diameters can produce a large ideal force ratio without a gear train. It also demonstrates the corresponding reduction in load travel for each handle revolution.
The calculator compares ideal geometry and effort. It does not establish rope selection, brake behavior, anchorage strength or a safe working load for a hoist.
Deriving the diameter-based equations
For one shaft revolution, rope taken up is πD1 and rope released is πD2. Net take-up is π(D1−D2), so load rise is π(D1−D2)/2.
The handle moves through 2πL per revolution. Equating ideal work gives F·2πL = W·π(D1−D2)/2. Therefore W/F = 4L/(D1−D2).
Input torque is FL = W(D1−D2)/4. A longer handle reduces hand force but does not change the required ideal shaft torque or rise per turn. Here W and F are forces in lbf, lengths are inches, and torque is lbf·in.
Default calculation
With an 18 in crank radius, 6 in and 5 in effective drum diameters and a 400 lbf load, the ideal mechanical advantage is 72:1. Hand force is approximately 5.56 lbf, rise is 1.571 in per turn and shaft torque is 100 lbf·in.
Increasing only the crank radius to 36 in halves hand force to 2.78 lbf. It leaves rise and shaft torque unchanged. Increasing the diameter difference increases rise per turn and required torque while reducing mechanical advantage.
Equal diameters take up and pay out the same rope length. The calculator flags this as an invalid lifting pair because it gives zero differential rise.
What the ideal model leaves out
Small diameter differences give high force multiplication and slow load motion. Rope buildup changes the effective winding diameters and therefore changes both travel and force ratio. Bearing losses and bending losses increase actual effort above the ideal result.
Neither self-locking nor a holding brake is implied by the force ratio. The drawing is a mechanism explanation, not a fabrication specification or lifting-equipment rating.
Chinese windlass questions
Why does the formula use 4L rather than 2L?
The inputs are drum diameters. In radius notation the ratio is 2L/(R1−R2); replacing each radius with half its diameter gives 4L/(D1−D2).
Does a longer crank lift the load farther each turn?
No. It reduces ideal hand force. Rise per shaft turn depends only on the drum diameter difference.
Why do equal drum diameters show an invalid pair?
Equal circumferences produce zero net rope take-up. This is not a useful lifting configuration and does not imply unlimited load capacity.
Do the two drums rotate in opposite directions?
No. They are fixed to one shaft. Opposed rope winding makes one drum take up while the other pays out.
Is the displayed hand force a rated hoist capacity?
No. It is an ideal work-balance result. Friction, braking, rope strength, supports and operating conditions are outside this calculation.
References & Further Reading
- Gardner D. Hiscox, Mechanical Movements, Powers, Devices and Appliances, printed page 25, figure 40: Chinese windlass and half-circumference-difference rise.
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