Hiscox figure 1028 shows a flange on a hoisting-drum shaft carrying three pivoted hooks. At ordinary speed they clear three studs fixed to the frame; when thrown outward they catch those studs. The animation distinguishes the rotating parts from the stationary frame.
Centrifugal Safety Catch for Hoisting Drums Interactive Calculator
Explore mass, radius and speed in the mrω² relationship. Compare the result with an entered radial reference force. The accompanying historical hook sequence illustrates clearance and engagement as a separate teaching sequence.
Equation Used
- Point mass at constant radius and steady angular speed.
- Reference force is an entered constant acting in the radial comparison.
- No hook-pivot moment balance, spring geometry, gravity, friction, transient acceleration or impact model.
- Historical engagement sequence uses illustrative proportions and timing, independent of calculated equal-force speed.
- Results do not set a hoist trip threshold, inspection interval or arrest capacity.
This historical teaching view shows pivoted hooks meeting fixed frame studs. The scalar force calculator is a separate ideal comparison, not a hoist safety-device design or calibration model. Actual engagement depends on moments, geometry and dynamics; arrest capacity is not calculated.
Rotating hooks and stationary studs
In Hiscox figure 1028, flange B is fastened to the drum shaft. Three hooks pivot on this flange. Studs D are attached to the stationary frame, so the hooks move past them during normal rotation.
The source describes the hooks hanging back at ordinary speed and moving outward under unusual acceleration of speed. Their noses then catch the studs. The three frame studs provide contact points for the hook noses. Their positions and the hook pivots determine the engagement geometry.
The animation follows that distinction: the frame and studs stay fixed, the flange carries the hooks, and the hook noses meet the studs in the illustrated caught position. The sequence pauses visually at engagement before repeating. It is a geometric demonstration; the stopping motion is not a prediction of impact deceleration.
Historical hoisting mechanism
The cited reference identifies this arrangement as a safety catch for hoisting drums. It illustrates a mechanically initiated stop using shaft rotation and pivoted hooks.
The historical hook-and-stud arrangement explains mechanically initiated engagement. A modern hoist must be assessed from its own device documentation and operating requirements; this illustration does not identify or rate its protection system.
What the force calculator actually calculates
A mass m at radius r rotating at N rpm requires inward force F = mr(2πN/60)². In a rotating-frame description, the corresponding outward centrifugal term has the same magnitude.
Setting that magnitude equal to an entered radial reference force Fref gives N_eq = 60/(2π)√(Fref/(mr)). The other outputs compare N_eq with the selected speed and subtract the calculated force from the reference.
For a pivoted hook, a force magnitude alone is insufficient: the force line and its moment arm about the hook pin matter, together with any spring force, gravity, friction and acceleration. Equal-force speed is therefore not a verified trip RPM. Hook or stud strength and the drum’s stopping energy are separate questions.
Example force comparison
For m = 2 kg, r = 120 mm and N = 300 rpm, angular speed is 31.416 rad/s and mrω² = 236.87 N.
With a radial reference force of 313.2 N, equal-force speed is about 344.97 rpm, or 114.99% of the comparison speed. Reference minus calculated force is 76.33 N.
Doubling speed multiplies force by four. Doubling mass or radius doubles force at the same speed and divides equal-force speed by √2. These are checks of the scalar equation, not instructions for selecting or adjusting a hoist spring.
Engagement and stopping are different problems
The illustration explains which parts meet. It does not calculate whether the hook can swing out, how quickly it does so, whether it rebounds or how much load the stud carries.
Stopping a rotating drum involves its inertia, the suspended load and the compliance of the mechanism. None of those quantities appears in mrω² alone. A real safety device cannot be selected, calibrated or returned to service using this teaching calculator.
Calculator questions
Which parts remain stationary?
The frame and its three studs remain fixed. The flange turns with the drum shaft and carries the pivoted hooks past the studs.
Does equal-force speed tell me when the catch will trip?
No. It is the solution of a radial point-mass force comparison. A pivoted hook needs a moment balance and a dynamic model, which are not supplied here.
Why does the animation catch even when I change the reference force?
The repeating sequence is a historical motion demonstration, independent of the scalar force comparison. Its timing and proportions are illustrative, not a simulation of a calibrated device.
What does a negative force difference mean?
It means mrω² exceeds the entered reference at the selected speed. It does not prove that an actual hook has engaged or that the device can stop the load.
Can the results be used to set a safety device?
No. Device-specific manufacturer procedures and qualified assessment are needed. The calculator supplies neither a trip setting nor an arrest-load rating.
References & Further Reading
- Gardner D. Hiscox, Mechanical Movements, Powers, Devices and Appliances (1901), printed page 259, figure 1028: three shaft-flange hooks engaging fixed frame studs.
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