This roller-closure mechanism moves two roller bearings along a fixed horizontal guide by translating a head containing two angled slots. As the head moves, the slot intersections bring the rollers together or separate them. The calculator connects slot angle to head travel and ideal force, and compares nip gap and tolerance.
Closure of Rollers Interactive Calculator
Adjust the roller gap and load, then change the slot angle to see how the translating head closes the rollers and trades head travel against ideal force.
Equation Used
- Both roller centers stay on the fixed horizontal guide.
- Equal angled slots move rigidly with the translating head.
- Each roller carries the entered inward load; the ideal head force is not a capacity rating.
- Roller size is illustrative; gap and head travel share one drawing scale.
- No friction, frame compliance, springback or material-force law is included.
The translating head’s angled slots move the rollers inward or outward along the fixed guide. Slot angle links head travel with ideal force, while the gap entries give a nominal thickness-reduction comparison. Roller diameter is illustrative; actual material reduction also depends on material response and machine stiffness.
How the angled-slot closure works
Hiscox figure 1376 shows two rollers A and B carried in the fixed guide C. The translating head D carries an opposed pair of angled slots. Each roller pin lies at the intersection of its moving slot and the fixed guide, so the rollers move horizontally while the head moves vertically.
In the displayed orientation, lowering D reduces the gap. Raising D opens it. The blue head and slot members form one rigid moving part; they are not hinged links. The visualization isolates the change in roller separation; a rotational drive is outside this model.
A closure mechanism, separate from the roller drive
The slot arrangement can set the separation of a pair of rollers. A complete feed or forming machine would also need a suitable rotational drive, bearings, structure and control system. Those parts are outside this closure model.
Travel and force equations
With slot angle β measured from vertical, a downward head movement y moves each roller inward by y tan β. The total gap change is therefore 2y tan β.
Head travel = |g_start − g_final|/(2 tan β).
If each roller carries an equal inward force F, ideal virtual work gives head force = 2F tan β. This omits friction, acceleration and elastic deformation.
The nip estimate is draft = max(entry thickness − final gap, 0), with percentage reduction equal to 100 × draft/entry thickness. Gap tolerance is expressed as a plus/minus amount; its relative value is 100 × tolerance/final gap, after converting micrometres to millimetres.
Worked closure example
At 1.20 mm entry thickness, 0.84 mm final gap and 25° slots, the nominal draft is 0.36 mm, or 30%. Each roller moves inward by 0.18 mm; the head moves downward by 0.386 mm.
For an entered 4,000 kN inward load on each roller, the ideal head force is 2 × 4,000 × tan 25° = 3,730.5 kN. This calculation does not establish that a mechanism can carry that load. A ±10 µm gap tolerance is ±1.19% of the 0.84 mm gap.
Changing the slot angle
A smaller angle from vertical requires more head travel for the same gap change and less ideal head force for a given roller load. A larger angle shortens the stroke and raises the head-force requirement. Slot contact forces, wear, friction and guide stiffness still need separate evaluation.
Questions about roller closure
Why do the rollers move horizontally?
Their bearing pins are constrained by fixed guide C. The translated angled slots set their horizontal positions.
Does the force control predict material reduction?
No. The load is an input. The draft output is a geometric estimate and does not solve a material deformation law.
What if the final gap is wider than the incoming material?
The tool reports zero thickness reduction. The mechanism still animates the requested opening motion.
Is the tolerance percentage a full tolerance band?
No. It is the entered plus/minus amount divided by the final gap. The full band would be twice that value.
References & Further Reading
- G. D. Hiscox, Mechanical Movements, Powers, Devices and Appliances, printed page 333, figure 1376: fixed guide C and translating slotted head D.
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