Direct-acting Centrifugal Governor: How It Works, Parts, Formula & Uses Explained

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A direct-acting centrifugal governor can use weights that slide radially instead of swinging on pendulum arms. In Hiscox mechanism 300, steel ribbons pass over pulleys and connect the sliding weights to a spring-loaded collar. Outward weight motion lifts the collar, which supplies an output for valve control.

Direct-acting Centrifugal Governor Interactive Calculator

Compare three imposed speeds for a direct-acting radial-ball governor. Steel ribbons lift the collar as the two weights move outward. This illustrative example uses 0.5 kg per ball, 120 mm inner radius, 10 N total preload, 2500 N/m spring stiffness and 60 mm travel.

0°

Collar lift at first speed
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Collar lift at middle speed
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Collar lift at last speed
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Last minus first lift
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Equation Used

2mω²(r₀+x) = F₀+kx. Thus x_free = (2mω²r₀−F₀)/(k−2mω²), with ω = 2πN/60. Illustrated travel x is limited to 0–60 mm by mechanical stops.
The three speeds are compared in a smooth prescribed sweep. The calculation is quasi-static and includes stop reactions, but no engine feedback, inertia, friction or acceleration lag. At low speed, the spring preload can hold the collar at its inner stop, giving zero lift.
  • Two equal 0.5 kg masses move symmetrically on radial guides.
  • Inner mass radius 120 mm; maximum radial displacement 60 mm.
  • Both inextensible ribbons lift one collar with a 1:1 travel ratio.
  • Total spring preload 10 N and stiffness 2500 N/m; collar weight is neglected.
  • The spring slope exceeds the centrifugal-force slope throughout 30–300 rpm, so the free equilibrium is stable.

The example dimensions and spring data are stated modeling assumptions, not historical specifications. The familiar pendulum-governor height equation g/ω² does not describe this ribbon-and-spring arrangement.

Same mechanism and inputs as the interactive calculator.

Radial travel becomes collar lift

The weights run along the rotating radial arms on rollers. Each weight pulls a ribbon over a pulley near the spindle. Increasing the horizontal ribbon span shortens its vertical span by the same amount, so the collar rises as the weights move outward.

The plan view shows rotation about the spindle. The enlarged lower view turns with that assembly, keeping the ribbons, pulleys and spring visible. Its visible collar extension represents the output connection; the engine throttle linkage is not simulated.

The spring resists collar lift. At an unconstrained equilibrium, the total centrifugal force of the two balls equals the spring force. Inner and outer stops supply any remaining force when the calculated free position lies beyond the available travel.

Speed-sensitive mechanical output

The historical mechanism connects the grooved collar to a lever and throttle valve. The governor is the speed-sensitive part of that control system. The engine, valve response and driven load determine how the complete system behaves.

This calculator isolates the radial weights, ribbons and spring. It compares three selected speeds for one stated example and does not predict hunting, settling time or regulated engine speed.

Spring equilibrium rather than pendulum height

For each ball of mass m at radius r₀+x, centrifugal force is mω²(r₀+x). Two ribbons transmit the combined force 2mω²(r₀+x) to the collar. A preloaded linear spring supplies F₀+kx.

Equating these gives x_free = (2mω²r₀−F₀)/(k−2mω²). Here m=0.5 kg, r₀=0.12m, F₀=10 N and k=2500 N/m. The calculator converts the resulting displacement to millimetres and applies the actual modeled 0–60 mm stops.

The denominator remains positive over the entered 30–300 rpm range. At an inner stop, preload exceeds the outward force; at an outer stop, the outward force exceeds spring force. The animation displays both forces so these stop conditions are distinguishable from free equilibrium.

Comparing 60, 120 and 200 rpm

At 60 rpm the outward force at the inner radius is below the 10 N preload, so the collar stays at its inner stop. At 120 rpm the ideal collar lift is approximately 3.82 mm. At 200 rpm it is approximately 20.68 mm.

The example reaches its 60 mm outer stop at about 285 rpm. Beyond that point additional speed raises stop reaction rather than increasing lift. Reversing the order of the first and last comparison speeds reverses the sign of their reported lift difference.

What the fixed example omits

The values are illustrative and do not specify a historical or commercial governor. Different masses, spring data, initial radii or ribbon ratios produce a different characteristic.

Transient motion would require inertia, damping, ribbon and spring compliance, bearing friction and a model of the engine and valve. The imposed speed sweep makes the equilibrium curve visible without claiming those dynamics have been calculated.

Direct-acting governor questions

Why does the collar sometimes remain at zero?

The spring preload holds the weights against their inner stops until centrifugal force is high enough.

Why does lift stop at 60 mm?

That is the outer travel stop of this illustrated example. Additional force is carried by the stop.

Why does the lower drawing not spin?

It is a section viewed in the rotating frame. The plan view above shows the actual direction of rotation at reduced display speed.

Does each speed input affect the result?

Each controls its own comparison output and part of the imposed speed sweep. Multiple speeds can share a stop position while producing different stop forces.

Is this a swinging-ball governor?

No. Its balls slide along radial guides and act through ribbons and a spring.

Reference

Gardner D. Hiscox, Mechanical Movements, Powers, Devices and Appliances, mechanism 300, printed page 84. The source describes sliding balls on rollers, steel ribbons over pulleys and a spring-linked grooved collar.

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