Two hinged weights act through links against springs to rotate the inner part of a compound eccentric. Changing its relative angle changes the resultant eccentric throw. Use the calculator to explore how linkage geometry, spring tension and centrifugal force set the eccentric angle and throw.
Crank-shaft Governor · Linked Compound Eccentric
Compare shaft speed, weights and springs in a linked compound-eccentric governor. Read a free balance position separately from a travel stop.
Equation Used
- Two point-mass weights and two tension springs.
- The listed linkage dimensions are illustrative and fixed in this model.
- Preload is defined at the 30° low stop.
- Inner travel is 30°–130° on one continuous branch.
- Massless links and output; no valve load, friction or transient dynamics.
Stop status requires a reaction; do not interpret it as free spring balance.
Weights, springs and a compound eccentric
Hiscox no. 317 shows weights and springs carried by the rotating shaft assembly. Links transmit their motion to the inner eccentric. The source identifies the resulting change in the main eccentric throw.
The animation first solves the steady position relative to the carrier, then rotates that settled assembly. Blue links join the weight arms to the inner disc; gold springs oppose outward weight movement. The output strap and a selected 280 mm connecting rod show the resulting slider stroke. The lower diagram separates the fixed first offset O–I from the adjustable second offset I–E.
The calculation uses the illustrative dimensions listed below, rather than dimensions for a particular engine. It is a steady rotating-frame model, not a simulation of overspeed, hunting or transient response.
Compare steady speed and spring settings
Change shaft speed, each weight mass, spring rate, preload and the second eccentric offset. The spring and centrifugal generalized torques determine the inner angle. At a travel stop, the remaining load is carried by the stop; the result explicitly says that it is not a free spring equilibrium.
Changing the second offset changes eccentric throw without changing the solved balance in this model, because valve load and eccentric inertia are omitted. It would not generally be independent in a loaded, dynamic governor.
Geometry and virtual-work balance
Lengths are in millimetres. O=(0,0), I=(12,0), and the inner angle q is limited to 30°–130°. For s=+1 and −1, each arm pivot is P=(−120s,−80s), its spring anchor is S=(120s,80s), and its inner-disc link pin is U=(12+36s cos q,36s sin q). The weight centre W is the selected continuous intersection of circles |W−P|=150 and |W−U|=125.
For each spring, L=|W−S| and Fₛ=F₀+k[L(q)−L(30°)]. With ω=2πn/60, centrifugal force magnitude is F𝚌=mω²|W|/1000 newtons. The generalized drive torque in N·mm is Q=Σ[mω²(W·dW/dq)/1000−Fₛ(dL/dq)]. The solver finds Q=0 with restoring slope, or identifies the appropriate travel stop.
The resultant eccentric vector is E=(12+e₂ cos q,e₂ sin q). Its magnitude satisfies e²=12²+e₂²+24e₂ cos q, and the illustrative in-line slider stroke is 2e. Forces are transmitted through lever geometry; adding centrifugal forces and dividing by spring rate would omit that geometry.
Read equilibrium and travel stops separately
The defaults use two 4 kg point weights, 20 N/mm springs with 50 N preload at the low stop, 12 mm second offset and 250 rpm shaft speed. The selected linkage settles at an interior spring-balanced position.
At 50 rpm the low travel stop holds it; at 500 rpm it reaches the high stop. Within the working range, increasing speed generally increases the inner angle and reduces the resultant throw for the displayed compound-eccentric orientation.
What the calculation includes
Both closed linkage loops, changing spring lengths, preload and centrifugal moments are included. All moving mass is concentrated at the two weight centres. Links, the eccentric and output parts are treated as massless, and valve load, friction, damping and gravity variation through a revolution are omitted.
The 30° and 130° travel stops bound the illustrated linkage motion. No spring stress, bearing load, allowable operating speed or governor stability under engine feedback is rated here. The two weight forces can differ because the compound eccentric has an offset centre.
Crank-shaft governor questions
Why does spring rate affect throw?
It changes the balance angle of the linked weights; that angle changes the vector sum of the two eccentric offsets.
Is a stop position an equilibrium?
It is constrained by a stop reaction. The springs alone do not balance the centrifugal generalized torque there.
Why are the two centrifugal forces different?
The selected linkage about the offset inner centre gives different weight radii. Each force uses its own radius.
Is the output motion a valve-timing prediction?
No. It illustrates eccentric throw with an in-line slider. Valve lap, lead, ports and load are not specified.
Primary reference
Gardner D. Hiscox, Mechanical Movements, Powers, Devices and Appliances, no. 317, printed page 89. The illustration and text identify spring-balanced weights moving an inner eccentric to vary the main throw. The calculator uses illustrative dimensions and a steady spring-and-weight balance.
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