A compensating pendulum bob offsets thermal changes that would otherwise alter a clock’s rate. In a mercury-jar arrangement, rod expansion lowers the jar while expansion of the liquid raises its center within the jar. Correct compensation depends on the complete pendulum. This calculator illustrates the opposing movements and estimates the rate change of an uncompensated point-bob model.
Compensating Pendulum Bob Interactive Calculator
Change equivalent rod length, thermal expansion and temperature rise to compare downward growth with the upward bob-center shift needed to cancel it. The enlarged jar comparison makes these otherwise microscopic movements visible.
Equation Used
- Massless, uniformly expanding equivalent rod with a point bob and small swing angle.
- Constant gravity and expansion coefficient; the clock is calibrated at the reference temperature.
- The liquid-center illustration assumes exact compensation and neglects jar/rod mass and jar expansion.
- A real compound pendulum requires its mass distribution, moment of inertia, suspension and temperature distribution; no mercury height is determined here.
Ideal point-bob thermal model; jar dimensions are illustrative. Compensation shown is the required target, not a sized mercury jar. Thermal displacements are magnified. No restoration or mercury-handling procedure is provided.
Opposing thermal movements
The upper illustration shows a suspended rod, a glass jar containing a liquid mass, and a threaded rating adjustment. It follows the type of mercury bob shown in Hiscox’s Mechanical Movements, figure 1147. The rating adjustment sets the pendulum length; it is distinct from thermal compensation.
The lower comparison separates thermal motion from swinging motion. Its reference liquid center is marked by a horizontal dashed line. On warming, the jar moves downward with the expanding rod. In the ideal compensated example, the liquid center rises by the same distance relative to that jar. The thermal movements are magnified; neither the jar dimensions nor its liquid quantity are a design calculation.
For a uniform liquid column, the center moves by half the change in column height relative to its base. A constrained liquid’s column-height expansion also depends on expansion of its container’s cross-sectional area. It is not obtained by simply dividing a liquid’s volumetric expansion coefficient by three. The former mercury-height formula on this page has therefore been removed.
Historical use and what this calculator can show
Mercury-compensated pendulums were used in precision regulator clocks. The Whipple Museum records a William Hardy astronomical regulator with a mercury pendulum and rating adjustment. Its purpose was to reduce temperature-related changes in timing.
Use the controls here to explore thermal growth and the sensitivity of an ideal pendulum’s rate. They do not establish the accuracy of a particular historical clock or specify a replacement bob.
Thermal growth and clock rate
Let L be equivalent rod length in meters, α its expansion coefficient in 1/°C, and ΔT the rise above the calibration temperature. The calculator converts the entered µm/(m·°C) coefficient by multiplying it by 10−6.
ΔL = αLΔT. For a small-angle point-bob pendulum, T = 2π√(L/g), so Thot/Treference = √(1 + αΔT). The clock’s loss over one actual day at that fixed temperature is 86400[1 − 1/√(1 + αΔT)] seconds. Its initial sensitivity at the reference temperature is 43200α s/day/°C.
The compensation target in this simplified model is an upward bob-center displacement equal to ΔL. A real distributed-mass pendulum cannot be fully sized from these three inputs alone.
Example: a one-meter equivalent rod
Inputs
L = 1 m, α = 11 µm/(m·°C), and ΔT = 20°C.
Calculate growth
The rod grows 11 × 1 × 20 = 220 µm. The ideal upward center-shift target is therefore also 220 µm.
Calculate uncompensated rate
The initial sensitivity is 0.4752 s/day/°C. The exact point-bob rate expression gives approximately 9.50 seconds lost per actual day at the warmer temperature. This is not a total seasonal error, and it is not a performance prediction for a compensated clock.
Limits of the simplified comparison
A lower rod expansion coefficient reduces the required compensating movement. A longer rod grows farther in absolute terms, but the fractional thermal length change αΔT is unchanged, so the ideal daily rate loss does not depend on L. Temperature rise changes both growth and loss.
Real pendulums have distributed mass, finite amplitude, suspension effects, and unequal heating of rod and bob. Keeping one liquid center fixed is not by itself a complete calculation of the center of oscillation. Those effects and restoration procedures are outside this model.
Questions about the model
Why does length change growth but not daily rate loss?
Absolute growth is αLΔT, while fractional growth is αΔT. The point-bob period ratio depends on fractional growth, so L cancels from that rate calculation.
Is the illustrated jar a specified mercury quantity?
No. It shows the direction and required size of an ideal compensating center shift. Container expansion, mass distribution and liquid quantity have not been entered.
Why is the thermal motion enlarged?
The calculated movement is measured in micrometers and would be nearly invisible at the pendulum’s drawing scale. The enlarged view uses a separate displacement scale.
Does zero temperature rise stop the pendulum?
No. It removes thermal growth and rate change while the normal pendulum swing continues.
References
- Gardner D. Hiscox, Mechanical Movements, Powers, Devices and Appliances, figure 1147, printed page 283: mercury jar, threaded rating adjustment and opposing thermal expansion.
- Whipple Museum: William Hardy astronomical regulator, description of its mercury-compensated pendulum and graduated rating unit.
The animated geometry is an explanatory drawing rather than a measured reconstruction of either example.
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