Clock Gear Train & Pendulum Calculator
Choose a beat rate or effective pendulum length, then search practical integer wheel and pinion counts for a clock train. The results show the closest trains, the beat rate they require, and the timing consequence of the remaining error.
Best wheel train24/6 × 24/6 × 30/8
- Overall train ratio
- 60
- Required beat rate
- 3,600
- Ratio error
- 0%
- Predicted clock error (seconds/day)
- 0
- Effective pendulum length (in)
- 39.14
- Alternative trains
- 24/6 × 24/6 × 45/12 (0.000 s/day); 24/6 × 24/6 × 60/16 (0.000 s/day); 24/6 × 27/7 × 35/9 (0.000 s/day); 24/6 × 28/7 × 30/8 (0.000 s/day); 24/6 × 28/7 × 45/12 (0.000 s/day); 24/6 × 28/7 × 60/16 (0.000 s/day); 24/6 × 30/8 × 32/8 (0.000 s/day); 24/6 × 30/8 × 36/9 (0.000 s/day); 24/6 × 30/8 × 40/10 (0.000 s/day); 24/6 × 30/8 × 44/11 (0.000 s/day); 24/6 × 30/8 × 48/12 (0.000 s/day); 24/6 × 30/8 × 52/13 (0.000 s/day); 24/6 × 30/8 × 56/14 (0.000 s/day); 24/6 × 30/8 × 60/15 (0.000 s/day); 24/6 × 30/8 × 64/16 (0.000 s/day); 24/6 × 32/8 × 45/12 (0.000 s/day); 24/6 × 32/8 × 60/16 (0.000 s/day); 24/6 × 35/9 × 54/14 (0.000 s/day); 24/6 × 36/9 × 45/12 (0.000 s/day); 24/6 × 36/9 × 60/16 (0.000 s/day)
Target beat rate: 3600.000 BPH. A positive daily error means the indicated clock runs fast.
Equivalent stage orderings are counted once; tooth form, center distance, strength, and clearance are not checked.
How to use this calculator
- Choose whether the target is a known beat rate or the effective length of a pendulum.
- Enter the escape-wheel tooth count and confirm how many beats occur per tooth.
- Set the number of wheel-and-pinion stages and the tooth-count limits you can make or use.
- Set an acceptable ratio error, then compare the best train with the listed alternatives.
How the clock wheel train is calculated
A clock train reduces the rotation of the center wheel through successive driving wheels and driven pinions until it reaches the escape wheel. This calculator searches every allowed integer wheel and pinion pair, rejects any pair whose driving wheel has no more teeth than its pinion, and keeps one ordering of equivalent stages.
For each candidate, the train ratio is:
R = (W₁ ÷ P₁) × (W₂ ÷ P₂) × … × (Wₙ ÷ Pₙ)
The required beat rate is B = k × E × R, where E is the escape-wheel tooth count and k is the selected number of beats per escape-wheel tooth. In beat-rate mode, that result is compared directly with your target. In pendulum-length mode, the target is calculated as B = 3600 × √(39.14 ÷ L), with L in inches.
The displayed ratio error is 100 × (B − target) ÷ target. To make the practical consequence easier to judge, the calculator also reports 86,400 × (target ÷ B − 1) seconds per day. A positive result means the hands would indicate fast if the oscillator actually holds the entered target rate.
What moves the result most
Each stage is multiplicative, so a small tooth-count change can matter more in a train with several stages. The escape-wheel count and beats-per-tooth assumption also scale the required BPH directly. A traditional anchor escapement commonly gives two beats for each escape-wheel tooth, but that must match the escapement you intend to use.
The effective pendulum length is calculated from the candidate BPH as L = 39.14 × (3600 ÷ B)². It is the equivalent simple-pendulum length from suspension point to center of oscillation, not the physical length of a rod. The search ranks matching trains first by absolute predicted seconds per day and then by the lowest largest single-stage ratio.
What this calculation does not design
A ratio that works on paper is only a starting point for a movement. This search does not check tooth form, module or diametral pitch, undercut, center distances, wheel clearance, plate space, torque, friction, backlash, or train layout. It also omits temperature compensation, circular error, suspension-spring effects, pendulum mass distribution, and escapement geometry. Check the chosen teeth against the actual mechanism before making parts.
Worked example
For a 30-tooth escape wheel with two beats per tooth, a three-stage train of 60/10, 50/10, and 48/12 has a ratio of 6 × 5 × 4 = 120. Its required rate is 2 × 30 × 120 = 7,200 BPH, so it is an exact match for a 7,200 BPH target and predicts 0 seconds per day of ratio-based error.
The corresponding ideal effective pendulum length is 39.14 × (3600 ÷ 7200)² = 9.785 inches. Reversing the stage order gives the same mathematics, so it appears only once in the search results.
Common questions
How do wheel and pinion tooth counts determine clock BPH?
Each driving-wheel tooth count divided by its driven-pinion leaf count is a stage ratio. Multiplying those ratios, then multiplying by the escape-wheel teeth and beats per tooth, gives the beat rate required for the clock to keep time.
What is the difference between a tick, a beat, and an oscillation?
A beat is one impulse or audible tick associated with the escapement. A full pendulum oscillation contains two beats in the usual clock convention, which is why a seconds pendulum has 3,600 beats per hour.
Why does an escape wheel often use two beats per tooth?
In a conventional anchor escapement, one tooth is released on each half swing of the pendulum. The escape wheel therefore advances one tooth over two beats, but escapement designs differ and the actual action should be confirmed.
Why is effective pendulum length different from rod length?
Effective length runs from the suspension point to the pendulum’s center of oscillation. The bob’s shape, its position, the rod, and the suspension all affect it, so a rod cut to the calculated length is not necessarily a seconds pendulum.
How much ratio error causes one second per day of timing error?
Near an exact match, one second per day is roughly 0.00116% of the rate. The relationship is reciprocal rather than perfectly linear, so the calculator reports the actual seconds-per-day result for each candidate instead of relying on that approximation.