A dead-beat escapement holds its escape wheel still while the pendulum continues swinging. Each pallet has a locking surface concentric with the anchor arbor, followed by an inclined impulse face. A tooth slides along the locking arc without forcing the wheel backward; the impulse face then lets the wheel advance and pass energy to the pendulum.
The animation separates entry lock, entry impulse, drop, exit lock and exit impulse. Both pallet contacts are solved against the same 30-tooth wheel. An enlarged view makes the working face visible.
Dead-beat Clock Escapement Interactive Calculator
Follow a 30-tooth wheel through lock, impulse and drop. Adjust the pendulum and wheel to compare ideal timing, tip force and impulse work.
Equation Used
- Simple pendulum with the entered effective length.
- Constant escape-wheel shaft torque.
- One tooth pitch per complete oscillation, two impulses.
- Each impulse advances 0.42 pitch; each drop advances 0.08 pitch.
- No friction or impact energy credited to useful impulse work.
- Illustrative proportions, not production escapement dimensions.
Hiscox 1171: concentric locking arcs hold the wheel still while the anchor keeps swinging.
Why the wheel remains still during lock
Hiscox figure 1171 illustrates a dead-beat escapement with two pallets carried by an anchor above the escape wheel. The locking faces follow circles centered on the anchor arbor. The entry pallet uses the outer surface and the exit pallet the inner surface. Because those locking arcs rotate about their own center, a tooth can stay at the same wheel position while the anchor moves.
When a tooth reaches an inclined impulse face, forward wheel motion turns the anchor. The tooth then clears that pallet and the wheel drops a short distance until another tooth reaches the opposite locking surface. A complete pendulum oscillation contains two impulses and advances this wheel by one tooth pitch.
The selected teaching geometry has 30 teeth, a 7.5-pitch pallet span, an arbor 1.3 wheel radii above the wheel center, and straight impulse faces spanning anchor angles −1.5° to +1.5°. Each impulse advances the wheel 0.42 tooth pitch; the following drop adds 0.08 pitch. These proportions illustrate the contact sequence; they are not dimensions for a commercial movement.
Use the calculator
Effective pendulum length sets the time scale. Half-swing angle is the angle on either side of vertical; the period includes the finite-amplitude correction. Escape-wheel diameter changes tip force at the selected shaft torque. Torque changes the ideal work supplied during each impulse.
The animation automatically plays and can be paused, reset or scrubbed through a full oscillation. The contact enlargement uses the current wheel and anchor positions. Green marks a loaded contact; amber identifies the brief drop between pallets.
Changing the pendulum amplitude does not redraw the pallet profiles: the same locking arcs and impulse faces remain in use throughout the 2°–5° half-swing range. The pendulum rod is shortened in the drawing, with a break mark, so the escapement remains large enough to inspect.
Pendulum timing and escape-wheel work
For effective length L in metres, gravitational acceleration g=9.80665 m/s² and half-swing angle a, the ideal simple-pendulum period is T=2π√(L/g)/AGM(1,cos(a/2)). AGM is the arithmetic–geometric mean; this is the complete-elliptic-integral expression for the finite-amplitude period.
The beat interval is T/2 and the beat rate is 120/T per minute. With z=30 teeth and one tooth advanced per full oscillation, wheel speed is 60/(zT) rpm.
Wheel pitch angle is δ=2π/z. In this selected geometry, each impulse advances the wheel by 0.42δ. At constant shaft torque M, ideal work per impulse is E=M×0.42δ, and ideal impulse power is 2E/T. When M is entered in mN·m, E is in mJ and power in mW.
Tangential force at the tooth tip is F=M/R, using torque in N·mm and tip radius R in mm. Numerically, one mN·m equals one N·mm. This is a tip-force reference, not the varying normal reaction on the inclined pallet face.
The remaining 0.08δ of wheel advance per beat occurs during drop. Its shaft work is not counted as useful pendulum impulse work. The calculation does not equate supplied energy with pendulum amplitude, since pendulum mass and losses are not specified.
Example: a one-metre effective pendulum
At L=1000 mm and a 3° half-swing, the ideal period is approximately 2.007 s. The beat interval is about 1.003 s, the beat rate about 59.798 per minute and the 30-tooth wheel speed about 0.997 rpm.
For a 100 mm tip diameter and 6 mN·m shaft torque, tangential tip force is 0.120 N. Each 5.04° wheel impulse receives approximately 0.528 mJ of ideal shaft work. Two impulses per period give approximately 0.526 mW of ideal impulse power.
Doubling effective length increases period by √2. Doubling shaft torque doubles tip force and impulse work without changing the ideal pendulum period. Doubling wheel diameter halves tangential tip force at the same torque.
Model boundaries
The wheel-to-pallet geometry is solved at each pose. During lock the wheel angle is constant. During impulse a straight pallet face intersects the tooth-tip circle, determining the wheel angle. During drop the wheel advances through the specified clearance to the opposite pallet.
The pendulum pose uses a sinusoidal display path; its cycle duration uses the finite-amplitude period. Drop is interpolated over a short anchor-angle interval to make the event visible. This is a contact and timing illustration, not an inertial simulation of tooth impact or a prediction of clock rate accuracy.
Tooth profiles, pallet thickness, clearances, arbor placement and swing limits are selected teaching dimensions. The calculator excludes friction, elastic deformation, recoil from imperfect construction, crutch clearance, suspension stiffness, air drag and temperature effects. It does not supply manufacturing dimensions or a guaranteed timekeeping accuracy.
Dead-beat escapement questions
Does the escape wheel move backward?
Not in the ideal geometry shown. The wheel holds still during lock, then advances during impulse and drop.
Why are the two locking faces different?
They contact teeth on opposite sides of the wheel. The entry pallet locks on an outer concentric surface and the exit pallet on an inner one.
Does a longer pendulum increase the impulse energy?
At a fixed escape-shaft torque and the same wheel impulse angle, work per impulse is unchanged. The longer period reduces impulse power.
Why does swing amplitude change the calculated period?
A finite-angle pendulum takes slightly longer than the infinitesimal-angle approximation. The calculation retains this effect.
Can these figures predict a clock’s accuracy?
No. Actual rate depends on the oscillator, construction, adjustment, environment and losses. This model explains contact, ideal timing and shaft work.
Primary references
Gardner D. Hiscox, Mechanical Movements, Powers, Devices and Appliances, figure 1171, printed page 288: original dead-beat escapement illustration.
Erwin Sattler / Uhrenbausatz M1 clock book, printed pages 70–73: entry and exit locking surfaces, impulse faces, lock, impulse and drop. The dimensions used in this calculator illustrate the principle; they do not specify the referenced clock movement.
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