Swiss Lever Escapement Impulse Duration
Calculates the exact duration in milliseconds during which the pallet jewel imparts kinetic energy to the balance impulse pin during each half-oscillation.
\[ t_{impulse} = \frac{\alpha_{lift}}{\omega_{avg}} = \frac{\alpha_{lift} \cdot T}{4 \cdot \theta_{amp}} \]
Plain Text Formula:
t_impulse = alpha_lift / omega_avg = (alpha_lift * T) / (4 * theta_amp)
📐 Variable Definitions & Parameters
| Symbol | Variable Name | Standard Unit | Physical Definition |
|---|---|---|---|
| \[ t_{impulse} \] | Impulse Duration | milliseconds (ms) | Contact duration during energy delivery |
| \[ \alpha_{lift} \] | Lift Angle | radians or degrees | Angular span of engagement |
| \[ T \] | Oscillation Period | ms | Full oscillation duration |
| \[ \theta_{amp} \] | Balance Amplitude | radians or degrees | Peak half-oscillation amplitude |
💡 Worked Engineering Example
Problem Statement:
For a 28,800 BPH movement (T = 250 ms) with 52° lift angle and 300° amplitude, find impulse duration.
Step-by-Step Solution:
- t_impulse = (52° * 250 ms) / (4 * 300°)
- t_impulse = 13,000 / 1,200 = 10.83 ms
🎯 Result: The pallet fork transfers energy to the balance for exactly 10.83 milliseconds per tick.