STEADY RESPONSE
Speed is not vibration.
Each point assumes constant speed and settled motion. The vertical line marks Ω/ωₙ = 1; damping shifts the unbalance-response maximum slightly above it.


ROTATING MACHINERY / AN INVISIBLE FORCE, MADE VISIBLE
Place the mass centre slightly off the shaft axis. Then accelerate through the rotor’s natural frequency and see how balance, damping and time change the motion.
A flexible shaft. One disc.
Two directions of motion.
The drive prescribes a smooth speed ramp. The model calculates the lateral motion; it does not simulate an engine governor.
STEADY RESPONSE
Each point assumes constant speed and settled motion. The vertical line marks Ω/ωₙ = 1; damping shifts the unbalance-response maximum slightly above it.
PHASE
At Ω/ωₙ = 1, displacement lags the rotating unbalance by 90°. Far above this frequency, the lag tends towards 180°. With zero unbalance, the plotted transfer-function phase has no observable displacement to describe.
CALCULATED TRANSIENT
The solid histories include ramp acceleration and the rotor’s memory. Dotted curves are the instantaneous constant-speed predictions, shown for comparison rather than used as the transient solution. Click the graph to move the playback cursor.
BALANCE VECTOR
Drag the correction endpoint, or use the magnitude and angle controls. Vectors are equivalent eccentricities. They are not instructions for adding a mass to a real engine.
THREE THINGS TO NOTICE
An offset mass centre creates a rotating force. At constant speed its magnitude grows with speed squared.
Mass and stiffness set the natural frequency. Damping limits the amplification and removes lateral mechanical energy.
Compare the two ramp histories. The transient peak need not occur at the natural frequency or match the steady peak.
THE MODEL BEHIND THE MOTION
m is rotor mass [kg], k is lateral stiffness [N/m], c is viscous damping [N s/m], and ζ is the damping ratio. The isotropic model has the same k and c in both directions.
e is residual eccentricity [m]; X is the settled orbit radius [m]. At r = 1, X = e/(2ζ). A balanced rotor has e = 0 and no unbalance response.
θ includes the residual vector’s initial angle. x and y are shaft-centre displacement from static equilibrium [m], positive right and up. Ω is prescribed spin speed [rad/s]. The model integrates these equations with fourth-order Runge–Kutta and checks lateral energy input against stored and dissipated energy.
READ THE RESULT AT THE RIGHT SCALE
A linear isotropic Jeffcott rotor: one disc, a massless flexible shaft and stationary supports. Gravity is absorbed into the equilibrium position. No gyroscopic splitting, bearing non-linearity, rub, clearance, fatigue, real engine vibration spectrum or maintenance acceptance limit is predicted.
The shaft outline is schematic. The orbital displacement is calculated and shown on a labelled scale. The rotor phase is sampled, so apparent animation speed does not measure physical RPM. A warning appears if the calculated excursion exceeds 2 mm: this is a teaching-model scale check, not a component limit.
Analytical limits, transient refinement and lateral energy balance are checked locally. These checks verify the equations; they do not establish experimental accuracy for an aircraft engine.
All parameter defaults are synthetic teaching choices. The sources support the mechanisms; no manufacturer geometry or measured performance is reproduced.