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How do I size a linear motor?

There is no mechanism: the forcer pushes the carriage directly, so the demand is a force per phase — the mass times the acceleration, plus friction (with the magnetic attraction of an iron-core motor adding to the normal force), gravity on an incline and any process force. The peak must fit the motor's peak force and the RMS over the cycle, dwell included, its continuous force.

What you need

The math: Move time on a trapezoidal profileopen — every formula, so anyone can check it

A point-to-point move accelerates at a, runs at the top speed vmax, and decelerates at d. The area under the velocity–time profile is the distance s.

s_accel = v_max² / (2·a)         distance the acceleration uses
s_decel = v_max² / (2·d)         distance the deceleration uses

if s_accel + s_decel ≤ s   (trapezoid — the top speed is reached)
    t_accel  = v_max / a
    t_cruise = (s − s_accel − s_decel) / v_max
    t_decel  = v_max / d
else                        (triangle — the acceleration runs straight into the deceleration)
    v_peak   = √( 2·s·a·d / (a + d) )
    t_accel  = v_peak / a,   t_cruise = 0,   t_decel = v_peak / d

t_total = t_accel + t_cruise + t_decel

The inverse — the time is given, what speed and acceleration does it need? — splits the time into acceleration, cruise and deceleration by two fractions and reads the peak speed off the area:

t_accel = f_accel·t_total,   t_decel = f_decel·t_total,   t_cruise = the rest
v_peak  = s / (t_accel/2 + t_cruise + t_decel/2)
a = v_peak / t_accel,   d = v_peak / t_decel

Worked example. 100 mm at 0.5 m/s with 3.5 m/s²: accelerating and decelerating need 35.7 mm each, so the move is a trapezoid — 0.143 s accelerating, 0.057 s cruise, 0.143 s decelerating, 0.343 s in all. Halve the distance and it turns triangular, peaking at 0.418 m/s in 0.239 s.

On the cyclogram this whole move — all three phases — is the bar's ramp-up (the stroke out); the body is the stay at the end position, and the ramp-down is the move back. A bar is travel over time, not speed.

Package motion-time (calcpkg/1, open format) — pure functions with tests; the browser version is held against it number for number.

The math: Linear motor — force demand per phaseopen — every formula, so anyone can check it
F_gravity  = (m − m_cw)·g·sin α
F_friction = μ · ( m·g·cos α + F_attraction )        iron-core motors pull toward the track
F_static   = F_gravity + F_friction + F_ext

F_accel = (m + m_cw)·a + F_static
F_cruise = F_static
F_decel = F_static − (m + m_cw)·d
F_peak  = max |F|
F_rms   = √( Σ F_i²·t_i / (Σ t_i + t_dwell) )

check:  F_peak ≤ F_peak,motor    F_rms ≤ F_continuous    v_peak ≤ v_max

There is no gearbox to hide behind: if the continuous force is short, the cycle needs a longer dwell, a gentler profile, or the next motor size.

Package mech-linear-motor (calcpkg/1, open format) — pure functions with tests; the browser version is held against it number for number.

Questions people ask

What is the attraction force?

Iron-core linear motors are pulled toward the magnet track by several times their continuous force. That pull is a normal force on the guides, so it adds friction. Ironless motors have none.

Why does the dwell matter?

The RMS force is what heats the motor; time at zero force lowers it. A move that fails on continuous force can pass with a longer dwell between moves.