The mechanics of drives

Most sizing calculations in this course are mechanics. Eight relations cover almost all of them.

Speed, torque and power

Motors are rated in rpm, but physics works in radians per second:

$$\omega = \frac{2\pi n}{60} \qquad 1500 \text{ rpm} = 157.1 \text{ rad/s}$$

Torque T (N·m) is a turning force: a force F at radius r gives $T = F r$. Rotating power is torque times angular speed, and linear power is force times speed:

$$P = T\,\omega \qquad P = F\,v$$

Lifting a mass m at speed v needs $F = m g$, so $P = m g v$. Lab IV's hoist lifts 5 000 kg at 0.25 m/s: $P = 5000 \times 9.81 \times 0.25 = 12.3$ kW.

Gearboxes

A motor is fast and weak; a hoist drum is slow and strong. A gearbox with ratio $i$ (motor speed ÷ load speed) trades one for the other:

$$\omega_\text{load} = \frac{\omega_\text{motor}}{i} \qquad T_\text{load} = T_\text{motor}\cdot i \cdot \eta$$

Motor speed

1,450 rpm

Motor torque

95 N·m

Gear ratio i

20 : 1

Efficiency η

0.90

Load speed

7.59 rad/s

Load torque

1,710 N·m

Power in

14.43 kW

Power out

12.98 kW

Lost as heat

1.44 kW

Set the ratio to 243 with a 1450 rpm motor. The drum turns at 0.625 rad/s, exactly the Lab IV hoist.

Inertia and acceleration

Newton's law for rotation: any torque surplus accelerates the inertia J (kg·m²).

$$J \frac{d\omega}{dt} = T_\text{motor}(\omega) - T_\text{load}(\omega)$$

This one equation is behind every dynamic simulation in the labs. The motor speeds up while its torque beats the load, and settles where the two are equal.

FoundationStart here if this is new to you

Think of pushing a merry-go-round. The harder you push compared with the friction, the faster it speeds up. The heavier it is, the slower it speeds up. Once your push only just balances the friction, it keeps turning at a steady speed. That is the whole equation.

Constant torque (conveyor, hoist)Fan or pump (T ∝ ω²)

Inertia J

0.50 kg·m²

Load torque

60 N·m

Runs at 1,461 rpm

Time to 95 % speed

1.21 s

Final slip

0.0257

Starting torque

84.0 N·m

J · dω/dt = T_motor(ω) − T_load(ω)

Try a constant load of 90 N·m: the motor's starting torque is only 84 N·m, so it never moves. Lab I asks you to notice exactly this margin.

A mass moving in a straight line, seen from a shaft where one radian of rotation moves it $r_\text{eff}$ metres, adds inertia $J = m\, r_\text{eff}^2$. Through a gearbox, a load inertia appears at the motor divided by $i^2$.

Which way does efficiency go?

Predict first

A hoist lowers its load. 12.26 kW of mechanical power comes from the falling load and the gearbox efficiency is 0.85. How much electrical power reaches the drive?

14.4 kW (12.26 / 0.85)12.26 kW10.4 kW (12.26 × 0.85)

Motoring (lifting)Generating (lowering)

Mechanical power

12.26 kW

Efficiency η

0.85

Drawn from the grid

14.42 kW

Losses

2.16 kW

P_elec = P_mech / η

ExplorerGo deeper: derivations and open questions

Where does $J = m r^2$ come from? Equate kinetic energies: $\tfrac{1}{2} m v^2 = \tfrac{1}{2} J \omega^2$ with $v = r\omega$.

Optimal gear ratio. For a pure inertia load $J_L$ driven by a motor of inertia $J_M$, the load acceleration $\alpha_L = T i / (J_M i^2 + J_L)$ is largest when $i = \sqrt{J_L / J_M}$. Try deriving it, then think about why servo designers care.