Hoist equilibrium speeds in four quadrants

Equilibrium (steady-state) speed is where the motor torque equals the load torque: $T = T_l$.

Data

• Characteristic 1 (quadrants I, II, IV): $T = 200 - 0.2N$
• Characteristic 2 (quadrants II, III, IV): $T = -200 - 0.2N$
• Loaded hoist: $T_l = 100$ N-m; unloaded hoist: $T_l = -80$ N-m (the counterweight is heavier than the empty cage, so the load torque is negative).

A hoist load is an active (gravity) load, so $T_l$ keeps its sign when the speed reverses.

Quadrant I: forward motoring (hoisting the loaded cage), characteristic 1, $T_l = 100$:

$$\begin{aligned} 200 - 0.2N &= 100 \\ N &= \frac{200-100}{0.2} = 500\ \text{rpm} \end{aligned}$$

$T = +100$, $N = +500$: both positive, quadrant I.

Quadrant II: forward braking (hoisting the empty cage; counterweight pulls), characteristic 1, $T_l = -80$:

$$\begin{aligned} 200 - 0.2N &= -80 \\ N &= \frac{200+80}{0.2} = 1400\ \text{rpm} \end{aligned}$$

$T = -80$, $N = +1400$: torque negative, speed positive, quadrant II (regenerative braking; speed above no-load speed of 1000 rpm).

Quadrant III: reverse motoring (lowering the empty cage), characteristic 2, $T_l = -80$:

$$\begin{aligned} -200 - 0.2N &= -80 \\ N &= \frac{-200+80}{0.2} = -600\ \text{rpm} \end{aligned}$$

$T = -80$, $N = -600$: both negative, quadrant III.

Quadrant IV: reverse braking (lowering the loaded cage), characteristic 2, $T_l = 100$:

$$\begin{aligned} -200 - 0.2N &= 100 \\ N &= \frac{-200-100}{0.2} = -1500\ \text{rpm} \end{aligned}$$

$T = +100$, $N = -1500$: torque positive, speed negative, quadrant IV (regenerative braking during lowering).

            N (rpm)
   II: 1400 |  I: 500
  (T=-80)   |  (T=+100)
 -----------+-----------> T
  III: -600 |  IV: -1500
  (T=-80)   |  (T=+100)

Answer: equilibrium speeds are 500 rpm (I), 1400 rpm (II), -600 rpm (III) and -1500 rpm (IV).