Actuators: Hydraulic and Pneumatic Cylinders, DC Motors, Stepper Motors and Servo Motors
1. Hydraulic and pneumatic actuators
A cylinder turns fluid pressure into force and flow into speed: F = pA and v = Q/A, and the power delivered is pQ. In a double-acting cylinder the rod side has only the annulus area A − a, so for the same pressure the retract force is smaller and for the same flow the retract speed is larger. With a 50 mm bore, a 25 mm rod and 10 MPa: extend F = 10⁷ × π × 0.025² ≈ 19.6 kN; retract F = 10⁷ × π(0.025² − 0.0125²) ≈ 14.7 kN. At 20 L/min (3.33 × 10⁻⁴ m³/s) the extend speed is 3.33 × 10⁻⁴/1.963 × 10⁻³ ≈ 0.17 m/s.
| Property | Hydraulic (oil) | Pneumatic (air) |
|---|---|---|
| Working pressure | High — tens of MPa | Low — under about 1 MPa |
| Force density | Very high | Low |
| Stiffness and position control | Stiff: oil is nearly incompressible | Soft and springy: air compresses |
| Typical use | Presses, heavy robots, excavators | Fast pick-and-place, clamping, grippers, end-to-end strokes |
| Drawbacks | Leaks, fire risk, pumps and filters | Poor intermediate positioning, noise |
Valves route the fluid: a directional control valve is named by ports and positions — a 4/3 valve has four ports and three positions, a 5/2 valve (common in pneumatics) five ports and two; flow control valves set speed, and pressure relief valves set the maximum force. Hydraulic actuators hold their force-per-kilogram advantage because the pressure is so high; pneumatic ones win on speed, cleanliness and cost wherever the motion is end-stop to end-stop.
2. DC motors and their torque–speed line
A permanent-magnet (or separately excited) DC motor obeys two relations with one constant K (in SI units the torque constant in N·m/A equals the back-EMF constant in V·s/rad): T = K I_a and E_b = K ω, with the armature circuit V = E_b + I_a R_a. Eliminating I_a gives the torque–speed characteristic, a straight line: ω = V/K − (R_a/K²) T. It meets the speed axis at the no-load speed ω₀ = V/K and the torque axis at the stall torque T_s = K V/R_a. Output power Tω is a parabola along that line, with its maximum P_max = T_s ω₀/4 at half the stall torque and half the no-load speed.
Worked: V = 24 V, R_a = 2 Ω, K = 0.1. Then ω₀ = 240 rad/s, T_s = 0.1 × 24/2 = 1.2 N·m and P_max = 1.2 × 240/4 = 72 W. At a load of 0.4 N·m: I_a = 4 A, E_b = 24 − 8 = 16 V, ω = 160 rad/s. Speed is controlled by armature voltage, which shifts the line parallel to itself; a wound-field motor can also be run above base speed by field weakening. A series motor has a hyperbolic characteristic with very high starting torque, and must never run unloaded.
3. Stepper motors
A stepper moves a fixed angle per input pulse, so it positions open-loop: count pulses, know the angle. For a variable-reluctance motor with N_s stator poles and N_r rotor teeth, the step angle is (N_s − N_r) × 360°/(N_s N_r) — 15° for 8 and 6. A two-phase hybrid with 50 rotor teeth gives the common 1.8°, 200 steps per revolution. Speed follows the pulse rate: n (rpm) = step angle × pulse rate × 60/360, so 1.8° at 1000 pulses/s is 300 rpm. Half-stepping alternates one and two energised phases and halves the step; microstepping proportions the phase currents to subdivide it further.
A stepper’s performance is read from two curves against pulse rate: the pull-in curve, the torque at which it can start and stop in step without ramping, and the higher pull-out curve, the torque it can sustain once running. Load beyond pull-out, or acceleration too steep, makes it lose steps silently, since there is no feedback; mechanical resonance at low speed does the same. Holding torque is the torque it resists when energised and stationary; detent torque is the smaller residual torque of a permanent-magnet or hybrid motor unpowered.
4. Servo motors and the choice of actuator
A servo motor is a motor run inside a closed loop: an encoder or resolver measures position or speed, a controller compares it with the command, and a drive corrects the current. The motor may be a brushed DC, a brushless DC (electronically commutated from Hall or encoder signals) or an AC synchronous machine; what makes it a servo is the feedback, which lets it hold position against load, run smoothly at any speed, and deliver short peaks of several times its continuous torque. The hobby "RC servo" is the same idea miniaturised, its angle commanded by the width of a pulse repeated about every 20 ms.
| Aspect | Stepper | Servo |
|---|---|---|
| Control | Open-loop, pulse counting | Closed-loop with feedback |
| Torque at speed | Highest at low speed, falls quickly | Nearly flat to rated speed |
| Failure mode | Loses steps without knowing | Detects and corrects error |
| Cost and complexity | Low | Higher: encoder, tuned drive |
5. Gearing a motor to its load
Motors are fast and weak; joints are slow and strong. A reduction of ratio n (input turns per output turn) with efficiency η gives output speed ω/n and output torque n η T. A motor giving 0.5 N·m at 3000 rpm through a 20:1 gearbox of 90% efficiency delivers 150 rpm and 0.5 × 20 × 0.9 = 9 N·m. Seen from the motor, a load inertia J_L is reduced to J_L/n² — 2 kg·m² through 20:1 looks like 0.005 kg·m² — and a load torque to T_L/n. The ratio that gives the load its greatest acceleration for a given motor torque is n = √(J_L/J_m), at which the reflected load inertia equals the motor’s own.
Key takeaways
- F = pA and v = Q/A; a double-acting cylinder retracts with less force and more speed because of the rod.
- Hydraulics for stiffness and force density, pneumatics for fast, cheap, end-to-end motion.
- DC motor: ω = V/K − R_a T/K²; ω₀ = V/K, T_s = KV/R_a, P_max = T_s ω₀/4.
- VR stepper step = (N_s − N_r)360°/(N_s N_r); rpm = step × pulse rate × 60/360; beyond pull-out torque it loses steps.
- A gearbox of ratio n multiplies torque by nη and divides speed by n; load inertia reflects as J_L/n².
Practice questions (14)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
A hydraulic cylinder of 50 mm bore is supplied at 10 MPa. The force it exerts while extending, correct to one decimal place, is ____ kN.
Numerical answer — type the value.
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Answer: 19.6
A = π × 0.025² = 1.9635 × 10⁻³ m², F = pA = 10⁷ × 1.9635 × 10⁻³ = 19 635 N ≈ 19.6 kN. Using the diameter as the radius gives four times as much, 78.5 kN.A double-acting hydraulic cylinder of 50 mm bore has a rod of 25 mm diameter. Supplied at 10 MPa on the rod side, the force it exerts while retracting, correct to one decimal place, is ____ kN.
Numerical answer — type the value.
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Answer: 14.7
Retraction acts on the annulus: A − a = π(0.025² − 0.0125²) = π × 4.6875 × 10⁻⁴ = 1.4726 × 10⁻³ m², so F = 10⁷ × 1.4726 × 10⁻³ ≈ 14.7 kN. Using the full bore area again gives 19.6 kN, the extension force; on the rod side the pressure acts only on the annulus.Compared with a hydraulic actuator, a pneumatic actuator is less suited to holding an intermediate position under varying load mainly because:
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Answer: A — air is compressible, so the actuator behaves like a spring
A change in load changes the air volume noticeably, so the piston moves — low stiffness. Oil is almost incompressible. Air is far less dense than oil, pneumatic pressures are lower, and double-acting pneumatic cylinders are standard.A permanent-magnet DC motor has K = 0.1 V·s/rad and armature resistance 2 Ω and is supplied at 24 V. Its no-load speed is ____ rad/s.
Numerical answer — type the value.
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Answer: 240
At no load T = 0, so I_a = 0 and E_b = V: ω₀ = V/K = 24/0.1 = 240 rad/s. R_a does not enter, since no current flows through it.A permanent-magnet DC motor has K = 0.1 V·s/rad (back-emf constant and torque constant) and armature resistance 2 Ω, and is supplied at 24 V. Driving a steady load torque of 0.4 N·m, its speed is ____ rad/s.
Numerical answer — type the value.
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Answer: 160
I_a = T/K = 0.4/0.1 = 4 A; E_b = V − I_a R_a = 24 − 8 = 16 V; ω = E_b/K = 160 rad/s. From the line: ω = 240 − (2/0.01) × 0.4 = 240 − 80 = 160.A permanent-magnet DC motor has K = 0.1 V·s/rad (back-emf constant and torque constant) and armature resistance 2 Ω, and is supplied at 24 V. The maximum mechanical output power it can deliver at this voltage is ____ W.
Numerical answer — type the value.
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Answer: 72
T_s = KV/R_a = 0.1 × 24/2 = 1.2 N·m and ω₀ = 240 rad/s. P = Tω along the line peaks at T_s/2 and ω₀/2: 0.6 × 120 = 72 W. Multiplying T_s by ω₀ gives 288 W, a point that is never reached since stall and no-load cannot occur together.At constant armature voltage, the torque–speed characteristic of a permanent-magnet DC motor is:
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Answer: A — a straight line falling from the no-load speed to zero speed at the stall torque
ω = V/K − (R_a/K²)T is linear in T. The hyperbola belongs to the series motor, whose flux rises with current; the parabola is the power curve, not the speed curve.A variable-reluctance stepper motor has 8 stator poles and 6 rotor teeth. Its step angle is ____ degrees.
Numerical answer — type the value.
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Answer: 15
Step angle = (N_s − N_r) × 360°/(N_s N_r) = 2 × 360/48 = 15°, i.e. 24 steps per revolution. 360/8 = 45° or 360/6 = 60° are the pole pitches, not the step.A stepper motor with a 1.8° step angle is driven at 1000 pulses per second. Its speed is ____ rpm.
Numerical answer — type the value.
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Answer: 300
1.8° × 1000 = 1800°/s = 5 rev/s = 300 rpm. Equivalently 200 steps per revolution at 1000 steps/s is 5 rev/s.Driving a stepper motor in half-step mode instead of full-step mode:
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Answer: A — halves the step angle, doubling the steps per revolution
Alternating one-phase-on and two-phase-on states places the rotor halfway between full-step positions, so a 1.8° motor steps 0.9° and makes 400 steps per revolution. It is still open-loop and still needs a driver.A motor delivering 0.5 N·m at 3000 rpm drives a load through a 20:1 speed-reducing gearbox of 90% efficiency. The output torque is ____ N·m.
Numerical answer — type the value.
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Answer: 9
T_out = n η T_in = 20 × 0.9 × 0.5 = 9 N·m, at 3000/20 = 150 rpm. Ignoring efficiency gives 10 N·m; dividing the torque by the ratio gives 0.0225 and would describe a speed-increasing box.A load of inertia 2 kg·m² is driven through a 20:1 speed reduction. The load inertia as seen at the motor shaft is ____ kg·m².
Numerical answer — type the value.
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Answer: 0.005
Reflected inertia = J_L/n² = 2/400 = 0.005 kg·m², from equating kinetic energies with ω_m = 20 ω_L. Dividing by n alone gives 0.1, the rule for torque, not inertia.Which statements comparing stepper and servo drives are true?
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Answer: A — A servo drive uses position or speed feedback; B — A stepper is commonly run open-loop by counting pulses; C — A stepper can lose steps without detecting it if the load exceeds its pull-out torque
(A) Feedback is what defines a servo. (B)–(C) With no feedback, a missed step is invisible to the controller. (D) False: in a series motor the field current is the armature current, so torque rises roughly with its square at start — very high starting torque, which is why series motors drive traction and hoists.For a double-acting hydraulic cylinder with a single rod, supplied at the same pressure and the same flow rate in both directions, which statements are true?
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Answer: B — The extension force is the pressure times the full bore area; C — The extension speed is the flow rate divided by the full bore area; D — Retraction is faster than extension
(A) False: retraction acts on the smaller annulus, so its force is smaller. (B) F = pA on the cap side. (C) v = Q/A. (D) The same flow into the smaller annulus area moves the piston faster.