Gyroscopic couple and its effect on vehicles

Why it matters

Every spinning part of a vehicle, the wheels, the engine flywheel, a turbine rotor, carries angular momentum. When the vehicle turns, pitches or leans, the direction of that angular momentum changes, and a couple is needed to change it. The reaction to that couple shifts load between the wheels of a car on a bend, changes the lean of a motorcycle, and makes a ship's bow rise or fall when it steers.

Key ideas

Spin, precession and the gyroscopic couple. A rotor spins at ω about its own axis, so its angular momentum is H = I·ω along the spin axis (right-hand rule). If the spin axis itself turns at angular velocity ω_p (precession) about a perpendicular axis, the direction of H changes at the rate ω_p, so a couple C = I·ω·ω_p must act on the rotor. Its axis is perpendicular to both the spin axis and the precession axis.

Active and reactive couple. The couple applied to the rotor (by its bearings) to make it precess is the active gyroscopic couple. The rotor pushes back on its bearings, and hence on the vehicle, with an equal and opposite reactive gyroscopic couple. Vehicle effects come from the reactive couple.

Direction rule. Represent spin and precession as vectors by the right-hand rule. The active couple vector is ω_p × H: rotate the spin vector through 90° in the sense of precession; the active couple turns the spin axis in that direction. The reactive couple on the frame is the opposite. Always state which one you mean.

Car on a curve (wheels and engine).

• Wheels spin with ω_w = v / r_w; the car precesses about the vertical at ω_p = v / R.
• The reactive couple from the four wheels tends to overturn the car outward: it adds load to the outer wheels and removes it from the inner wheels, just like the centrifugal couple.
• The engine flywheel turns at G·ω_w (G = engine speed / wheel speed). If it spins in the same sense as the wheels, its couple adds; if opposite, it subtracts.
• Centrifugal couple m·v²·h / R also lifts the inner wheels. The inner wheels leave the road when the combined reduction in reaction equals their static load.

Two-wheeler leaning on a bend. The wheel spin axes are tilted by the lean angle θ, so the gyroscopic couple becomes (2I_w + G·I_e)·(v²/(R·r_w))·cos θ. Together with the centrifugal couple it must be balanced by the overturning moment of the weight, which sets the required lean angle.

Ship with a turbine rotor (axis along the ship).

• Steering (turning about the vertical): couple about the transverse axis, which raises or lowers the bow (pitching).
• Pitching (bow moving up and down): the precession is about the transverse axis, so the couple is about the vertical axis and the ship tends to turn (yaw) left or right.
• Rolling: the precession axis is parallel to the spin axis, so there is no gyroscopic effect.

Aeroplane. An engine and propeller rotating clockwise when viewed from the rear, in an aeroplane turning left, produce a reactive couple that raises the nose and lowers the tail; turning right lowers the nose.

Formulas

C = I·ω·ω_p

• C: gyroscopic couple (N·m); I: moment of inertia of the rotor about its spin axis (kg·m²); ω: spin speed (rad/s); ω_p: precession speed (rad/s). Spin and precession axes perpendicular.

ω_w = v / r_w and ω_p = v / R

• v: vehicle speed (m/s); r_w: wheel radius (m); R: radius of the curve (m).

C_g = (4·I_w ± G·I_e)·v² / (r_w·R) (four-wheeler)

• I_w: inertia of each wheel (kg·m²); I_e: inertia of the engine rotating parts (kg·m²); G: engine-to-wheel speed ratio; + if the engine rotates in the same sense as the wheels.

P_g = C_g / (2x) and P_c = m·v²·h / (2·R·x)

• Change in vertical reaction at each wheel due to the gyroscopic and centrifugal couples (N): + at the outer wheels, − at the inner wheels; x: track width (m); h: height of the centre of gravity (m); m: vehicle mass (kg).

tan θ = (v² / (R·g))·[1 + (2·I_w + G·I_e) / (m·h·r_w)] (two-wheeler lean)

• θ: lean angle from the vertical; other symbols as above, m and h for the rider plus vehicle.

Worked examples

Example 1 (standard: ship). A ship's turbine rotor has a mass of 2000 kg and a radius of gyration of 0.4 m, and runs at 3000 rpm clockwise when viewed from the stern. The ship turns to the left (port) on a curve of 200 m radius at 8 m/s. Find the gyroscopic couple and its effect.

1. I = m·k² = 2000 × 0.4² = 320 kg·m².
2. ω = 2π × 3000 / 60 = 314.16 rad/s; ω_p = v / R = 8 / 200 = 0.04 rad/s.
3. C = 320 × 314.16 × 0.04 = 4021 N·m.
4. Direction: clockwise viewed from the stern means the spin vector points forward, towards the bow. Turning left is anticlockwise viewed from above, so the precession vector points up. Rotating the spin vector 90° in the precession sense takes it towards port, so the active couple acts about the port axis and tends to dip the bow.
5. The reactive couple on the ship is opposite: it raises the bow and lowers the stern.

Example 2 (GATE level: car on a curve). A car of mass 2000 kg has a track width of 1.5 m and its centre of gravity 0.6 m above the road; the load is shared equally by the four wheels. Each wheel has I_w = 1.5 kg·m² and radius 0.35 m. The engine rotating parts have I_e = 1.0 kg·m², rotate in the same sense as the wheels, and turn 4 times as fast as the wheels. The car rounds a curve of 100 m radius at 72 km/h. Find the reaction on each inner wheel and on each outer wheel, and the speed at which the inner wheels would lift.

1. v = 72 / 3.6 = 20 m/s. Static load per wheel = 2000 × 9.81 / 4 = 4905 N.
2. Gyroscopic couple: C_g = (4 × 1.5 + 4 × 1.0) × 20² / (0.35 × 100) = 10 × 400 / 35 = 114.3 N·m. Per wheel: P_g = 114.3 / (2 × 1.5) = 38.1 N.
3. Centrifugal couple: C_c = 2000 × 20² × 0.6 / 100 = 4800 N·m. Per wheel: P_c = 4800 / 3.0 = 1600 N.
4. Inner wheels: 4905 − 38.1 − 1600 = 3267 N each. Outer wheels: 4905 + 38.1 + 1600 = 6543 N each.
5. Lift-off speed: 4905 = v² × [(10 / (0.35 × 100)) + (2000 × 0.6 / 100)] / 3.0 = v² × (0.2857 + 12) / 3.0 = 4.095·v², so v² = 1197.7 and v = 34.6 m/s = 125 km/h (ignoring tyre side-slip, which in practice occurs first).

The gyroscopic share is small for a car (38 N against 1600 N), which is why it is usually neglected in vehicle design, but it is not negligible for motorcycles, aircraft and ships.

Common mistakes

• Mixing up the active couple (on the rotor) and the reactive couple (on the vehicle). Vehicle effects come from the reactive one.
• Using the wheel speed in rpm or forgetting ω_w = v / r_w.
• Adding the engine's couple when it rotates opposite to the wheels.
• Forgetting cos θ for a leaning two-wheeler.
• Saying rolling of a ship produces a gyroscopic effect. The spin and precession axes are parallel, so it does not.
• Dividing the couple by the track width x instead of 2x when finding the reaction change at each of two wheels on one side.

For GATE ME

Questions ask for the magnitude of a gyroscopic couple (often for a ship's rotor or an aircraft propeller), its effect (bow rises or falls, nose rises or dips), the reaction changes at the wheels of a car on a curve, or the limiting speed for wheel lift-off. Practise the vector rule with three or four sketches of spin and precession directions until you can state the effect without hesitating.

Quick check

1. I = 0.5 kg·m², ω = 200 rad/s, ω_p = 0.5 rad/s. Couple?
2. Does rolling of a ship produce a gyroscopic couple from a rotor along its length?
3. On a bend, does a car's wheel gyroscopic couple load the inner or the outer wheels more?
4. How do you get the precession speed of a car on a curve?
5. Engine rotating opposite to the wheels: does its gyroscopic couple add to or subtract from that of the wheels?

Answers: 1. 50 N·m. 2. No. 3. The outer wheels. 4. ω_p = v / R. 5. It subtracts.