Independent suspension: MacPherson strut and double wishbone

Why it matters

Almost every passenger car sold today has independent front suspension, usually a MacPherson strut, and performance cars, SUVs and pickups often use double wishbones. How the links are laid out decides camber change, roll centre height, packaging space and cost — the trade-offs an automotive engineer is expected to explain and calculate.

Key ideas

Independent suspension lets each wheel on an axle move vertically without directly moving the other. Benefits over a rigid axle: lower unsprung mass (the differential, if any, is body-mounted), less interaction between the wheels, more freedom to tune camber and roll-centre height, no axle beam under the engine, so lower bonnet and engine. Costs: more joints and bushes, more alignment settings, and (in most layouts) camber changes relative to the road as the body rolls.

MacPherson strut.

• A telescopic strut — damper tube with a coil spring around it — is rigidly bolted to the steering knuckle at the bottom and mounted to the body at the top through a rubber-isolated top mount with a bearing. A single lower arm (wishbone or transverse link with tension/compression rod) locates the bottom of the knuckle through a ball joint. An anti-roll bar link often attaches to the strut.
• The steering axis runs from the top-mount bearing to the lower ball joint, so the strut swivels when the wheel is steered.
• The strut is a structural member: it carries bending loads from the tyre forces, which raises friction in the damper (reduced by offsetting the spring axis).
• Advantages: few parts, cheap, light, compact sideways — leaves room for a transverse engine and drive shafts; loads are spread into the body over a large height (top mount to lower arm).
• Disadvantages: tall (needs a high bonnet line/strut tower), limited camber control (camber gain in bump is small, so roll causes positive camber on the outer wheel), strut friction, road noise transmitted through the top mount.

Double wishbone (short–long arm, SLA).

• Two transverse A-shaped arms, upper and lower, each pivoted on the body at two points and carrying the knuckle (upright) through a ball joint at its outer end. Spring and damper usually act on the lower arm, or on the upper arm, or through a pushrod in racing cars.
• Equal and parallel arms keep the wheel parallel to the body: camber relative to the road changes by the full body roll angle, and the track changes in bump — not good.
• Unequal arms with the upper arm shorter make the top of the wheel move inward as it rises, giving negative camber gain in bump. In a corner this counteracts the positive camber caused by body roll on the loaded outer wheel, keeping the tyre flatter. The arm lengths and angles also place the instantaneous centre and hence the roll centre.
• Advantages: best kinematic control (camber gain, roll centre, anti-dive), low height, stiff location, freedom of tuning.
• Disadvantages: more parts and cost, wider package (intrudes on engine or boot space), more alignment adjustments.

Other layouts. Multi-link (four or five separate links — the modern refinement of the double wishbone), trailing arm and semi-trailing arm (rear), swing axle (large camber change; obsolete), and the twist (torsion) beam — semi-independent, common at the rear of front-wheel-drive cars.

Key geometric ideas.

• Instantaneous centre (IC) of the knuckle in front view: where the lines of the two arms (or, for a strut, the lower arm line and the line perpendicular to the strut axis at the top mount) meet.
• Front-view swing-arm length (fvsa) — distance from the tyre contact centre to the IC; camber gain per unit bump ≈ 1 / fvsa.
• Roll centre — where the line from each contact centre to its IC crosses the vehicle centreline; the body rolls about this point under lateral load.
• Motion (installation) ratio — spring travel per unit wheel travel; it sets the wheel rate from the spring rate.

Formulas

MR = d_s / d_w

• MR = motion ratio, d_s = distance from the arm's inner pivot to the spring seat, d_w = distance from the pivot to the ball joint (wheel), both measured along the arm (m). Spring perpendicular to the arm.

k_w = k_s · (MR · cos α)²

• k_w = wheel rate (N/m), k_s = spring rate (N/m), α = inclination of the spring axis from the direction of motion of its seat. Small motions.

F_s = F_w / (MR · cos α)

• F_s = spring force (N) for a vertical wheel load F_w (N), same geometry.

f_n = (1/2π) · √(k_w / m_s)

• f_n = ride frequency (Hz), m_s = sprung mass on that corner (kg).

h_rc = (B / 2) · h_ic / fvsa

• h_rc = roll centre height (m), B = track (m), h_ic = height of the IC above the ground (m), fvsa = horizontal distance from the contact centre to the IC (m). Symmetric suspension, straight-line construction.

Δγ ≈ z / fvsa (radians)

• Δγ = camber change for a wheel travel z (m), small motions.

Worked examples

Example 1 (standard) — wheel rate and ride frequency. A double-wishbone lower arm is 0.40 m from inner pivot to ball joint; the coil spring sits 0.28 m from the pivot and is inclined at 15° to the direction of its seat's motion. The spring rate is 45 N/mm, and the static wheel load is 3.2 kN. Find the wheel rate, the spring force and the ride frequency.

1. MR = d_s / d_w = 0.28 / 0.40 = 0.70.
2. k_w = k_s · (MR · cos α)² = 45 × (0.70 × 0.9659)² = 45 × 0.4572 = 20.6 N/mm.
3. F_s = F_w / (MR · cos α) = 3200 / (0.70 × 0.9659) = 4733 N.
4. Sprung mass on the corner (taking all of the wheel load as sprung, for simplicity) m_s = 3200 / 9.81 = 326 kg.
5. f_n = (1/2π)·√(k_w / m_s) = 0.1592 × √(20 570 / 326) = 0.1592 × 7.94 = 1.26 Hz. k_w ≈ 20.6 N/mm, F_s ≈ 4.73 kN, f_n ≈ 1.26 Hz. The spring carries about 1.5 times the wheel load because of the lever ratio.

Example 2 (GATE level) — roll centre and camber gain. A symmetric double-wishbone front suspension has a track of 1.5 m. In the front view, the lines of the upper and lower arms meet at an instantaneous centre 2.4 m inboard of the tyre contact centre and 0.25 m above the ground. Find the roll centre height and the camber change for 30 mm of bump travel.

1. h_rc = (B/2)·h_ic / fvsa = 0.75 × 0.25 / 2.4 = 0.0781 m.
2. Δγ ≈ z / fvsa = 0.030 / 2.4 = 0.0125 rad = 0.72°.
3. Because the IC is inboard of the wheel, the top of the wheel moves inward in bump: the change is towards negative camber. h_rc ≈ 78 mm; Δγ ≈ −0.72° (negative camber gain) for 30 mm bump.

Common mistakes

• Calling a MacPherson strut "just a damper": it is a structural member that locates the wheel and carries bending.
• Thinking equal, parallel wishbones keep the tyre upright in a corner — they keep it parallel to the body, so it leans with the body roll.
• Using MR instead of MR² for wheel rate (the lever ratio acts on both force and displacement).
• Mixing the roll centre (front view of both sides) with the IC (one side).
• Assuming independent suspension has no camber change; most layouts have plenty.

For GATE ME

Expect conceptual comparisons of suspension types and simple kinematics: lever (motion) ratios and wheel rate, spring force from wheel load, ride frequency, instantaneous centre and roll-centre construction, camber change. Practise the IC construction by sketching both arms and extending their lines.

Quick check

1. What forms the steering axis in a MacPherson strut?
2. Which wishbone is made shorter in an SLA suspension, and why?
3. A spring of 30 N/mm acts at a motion ratio of 0.6 with no inclination. What is the wheel rate?
4. Name one packaging advantage of the MacPherson strut.
5. Where is the roll centre found?

Answers: 1. The line from the top-mount bearing to the lower ball joint. 2. The upper, to give negative camber gain in bump. 3. 30 × 0.36 = 10.8 N/mm. 4. It needs little lateral space, leaving room for a transverse engine and drive shafts. 5. Where the lines from the tyre contact centres to their instantaneous centres cross the vehicle centreline.