Interactive Lecture
Understanding Suspension Systems and Ride Comfort
by Dr Arnaldo Delli Carri
Ride quality refers to how well a vehicle isolates occupants from road disturbances, while handling describes the vehicle's responsiveness to driver inputs. These objectives often conflict: soft suspensions improve ride comfort but reduce handling precision, creating a fundamental trade-off in suspension design.
The suspension must do four things at once: isolate occupants from road disturbances, maintain tyre contact for traction, control body motion for handling, and support the vehicle weight. These objectives pull in different directions — that tension is what suspension design is all about.
The quarter-car model represents one corner of the vehicle as a two-degree-of-freedom system:
Sprung Mass (Vehicle Body) Equation:
\[m_s \ddot{z}_s = -k(z_s - z_u) - c(\dot{z}_s - \dot{z}_u)\]
Unsprung Mass (Wheel/Tyre) Equation:
\[m_u \ddot{z}_u = k(z_s - z_u) + c(\dot{z}_s - \dot{z}_u) - k_t(z_u - z_r)\]
where:
The suspension system has two natural frequencies corresponding to sprung and unsprung mass modes:
Sprung Mass Natural Frequency (Body Bounce):
\[f_s = \frac{1}{2\pi}\sqrt{\frac{k}{m_s}} \approx 1-1.5 \text{ Hz (typical)}\]
This body-bounce frequency \(f_s\) is the natural frequency of the sprung mass. It is the same quantity written \(f_n\) in the summary table and the problems, with natural angular frequency \(\omega_n = 2\pi f_s = \sqrt{k/m_s}\).
Unsprung Mass Natural Frequency (Wheel Hop):
\[f_u = \frac{1}{2\pi}\sqrt{\frac{k + k_t}{m_u}} \approx 10-15 \text{ Hz (typical)}\]
The large frequency separation ensures that wheel hop doesn't couple strongly with body motion.
The damping ratio controls how quickly oscillations decay:
\[\zeta = \frac{c}{2\sqrt{km_s}}\]
Typical Values:
Underdamping (\(\zeta < 1\)) is always used because critical damping (\(\zeta = 1\)) transmits too much high-frequency road harshness.
The transmissibility function describes how road inputs are transmitted to the vehicle body:
\[T(\omega) = \left|\frac{Z_s(\omega)}{Z_r(\omega)}\right| = \frac{\sqrt{1 + (2\zeta r)^2}}{\sqrt{(1-r^2)^2 + (2\zeta r)^2}}\]
where \(r = \omega/\omega_n\) is the frequency ratio, \(\omega = 2\pi f\) is the road excitation frequency, and \(\omega_n = 2\pi f_s = \sqrt{k/m_s}\) is the natural angular frequency of the sprung mass.
Key Observations:
During cornering, the vehicle body rolls about its roll axis:
Roll Angle:
\[\phi = \frac{m h a_y}{K_\phi}\]
where:
Anti-Roll Bars: Additional torsional springs connecting left/right wheels to increase roll stiffness without affecting ride stiffness.
The Fundamental Compromise:
This simulation shows a quarter-car model responding to a road bump. Adjust suspension parameters to see the trade-off between ride comfort and wheel control. It covers the ride (vertical bounce) behaviour only; the handling side — body roll and anti-roll bars — is treated analytically in the theory above and in Problem 3.
Ride and handling engineering impacts vehicle design across all segments:
| Concept | Equation | Description |
|---|---|---|
| Sprung Mass EOM | \(m_s \ddot{z}_s = -k(z_s - z_u) - c(\dot{z}_s - \dot{z}_u)\) | Vehicle body motion |
| Unsprung Mass EOM | \(m_u \ddot{z}_u = k(z_s - z_u) + c(\dot{z}_s - \dot{z}_u) - k_t(z_u - z_r)\) | Wheel/tyre motion |
| Natural Frequency | \(f_n = \frac{1}{2\pi}\sqrt{\frac{k}{m_s}}\) | Body bounce frequency |
| Damping Ratio | \(\zeta = \frac{c}{2\sqrt{km_s}}\) | Oscillation decay rate |
| Transmissibility | \(T = \frac{\sqrt{1 + (2\zeta r)^2}}{\sqrt{(1-r^2)^2 + (2\zeta r)^2}}\) | Road-to-body amplitude ratio |
| Roll Angle | \(\phi = \frac{m h a_y}{K_\phi}\) | Body roll during cornering |
| Roll Stiffness | \(K_\phi = k_f t_f^2 + k_r t_r^2\) | Resistance to roll |
A quarter-car model has sprung mass ms = 400 kg and spring stiffness k = 25,000 N/m. Calculate: (a) the natural frequency in Hz, (b) the damping coefficient for ζ = 0.35, and (c) the period of oscillation.
Show Solution
(a) Natural frequency:
\[\omega_n = \sqrt{\frac{k}{m_s}} = \sqrt{\frac{25000}{400}} = \sqrt{62.5} = 7.91 \text{ rad/s}\]
\[f_n = \frac{\omega_n}{2\pi} = \frac{7.91}{2\pi} = 1.26 \text{ Hz}\]
(b) Damping coefficient:
\[c = 2\zeta\sqrt{km_s} = 2 \times 0.35 \times \sqrt{25000 \times 400}\]
\[c = 0.7 \times \sqrt{10,000,000} = 0.7 \times 3162 = 2213 \text{ N·s/m}\]
(c) Period of oscillation:
\[T = \frac{1}{f_n} = \frac{1}{1.26} = 0.79 \text{ seconds}\]
Interpretation: This 1.26 Hz frequency is typical for comfort-oriented vehicles. The body completes one oscillation in 0.79 seconds, which feels smooth to occupants.
Answer: (a) \(\omega_n = 7.91\,\text{rad/s}\), \(f_n = 1.26\,\text{Hz}\); (b) \(c = 2213\,\text{N·s/m}\); (c) period \(T = 0.79\,\text{s}\).
For the suspension in Problem 1 with ζ = 0.35, calculate the transmissibility at: (a) low frequency (0.5 Hz), (b) resonance (1.26 Hz), and (c) high frequency (10 Hz). At what frequency does the suspension start providing isolation (T < 1)?
Show Solution
(a) At f = 0.5 Hz:
\[r = \frac{f}{f_n} = \frac{0.5}{1.26} = 0.40\]
\[T = \frac{\sqrt{1 + (2 \times 0.35 \times 0.40)^2}}{\sqrt{(1-0.40^2)^2 + (2 \times 0.35 \times 0.40)^2}}\]
\[T = \frac{\sqrt{1 + 0.078}}{\sqrt{0.706 + 0.078}} = \frac{1.038}{0.885} = 1.17\]
(b) At resonance, f = fn = 1.26 Hz (r = 1):
\[T = \frac{\sqrt{1 + (2 \times 0.35)^2}}{\sqrt{0 + (2 \times 0.35)^2}} = \frac{\sqrt{1.49}}{\sqrt{0.49}} = \frac{1.22}{0.70} = 1.74\]
(c) At f = 10 Hz:
\[r = \frac{10}{1.26} = 7.94\]
\[T = \frac{\sqrt{1 + (2 \times 0.35 \times 7.94)^2}}{\sqrt{(1-7.94^2)^2 + (2 \times 0.35 \times 7.94)^2}}\]
\[T = \frac{\sqrt{1 + 30.9}}{\sqrt{3849 + 30.9}} = \frac{5.65}{62.3} = 0.091\]
Isolation threshold: T < 1 when \(r > \sqrt{2} = 1.41\)
\[f_{isolation} = 1.41 \times f_n = 1.41 \times 1.26 = 1.78 \text{ Hz}\]
Above 1.78 Hz, the suspension provides isolation. The 10 Hz result shows strong isolation (about a 91% reduction) at high frequencies.
Answer: (a) \(T = 1.17\); (b) \(T = 1.74\); (c) \(T = 0.091\). The suspension provides isolation (\(T < 1\)) above 1.78 Hz.
A 1500 kg vehicle has CG height h = 0.55 m above the roll axis, front track tf = 1.5 m, rear track tr = 1.5 m, and equal spring rates kf = kr = 25,000 N/m. Calculate the roll angle during a 0.8g lateral acceleration corner. What would be the effect of adding anti-roll bars that double the effective roll stiffness?
Show Solution
Step 1: Calculate roll stiffness:
\[K_\phi = k_f t_f^2 + k_r t_r^2 = 25000 \times 1.5^2 + 25000 \times 1.5^2\]
\[K_\phi = 2 \times 25000 \times 2.25 = 112,500 \text{ N·m/rad}\]
Step 2: Calculate roll angle:
\[\phi = \frac{m h a_y}{K_\phi} = \frac{1500 \times 0.55 \times 0.8 \times 9.81}{112,500}\]
\[\phi = \frac{6474}{112,500} = 0.0575 \text{ rad} = 3.30°\]
Step 3: With doubled roll stiffness:
\[K_\phi' = 2 \times 112,500 = 225,000 \text{ N·m/rad}\]
\[\phi' = \frac{6474}{225,000} = 0.0288 \text{ rad} = 1.65°\]
Answer: \(K_\phi = 112\,500\,\text{N·m/rad}\); roll angle \(\phi = 3.30°\); with anti-roll bars doubling the roll stiffness, \(\phi = 1.65°\).
Conclusion: Anti-roll bars halve the roll angle from 3.3° to 1.65°, improving handling feel and reducing lateral load transfer. However, they don't affect the ride frequency (ride comfort) since they only act on asymmetric (left-right) motion, not symmetric (both wheels together) bounce.