Embedded Files
How can we adjust the compromise between good ride and precise handling to ensure a vehicle characteristics suit its market and the roads where it will be used?
Introduction
This article covers material studied by the author at the University of Hertfordshire, with the aim to analytically evaluate the spring stiffness and damping coefficient for a vehicle using a four-post rig.
The vehicle used was a Ford Focus 1.8 Mk 1.
A graph is shown here of suspension natural frequency versus sprung mass and vertical stiffness.
For example with a vertical stiffness of 100 lb/in (~17.5 N/mm) for a sprung mass of 600 lbs (~275 kg) the frequency is ~77 cpm (~1.283 Hz).
Nomenclature and Vehicle Data
a: distance from centre of gravity to front axle = 1.3m
b: distance from centre of gravity to rear axle = 1.32m
c: damping coefficient (N*s/m)
I: second moment of mass = 1720 kg*m^2
k: spring stiffness (N/m)
kt: tyre stiffness = 170,000 N/m
m: total vehicle mass = 1122 kg
muf: mass unsprung front (single wheel) = 27 kg
mur: mass unsprung front (single wheel) = 24 kg
Background Theory
<insert graphs and equations>
When damping is present, the resonance of the system occurs ad the damped natural frequency, given by:
omega_d = omega_n * sqrt (1 - zeta^2)
kf = ( omega_n^2 * mf * kt ) / ( kt - omega_n^2 * mf )
kr = ( omega_n^2 * mr * kt ) / ( kt - omega_n^2 * mr )
where mf = 276 kg; mr = 285 kg; ky = 170,000 N/m
Stiffness and Damping Calculation
The following equations are used to calculate front and rear N/m and Ns/m values:
NOTE that the comma is used for thousands, and the dot is used as decimal separator
damping const c= 2*zeta*sqrt(k*m)
Quarter Vehicle Model (QVM)
Front corner sprung mass = 276 kg - 27 kg = 249 kg
assuming suspension stiffness 16,000 N/m
<insert graphs and equations>
we solve to obtain: omega_1 = 83.1 rad/s, omega_2 = 7.29 rad/s
hence wheel frequency = 13.22 Hz, body frequency = 1.16 Hz
[download article link]
======================================================================================
1/4 Car Model
The diagrams below show the basic theory, where we derive the equations via Newton's 2nd Law:
F_net = m * a
m*a + c*v + k*x = F
a + c/m*v + k/m*x = F/m (after dividing everything by mass)
In Simulink we can represent it this way with 3 gains of 1/mass, Damping Coeff, Spring Stiffness:
However our suspension is subtly different because the force is input at the tyre contact patch.
Further reading: 1/4 Car Suspension - University of Michigan , 1/4 Car - Intl Journal of Eng Research
1/2 Car Model
Further reading: Mathworks , FileExchange ,
Kinematics
A useful discussion about influence on tyres: Dr Racing Blog
Some interesting and useful information from the McLaren F1:
wheel travel = +80 -90 F&R
Ride Frequency = 80/92 cpm
Roll Axis Centroid = roll center 60 mm a. ground @ F. 90 mm R.
Anti-Roll Bars= Front only
Roll Stiffness = 3 degrees per g lat
Anti-Dive = Yes (40-50%) investigate farb
Anti-Squat = ?
Damping = aluminum shocks[?] - ride ht adj?
velocity ratios = as near 1:1 as poss. (same F&R) + rising rate
Camber Change = 75% of body roll = 4.4 degrees for 80 mm travel
Wheel size = 18" dia x 8.5" F x 11.5" r
Tyre size = 245x40x18
Wheel base = 104" (2640)
Track = 1800 over tyres F&R [corrected to 1554]
Front Offset and Trail = Offset 14 mm @ gr, trail 30 mm
KPI = 10 degrees [corrected to 9 degrees]
Castor = 4 degrees
Static Camber = -45' F&R
Toe In = 3mm neg. F O F
Bump Steer = None
Center of pressure = 57% of wheel base
CD = .330
CL Front and Rear = .150 F .150 R (stable in yaw)
[something off the page] torsional stiffness = 10,000 ft lb/degree axle to axle min
Steering Wheel Diam = 13" dia
Unsprung Weight = 37 kg F 49 R
Frontal Area = 1.62 m^2
CDA = .535
Ground Clearance = 120mm
Ackermann Characteristics = 0-5 degrees / 0% 5-10 degrees / 35% 10-20 degrees / 80% -> 100% From [?] 25 degrees
References: Susp Design
Interesting article by Willem Toet: thinking-kinematics-led-optimisation-willem-toet
Page updated
Google Sites
Report abuse