Vehicle Ride
Q1.
Ans: Wheel rate is different from spring rate because of motion ratio/installation ratio which defines the geometeric relationship between the change in position of force. Motion ratio aries due to the fact spring is mounted on lower control arm and will be having a offset from the wheel center.
Kw = Ks*(MR)^2
we can say that spring rate is always less than spring rate.
Q2.
Ans: wheel rate is defined as the vertical force acting on the wheel per unit vertical displacement of wheel along the spindle realtive to the chassis. Ride rate is defined as the verica force acting on the wheel at the tire contact patch per unit dispalcement of the sprung mass relative to the chassis.
ride rate
Kr = (Kw*Ks)/(Kw+ks)
but we know that,
Ks =Kw/(MR)^2
we get
Kr = Kw/[1+(MR)^2]
since denominator is greater than 1, ride rate is always less than wheel rate.
Q3.
Ans: i) No we cant use the same amount of daming of low and high frequency region. For low frequency region high damping is preferred while for high frequency region low damping is preferred.
ii) No we cannot use the same damping for compression and rebound. Compression damping is less than rebound damping by a factor of 2-3. Because in comperssion, spring aborbs energy and hence the damping needs to be lower. In rebound, spring releases energy and this rate at which energy is released must be controlled by damping .
Q4.
Ans: If both rear and front ride frequence are same, out of phase motion wil cause between the front and rear suspesnion because of the wheelbase seperation. This leads to pitching of vehicle .
Q5.
Ans: High unsprung mass is preferred, lowing damping is preferred.
Q6.
Ans: Ride frequency point of view air suspension is beneficial,
i) air suspension have better compliance tha steel suspension
ride frequency is,
fs = (1/2pi)*`square root`(Kr/Ms)
air suspension stiffness is lower value, ride rate(Kr) value goes down.
ii) The driver can increase the firmness when towing to account for heavier loads.
Q7.
Ans: Ms = 2000 kg
50/50 weight distribution
on each corner, Ms = 2000/4 = 500 kg
Mus = 50 kg
Ks = 25 N/mm
Kw = 250 N/mm
g = 10 m/s^2
i) fs = (1/2pi)*'Square root'(kr/Ms)
Ride rate,
Kr = Kw*Ks/(Kw+Ks)
Kr = 25*250/(25+250) = 6250/275
Kr = 22.72 N/mm = 22727.27 N/m
fs = 0.1591*'Square root'(22727.22/500)
fs = 1.07 Hz
ii) fus = (1/2pi)*'square root'(Keq/Mus)
Equivalent stiffness,
Keq = Kw +Ks = 250 + 25
Keq = 2750000 N/m
fus = 0.1591*'square root'(275000/50)
fus = 11.8 Hz
Q8.
Ans: wheelbase = 2.5 m
speed = 16.67 m/s
T(lag) = wheelbase/speed
T(lag) = 2.5/16.67
T(lag) = 0.15 s
to get flat ride,
T(front) = T(rear) + T(lag)
Case I: f1 = 1.2 Hz
T(front) = (1/1.2) = 0.833 s
f2 = 0.8 Hz
T(rear) = (1/0.8) =1.25 s
T(front) = T(rear) + T(lag)
but,
0.833< 1.25 + 0.15
Cae II: f1 = 1.2 Hz
T(front) = 0.83 s
f2 = 1.45 Hz
T(rear) = 0.68 s
T(front) = T(rear) + T(lag)
0.83 = 0.68 + 0.15
thus CaseII works best for flat ride perspective