Vehicle Ride
ANS_1:-it is because of the motion ratio, where the value of (b) always grater than (a), so the motion ratio is always less than 1.
Because of this the wheel rate(kw) is always different from spring rate(ks).
kw=ks(a/b)^2
b=distance between trailing arm pivot to the wheel mtg
a=distance between trailing arm pivot to spring pivot
ANS_2:- Reason for its difference,because ride rate is the function of the spring stiffness(Ks) & tyre stiffness(Kt), but the wheel rate depends on spring stiffness(Ks).
ANS_3:- No, we cannot use same amount of damping because,
1.1) Low frequency required high damping (needs to act slower)
1.2) High frequency required low damping (needs to act faster)
2.1) compression required high damping (store the bump energy)
2.2) rebound required low damping (release the spring energy faster)
ANS_4:- Because the rear suspension always hits a bump later on the first wheel, so to meet with the front wheel frequency the rear wheel frequency needs to be higher.
ANS_5:- For Vibration Isolation:-
a) Lower unsprung mass good for isolation because the higher unsprung mass natural frequency is high which is not good.
b) Low damping is good, because it can overcome the frequency earliest.
ANS_6:- Benefit of air suspension in heavy trucks.
a) we can manage the ride height, results low ride height means better ride & cornering performance.
b) we can also manage the ride rate.
ANS_7:-
M=2000Kg
sprung mass per corner=2000/4=500Kg
unsprung mass per corner=50Kg
spring stiffness(Ks)=25 N/mm
tire stiffness(Ktr)=250 N/mm
Ride rate(Kr)=(Ks*Ktr)/(Ks+Ktr)=22.
Sprung mass natural frequency= 1/2pi(sqrt(Kr/sprung mass)=1.07Hz
unsprung mass natural frequency= 1/2pi(sqrt((Ks+Ktr)/sprung mass))=11.80Hz
ANS_8:-
Wheel Base=98.42 inch=2.5m
Speed=37.42mph=16.72m/s
Time Lag= Wheel base/speed=2.5/16.72= 0.15s
1) Front ride frequency=1.2Hz
time period(ts)=1/1.2=0.833s
Rear ride frequency=0.8Hz
time period(ts)=1/0.8=1.25s
TIME LAG=1.25-0.833= 0.417
2) Front ride frequency=1.2Hz
time period(ts)=1/1.2=0.833s
Rear ride frequency=1.45Hz
time period(ts)=1/1.45=0.68s
TIME LAG=0.68-0.833= -0.153s
When we compare the wheelbase time lag with both the sets the set-2 shows better result for flat ride perspective.
Because of this the wheel rate(kw) is always different from spring rate(ks).
kw=ks(a/b)^2
b=distance between trailing arm pivot to the wheel mtg
a=distance between trailing arm pivot to spring pivot
ANS_2:- Reason for its difference,because ride rate is the function of the spring stiffness(Ks) & tyre stiffness(Kt), but the wheel rate depends on spring stiffness(Ks).
ANS_3:- No, we cannot use same amount of damping because,
1.1) Low frequency required high damping (needs to act slower)
1.2) High frequency required low damping (needs to act faster)
2.1) compression required high damping (store the bump energy)
2.2) rebound required low damping (release the spring energy faster)
ANS_4:- Because the rear suspension always hits a bump later on the first wheel, so to meet with the front wheel frequency the rear wheel frequency needs to be higher.
ANS_5:- For Vibration Isolation:-
a) Lower unsprung mass good for isolation because the higher unsprung mass natural frequency is high which is not good.
b) Low damping is good, because it can overcome the frequency earliest.
ANS_6:- Benefit of air suspension in heavy trucks.
a) we can manage the ride height, results low ride height means better ride & cornering performance.
b) we can also manage the ride rate.
ANS_7:-
M=2000Kg
sprung mass per corner=2000/4=500Kg
unsprung mass per corner=50Kg
spring stiffness(Ks)=25 N/mm
tire stiffness(Ktr)=250 N/mm
Ride rate(Kr)=(Ks*Ktr)/(Ks+Ktr)=22.
Sprung mass natural frequency= 1/2pi(sqrt(Kr/sprung mass)=1.07Hz
unsprung mass natural frequency= 1/2pi(sqrt((Ks+Ktr)/sprung mass))=11.80Hz
ANS_8:-
Wheel Base=98.42 inch=2.5m
Speed=37.42mph=16.72m/s
Time Lag= Wheel base/speed=2.5/16.72= 0.15s
1) Front ride frequency=1.2Hz
time period(ts)=1/1.2=0.833s
Rear ride frequency=0.8Hz
time period(ts)=1/0.8=1.25s
TIME LAG=1.25-0.833= 0.417
2) Front ride frequency=1.2Hz
time period(ts)=1/1.2=0.833s
Rear ride frequency=1.45Hz
time period(ts)=1/1.45=0.68s
TIME LAG=0.68-0.833= -0.153s
When we compare the wheelbase time lag with both the sets the set-2 shows better result for flat ride perspective.