Vehicle Ride

difference between wheel rate and spring rate is the wheel rate is what the driver feels and the tires deal with while driving

2) İt is nearly the same but in ride rate the vertical displacement is between the contact patch and the from sprung mass but for wheel rate it is from wheel center.

3)

a) To reduce the transibilty ratio it is prefered to use high damping in lower frequency and low damping for high frequency.

b) When the spring compress the spring will store an energy inside the spring that's why it is not necessary to use high damipng in compression whereas in rebound the spring will release energy that's why it is necessary to use high damipng in rebound.

4) There is a timelag between front and rear wheel according to the vehilce speed and wheelbase lenght. Therefore, the front and rear wheel will oscilate differently. To solve this problem the rear ride frequency need to be higher than the front to have an simultaneous oscilate.

5) a lower unsprung mass and low damping for having a better viration isolation.

6) Because air suspansion can reduce the effect on the pitchin depending on the load varies on wheels

7)

Sprung mass pre corner (ms)=500 kg

Sprung mass ride rate (ks)= 25*250/(25+250)=22.73 N/mm

sprung mass natural frequency=1/(2*pi)*sqrt(ks/ms)=1.0731 hz

unsprung mass pre corner (mus)=50 kg

unsprung mass ride rate (kus)= (25+250)=275 N/mm

unsprung mass natural frequency=1/(2*pi)*sqrt(kus/mus)=11.8 hz

8)

Wheelbase=2490 mm

speed= 16.73m/s

time lag=wheel base/ speed = 0.015 sec

frequency

Front ride frequency of 1.2Hz, Rear ride frequency of 0.8Hz

time period> T(f)= 1/ f(f) = 0.83 T(r)= 1/f(r)= 1.25sec

T(f) < Tr + time lag flat ride is not possibe

Front ride frequency of 1.2Hz, Rear ride frequency of 1.45Hz

T(f) = 0.83 sec T(r) = 0.69 sec

T(f) >= T(r) + time lag flat ride is possible