Vehicle Ride
Aim:
1.Why is the wheel rate usually different from the spring rate?
Ans: Wheel rate is a measure of the effective stiffness of the suspension system measured at the wheel centre. Spring's rate is the amount of weight required to compress itself a single inch. Wheel rate is the product of spring rate and motion rate square where as spring rate depends on the spring stiffness only.
2.Why is the ride rate different from the wheel rate?
Ans: Both ride rate and wheel rate are measure of vertical distance. Ride rate is the vertical deviation measured from sprung mass to contact patch. Wheel rate is the vertical deviation measured from sprung mass and wheel centre. In both the reference point is different i.e. for ride rate contact patch is reference and for wheel rate wheel centre is reference.
3.Can you use the same amount of damping for
I.Low and high frequency regions
II.Compression and rebound
Ans: Both the above two condition is not possible, because at low frequency region high damping is required and at high frequency low damping. For compression spring absorb energy so low damping is required and at the time of rebound spring releases energy so high damping is required.
4.Why does the rear ride frequency need to be higher than the front?
Ans: Rear wheel needs higher frequency than front to reduce the pitching motion because rear wheel hits the bump after some time of the front wheel.
5.For improved vibration isolation in the wheel hop region, would you prefer:
I.A higher or lower unsprung mass?
II.Higher or lower damping?
Ans: For better vibration isolation we need higher value of natural frequency.
Natural frequency is inversely proportional to the unsprung mass, so lower the mass higher will bethe natural frequency better the isolation. Same way natural frequency is inversely proportional to the damping, so lower the value of damping lower will be the natural frequency better the isolation.
6.From a ride frequency standpoint, why are air suspensions beneficial, especially in heavy vehicles (do some research to answer this)
Ans: Air suspensions beneficial for heavy vehicle because depending on the load air pressure in the suspenssion can change so it avoid pitching effect.
7.Calculate sprung mass and unsprung mass natural frequencies for a vehicle with a 2000 kg sprung mass, 50/50 weight distribution, 50 kg of unsprung mass in each corner, spring stiffness of 25 N/mm and tire stiffness of 250 N/mm. Assume gravity to be 10 m/sec^2 to simplify calculations. Report your answer in hertz
Ans: Sprung Mass (Mts) = 2000Kg
Sprung Mass at Each wheel (Ms) = (Mts)/4
= 500kg
Unsprung Mass (Mus) = 50Kg
Ks = 25N/mm
Kt = 250N/mm
Now for Sprung Mass: both this stiffness are in series.
K1 = `(Ks*Kt)/(Ks+Kt)`
= `(25*250)/(25+250)`
= 22727.27 N/m
Natural frequency:
f1 = `(1/(2*pi))*sqrt((K1)/(Ms))`
= ``(1/(2*pi))*sqrt((22727.77)/(500))``
= 1.073 Hz
Natural frequency for sprung mass is 1.073 Hz
Now for UnSprung Mass: both this stiffness are in Parallel.
K2 = `(Ks+Kt)`
= `(25+250)`
= 275000 N/m
Natural frequency:
f2 = `(1/(2*pi))*sqrt((K2)/(Mus))`
= ``(1/(2*pi))*sqrt((275000)/(50))``
= 11.80 Hz
Natural frequency for Unsprung mass is 11.80 Hz.
8.Calculate the time lag between the front and rear suspensions for:
1.Front ride frequency of 1.2Hz, Rear ride frequency of 0.8Hz
2.Front ride frequency of 1.2Hz, Rear ride frequency of 1.45Hz
Now calculate the time lag due to wheelbase. Wheelbase = 98.42 inches, speed = 37.42 mph.
From these numbers, determine which ride frequency set (1 or 2) works best from a flat ride perspective.
Ans: Time period is inverse of the frequency.
Front frequency = 1.2 Hz
Time period = 1/1.2
= 0.833 sec
Rear frequency = 0.8 Hz
Time period = 1/ 0.8
= 1.25 sec
For flat ride front time period should be more than rear which is not satisfied in this case.
For second set of values,
Front frequency = 1.2 Hz
Time period = 1/1.2
= 0.833 sec
Rear frequency = 1.45 Hz
Time period = 1/1.45
= 0.689 sec
For flat ride front time period should be more than rear which is satisfied in this case. Hence flat ride is possible with second set of values.
Time lag due to wheelbase:
Wheel base = 98.42 in (converting to meters)
= 2.499 m
Speed = 37.42 mph (converting to mps)
= 16.72 m/s
Time lag = Wheelbase/speed
= 2.499/16.72
= 0.149 sec