BillaNichols
Electrical
Is the correct?
Below is the calculation I used to determine my spring rates for auto cross racing. My target natural frequency is 2.2 in the front and 2.5 in the rear. The calculation is three steps, finding the wheel rate, find the wheel rate in series with the spring rate of the tire and using this value to calculate the natural frequency.
The equations:
The wheel rate Kw
Kw=Ks*(MR)^2
Where:
Ks = the coilover Spring rate
MR = the Motion Ratio
The wheel rate with the contribution of the tire spring rate
Kw' = (Kw*Kt)/(Kw+Kt)
Where:
Kw' = combined spring rate
Kt = the spring rate of the tire
Kw = the spring of the wheel (the contribution of the coilover spring acting through the suspension)
The Natural Frequency
NF = 1/2π(Kw'/Msm)^1/2
Where:
NF = the Natural Frequency
Kw' = the Spring Rate at the wheel
Msm = the Sprung Mass of corner of interest
The front unsprung mass = 312 kilograms
The rear unsprung mass = 294 kilograms
Front Corner NF Calcs
Selecting a front spring rate of 140,101 Newtons per Meter (N/M) (800 lbs/in)
Kw=Ks*(MR)^2
Kw = [(140,101 N/M)*((.7334)^2)]
Kw = 75,357 N/M
Kw' = (Kw*Kt)/(Kw+Kt)
Kw' = (75,357 N/M * 318,030 N/M)/(75,357 N/M + 318,030 N/M)
Kw' = 60,922 N/M
NF = 1/2π(Kw'/Msm)^1/2
NF = (.1592)*[(60,922 N/M)/312 Kg]^1/2
NF = 2.22 Hz
Rear Corner NF Calcs
Selecting a rear spring rate of 210,152 Newtons per Meter (N/M) (1200 lbs/in)
Kw=Ks*(MR)^2
Kw = [(210,152 N/M)*((.6822)^2)]
Kw = 97,804 N/M
Kw' = (Kw*Kt)/(Kw+Kt)
Kw' = (97,804 N/M * 318,030 N/M)/(97,804 N/M + 318,030 N/M)
Kw' = 74,801 N/M
NF = 1/2π(Kw'/Msm)^1/2
NF = (.1592)*[74,801 N/M)/294 Kg]^1/2
NF = 2.54 Hz
Artificer of control system engineering
Below is the calculation I used to determine my spring rates for auto cross racing. My target natural frequency is 2.2 in the front and 2.5 in the rear. The calculation is three steps, finding the wheel rate, find the wheel rate in series with the spring rate of the tire and using this value to calculate the natural frequency.
The equations:
The wheel rate Kw
Kw=Ks*(MR)^2
Where:
Ks = the coilover Spring rate
MR = the Motion Ratio
The wheel rate with the contribution of the tire spring rate
Kw' = (Kw*Kt)/(Kw+Kt)
Where:
Kw' = combined spring rate
Kt = the spring rate of the tire
Kw = the spring of the wheel (the contribution of the coilover spring acting through the suspension)
The Natural Frequency
NF = 1/2π(Kw'/Msm)^1/2
Where:
NF = the Natural Frequency
Kw' = the Spring Rate at the wheel
Msm = the Sprung Mass of corner of interest
The front unsprung mass = 312 kilograms
The rear unsprung mass = 294 kilograms
Front Corner NF Calcs
Selecting a front spring rate of 140,101 Newtons per Meter (N/M) (800 lbs/in)
Kw=Ks*(MR)^2
Kw = [(140,101 N/M)*((.7334)^2)]
Kw = 75,357 N/M
Kw' = (Kw*Kt)/(Kw+Kt)
Kw' = (75,357 N/M * 318,030 N/M)/(75,357 N/M + 318,030 N/M)
Kw' = 60,922 N/M
NF = 1/2π(Kw'/Msm)^1/2
NF = (.1592)*[(60,922 N/M)/312 Kg]^1/2
NF = 2.22 Hz
Rear Corner NF Calcs
Selecting a rear spring rate of 210,152 Newtons per Meter (N/M) (1200 lbs/in)
Kw=Ks*(MR)^2
Kw = [(210,152 N/M)*((.6822)^2)]
Kw = 97,804 N/M
Kw' = (Kw*Kt)/(Kw+Kt)
Kw' = (97,804 N/M * 318,030 N/M)/(97,804 N/M + 318,030 N/M)
Kw' = 74,801 N/M
NF = 1/2π(Kw'/Msm)^1/2
NF = (.1592)*[74,801 N/M)/294 Kg]^1/2
NF = 2.54 Hz
Artificer of control system engineering