Problem Set-1b Week 11Challenge

A car has a weight of 1900 lb front axle and 1552 lb on the rear with a wheelbase of 100.6 inches. 
The tires have the following cornering stiffness values

Given Date:-
mf=1900lbs
mr=1552lbs
l=1006.6in

wf=8452.7865N
wr=6904.591N
l=2.555m

1. Ackermann steer angles for 500, 200 and 50 ft radius
ackermann steer angle = l/R*(180/3.14)+(k*ay) in deg
k=understeer gradient=(wf/Caf-wr/car) in deg/g
ay=lateral  acceleration

a) Assuming neutral steer condition, k=0,i.e(wf/car=wr/car)
For, R=500ft, 
R=500*12 = 6000in
ackermann steer angle = (l/R)*(180/3.14)
=(100.6/6000)*(180/3.14)
= 0.96073 deg/g

For, R=200ft, 
R=200*12 = 2400in
ackermann steer angle = (l/R)*(180/3.14)
=(100.6/2400)*(180/3.14)
= 2.4018 deg/g

For, R=50ft, 
R=50*12 = 600in
ackermann steer angle = (l/R)*(180/3.14)
=(100.6/600)*(180/3.14)
= 9.6073 deg/g

2) Understeer Gradient
k=understeer gradient=(wf/car)-(wr/car)
wf/tire=950lbs
wr/tire=776lbs

caf=[(257-225/1125-900)]+225
   = 232.11 lbs/deg
or
   = 232.11*0.4535*9.82
   = 1032.619 N/deg

car=[225-171/900-675]+171
   = 195 lbs/deg
or
   = 195*0.4535*9.82
   = 876.522 N/deg 

k=(8452.786*0.5)/1032.619-(6904.591*0.5)/876.522) = 0.1542 deg/g

3) Characterstics Speed
=2*ackermann
u^2=l*g*57.3=2.555*9.81*57.3/0.1542
=9314.744(m/s)^2
u=96.512m/s

4) Lateral acceleration gain at 60 mph
at ay=1,R*g=u^2,u=(R*g)^(1/2)
for R=500ft = 6000in 
u=38.665m/s
for R=200ft = 2400in 
u=24.4543 m/s
for R=50ft = 600in 
u=12.227 m/s

lateral acceleration gain = 60-given
=u^2/(57.3*l*g)/1+(k*u^2/57.3*l*g)
=0.4452 g/deg

5. Yaw velocity gain at 60 mph
yaw veloctiy = yaw rate = r=57.3*u/R, u=60mph
yaw velocity gain = =(u/l)/(1+k*u^2/57.3*l*g)
=9.383 (m/s)/deg

6. Slideslip angle at the CG on an 800 ft radius turn at 60 mph.
W=1900+1552=3452lbs, wf=0.55wr=0.45
a=0.45*100.6=45.27in
b=0.55*100.6=55.33in

At low speed, neutral speed setup
body side slip angle = (lr/l)*ackermann
For R=500ft, ackermann = 0.96073deg/g
slip angle=(55.33/100.6)*0.96073=0.528deg

For R=200ft, ackermann = 2.4018deg/g
slip angle=(55.33/100.6)*2.4018=1.32 deg

For R=50ft, ackermann = 9.6073deg/g
slip angle=(55.33/100.6)*9.6073=5.28deg

At high speeds,
body side slip angle,(lr/R)-Br, Br=ar=Wr/Car*ay
For R=500ft and U=60mph
Br=ar=(6904.591/1498.827)*0.4756=2.192deg
B=0.009221-2.1912=-2.1819(deg)
-sign indicates direction of slip of the rear wheel

R=200ft,Br=ar=(6904.591/1498.824)*0.47567=5.477 deg
B=-0.0023045-5.477=-5.4547deg

R=50ft,Br=ar=(6904.591/1498.824)*4.75644=21.911 deg
B=-0.09221-21.911=-21.8187deg

7. Why is the calculated understeer gradient quite low (usually it is around 3-4 deg/g)
High cornering stiffness in the rear of the vehicle.
The accuracy of the cornering results crucial and instantaneos results are required to get the accurate realistic results.