Handling and steering system calculations for a given car model

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

Determine the following: (2 points each except for the first question)

1. Ackermann steer angles for 500, 200 and 50 ft radius(3 points)
2. Under-steer gradient
3. Characteristic speed
4. Lateral acceleration gain at 60 mph
5. Yaw velocity gain at 60 mph
6. Side-slip angle
7. Why is the calculated under-steer gradient quite low (usually it is around 3-4 deg/g)

Given data:

mf = 1900 lbs, mr = 1552 lbs, l = 100.6 inches

Wf = 1900*0.4535*9.81 = 8452.7865 N or 861.65 Kg

Wr = 703.832 Kg or 6904.5919 N.

l = 100.6*25.4/1000 =2.555 m

1. Ackermann steer angles for 500, 200 and 50 ft radius:

Ackermann steer angle = δ = l/R * 180/3.14 + (k*ay) [degrees]

k = Under-steer gradient = (W_f /C_αf – W_r/C_αr) in [deg/g], and

ay = Lateral acceleration = 1[g] (assuming)

a) Assuming neutral steer conditions, k = 0, such that, W_f/C_αf = W_r/C_αr

for R = 500ft = 500*12 = 6000 [inches], δ = l/R * 180/3.14 = 0.96073 [deg/g]

R = 200ft = 2400 inches, δ = l/R * 180/3.14 = 2.4018 [deg/g]

R = 50ft = 600 inches, δ = l/R * 180/3.14 = 9.6073 [deg/g]

2. Under-steer gradient:

k = under-steer gradient = (W_f /C_αf – W_r/C_αr)

wf/tire = 950lbs, and wr/tire =776lbs

hence, from the table,

C_αf = [(257-225/1125-900)*50]+225 = 232.11 [lb/deg] = 232.11*0.4535*9.81=1032.6190 [N/deg], and

C_αr = [(225-171/900-675)*101]+171 = 195 [lb/deg] = 876.5228 [N/deg]

k = ((8452.7865*0.5)/1032.6190 – (6904.5919*0.5)/876.5228) = 0.1542 [deg/g]

3. Characteristic speed:

Characteristic speed = 2* Ackerman angle

u² = l*g*57.3/k = 2.55524*9.81*57.3/0.1542 = 9,314.744 (m/s)², u = 96.512 [m/s]

4. Lateral acceleration gain at 60 mph:

at ay = 1, R*g = u², u = (R*g)^1/2

for R = 500ft = 500*12 = 6000 inches, u = 38.6657 [m/s]

R = 200ft = 2400 inches, u = 24.4543 [m/s]

R = 50ft = 600inches, u = 12.2271 [m/s]

Lateral acceleration gain = ay/δ, and u = 60 mph – given,

after solving for both,

ay/δ = u²/(57.3*l*g) / 1+(k*u²/57.3*l*g)

ay/δ = 0.445286 [g/deg]

5. Yaw velocity gain at 60 mph

Yaw velocity = yaw rate = r = 57.3*u/R, u = 60 mph

after solving for both,

Yaw velocity gain = r/δ = (u/l) / (1+k*u²/57.3*l*g)

r/ δ = 9.3839 [(deg/sec)/deg]

6. Side-slip angle

W = 1900 + 1552 = 3452 lbs, Wf = 0.55 Wr = 0.45

a = 0.45*100.6 = 45.27 inches, b = 0.55*100.6 = 55.33 inches.

At low speeds, Neutral steer setup,

Body side slip angle β = (lr/l)* δ

for R = 500ft, δ = 0.96073 [deg/g],

β = (55.33/100.6) * δ = 0.5284015 [deg]

R = 200ft, δ = 2.4018 [deg/g],

β = (55.33/100.6) * δ = 1.32099 [deg]

R = 50ft = δ = 9.6073 [deg/g],

β = (55.33/100.6) * δ = 5.284015 [deg]

at high speeds,

Body side slip angle β = (lr/R) - βr, βr = αr = W_r / C_αr * ay

for R = 500ft, and u = 60mph,

βr = αr = (6904.5919 / 1498.8274)* 0.47567 = 2.1912 [deg],

β = 0.009221 - 2.1912 = -2.1819 deg = 2.1819 [deg] in opposite direction to the front wheels or in the direction of slip of rear wheels.

Similarly,

R = 200ft, βr = αr = (6904.5919 / 1498.8274)* 1.18911 = 5.4777 [deg]

β = 0.023054 – 5.4777 = (-) 5.4547 [deg]

R = 50ft, βr = αr = (6904.5919 / 1498.8274)*4.75644 = 21.911056 [deg]

β = 0.09221 – 21.9110 = (-) 21.8187 [deg]

7. Why is the calculated under-steer gradient quite low ?

a) One of the reasons for the low under-steer gradient is due to the high cornering stiffness in the rear of the vehicle. Other than that, the accuracy of the cornering stiffness figures is also crucial and instantaneous figures are required to get realistic results.