Problem Set-1b Week 11Challenge : Basic Handling parameters

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

Solution:

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

Ackerman steer angle, `delta = L/R*180/pi`

a) For R = 500ft

`delta = (100*6)/(500*12)*180/pi = 0.961deg`

b) For R = 200ft

`delta = (100*6)/(200*12)*180/pi = 2.4deg`

c) For R = 50ft

`delta = (100*6)/(50*12)*180/pi = 9.61deg`

2. Understeer gradient

The weight distribution on each front and rear wheel shall be 50/50 of the equivalent weights at front and rear.

Therefore, weight of each wheel:

`w_f = 950lb,\ w_r =776lb`

To calculate the understeer gradient, we need to know the cornering stiffness at these loads. To get the values from the supplied table, we utilise the interpolation formula.

Therefore,

`C_{alpha_f} = 225+(950-900)*((257-225)/(1125-900)) = 232.11 {lb}/{deg}`

`C_{alpha_r} = 171+(776-675)*((225-171)/(900-675)) = 195.24 {lb}/{deg}`

Understeer gradient, `K = {w_f}/{C_{alpha_f}}-{w_r}/C_{alpha_r} = 950/232.11-776/195.24 = 0.118{deg}/g`

3. Characteristic speed

The characteristic speed is the speed at which the required steering angle to negotiate the turn is twice the ackerman angle.

It is given by:

`V_{char} = sqrt{{57.3Lg}/K = sqrt{{57.3*100.6*32.2}/{12*0.118}`

`V_{char} = 362.05 {ft}/{sec} = 246.85mph`

4. Lateral acceleration gain at 60 mph

60mph = 88 ft/sec

yaw rate

`a_y/delta = {V^2/{57.3Lg}}/{1+{KV^2}/{57.3Lg} } = {88^2/{57.3*100.6*32.2/12}}/{1+{0.118*88^2}/{57.3*100.6/12*32.2}} = 0.473 {g}/{deg}`

5. Yaw velocity gain at 60 mph

60mph = 88 ft/sec

yaw rate

`r/delta = {V/L}/{1+{KV^2}/{57.3Lg} } = {88/{100.6/12}}/{1+{0.118*88^2}/{57.3*100.6/12*32.2}} = 9.911 {{deg}/{sec}}/{deg}`

6. Slideslip angle at the CG on an 800 ft radius turn at 60 mph.

`beta = 57.3c/R-{W_rV^2}/{C_{alpha_r}Rg}`

also, `c = {W_fL}/W_{t\otal} = {1900*100.6}/{(1900+1552)*12} = 4.614ft`

`C_{alpha_r} = 257+(1552-1125)*{300-257}/{1350-1125} = 338.60 {lb}/{deg}`

`beta = 57.3*4.614/800-{1552*88^2}/{338.60*800*32.3 = -1.047deg`

7. Why is the calculated understeer gradient quite low (usually it is around 3-4 deg/g) 

We calculated the understeer gradient using a 2DOF model and solely considered the cornering stiffness, which is a tyre property. However, in practise, other systems like as suspension, steering, and so on will also contribute to the understeer gradient.