Problem Set-1b Week 11Challenge : Basic Handling parameters
Problem:
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. Understeer gradient
3. Characteristic speed
4. Lateral acceleration gain at 60 mph
5. Yaw velocity gain at 60 mph
6. Slideslip angle at the CG on an 800 ft radius turn at 60 mph.
7. Why is the calculated understeer gradient quite low (usually it is around 3-4 deg/g)
Solution:
Given:
Front axle weight, Wf = 1900 lb
Rear axle weight, Wr = 1552 lb
Wheelbase, L = 100.6 in
1. Ackermann steer angle = `(L/R) *(180/pi)`
Ackermann steer angle for 500 ft radius = `(100.6/(500*12))*(180/pi)`
= 0.96 deg
Ackermann steer angle for 200 ft radius = `(100.6/(200*12))*(180/pi)`
= 2.4 deg
Ackermann steer angle for 50 ft radius = `(100.6/(50*12))*(180/pi)`
= 9.6 deg
2. Understeer gradient:
The weight distribution on each wheel at front and rear will be 50/50 of the corresponding weights at front and rear respectively.
Therefore weight of each wheel:
Wf = 950 lb
Wr = 776 lb
We use the interpolation formula to determine cornering stiffness values at these loads
Cf = `225+(950-900)*((257-225)/(1125-900))`
Cr = `171 + (776-675)*((225-171)/(900-675))`
`Understeer gradient: `((Wf)/(Cf))-((Wr)/(Cr))`
: 0.118 deg/g
3. Characteristic speed is the speed at which the steering angle requied to negotiate the turn is twice the ackerman angle.
It is given by,
`V = sqrt((57.3*Lg)/K)`
= `sqrt[(57.3*100.6*32.2)/(12*0.118)`
= 362.09 ft/sec
4. lateral acceleration gain: `[(V^2/(57.3Lg))]/[1+((KV^2)/57.3Lg)]`
= `[(88^2/(57.3*(100.6/12)*32.2))]/[1+(0.118*88^2/(57.3*(100.6/12)*32.2))]`
= 0.47 g/deg
5. Yaw Velocity Gain: `(V/L)/[1+(KV^2)/(57.3Lg)]`
= `(88/(100.6/12))/[1+(0.118*88^2)/(57.3*(100.6/12)*32.2)]`
= 9.92 deg/sec/deg
6. Slide slip angle is given by: `57.3*c/R-[(Wr*V^2)/(Cr*Rg)]`
where, c is the distance of CG from the rear axle.
We know, `Wf = W *c/L`
c = `[1900*100.6]/[(1900+1552)*12]`
= 4.61 ft
Cr = `257+(1552-1125)*[(300-275)/(1350-1125)]`
= 338.60
Slide slip angle : `57.3*(4.61/800)-[(1552*88^2)/(338.60*800*32.5)]`
: -1.04 deg
7. We determined the understeer gradient from the study of a 2DOF model and we only considered the cornering stiffness which is tire property. But in real scenario, other systems such as the suspension, steering etc. will also contribute to the understeer gradient.