Problem Set-1b Week 11Challenge : Basic Handling parameters
Given Data:
weight on front axle `w_f` = 1900 lb
weight on rear axle `w_r` = 1552 lb
wheelbase `l` = 100.6 in
The tires have the following stiffness values:
Deliverables:
1) Ackermann steer angles for 500, 200 and 50
| Ackerman Steering Angle `delta = (180/pi)*L/R` |
For the given values of radius:
| Radius (ft) | Radius (in) | Ackerman Angle `delta` |
| 500 | 6000 | 0.9607 |
| 200 | 2400 | 2.4016 |
| 50 | 600 | 9.6066 |
2) Understeer Gradient
| Understeer gradient `k = W_f/C_(alphaf)-W_r/C_(alphar` |
`W_f` is the weight on the front axle (one tire) `W_r` is the weight on the rear axle (one tire) `C_(alphaf` is the front cornering stiffness `C_(alphar` is the rea cornering stiffness |
`W_f = 1900/2 = 950`lb
`W_r = 1552/2 = 776`lb
Calculating `C_(alphaf` using interpolation:
`(1125-950)/(1125-900) = (257-C_(alphaf))/(257-225)`
`=>` `C_(alphaf` = 232.11 lb/deg
Calculating `C_(alphar` using interpolation:
`(900-776)/(900-675)=(225-C_(alphar))/(225-171)`
`=>` `C_(alphar` = 195.24 lb/in
Implementing the formula for understeer budget:
`k = 950/232.11-776/195.24`
`k = 0.1183` deg/g
3) Characteristic speed
| Characteristic Speed `u_(char) = sqrt(57.3*L*g/k)` |
`g = 386.08858267562` in/sec^2
`u_(char) = sqrt(57.3*100.6*386.08858267562/0.1183)`
`u_(char` = 4337.4 in/sec = 110.17 m/s
4) Lateral acceleration gain at 60 mph
v = 60 mph = 88 ft/sec
g = 32.2 ft/sec^2
| Lateral Acceleration Gain `a_y/delta = ((v^2)/(57.3Lg))/(1+kv^2/(57.3Lg)` |
`a_y/delta = (88/(57.3*100.6/12*32.2))/(1+(0.118*88^2/(57.3*100.6/12*32.2)))`
`a_y/delta = 0.473` g/deg
5) Yaw velocity gain at 60 mph
| Yaw Velocity Gain `r/delta = (y/L)/(1+kv^2/(57.3Lg))` |
`r/delta =(88/(100.6/12))/(1+0.118*88^2/(57.3*100.6/12*32.2))`
`r/delta = 9.911` deg/sec/deg
6) Slideslip angle at the CG on an 800 ft radius turn at 60 mph.
| Body Side Slip `beta= 57.3c/R-(w_rv^2)/(C_(alphar)gR)` |
R = 800 tf
`w_f = (w_f+w_r).c/L`
`=> c = (1900.100.6)/((1900+1552)*12) = 4.614`ft
Calculating `C_(alphar` using interpolation:
`(1552-1350)/(1350-1125)=(C_(alphar)-300)/(300-257)`
`=>` `C_(alphar)= 338.60` lb/deg
`beta = 57.3*4.614/800-(1552*88^2)/(338.60*800*32.2)`
`beta = -1.047`deg
7) Why is the calculated understeer gradient quite low (usually it is around 3-4 deg/g)
There are various factors that affect the understeer gradient including the steering and suspension system but here the understeer gradient is being calculated using only a 2 degree of freedom model and cornering stiffness.