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Directional Stability During Straight Line Braking for Heavy Vehicles

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Directional Stability During Straight Line Braking for Heavy Vehicles Directional stability during straight line braking for heavy vehicles: Malte Rothhämel 1) 2) Jolle Ijkema 1) 1) Scania Truck Development, Vehicle Dynamics, Södertälje Sweden 2) Royal Institute of Technology, Department of Vehicle Engineering, Stockholm, Sweden Contact: [email protected] All vehicles show some degree of directional instability during straight line braking since no real vehicles are truly symmetric. If not corrected by the driver, the vehicle may pull to a certain side of the road, so called brake pull. There are numerous sources of this behavior: e.g. kinematical brake steer, asymmetric mass distribution or brake imbalance. One possible approach to eliminate the negative effects of the inevitable causes for brake pull is active steering. This poster investigates the stability limits of the resulting vehicle motion when brake pull is compensated for by means of a steering correction in a theoretical way. The situation is described by means of a bicycle model with an external yaw moment. Its equations of motion are used to find states for the body slip angle (β) and steering wheel angle (δ) that give stable and translatory motion free from yaw velocity and yaw acceleration. The Model m⋅a ⋅d ⎛ cyF + cyR − ax ⎞ δ = − x ⎜ ⎟ (3) The analysis is based on a bicycle-model which c ⋅ L ⎜ c + 2⋅λ ⋅c − λ ⋅ a ⎟ describes the vehicle adequately (Fig. 1). Unlike yF ⎝ yR yR x ⎠ the typical bicycle-model the center of gravity c β = − yF δ (4) (COG) height is assumed to be a − c − c β x yF yR hCOG = 0.9m above the road surface to include the 0 changed wheel load distribution while braking. The Sensitivity v / m &x s 2 brake force distribution κ at the wheels are In addition to the longitudinal deceleration δ ble steering angle and body slip angle are sta assumed as proportional to the vertical wheel-load -4 (Fig. 2) with reference to the friction circle. The dependent of wheel base, centre of gravity ble sta behavior of the tires is simplified to a linear longitudinal and vertical position. These un relationship between the side force and the tire slip parameters have different influences to the κ vehicle’s behavior. Fig. 5 shows the -8 angle α with a maximum side force at α = 8° as the 0 02 04 06 08 1 0 0,2 0,4 0,6 0,8 1,0 upper limit (Fig. 3). The cornering stiffness, cy, is sensitivity of different parameters based on Fig. 4 Stability limit of the proportional to the vertical wheel load. a typical tractor with lateral deviation of vehicle as deceleration to F The equations of COG equal to 0.225m while braking with brake force distribution κ κ = xR m vx = 4 /s² with μ = 0.8. While varying one 10° FxR + FxF motion can be simplified with the Fig. 1 Simplified bicycle parameter the others remain at their β λL (1-λ)L model with yaw assumption that β standard value. A long wheelbase extends δ moment as result of the moment arm of the tire lateral forces COG and δ are small deviation of COG 5° which leads to the thus making the stabilizing yaw moments h COG linearization of trigonometrical functions. bigger. A rotating moment because of With the kinematic relation between asymmetric mass distribution cannot cause L hCOG/m FzF FzR as much rotation as a vehicle with short 0° steering angle, body slip angle and tire 0 0,8 1,6 Fig. 2 Model of the position of slip angles, the side-stiffness’ wheel base. As well, the rear axle’s release center of gravity dependency of the tire slip angle is decreases which also stabilizes the vehicle. 10° Maximum tire slip y The same result has a good longitudinal F angle α substituted. Combination and β max relative mass distribution λ which can rearranging of the equations creates the δ y max overall relation. influence the load of the rear axle. The The force and moment equilibrium can centre of gravity height will, of course, 5° Lateral tire force tire Lateral worsen the truck’s behavior because a Fy max be expressed as linear equations in β λ c = Maximum lateral tire force F y α max and δ (1) where the matrix A describes higher COG unloads the rear axle while 0° 0 5 10 15 0,1 0,5 0,9 the factors of the angles. braking. Tire slip angle [°] Truck data: Fig. 3 Simplified relation of lateral Conclusion m = 7000kg Lstandard = 3.7m β −1 M asym tire force and slip angle = A (1) In general active steering can accomplish hCOG,standard = 0.9m λstandard = 0.21 δ 0 lateral deviation of COG d = 0.225m Limitations the problem of brake pull but the possibilities Fig. 5 Sensitivity of the vehicle λLcyF − (1− λ)LcyR − λLcyF The determinant of A shows how the A = are limited by the decereleration. As long as to wheelbase and position − c − c − mv c system of equations is solvable (2). yF yR &x yF there is a yaw moment-imbalance due to an of center of gravity. This leads to two solutions: unbalanced COG, neither β nor δ will become ⎧ cyF = 0 equal to zero. This means that a tractor which is 2.55m wide and 5m If cyF is equal with zero, which det(A) = 0 ⇔ ⎨ (2) c + λmv = 0 means that the front wheels cannot ⎩ yR &x long, can at most pull up to a body slip angle of 5° before it exceeds the generate lateral force (e.g. locked wheels), det(A) will be equal with zero. border of a 3m wide lane. The first limitation is given by the front wheels’ The equation system is unsolvable which results in an unstable state. side-stiffness. If the front wheels cannot transmit side-force there will be The second solution is more complex: If (2) is true, the state will also be no correction possibility. The second limitation is the rear wheels’ side- stiffness. If the rear wheels cannot transmit enough side-force even active unstable. cy is always positive while vx is always negative during braking. This means that there is a state of λ·m that makes the equation equal to steering cannot correct the vehicle’s direction. zero. Moreover, (2) can be filled with vehicle specific data which results However, active steering can make the steering wheel angle (δSW) the limiting deceleration as function of the brake distribution. Fig. 4 shows become equal with zero which unburdens the driver’s efforts by providing this border of stability for different values of static brake force distribution. the active steering correction. But because of the limitations, in the product design, the lateral deviation of COG has to be further more Solving equation (1) results the steering angle (3) and the vehicles body avoided or minimized while also the longitudinal and vertical position of slip angle (4) that can be calculated by means of vehicle specific COG and the wheelbase length must be carefully considered. parameters and the deceleration within the stability limit..
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