Simple Car Dynamics Claude Lacoursière HPC2N/VRlab, Umeå Universitet, Sweden, and CMLabs Simulations, Montréal, Canada May 18, 2005 Typeset by FoilTEX May 16th 2005 Outline basics of vehicle dynamics different models for different applications analysis of the different components building a simple car simulation Typeset by FoilTEX May 16th 2005 1
Basic definition Ground vehicles consist of: a frame: either one rigid part or some articulated components ground support: wheels, tracks, or pods, and suspension steering mechanism (sometimes coupled to suspension) power plant (engine) drive train (transmission, gear box, differential, brakes) Ground vehicle dynamics addresses all these issues by providing models for each of the components. Track vehicles will not be considered in this (short) lecture. From now on, ground vehicle = car. Typeset by FoilTEX May 16th 2005 2 Driving conditions Race Track perfectly flat road, absolutely no obstacle Well Maintained Street very flat surface, hills, no obstacles Bumpy Road mostly flat surface, some holes and bumps, hills, small obstacles Rough Road not very flat, many holes, small and big obstacles Off Road arbitrary terrain conditions, arbitrary obstacles These are increasingly difficult when it comes to finding the contact points on the wheels or track, and computing the forces between the ground and the ground support. Off road will require full rigid body model as well as detailed geometric inteference computation (collision detection). Race track driving requires very little geometric information. We will concentrate race track conditions but build a model that is suitable for bumpy road conditions. Typeset by FoilTEX May 16th 2005 3
Some basic numbers Car driving at 100 Km per hour is : 28 metres per second Typical wheel: 35cm radius 12 revolutions per second At 60 frames per second, that s 1/5 revolution per frame or 1.2 rad per frame. Small angle formula i.e., sin(θ) θ works only for θ 0.2. We re way off. Racing car: 200km per second, 25 revolutions per second, 2.5 rads per frame! Using real wheels to simulate race cars is difficult! This is why we concentrate on a point approximation for the wheels i.e., the car is now a frame and at most one contact point per wheel. Typeset by FoilTEX May 16th 2005 4 Basic physics of a simple car The forces acting on a car N r F w N f F r W F f N r N f F r F f Normal force on the rear wheel Normal force on the front wheel Traction (friction) force on the rear wheel Traction (friction) force on the front wheel W Gravitational force on Center of Mass (CM) F w Drag resistance on CM The diagram is that of a rear traction car which is accelerating. Typeset by FoilTEX May 16th 2005 5
Basic physics of a simple car Note carefully: gravity acts on the CM but normal, traction, and braking forces act at the contact area of the wheels. Traction, braking and normal forces apply torques on the CM. Because of this, we need a rigid body model for the frame, at least. Wind drag also applies torques but that will be neglected. Typeset by FoilTEX May 16th 2005 6 Basic physics of circles v 2 r 2 v 1 F 2 F 1 r 1 ω 2 ω 1 Essential relationships: Linear velocity v = ω/r Centripetal force F = mv 2 /r, without this force, you don t turn! Velocity ratios for common angular velocity: v 2 = v 1 r 2 /r 1 Typeset by FoilTEX May 16th 2005 7
What does that mean for cars? Inside and outside wheels are on different radii during cornering. To turn without skidding, need different linear velocities for inside or outside. Wheels are subject to tangential force of mv 2 /r which can be high. π β π α α β Typeset by FoilTEX May 16th 2005 8 Basic Wheel Physics A perfectly cylindrical wheel cannot turn without slipping! ω 2 ω 1 v 2 I.e., ω 1 = ω 2 and v 1 = v 2 so we can only have straight line motion. This is why we use tires which are deformable. A tire under stress will deform and slip i.e., direction of travel will differ from that of a perfectly rigid cylinder. Despite the abundand folklore, tire slip has nothting to do with the contact patch slipping on the ground. Tire slip is the net kinematic result of bulk tire deformations. v 1 Typeset by FoilTEX May 16th 2005 9
Details of wheel-ground contact Longitudinal slip Tangential slip Actual direction of travel R e R 0 Slip angle α Direction of the wheel When the tire is under a load, it compresses and the effective radius is R e < R 0, i.e., the velocity of the center of the wheel is not v = R 0 ω but instead, v = R e ω. This generates a friction force because work is done to compress the tire, not because the contact patch slips on the ground!!!! When there is a tangential force on a tire, it compresses tangentially and doesn t go along a line of travel that is in the vertical plane. Typeset by FoilTEX May 16th 2005 10 Longitudinal slip Wheel center of mass moves at linear velocity v x, angular velocity ω Point on contact patch is actually moving at v r = ωr e = v x The slip is defined as: κ = ωr 0 v x vx, and σ = κ 1+κ. longitudinal force (relative units) Longitudinal force as a function of slip ratio 1.5 1 0.5 0-0.5-1 -1.5-20 -15-10 -5 0 5 10 15 longitudinal slip ratio in percent The relative units are actually ratios to the normal force. We can approximate the slip with F s = σc l C l 1/0.06. 20 and clip this at F s = 1, and To use these relations, we need to integrate the angular velocities of the wheels. Typeset by FoilTEX May 16th 2005 11
Tangential slip Here, the line of travel of the tire differs from the plane by angle α. If the forward velocity of the wheel is v x and the tangential velocity of the contact patch is v y, the slip is defined as: σ t = v y v x = tan(α). Again, this shows up in the force ratio i.e., the ratio of the tangential force, F y, to the normal load, F z, and we essentially have: F y = σ t C t F z Of course, we need to clip σ to the friction coefficient but for good tires, this can be high i.e., µ 1.5. The relationship between slip angle and tangential force is very similar to that of the longitudinal slip. Typeset by FoilTEX May 16th 2005 12 The Power Plant Engines have rating curves which describes how much torque they can deliver for a given RPM. Typically, two pairs of numbers are given: max power rating: given max power achieved at given rpm max torque rating: given max torque achieved at given rpm a typical figure is given here: Non-descript Actual Diesel engine Typeset by FoilTEX May 16th 2005 13
The Power Plant For most cases, it s enough to assume that power delivered is proportional to engine RPM up to the max. This means that P = C p Ω 0 is the power, and that: τ max = P/Ω 0 C p is the constant max torque. Of course, using curve lookup is a better idea. Typeset by FoilTEX May 16th 2005 14 The throttle, and the RPM The gas pedal controls the acceleration of the engine, we know that! Simple model: torque delivered = throttle max torque The RPM is estimated by the current velocity of the drive wheels i.e., Ω 0 = ω d r f where ω d is the average angular velocity of the drive wheels, Ω 0 is the engine angular velocity, and r f is the final gear ratio, i.e., the ratio from the gear box times the ratio from the differential. Note: this is working in open loop fashion i.e., there is no self inertia and no feedback mechanism. A good drive train model would have all these inertias and feedback loops simulated but that s too complicated for our purposes. Typeset by FoilTEX May 16th 2005 15
Differentials When traction is delivered to the wheels, it goes through a differential which is a device that enforces the following constraint: ω s = r d (ω l + ω r ) where ω s is the drive shaft angular velocity, ω l is the angular velocity of the left wheel, ω r is the angular velocity of the right wheel, and r d is the differential gear ratio. In practice, this constraint means that the torque applied on each wheel is equal: that s why you can get stuck in the snow if one wheel is free. Therefore, this can be ignored i.e., after computing the torque on each wheel based on wheel slip, one can simply average the two and use that value instead to model a differential. Typeset by FoilTEX May 16th 2005 16 Steering A typical steering mechanism does a lot of fancy things i.e.,: set different steering angles on each wheels set an inclination angle which depends on the steering angle adjust differential parameters (for front wheel driving) But the difference in steering angle is very small and the other adjustments are also high order corrections. Therefore, use the simplest possible model: steering angle = angle of the wheel with respect to the chassis direction. To really model steering, we need to also account the self-aligning torque on the wheel i.e., tangential forces on the wheel tend to set them back to a straight line. We ignore that. Typeset by FoilTEX May 16th 2005 17
Brakes A brake delivers a force roughly proportional to the brake pedal input times some maximum torque. The result is applied directly to the wheels but in the direction opposite to the motion. Give max braking torque τ b, and a brake pedal signal d b [0, 1], a simple model is this: ω τ = d b τ b ω + ɛ, ɛ > 0. Of course, this will have problems when ω 0 and there are other forces on the car: the wheels will jitter. This can be clamped near zero with smart control but it s not trivial. Typeset by FoilTEX May 16th 2005 18 The simplest model: soap box car The frame is a box: use a rigid body. Use symmetric inertia tensor. Wheels are also rigid bodies but: only integrate angular velocities ω (i) about axle inertia tensor: I (i) = m (i) R 2 0 /2, and m(i) m/50 Suspension is modelled using springs: aim at 5% of R 0 under normal weight so that k = (gm/4)/(r 0 /20) Ground forces are applied at the contact points. The drive train is modeled by feeding back the wheels angular velocities. Typeset by FoilTEX May 16th 2005 19
The simplest model: soap box car Dynamics model: location of contact points compute traction forces compute cornering forces compute drive torques compute brake torques distribute forces on frame and torques on wheels gravity on CM wind drag on CM Typeset by FoilTEX May 16th 2005 20 Locating the contact points Fixed locations of wheel attachments in chassis frame: a (i), i = 1, 2, 3, 4. World coordinates: A (i) = X + Ra (i), X is the CM position, R rotation matrix. At each of these points, find the point of the ground that is closest. For flat ground, it s going to be the intersection of the ray: A (i) λrz with ground plane: z T [A (i) λ (i) Ra (i) ] = 0 λ (i) = zt A (i) z T [Rz] where z = (0, 0, 1) T is the unit vector pointing upwards. If the ground is not flat, one must compute the quantities: λ (i,j) = n(j)t A (i) n (j)t [Rz] for each triangle j which is in the neighborhood of point a (i), and where n (j) is the normal of the polygon. One must then find out which triangle contains the intersection and from this, recover λ (j). This will not be covered here but it s not difficult. Look for ray and polygon intersection on the web. Typeset by FoilTEX May 16th 2005 21
Computing normal loads Once we know the distance from the ground λ (i), we get the normal force magnitudes: ( 0 if λ N (i) (i) > l 0 = k(l 0 λ) otherwise Normal forces get applied at the location of the contact points: A (i) λ (i) Rz, and in the direction of n (i), the normal to the ground contact at point i. It s a good idea to compute a damping force for the suspension spring as well. To do this, you need to project the velocity of the attachment point on the axis of the suspension. Typeset by FoilTEX May 16th 2005 22 Computing traction and cornering torques Geometry. First, for each contact point, compute local contact plane basis vectors. Compute velocity in the contact plane i.e., v (i) = v + ω [Ra (i) λ (i) Rz] and v (i) z = n (i) v (i), v (i) x = s(i) v (i), v (i) y = t (i) v (i), and s (i), t (i), n (i) form a right handed coordinate system at the point of contact. Also, ω is the angular velocity of the frame, R is the current rotation matrix, a (i) is the body fixed position of the i th attachment point, z is the upward unit vector, and λ (i) was computed previously. For the driving wheels, s (i) should be in the plane of the drive wheel i.e., given zero slip, s (i) would be the direction of travel. Compute the longitudinal velocity of the car i.e., v x = (Rx) v where v is the velocity of the car, and x is a normal vector pointing forward. Typeset by FoilTEX May 16th 2005 23
Computing traction and cornering torques Each wheel will get a net torque which is the sum of the drive torque, the longitudinal rolling resistance torque, and a brake torque. The corresponding net force is applied to the frame at the contact point. Tangential slip Compute tan(α (i) ) = v (i) y /v x v x v (i) y /( v x +ɛ) 2 = σ (i) t (avoid dividing by zero). Set F (i) t = σ (i) t C t N (i) and clamp this with µn (i) where µ is the friction coefficient. Apply this force on the cassis at the contact point, ignore self-aligning torque. Longitudinal Slip Compute the longitudinal slip: κ (i) = ω(i) R 0 vx vx +ɛ, and σ (i) l = κ (i) /(1 + κ (i) ) (avoid dividing by zero!), and from this, compute the traction force: F (i) l = C l σ (i) l N (i). Apply this as a torque on wheel i: τ (i) l = F (i) l R 0. Brake Compute the brake torque τ (i) b = d b τ b ω (i) and corresponding brake ω (i) +ɛ force F (i) b = τ (i) b s(i) /R 0, where d b is the break pedal value (could have a different one for each wheel). Typeset by FoilTEX May 16th 2005 24 Computing drive torque Estimate engine angular velocity Ω 0 ω d /(r g r d ) where ω d is the average driving wheel angular velocity (last time step), and r g, r d are the gear and differential gear ratios, respectively. Get the gas pedal input: d g Get max torque from table and RPM value (Ω 0 ) Compute: τ e = d g τ max (Ω 0 ) Compute: τ d = r g r d τ e f l where f l is a loss factor (if you like) Apply τ (i) ω d = τ (i) d 2ω to each drive wheel, where ω d is the average drive-wheel d velocity: this will balance the power delivered on each wheel. Typeset by FoilTEX May 16th 2005 25
Wind drag Simple wind drag of F w = C w v x v x, add to CM Could add up a number of drag forces, including downward pressure and lifts, depending on the shape of the car... not today! Typeset by FoilTEX May 16th 2005 26 Summing up and integrating Sum all the torques on the drive wheels. Sum up all the forces on the contact points and compute torques on CM. Integrate using symplectic Euler: v n+1 = v n + hm 1 F n (x n, v n ) (1) ω n+1 = ω n + hi 1 τ n (x n, v n ) (2) q n+1 = q n + h 2 ET (q n )ω n+1, q n+1 = q n+1 / q n+1 (3) x n+1 = x n + hv n+1 (4) ω (i) n+1 = ω(i) n + hi(i) 1 τ (i) (5) and 2 q 1 q 0 q 3 3 q 2 E(q) = 4q 2 q 3 q 0 q 1 5 q 3 q 2 q 1 q 0 Typeset by FoilTEX May 16th 2005 27
Open issues Multiple contacts on a wheel e.g., obstacles Problems at low velocity: κ misbehaves near v x 0 brakes can go unstable Typeset by FoilTEX May 16th 2005 28 Conclusions Cars are very sophisticated things! A good model needs many coupled differential equations and algebraic conditions. Much work to do on tire models to get nice stable contacts and good handling. Typeset by FoilTEX May 16th 2005 29
References Pedestrian s references: Basic game physics car tutorial: http://home.planet.nl/ monstrous/tutcar.html Another interesting set of tutorials: Advanced references: http://www.miata.net/sport/physics/index.html Jo Yung Wong, Theory of Ground Vehicles, 3rd ed., Wiley and Sons, 2001. Tyre models: Hans B. Pacejka, Tyre and Vehicle Dynamics, Butterworth- Heinemann, Oxford, 2002. Milliken and Milliken, Race Car Vehicle Dynamics, SAE, 1995 Milliken and Milliken, Chassis Design Principle and Analysis, SAE, 1995 Typeset by FoilTEX May 16th 2005 30