Vehicles FISICA

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Simple Car Dynamics

Claude Lacoursiere

HPC2N/VRlab, Umea Universitet, Sweden,

and CMLabs Simulations, Montreal, 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”.


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 thewheels 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.




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.


Basic physics of a simple car

F w N f

F r F f

N r

The forces acting on a car

W

N r Normal force on the rear wheel

N f Normal force on the front wheel

F r Traction (friction) force on the rear wheel

F f 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.




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.


Basic physics of circles

r1

r2

ω

2

ω1

F 1

F 2

v2

v1

Essential relationships:

Linear velocity v = ω/r

Centripetal force F = mv2/r, without this force, you don’t turn!

Velocity ratios for common angular velocity: v2 = v1r2/r1




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 mv2/r which can be high.

α β

π − απ − β


Basic Wheel Physics

A perfectly cylindrical wheel cannot turn without slipping!

ω1

ω2

v1

v2

I.e., ω1 = ω2 and v1 = v2 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.




Details of wheel-ground contact

Longitudinal slip Tangential slip

R0

Re

Actual direction of travel

Slip angle

Direction of the wheel

α

When the tire is under a load, it compresses and the effective radius is Re < R0,

i.e., the velocity of the center of the wheel is not v = R0ω but instead, v = Reω.

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 goalong a line of travel that is in the vertical plane.


Longitudinal slip

Wheel center of mass moves at linear velocity vx, angular velocity ω

Point on contact patch is actually moving at vr = ωRe = vx

The slip is defined as: κ =ωR0−vx|vx|

, and σ = κ1+κ

.

Longitudinal force as a function of slip ratio

longitudinal slip ratio in percent l o n g i t u d i n a l f o r c e

( r e l a t i v e

u n i t s )

20151050-5-10-15-20

1.5

1

0.5

0

-0.5

-1

-1.5

The relative units are actually ratios to the normal force.

We can approximate the slip with F s = σC l and clip this at |F s| = 1, and

C l ≈ 1/0.06.

To use these relations, we need to integrate the angular velocities of the wheels.




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 vx and the tangential velocity of the contact

patch is vy, the slip is defined as:

σt =vy

vx= 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 = σtC tF 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.


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:

Actual Diesel engineNon-descript




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 pΩ0 is the power, and that:

τ max = P/Ω0 ≈ C p is the constant max torque.

Of course, using curve lookup is a better idea.


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.,

Ω0 =ωd

rf

where ωd is the average angular velocity of the drive wheels, Ω0 is the engine angularvelocity, and rf 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.




Differentials

When traction is delivered to the wheels, it goes through a differential which is a

device that enforces the following constraint:

ωs = rd(ωl + ωr)

where ωs is the drive shaft angular velocity, ωl is the angular velocity of the leftwheel, ωr is the angular velocity of the right wheel, and rd 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.


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.




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 db ∈ [0, 1], a simple model is

this:τ = −dbτ b

ω

|ω| + , > 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.


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)R20/2, and m(i) ≈ m/50

Suspension is modelled using springs: aim at 5% of R0 under normal weight so that

k = (gm/4)/(R0/20)

Ground forces are applied at the contact points.

The drive train is modeled by feeding back the wheels angular velocities.




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


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:

zT [A(i) − λ(i)Ra(i)

] = 0 ⇒ λ(i)=

zT A(i)

zT [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.




Computing normal loads

Once we know the distance from the ground λ(i), we get the normal force

magnitudes:

N (i)

=

(0 if λ(i) > l0

k(l0 − 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.


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 ith

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., vx = (Rx) · v where v is the

velocity of the car, and x is a normal vector pointing forward.




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

/vx ≈ vxv(i)

y

/(|vx|+)2 = σ(i)

t

(avoid

dividing by zero). Set F (i)t = −σ

(i)t C tN (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)R0−vx|vx|+

, and

σ(i)l

= κ(i)/(1 + κ(i)) (avoid dividing by zero!), and from this, compute the

traction force: F (i)l

= C lσ(i)l

N (i). Apply this as a torque on wheel i:

τ (i)l

= F (i)l

R0.

Brake Compute the brake torque τ (i)b

= −dbτ bω(i)

|ω(i)|+and corresponding brake

force F (i)b

= τ (i)b

s(i)/R0, where db is the break pedal value (could have a

different one for each wheel).


Computing drive torque

Estimate engine angular velocity Ω0 ≈ ωd/(rgrd) where ωd is the average driving

wheel angular velocity (last time step), and rg, rd are the gear and differential gear

ratios, respectively.

Get the gas pedal input: dg

Get max torque from table and RPM value (Ω0)

Compute: τ e = dgτ max(Ω0)

Compute: τ d = rgrdτ ef l where f l is a loss factor (if you like)Apply τ

(i)d

= τ dω(i)

2ωdto each drive wheel, where ωd is the average drive-wheel

velocity: this will balance the power delivered on each wheel.




Wind drag

Simple wind drag of F w = −C wvx|vx|, 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!


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:

vn+1 = vn + hm−1F n(xn, vn) (1)

ωn+1 = ωn + hI −1

τ n(xn, vn) (2)

qn+1 = qn +h

2E T (qn)ωn+1, qn+1 = qn+1/|qn+1| (3)

xn+1 = xn + hvn+1 (4)

ω(i)n+1 = ω(i)

n + hI (i)−1

τ (i) (5)

and

E (q) =

2

4

q1 q0 −q3 q2q2 q3 q0 −q1q3 −q2 q1 q0

3

5– Typeset by FoilTEX – – May 16th 2005 27



Open issues

Multiple contacts on a wheel e.g., obstacles

Problems at low velocity:

• κ misbehaves near vx ≈ 0

• brakes can go unstable


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.




References

Pedestrian’s references:

Basic game physics car tutorial: http://home.planet.nl/ monstrous/tutcar.html

Another interesting set of tutorials: http://www.miata.net/sport/Physics/index.html

Advanced references:

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


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